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# How to remove heartbeat signal from blood pressure signal?
I am working on a medical engineering project that involves processing signals received from a human body. We use a sensor to record the blood pressure, $$B(t)$$.
However, if you put your finger on many parts of your skin, you can notice your heartbeat. Accordingly, the recorded signal also has some weak heartbeat signal added, $$A'(t)$$. To obtain accurate blood pressure signal, we have a sensor that records the heartbeat signal, $$A(t)$$. However, $$A'(t)$$ is a delayed and weak version of $$A(t)$$, where the phase lag is unknown.
Knowing $$A(t)$$, how can we find accurate $$B(t)$$?
• @rodrigo-de-azevedo , as mentioned we have recorded h(t) for a period of time and have some recorded signal like B'(t) in the same period, which has B(t) and h'(t). h'(t) has lower amplitudes than h(t) (as the sensor captures it weakly). However, it is not something like multiplying h(t) by some constant. Usually, h(t) has higher frequencies than B(t) so a low pass filter may be used. However, as h'(t) may change from person to person and B(t) may have high frequency components, I am looking for a better solution. for instance some sort of decomposing B'(t) and finding components related to.. Sep 12, 2021 at 14:28
• h'(t) is unknown and we wish to find it, If you mean B'(t), I dont know if cross correlation would help for finding delay. besides, while h' is much similar to h at each point t we have something like h'(t)=ah(t-v) for some noisy a. so finding only the delay would not solve the problem Sep 12, 2021 at 14:48
• Are you trying to record the blood pressure averaged over time, or are you trying to record systolic and diastolic blood pressure? If the latter, I'm not sure how you're going to separate these out, because the instantaneous blood pressure varies in sync with the heartbeat. Sep 12, 2021 at 14:55
• Just as a note -- you may want to review your variable names. In signal processing, $h(t)$ is almost universally used to denote a system impulse response, so unless $h(t)$ is already used to denote variations due to heartbeat in biomechanical circles, you'll probably confuse more than you clarify by that choice of signal name. I'd see if I can find what people tend to use in papers; if that doesn't work I'm not sure what to suggest -- $x_h(t)$, or $p_h(t)$, maybe. Sep 12, 2021 at 14:57
• @TimWescott , I am not only recording the blood pressure, It was just one sample. There are other types of signals which are mixed with heart beat. The choice of h(t) was poor and I am changing it. Sep 12, 2021 at 15:53
If $$A(t)$$ is known, it can be zeroed in the synchrosqueezed representation - the remainder is then $$B(t)$$, recovered by inversion. $$A(t)$$ need not be known perfectly - just enough to indentify its time-frequency ridges.
• h(t) is not the impulse response it was just a poor naming, I changed it to A(t). thanks for the program I am working on it. Sep 12, 2021 at 16:01
Problem
Your model is: $$B'(t) = B(t) + A'(t)$$ where $$B'(t)$$ is your observed blood pressure corrupted with the heartbeat signal, $$A'(t)$$. You want to find $$B(t)$$ but you don't know $$A'(t)$$, instead you know $$A(t) = \gamma A'(t-d)$$.
Option 1
A similar problem led Widrow to invent the Least-mean-squares (LMS) filter.
We want to find a time-varying filter $$w(t)$$ that when multiplied with the heartbeat signal $$A(t)$$ approximates the blood pressure signal $$B(t)$$, i.e., $$B(t) = w(t)A(t)$$. To do so, we minimize the mean squared error between the measured signal and the model wrt $$w$$.
The error signal is $$\epsilon(t) = B'(t) - B(t) = B'(t) - w(t)A(t)$$
The LMS cost function is,
$$J(w) = \mathbb{E}(\epsilon^2(t)) \\ = J(w) = \mathbb{E} [(B'(t) - w(t) A(t))^2] \\ = J(w) \mathbb{E} [B'^2(t)] + w(t)^2 \mathbb{E}[A^2(t)] - 2w(t) \mathbb{E} [B(t)A(t)] \\ \frac{\partial J}{\partial w(t)} = 2w(t) \mathbb{E}[A^2(t)] - 2 \mathbb{E}[B'(t)A(t)]$$
Now, Widrow suggested an adaptive update to the filter weights to account for changes in the signal over time:
\begin{align} w(t) &= w(t-1) + 0.5 \mu \frac{\partial J}{\partial w(t)} \\ &= w(t-1) + \mu \left(w(t)A^2(t) - B'(t)A(t) \right) \\ &= w(t-1) + \mu \left[A(t) (w(t)A(t) - B'(t))\right] \\ &= w(t-1) + \mu A(t) \epsilon(t). \end{align}
Your best estimate of the blood pressure is $$w(t)A(t)$$.
Option 2
Let's take the Fourier transform of $$B'(t)$$ and $$A(t)$$ and look at them in the frequency domain,
$$B'(\omega) = B(\omega) + A'(\omega) \\ A(\omega) = \gamma \exp(-j\omega d) A'(\omega)$$
If we look at the cross-power spectrum of the signals, we get
$$B'^*(\omega)A(\omega) = \left(B^*(\omega)+A'^*(\omega)\right) \gamma \exp(-j\omega d)A'(\omega)$$
We can make the assumption that the blood pressure and the heartbeat signal are uncorrelated (I have a feeling a biologist would not be happy with this). Then, $$B^*(\omega)A'(\omega) = 0$$ and \begin{align} B'^*(\omega)A(\omega) &= \gamma \exp(-j\omega d) |A'(\omega)|^2 \\ &= \gamma^3 \exp(-j\omega d) |A(\omega)|^2 \\ \frac{B'^*(\omega)A(\omega) } {|A(\omega)|^2} & = \gamma^3 \exp(-j\omega d) \end{align} The inverse Fourier transform of this is some sort of a modified cross-correlation function,
$$\hat{R}_{B',A}(t) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{B'^*(\omega)A(\omega)} {|A(\omega)|^2} e^{j\omega t} d\omega = \gamma^3 \delta(t-d).$$
From this function, you can find $$\gamma$$ and $$d$$ and thus, $$A'(t)$$ and simply subtract it from $$B'(t)$$ to get $$B(t)$$.
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Calculus 8th Edition
$v(a) = -\frac{2}{a^3}$ $v(1) = -2$ $v(2) = -\frac{1}{4}$ $v(3) = -\frac{2}{27}$
First find the derivative of the function $s = \frac{1}{t^2}$ $v(a) = \lim\limits_{h \to 0}\frac{s(a+h)-s(a)}{h}$ => $\lim\limits_{h \to 0}\frac{\frac{1}{(a+h)^2}-\frac{1}{a^2}}{h}$ => $\lim\limits_{h \to 0}\frac{\frac{a^2 - (a+h)^2}{a^2(a+h)^2}}{h}$ => $\lim\limits_{h \to 0}\frac{a^2 - (a+h)^2}{ha^2(a+h)^2}$ => $\lim\limits_{h \to 0}\frac{a^2 - (a^2 +2ah + h^2)}{ha^2(a+h)^2}$ => $\lim\limits_{h \to 0}\frac{- 2ah - h^2}{ha^2(a+h)^2}$ =>$\lim\limits_{h \to 0}\frac{- 2a - h}{a^2(a+h)^2}$ => $-\frac{2a}{a^2a^2}$ => $-\frac{2}{a^3}$ $v(a) = -\frac{2}{a^3}$ Now use the derivative of $v(a)$ to find $v(1)$, $v(2)$, and $v(3)$ $v(1) = -\frac{2}{1^3}$ => $-\frac{2}{1}$ => $-2$ $v(2) = -\frac{2}{2^3}$ => $-\frac{2}{8}$ => $-\frac{1}{4}$ $v(3) = -\frac{2}{3^3}$ => $-\frac{2}{27}$
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5
A 121-9 ball Is traveling to thc Ieft with speed of 30.0 mis when Il Is struck by J racket The farce on Ine ball directad {0 ienstiera epplled over 21,0 ms of conta...
Question
A 121-9 ball Is traveling to thc Ieft with speed of 30.0 mis when Il Is struck by J racket The farce on Ine ball directad {0 ienstiera epplled over 21,0 ms of contact lime shown In the graph; What is the speed of the ball Immediatety eicr I Iaaves tne FechitF(N}60040Z00
A 121-9 ball Is traveling to thc Ieft with speed of 30.0 mis when Il Is struck by J racket The farce on Ine ball directad {0 ienstiera epplled over 21,0 ms of contact lime shown In the graph; What is the speed of the ball Immediatety eicr I Iaaves tne Fechit F(N} 600 40 Z00
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# Borel $\sigma$-algebra of a Borel subset
This is a question from math.stackexchange, which was not answered for a month now. I don't feel comfortable to post it on mathoverflow, but I am somehow blind to see the mistake in the argumentations below.
Let $$(X, \tau)$$ be a topological space. Then $$\sigma(\tau)$$ is the Borel $$\sigma$$-algebra on $$(X, \tau)$$. For any subset $$Y \subseteq X$$ the subspace topology on $$Y$$ is $$\tau|Y = \{ G \cap Y \mid G \in \tau \}$$ and the trace $$\sigma$$-algebra on $$Y$$ is $$\sigma(\tau)|Y = \{ B \cap Y \mid B \in \sigma(\tau) \}$$. It holds $$\sigma(\tau|Y) = \sigma(\tau)|Y$$. If $$Y \in \sigma(\tau)$$ then $$\sigma(\tau)|Y \subseteq \sigma(\tau)$$, hence $$\sigma(\tau|Y) \subseteq \sigma(\tau)$$.
Consider $$X = \mathbb{R}^2$$, $$\tau_e$$ the Euclidean topology and $$\tau_S$$ the Sorgenfrey plane topology (generated by semi-open rectangles $$[a, b) \times [c, d)$$). Then
• $$\tau_e \subsetneq \tau_S$$ (open rectangles $$(a,b) \times (c,d)$$ can be written as a union of semi-open rectangles)
• but $$\sigma(\tau_e) = \sigma(\tau_S)$$ (since $$[a, b) \times [c, d) \in \sigma(\tau_e)$$).
Consider the antidiagonal $$Y := \{ (x, -x) \mid x \in \mathbb{R} \}$$. Then $$Y$$ is a $$\tau_e$$-closed subset of $$X$$, hence a $$\tau_S$$-closed subset. For any $$x \in \mathbb{R}$$ it holds $$\{ (x, -x) \} = ([x, x+1) \times [-x,-x+1)) \cap Y \in \tau_S|Y$$, i.e. every point in $$Y$$ is $$\tau_S|Y$$-open in $$Y$$. Therefore, $$\tau_S|Y = \mathcal{P}(Y)$$ is the discrete topology, hence $$\sigma(\tau_S|Y) = \mathcal{P}(Y)$$.
Now, since $$Y$$ is $$\tau_S$$-closed in $$X$$, we have $$Y \in \sigma(\tau_S)$$ and therefore $$\sigma(\tau_S|Y) \subseteq \sigma(\tau_S) = \sigma(\tau_e)$$, hence $$\mathcal{P}(Y) \subseteq \sigma(\tau_e)$$. But this is a contradiction (e.g. by comparing the cardinalities: $$|Y| = \frak{c}$$, hence $$|\mathcal{P}(Y)| = 2^{\frak{c}}$$ while $$|\sigma(\tau_e)| = \frak{c}$$ because $$\sigma(\tau_e)$$ is generated by countably many sets (the open rectangles with rational endpoints); see also here).
What am I missing?
The problem is that you have to take uncountable unions of sets of the form $$[a,b) \times [c,d)$$ to get every open set in the Sorgenfrey plane, so the $$\sigma$$-algebra generated by $$[a,b) \times [c,d)$$ is strictly smaller than the Borel $$\sigma$$-algebra.
Which is to say, in your notation, $$\sigma(\tau_e) \subseteq \sigma(\tau_S)$$, but $$\sigma(\tau_S) \not\subseteq \sigma(\tau_e)$$.
Interestingly, it is the case that the Borel $$\sigma$$-algebra of the Sorgenfrey line agrees with the Borel $$\sigma$$-algebra of the usual topology on $$\mathbb{R}$$, and it is easy to give a false proof of this. The correct proof uses the hereditary Lindelöfness of the Sorgenfrey line (something not true of the Sorgenfrey plane).
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# Diagonalize a matrix
We have a matrix $A = \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \right)$.
How do you find a diagonal matrix $D$ and an orthogonal matrix $P$ so that:
$D=P^tAP$?
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Which theory do you have a avaliable? Do you know about eigenvalues and how to find eigenvectors? – Henning Makholm Feb 4 '12 at 14:18
Yeah, i tried it but it always gets wrong somewhere in the middle – Some1 Feb 4 '12 at 14:23
You should show what you've tried and what you're saying goes wrong with your attempt, you know... that would make it more interesting for the rest of us. – Ｊ. Ｍ. Feb 4 '12 at 14:25
I've tried to find the eigenvalues using $tI-A=0$ so it's: $\left( \begin{array}{ccc} (-t) & 1 & 1 \\ 1 & (-t) & 1 \\ 1 & 1 & (-t) \\ \end{array} \right)=0$. I get t=2,-1,1 im kind of stucked here... – Some1 Feb 4 '12 at 14:29
I think the solution is -1 and 2 only. – alpha.Debi Feb 4 '12 at 14:36
$A = \left( \begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{array} \right)$.
You know $D = \left( \begin{array}{ccc} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{array} \right)$.
Now, $Ax = \lambda x$
x = 2 (Case 1) $$Ax = 2x$$ $$Ax - 2x = 0$$ $$(A - 2I)x = 0$$ $$A -2I= \left( \begin{array}{ccc} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{array} \right)$$ $$\left( \begin{array}{ccc} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \\ \end{array} \right)x=0$$ Take it to echelon form you get
$$\left( \begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \\ \end{array} \right)x=0$$ Thus, if x = (a b c)'
a = b = c Similarly, do for the rest.
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$A$ is symmetric, then you can find an orthonormal basis of ${\bf R}^3$ with eigenvectors. Solve $\det(A-tI)=0$, find the proper values, and for each one the eigenspace. If one of the spaces has dimension 2 you have to apply Gram-Schmidt to get an orthonormal basis, and the other basis will have one vector that you can normalize. Then the union of the two bases is an orthonormal basis (eigenvectors of a symmetric matrix corresponding to different eigenvalues are orthogonal). Put this basis as columns of $P$..
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Still not working :\ – Some1 Feb 4 '12 at 15:01
But if you built the orthonormal basis of eigenvectors of A , and used it as the columns of P it must work. – alpha.Debi Feb 4 '12 at 15:45
I'm still learning this, I guess I have a mistake somewhere because your answer seems right... can anyone please show me how its done so ill know for other similar questions? – Some1 Feb 4 '12 at 15:49
My answer is a method used in linear algebra to diagonalize symetric matrices by an ortogonal matrix P,that is P^T=P^-1 – alpha.Debi Feb 4 '12 at 16:08
Some1, you're not helping. Can you find the eigenvalues? Can you find a basis for each eigenspace? Can you use Gram-Schmidt to get an orthonormal basis for each eigenspace? Can you make a matrix whose columns are the eigenvectors? It's not enough to say you don't understand - you have to show us what you do understand, and then we can help you over the rough spots. – Gerry Myerson Feb 4 '12 at 23:43
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## måndag 16 december 2019
### Prescription vs Prediction in CFD
Recent posts compare the standard methods of CFD based on turbulence and wall models (RANS, LES and DES as a combination), with DFS Direct Finite Element Simulation without turbulence and wall models as best possible solution of Euler's equations.
DFS has shown to accurately predict complex aerodynamics such as the stall of an airplane by capturing both turbulence and flow separation from first principle physics. DFS thus predicts the full flight characteristics of an airplane with the only input being the shape of the airplane. DFS not only predicts flow separation but also makes it understandable as 3d rotational slip separation with point or line stagnation.
This is a stunning example of the ideal according to Einstein of a mathematical model capable of predicting true physics without input of physical parameters. It is like predicting the circumference of a circle with radius 1 to be $2\pi$, just much more complicated and surprising.
With the standard methods of RANS-LES prediction is replaced by prescription mediated through the turbulence and wall models containing many parameters. As a result RANS-LES cannot truly predict flow separation since that has to be built into the wall model, by either prescribing the flow to stay attached to a smooth solid wall and separate at a corner, or separate under influence of an "adverse pressure gradient".
The main novelty of DFS is thus the possibility of true prediction, which is not possible with RANS-LES which include prescription. This connects to Bohr's comment to Einstein's claim that God does not trow dice, in the form:
• Einstein, stop telling God what to do!
RANS-LES tells physics what to do. DFS predicts what physics does.
DFS prediction of stall of a jumbojet with flow separation on top of the inner part of the wing, in close agreement with observation.
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# Knots and Applications
## In collaboration with INdAM and the Dipartimento di Matematica e Applicazioni, Milano Bicocca
1 May 2011 - 31 July 2011
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### Home > INT1 > Chapter A > Lesson A.1.4 > ProblemA-46
A-46.
Consider the rectangle at right.
1. Determine the perimeter of the large outer rectangle shown at right.
You can find the missing side length using the area and side length given for the upper left inner rectangle.
$5$ $7$ $8$ $40$ $8$ $6$ $30$ $6$ $7$
52 units
2. Notice that the areas of two of the parts have been labeled inside the rectangle. Calculate the total area. Remember to show all work leading to your solution.
$168$ square units
$5$ $40$ $6$ $30$ $7$
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## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
Published by Pearson
# Chapter 5 - Polynomials and Factoring - 5.7 Solving Quadratic Equations by Factoring - 5.7 Exercise Set - Page 352: 40
#### Answer
$x=\left\{ 0,4 \right\}$
#### Work Step by Step
Factoring the given equation, $12t=3t^2 ,$ results to \begin{array}{l}\require{cancel} -3t^2+12t=0 \\\\ -3t(t-4)=0 .\end{array} Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $-3t(t-4)=0 ,$ are \begin{array}{l}\require{cancel} -3t=0 \\\\ t=\dfrac{0}{-3} \\\\ t=0 ,\\\\\text{OR}\\\\ t-4=0 \\\\ t=4 .\end{array} Hence, $x=\left\{ 0,4 \right\} .$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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# In the figure below, ABCD is a quadrilateral with vertices as shown. (Figure can't copy) Why is $\overline{A B} \| \overline{D C} ?$...
## Question
###### In the figure below, ABCD is a quadrilateral with vertices as shown. (Figure can't copy) Why is $\overline{A B} \| \overline{D C} ?$
In the figure below, ABCD is a quadrilateral with vertices as shown. (Figure can't copy) Why is $\overline{A B} \| \overline{D C} ?$
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##### 161maw/V]Pre-Lab Questions What Bronsted-Lowry acid? What Bronsted-Lowty base?Provide the formula of the conjugate base for each of the following acids: HCIH,CO,21H;POasaliva is 6.4. Which of the two is more basic? 3. The pH of blood is 7.4 and that for 1 3
16 1maw/V] Pre-Lab Questions What Bronsted-Lowry acid? What Bronsted-Lowty base? Provide the formula of the conjugate base for each of the following acids: HCI H,CO, 2 1 H;POa saliva is 6.4. Which of the two is more basic? 3. The pH of blood is 7.4 and that for 1 3...
##### PartAHolowalki HIQHIU= 3t25"0 [OH Galcual [1o'in Ite Lxmcst YouI anstc [ UaIQ Iwo signilicont IiquteS1.3*10[H,o'] =SubmnitRexnrst AnsratPan BCalculate [HsO' | In Iha bollowlng uqou s sohation 25"@ [OH Expaos# Yout anewut ueing two #Iqulllcant Alqutae.20410 ? M[ot |
PartA Holowalki HIQHIU= 3t25"0 [OH Galcual [1o'in Ite Lxmcst YouI anstc [ UaIQ Iwo signilicont IiquteS 1.3*10 [H,o'] = Submnit Rexnrst Ansrat Pan B Calculate [HsO' | In Iha bollowlng uqou s sohation 25"@ [OH Expaos# Yout anewut ueing two #Iqulllcant Alqutae. 20410 ? M [ot |...
##### 0.6 L300020021009 v (cm-4)Figure 2. FT-IR Spectrum of Unknown A
0.6 L 3000 2002 1009 v (cm-4) Figure 2. FT-IR Spectrum of Unknown A...
##### Thls extreme Uae problem has solutlon with Dothtnderntn Vaiue andminimum Valve Use Lagrange multipllersfind the extreme Valvesthe function subjectyiven constraint;fx; Y, 2) = 10X 10ySx2 Sy2maximum Valueminlmum value
Thls extreme Uae problem has solutlon with Doth tnderntn Vaiue and minimum Valve Use Lagrange multipllers find the extreme Valves the function subject yiven constraint; fx; Y, 2) = 10X 10y Sx2 Sy2 maximum Value minlmum value...
##### Are the functions f(z)and g(z) = I = the same? Why? Explain in full sentencesType your answer in the box below.Edit - InserFormatsx 4 " A D M- 2 2 4
Are the functions f(z) and g(z) = I = the same? Why? Explain in full sentences Type your answer in the box below. Edit - Inser Formats x 4 " A D M- 2 2 4...
##### Arectangular tank with square base, an open top, and volume of 8,788 ft? is to be constructed of sheet steel . Find the dimensions of the tank that has the minimum surface area_Let s be the length of one of the sides of the square base and let A be the surface area of the tank. Write the objective function:A= (Type an expression: )
Arectangular tank with square base, an open top, and volume of 8,788 ft? is to be constructed of sheet steel . Find the dimensions of the tank that has the minimum surface area_ Let s be the length of one of the sides of the square base and let A be the surface area of the tank. Write the objective ...
##### The numbers of regular season wins for 10 football teams in a given season are given and standard deviation of the population data below: Determine the range, mean; variance set I 2,,6,15,4,15,10,14,10,4,6 2The range is (Simplify your answer )ureThe population mean is Simplify your answer: Round to the nearest tenth as needed )clickrThe populalion variance iS (Simplily your answer: Round (o the nearost hundredlh as needed )AdvisorThe populalion standard dovialion iS (Simplily your answer: Round
The numbers of regular season wins for 10 football teams in a given season are given and standard deviation of the population data below: Determine the range, mean; variance set I 2,,6,15,4,15,10,14,10,4,6 2 The range is (Simplify your answer ) ure The population mean is Simplify your answer: Round ...
##### Define reaction quotient: How does differ from equilibrium constant? If 12 10-42 a 500 K for the reaction shown below_ Hztg) = 2 Hg) IC[H2] = 10-2 M and [HJ-L2*10 22 M, in order t0 ilehieve equilibrium which way will the net reaetion proceed? Umtil when?
Define reaction quotient: How does differ from equilibrium constant? If 12 10-42 a 500 K for the reaction shown below_ Hztg) = 2 Hg) IC[H2] = 10-2 M and [HJ-L2*10 22 M, in order t0 ilehieve equilibrium which way will the net reaetion proceed? Umtil when?...
##### Solve each equation. Approximate the solutions to the nearest hundredth when appropriate.$$-x^{2}+10 x=18$$
Solve each equation. Approximate the solutions to the nearest hundredth when appropriate. $$-x^{2}+10 x=18$$...
##### Solve each problem. See Example 3.Tram works due north of home. Her husband Alan works due east. They leave for work at the same time. By the time Tram is 5 mi from home, the distance between them is 1 mi more than Alan's distance from home. How far from home is Alan?
Solve each problem. See Example 3. Tram works due north of home. Her husband Alan works due east. They leave for work at the same time. By the time Tram is 5 mi from home, the distance between them is 1 mi more than Alan's distance from home. How far from home is Alan?...
##### Use $E q .(3)$ to evaluate the curvature at the given point. $$\mathbf{r}(t)=\left\langle\cos t, \sin t, t^{2}\right\rangle, \quad t=\frac{\pi}{2}$$
Use $E q .(3)$ to evaluate the curvature at the given point. $$\mathbf{r}(t)=\left\langle\cos t, \sin t, t^{2}\right\rangle, \quad t=\frac{\pi}{2}$$...
##### 3 Ey wing cbe ceinitics (chac "(z) Sfc) = 2-62fz+0 - Linng je), Ae1 find the deljvetive of
3 Ey wing cbe ceinitics (chac "(z) Sfc) = 2-62 fz+0 - Linng je), Ae1 find the deljvetive of...
##### Solve the following integral equation y(t) K sinht T)u(r) dr = /.
Solve the following integral equation y(t) K sinht T)u(r) dr = /....
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# continuous random variable (stats)
by francisg3
Tags: continuous, random, stats, variable
P: 32 The probability density function of the time customers arrive at a terminal (in minutes after 8:00 A.M) is f(x)= (e^(-x/10))/10 for 0 < x c) Determine the probability that: two or more customers arrive before 8:40 A.M among five that arrive at the terminal. Assume arrivals are independent my logic is the following: Probability= 1-Probability 0 or 1 customers arrive before 8:40 A.M the answer is the following: P(X1>40)+ P(X1<40 and X2>40)= e-4+(1- e-4) e-4= 0.0363 from what is written above, it seems to be the probability that no one arrives before 8:40 P(X1>40) and the probability that one arrives before 8:40 (X1<40) and another arrives after 8:40 (X2>40). i tihnk i just need some help on understanding why X2 is brought in. Thanks!
P: 43 Your logic is correct. But the solution is wrong. Let p=P(a customer arrives at the terminal before 08:40)=P(X<40)=1-e^(-4) Then, P(0 or 1 customers arrive at the terminal before 08:40) = (1-p)^5 + nchoosek(5, 1)*p*(1-p)^4 = 5.5255*10^(-7) (approximately zero). Hence, P(2 or more customers arrive at the terminal before 08:40) = 1-5.5255*10^(-7) (approximately 1)
Related Discussions Calculus & Beyond Homework 0 Calculus & Beyond Homework 1 Precalculus Mathematics Homework 6 Calculus & Beyond Homework 1 Calculus & Beyond Homework 3
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## Evolving viable systems
It got cold really, really quickly today… Winter is coming –GRRM
In my last two posts I wrote about holey adaptive landscapes and some criticisms I had for the existing model. In this post, I will motivate the hijacking of the model for my own purposes, for macroevolution :-) That’s the great thing about models, by relabeling the ingredients something else, as long as the relationship between elements of the model holds true, we may suddenly have an entirley fresh insight about the world!
As we might recall, the holey adaptive landscapes model proposes that phenotypes are points in n-dimensional space, where n is a very large number. If each phenotype was a node, then mutations that change one phenotype to another can be considered edges. In this space for large enough n, even nodes with very rare properties can percolate and form a giant connected cluster. Gavrilets, one of the original authors of the model and its main proponent, considered this rare property to be high fitness. This way, evolution never has to cross fitness valleys. However, I find this unlikely; high fitness within a particular environment is a very rare property. If the property is sufficiently rare, then even if the nodes form a giant connected cluster, if the connections between nodes are sufficiently tenuous, then there is not enough time and population size for their exploration.
Think of a blind fruitfly banging its head within a sphere. On the wall of the sphere is pricked a single tiny hole just large enough for the fruitfly to escape. How much time will it take for the fruitfly to escape? The question clearly depends on the size of the sphere. In n-dimensions, where n is large, the sphere is awfully big. Now consider the sphere to be a single highly fit phenotype, and a hole is an edge to another highly fit phenotype. The existence of the hole is not sufficient to guarantee that the fruitfly will find the exit in finite time. In fact, even a giant pack of fruitflies — all the fruitflies that ever existed — may not be able to find it, given all the time that life has evolved on Earth. That’s how incredibly large the sphere is — the exit must not only exist, it must be sufficiently common.
The goal of this post is to detail why I’m so interested in the holey adaptive landscapes model. I’m interested in its property of historicity, the capacity to be contingent and irreversible. I will define these terms more carefully later. Gavrilets has noted this in his book, but I can find no insightful exploration of this potential. I hope this model can formalize some intuitions gained in complex adaptive systems, particularly those of evo-devo and my personal favorite pseudo-philosophical theory, generative entrenchment (also here and here). Gould had a great instinct for this when he argued for the historical contingency of evolutionary process (consider his work on gastropods, for example, or his long rants on historicity in his magnus opus — although I despair to pick out a specific section).
Before I go on, I must also rant. Gould’s contingency means that evolution is sensitive to initial conditions, yes, this does not mean history is chaotic, in the mathematical sense of chaos. Chaos is not the only way for a system to be sensitive to initial conditions, in fact, mathematical chaos is preeminently ahistorical — just like equilibriating systems, chaotic systems forget history in a hurry, in total contrast to what Gould meant, which is that history should leave an indelible imprint on all of future. No matter what the initial condition, chaotic systems settle in the same strange attractor, the same distribution over phase space. The exact trajectory depends on initial condition, yes, but because the smallest possible difference in initial condition quickly translates to a completely different trajectory, it means that no matter how you begin, you future is… chaotic. Consider two trajectories that began with very differently, the future difference between those two trajectories is no greater than two trajectories that began with the slightest possible difference. The difference in the difference of initial conditions is quickly obliviated by time. Whatever the atheist humanists say, chaos gives no more hope to free will than quantum mechanics. Lots of people seem to have gone down this hopeless route, not least of which is Michael Shermer, who, among many different places, writes here:
And as chaos and complexity theory have shown, small changes early in a historical sequence can trigger enormous changes later… the question is: what type of change will be triggered by human actions, and in what direction will it go?
If he’s drawing this conclusion from chaos theory, then the answer to his question is… we don’t know, we can’t possibly have an idea, and it doesn’t matter what we do, since all trajectories are statistically the same. If he’s drawing this conclusion from “complexity theory” — not yet a single theory with core results of any sort, then it’s a theory entirely unknown to me.
No, interesting historical contingency is quite different, we will see if the holey landscapes model can more accurately capture its essence.
Things in the holey landscapes model get generally better if we consider the rare property to be viability, instead of high fitness. In fact, Gavrilets mixes use of “viable” and highly fit, although I suspect him to always mean the latter. By viable, I mean that the phenotype is capable of reproduction in some environment, but I don’t care how well it reproduces. For ease of discussion, let’s say that viable phenotypes also reproduce above the error threshold, and there exist an environment where it is able to reproduce with absolute fitness >1. Else, it’s doomed to extinction in all environments, and then it’s not very viable, is it?
It turns out that the resultant model contains an interesting form of irreversibility. I will give the flavor here, while spending the next post being more technical. Consider our poor blind fruitfly, banging its head against the sphere. Because we consider viability instead of “high fitness”, there are now lots of potential holes in the sphere. Each potential hole is a neighboring viable phenotype, but the hole is opened or closed by the environment, which dictates whether that neighboring viable phenotype is fit.
Aha, an astute reader might say, but this is no better than Gavrilets’ basic model. The number of open holes at any point must be very small, since it’s also subject to the double filter of viability and high fitness. How can we find the open holes?
The difference is that after an environmental change, the sphere we are currently in might be very unfit. Thus, the second filter — high fitness — is much less constrictive, since it merely has to be fitter than the current sphere, which might be on the verge of extinction. A large porportion of viable phenotypes may be “open” holes, as opposed to the basic model, where only the highly fit phenotypes are open. Among viable phenotypes, highly fit ones may be rare, but those that are somewhat more fit than an exceedingly unfit phenotype may be much more common — and it’s only a matter of time, often fairly short time on the geological scale, before any phenotype is rendered exceedingly unfit. So you see, in this model evolution also did not have to cross a fitness valley, but I’m using a much more classical mechanism — peak shifts due to environmental change, rather than a percolation cluster of highly-fit phenotypes.
Now that our happier fruitfly is in the neighboring sphere, what is the chance that it will return to its previous sphere, as opposed to choosing some other neighbor? The answer is… very low. The probability of finding any particular hole has not improved, and verges on impossibility; although the probability of finding some hole is much better. Moreover, the particular hole that the fruitfly went through dictate what holes it will have access to next — and if it can’t return to its previous sphere, then it can’t go back to remake the choice. This gives the possibility of much more interesting contingency than mere chaos.
This mechanism has much in common with Muller’s Ratchet, or Dollo’s Law, and is an attempt to generalize the ratchet mechanism while formalizing what we mean, exactly, by irreversibility. I will tighten the argument next Thursday.
## Criticisms of holey adaptive landscapes
:-) My cats say hello.
In my last post I wrote about holey adaptive landscapes, a model of evolution in very high dimensional space where there is no need to jump across fitness valleys. The idea is that if we consider the phenotype of organisms to be points in n-dimensional space, where n is some large number (say, tens of thousands, as in the number of our genes), then high fitness phenotypes easily percolate even if they are rare. By percolate, I mean that most high-fitness phenotypes are connected in one giant component, so evolution from one high fitness “peak” to another does not involve crossing a valley, rather, there are only high fitness ridges that are well-connected. This is why the model is consider “holey”, highly fit phenotypes are entirely connected within the fitness landscape, but they run around large “holes” of poorly fit phenotypes that seem to be carved out from the landscape.
This is possible because as n (the number of dimensions) increases, the number of possible mutants that any phenotype can become also increases. The actual rate of increase depends on the model of mutation and can be linear, if we consider n to be the number of genes, or exponential, if we consider n to be the number of independent but continuous traits that characterize an organism. Once the number of possible mutants become so large that all highly fit phenotypes have, on average, another highly fit phenotype as neighbor, then percolation is assured.
More formally, we can consider the basic model where highly fit phenotypes are randomly distributed over phenotype space: any phenotype has probability $p_\omega$ of being highly fit. Let $S_m$ be the average size of the set of all mutants of any phenotype. For example, if n is the number of genes and the only mutations we consider are the loss-of-function of single genes, then $S_m$ is simply n, since this is the number of genes that can be lost, and therefore is the number of possible mutants. Percolation is reached if $S_m>\dfrac{1}{p_\omega}$. Later extensions also consider cases where highly fit phenotypes exist in clusters and showed that percolation is still easily achievable (Gavrilets’ book Origin of Species, Gravner et al.)
I have several criticisms of the basic model. As an aside, I find criticism to be the best way we can honor any line of work, it means we see a potential worthy of a great deal of thought and improvement. I’ll list my criticisms in the following:
1) We have not a clue what $p_\omega$ is, not the crudest ball-park idea. To grapple with this question, we must understand what makes an admissible phenotype. For example, we certainly should not consider any combination of atoms to be a phenotype. The proper way to define an admissible phenotypes is by defining the possible operations (mutations) that move us from one phenotype to another, that is, we must define what is a mutation. If only DNA mutations are admissible operations, and if the identical DNA string produces the same phenotype in all environments (both risible assumptions, but let’s start here), then the space of all admissible phenotypes are all possible strings of DNA. Let us consider only genomes of a billion letters in length. This space is, of course, $4^{10^9}$. What fraction of these combinations are highly fit? The answers must be a truly ridiculously small number. So small that if $S_m\approx O(n)$, I would imagine that there is no way that highly fit phenotypes reach percolation.
Now, if $S_m\approx O(a^n)$, that is a wholly different matter altogether. For example, Gravner et al. argued that $a\approx 2$ for continuous traits in a simple model. If n is in the tens of thousands, my intuition tells me it’s possible that higly fit phenotypes reach percolation, since exponentials make really-really-really big numbers really quickly. Despite well known evidence that humans really are terrible intuiters at giant and tiny numbers, the absence of fitness valleys becomes at least plausible. But… it might not matter, because:
2) Populations have finite size, and evolution moves in finite time. Thus, the number of possible mutants that any phenotype will in fact explore is linear in population size and time (even if those that it can potentially explore is much larger). Even if the number of mutants, $S_m$ grows exponentially with n, it doesn’t matter if we never have enough population or time to explore that giant number of mutants. Thus, it doesn’t matter that highly fit phenotypes form a percolating cluster, if the ridges that connect peaks aren’t thick enough to be discovered. Not only must there be highly-fit neighbors, but in order for evolution to never have to cross fitness valleys, highly-fit neighbors must be common enough to be discovered. Else, if everything populations realistically discover are low fitness, then evolution has to cross fitness valleys anyway.
How much time and population is realistic? Let’s consider bacteria, which number in the $5\times 10^{30}$. In terms of generation time, let’s say they divide once every twenty minutes, the standard optimal laboratory doubling time for E. Coli. Most bacteria in natural conditions have much slower generation time. Then if bacteria evolved 4.5 billion years ago, we have had approximately 118260000000000, or ~$1.2\times 10^{14}$ generations. The total number of bacteria sampled across all evolution is therefore on the order of $6\times 10^{44}$. Does that sound like a large number? Because it’s not. That’s the trouble with linear growth. Against $4^{10^9}$, this is nothing. Even against $2^{10000}$ (where we consider $10000$ to be n, the dimension number), $6\times 10^{44}$ is nothing. That is, we simply don’t have time to test all the mutants. Highly fit phenotypes better make up more than $\dfrac{1}{6\times 10^{44}}$ of the phenotype space, else we’ll never discover it. Is $\dfrac{1}{6\times 10^{44}}$ small? Yes. Is it small enough? I’m not sure. Possibly not. In any case, this is the proper number to consider, not, say, $2^{10000}$. The fact that $S_m\approx O(a^n)$ is so large is a moot point.
3) My last criticism I consider the most difficult one for the model to answer. The holey adaptive landscapes model does not take into account environmental variation. To a great extent, it confuses the viable with the highly fit. In his book, Gavrilets often use the term “viable”, but if we use the usual definition of viable — that is, capable of reproduction, then clearly most viable phenotypes are not highly fit. Different viable phenotypes might be highly fit under different environmental conditions, but fitness itself has little meaning outside of a particular environment.
A straightforward inclusion of environmental conditions into this model is not easy. Let us consider the basic model to apply to viable phenotypes, that is, strings of DNA that are capable of reproduction, under some environment. Let us say that all that Gavrilets et al. has to say are correct with respect to viable phenotypes, that they form a percolating cluster, etc. Now, in a particular environment, these viable phenotypes will have different fitness. If we further consider only the highly fit phenotypes within a certain environment, for these highly fit phenotypes to form a percolating cluster, it would mean we would have to apply the reasoning of the model a second time. It would mean that all viable phenotypes must be connected to so many other viable phenotypes that among them would be another highly fit phenotype. Here, we take “highly fit” to be those viable phenotypes that have relative fitness greater than $1-\epsilon$, where the fittest phenotype has relative fitness $1$. This further dramatizes the inability of evolution to strike on “highly fit” phenotypes through a single mutation in realistic population size and time, since we must consider not $p_\omega$, but $p_v\times p_\omega$, where $p_v$ is the probability of being viable and $p_\omega$ is the probability of being highly fit. Both of these probabilities are almost certainly astronomically small, making the burden on the impoverishingly small number of $6\times 10^{44}$ even heavier.
It’s my belief, then, that in realistic evolution with finite population and time, fitness valleys nevertheless have to be crossed. Eithere there are no highly fit phenotypes a single mutation away, or if such mutations exist, then the space of all possible mutations is so large as to be impossible to fully sample with finite population and time. The old problem of having to cross fitness valleys is not entirely circumvented by the holey adaptive landscapes approach.
Next Thursday, I will seek to hijack this model for my own uses, as a model of macroevolution.
Hello world :-)
My research interests has veered off pure EGT, but my questions still center around evolution — particularly the evolution of complex systems that are made up of many small components working in unison. In particular, I’ve been studying Gavrilets et al. ‘s model of holey fitness landscapes, I think it’s a model with great potential for studying macroevolution, or evolution on very long timescales. I’m not the first one to this idea, of course — Arnold and many others have seen the possible connection also, although I think of it in a rather different light.
In this first post, I will give a short summary of this model, cobbled together from several papers and Gavrilets’ book, the Origin of Species. The basic premise is that organisms can be characterized by a large number of traits. When we say large, we mean very large — thousands or so. Gavrilets envisions this as being the number of genes in an organism, so tens of thousands. The important thing is that each of these traits can change independently of other ones.
The idea that organisms are points in very high dimensional space is not new, Fisher had it in his 1930 classic Genetical Theory of Natural Selection, where he used this insight to argue for micromutationism — in such high dimensional space, most mutations of appreciable size are detrimental, so Fisher argued that most mutations must be small (this result was later corrected by Kimura, Orr and others, who argued that most mutations must be of intermediate size, since tiny mutations are unlikely to fix in large populations).
However, even Fisher didn’t see another consequence of high-dimensional space, which Gavrilets exploited mercilessly. The consequence is that in high-enough dimensional space, there is no need to cross fitness valleys to move between one high fitness phenotype to another; all high fitness genotypes are connected. This is because connectivity is exceedingly easy in high dimensional space. Consider two dimensions, to get from one point to another, there are only two directions to move in. Every extra dimension offers a new option for such movement, that’s why there’s a minimum dimensionality to chaotic behavior — we can’t embed a strange attractor in a two dimensional phase plane, since trajectories can’t help but cross each other. Three dimensions is better, but n-dimensional space, where n is in the tens of thousands — that’s really powerful stuff.
Basically, every phenotype — every point in n-D space, is connected to a huge number of other points in n-D space. That is, every phenotype has a huge number of neighbors. Even if the probability of being a highly fit organism is exceedingly small, chances are high that one would exist among this huge number of neighbors. We know that if each highly fit phenotype is, on average, connected to another highly fit phenotype (via mutation), then the percolation threshold is reached where almost all highly fit phenotypes are connected in one giant connected component. In this way, evolution does not have to traverse fitness minima.
If we consider mutations to be point mutations of genes, then mutations can be considered to be a Manhattan distance type walk in n-D space. That’s just a fancy way of saying that we have n genes, and only one can be changed at a time. In that case, the number of neighbors any phenotype has is n, and if the probability of being highly fit is better than 1/n, then highly fit organisms are connected. This is even easier if we consider mutations to be random movements in n-D space. That is, if we consider an organism to be characterized by $\mathbf{p}=(p_1, p_2, ... p_n)$, where $p_i$ is the $i^{th}$ trait, and a mutation from $\mathbf{p}$ results in $\mathbf{p_m}=(p_1+\epsilon_1, ... p_n+\epsilon_n$), such that $\epsilon_i$ is a random small number that can be negative, and the Euclidean distance between $\mathbf{p_m}$ and $\mathbf{p}$ is less than $\delta$, where $\delta$ is the maximum mutation size, then the neighbors of $\mathbf{p}$ fill up the volume of a ball of radius $\delta$ around $\mathbf{p}$. The volume of this ball grows exponentially with n, so even a tiny probability of being highly fit will find some neighbor of $\mathbf{p}$ that is highly fit, because of the extremely large volume even for reasonably sized n.
The fact that evolution may never have to cross fitness minima is extremely important, it means that most of evolution may take place on “neutral bands”. Hartl and Taube had foreseen this really interesting result. Gavrilets mainly used this result to argue for speciation, which he envisions as a process that takes place naturally with reproductive isolation and has no need for natural selection.
Several improvements over the basic result have been achieved, mostly in the realm of showing that even if highly fit phenotypes are highly correlated (forming “highly fit islands” in phenotype space), the basic result of connectivity nevertheless holds (i.e. there will be bridges between those islands). Gavrilets’ book summarizes some early results, but a more recent paper (Gravner et al.) is a real tour-de-force in this direction. Their last result shows that the existence of “incompatibility sets”, that is, sets of traits that destroy viability, nevertheless does not punch enough holes in n-D space to disconnect it. Overall, the paper shows that even with correlation, percolation (connectedness of almost all highly fit phenotypes) is still the norm.
Next Thursday, I will detail some of my own criticisms to this model and its interpretation. The week after next, I will hijack this model for my own purposes and I will attempt to show that such a model can display a great deal of historical contingency, leading to irreversible, Muller’s Ratchet type evolution that carries on in particular directions even against fitness considerations. This type of model, I believe, will provide an interesting bridge between micro and macroevolution.
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# function table calculator
There are two kinds of tables you can create: an automatically generated table and a user-generated table. It is common to want to apply a function to a variety/sequence/list of values, such as an Excel column. y = x x2 − 6x + 8. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). How to Use a Function Calculator? A variable is a value (usually a number) that you don't know or that could be a bunch of different numbers. （input by clicking each white cell in the table below）. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. This calculator uses provided target function table data in the form of points {x, f (x)} to build several regression models, namely: linear regression, quadratic regression, cubic regression, power regression, logarithmic regression, hyperbolic regression, ab-exponential regression and exponential regression. Example input. These are just the $$x$$ and $$y$$ values that are true for the given line. Our limit calculator with steps helps users to save their time while doing manual calculations. RapidTables. For example, let's say you just took a test. （ex. Interactive, free online graphing calculator from GeoGebra: graph functions, plot data, drag sliders, and much more! TI-84: Using Tables TI-84 Video: ... Go to: [Y=] to input a second function. Practice, practice, practice. If you're using a function, make sure it's taking the set as an input. L’expression utilisée comme premier paramètre équivaut essentiellement à une mesure.The expression used as the first parameter is essentially the same as a measure. Thank you for your questionnaire.Sending completion. matrix calculator Add, multiply, transpose matrices and more. $f\left (x\right)=\cos\left (2x+5\right)$. Home›Calculators›Math Calculators› Tangent calculator Tangent Calculator. The essential thing to know about these particular functions is that they can change the context of a calculation. Some functions are limited now because setting of JAVASCRIPT of the browser is OFF. You can also choose from calendar function rule calculator, as well as from plastic function rule calculator, and whether function rule calculator is scientific, graphic calculator, or general purpose calculator. Function is represented as Y = F(A,B,C,) where A,B,C, are inputs and the Y is the output. Mode: . This graphing calculator provides functionality similar to handheld graphing calculators like the TI-83/TI-84 in a convenient web-based format. 3*x+9）. If the function is one-to-one, there will be a unique inverse. However, a Boolean expression can use any function that looks up a single value, or that calculates a scalar value. In other words, a table of values is simply some of the points that are on the line. In your case, x = 83. Type in any function above then use the table below to input any value to determine the output: Use this free online constant of variation calculator to generate equation based on the given x and y values. Limit calculator is an online tools which is developed by Calculatored to make these calculations easy. f(x) is inputed as "expression". A data table is a range of cells in which you can change values in some of the cells and come up with different answers to a problem. Operations on Functions - Graphing Calculator Input two functions f and g and carry out operations such as adding, subtracting, multiplying, dividing and composing functions. An electronic calculator is typically a portable electronic device used to perform calculations, ranging from basic arithmetic to complex mathematics.. Wolfram|Alpha is a great tool for finding the domain and range of a function. Use the following guidelines to enter functions into the calculator. In economics, a production function represents the relationship between the output and the combination of factors, or inputs, used to obtain it. Calculate Reset: Result: * Use e for scientific notation. Search for additional learning materials, such as related worksheets and video tutorials. For example, if you have "x_1" and "2x_1+1" as your headings, we'll fill … It is used to find the area between z = 0 and any positive value, and reference the area to the right-hand side of the standard deviation curve. Table calculation functions allow you to perform computations on values in a table. The procedure to use the function calculator is as follows: Step 1: Enter the function f(x) in the given input field. Inverse Function Calculator. More than just an online function properties finder. When used as filters in CALCULATE, ALLxxx functions might display unexpected behaviors. Since, as we just wrote, every linear equation is a relationship of x and y values, we can create a table of values for any line. Change your answer find from a table, choose your platform! The calculator will find the average rate of change of the given function on the given interval, with steps shown. L’expression utilisée comme premier paramètre doit être une fonction qui retourne une table.The expression used as the first parameter must be a function that returns a table. The function approximation problem is how to select a function among a well-defined class that closely matches ("approximates") a target unknown function. Engineering functions; Complex number functions; Matrix functions; Statistical functions; Statistical functions on variables; Statistical test functions; Probability distribution functions; Date and time functions; Logical functions; String functions; Miscellaneous functions; Appendices. When you enter a function, the calculator will begin by expanding (simplifying) it. How to Use a Function Calculator? variable data table. Print the values of the table index while the table is being generated: Monitor the values by showing them in a temporary cell: Relations to Other Functions (5) Thank you for your questionnaire.Sending completion. Limit function belongs to difficult concepts of mathematics. Then hit the options button and click the table symbol that appears over the pair. $f\left (x\right)=\sqrt {x+3}$. tan -1: Calculate Reset: Degrees: ° Radians: rad: Tangent table. The expression c… How can I import table in excel and calculate each cells by a function I've write? This is also called as direct proportion and constant of variation (k). Expanded tables in DAX. Par exemple, vous pouvez calculer le pourcentage du total que représente une vente individuelle pour l'année, ou pour plusieurs années. GraphCalc is the best free online graphing calculator that almost completely replaces the TI 83 and TI 84 plus calculators. Then you pair them up as you would with coordinates. Notice that neither function is the example above, the value for this story, choose your platform! Fonctions de calcul de table disponibles dans Tableau Online Domain and Range Calculator Find the domain and range of a function with Wolfram|Alpha. Step 3: The graph of the function will be displayed in a new window. 5. Cobb-Douglas Production Function Calculator helps calculating the quantity of products, the marginal product of Labor and the marginal product of capital, given Cobb-Douglas Production Function. Add and . In algebra, a function is defined as an equation where it assigns exactly one output for the every given specified … You start by defining your set "b" along with a function or another list. Table 8-1 lists each Calculator function with its keyboard equivalent. The expression cannot reference a measure. Clicking on the sliders until you guess is available for this table, what would you get the button. Some functions are limited now because setting of JAVASCRIPT of the browser is OFF. The expression cannot use any function that scans a table or returns a table, including aggregation functions. Show Instructions. The expression cannot use a nested CALCULATE function. The calculator will find the domain, range, x-intercepts, y-intercepts, derivative, integral, asymptotes, intervals of increase and decrease, critical points, extrema (minimum and maximum, local, absolute, and global) points, intervals of concavity, inflection points, limit, Taylor polynomial, and graph of the single variable function. They are very similar in some aspects but also very different in others. Function Calculator A Function Calculator is a free online tool that displays the graph of the given function. Calculate the value of using each value in the relation and compare this value to the given value in the relation. Whats people lookup in this blog: Find The Function Rule For Input Output Table Calculator. （ex. Your feedback and comments may be posted as customer voice. What are the Functions? Trigonometry Calculator. To recall, an inverse function is a function which can reverse another function. I want to go through these with the accompanying tutorials during this post. f x = x 2 + x − 4. If you are familiar with graphing algebraic equations, then you are familiar with the concepts of the horizontal X-Axis and the Vertical Y-Axis. More than just an online function properties finder. Variables. 3. A good example of a data table employs the PMT function with different loan amounts and interest rates to calculate the affordable amount on a home mortgage loan. Graph any equation, find its intersections, create a table of values. The calculator will find the inverse of the given function, with steps shown. See the manual page for calctool for a more complete description of each function. It is also called an anti function. 4. For your friend, x might equal 97. Inverse tangent calculator. Exponents There is always a pattern in the way input values (x) and the output values (y) are related which is given by the function rule. Wolfram|Alpha is a great tool for finding the domain and range of a function. [1] 2020/11/10 23:53 Female / Under 20 years old / Elementary school/ Junior high-school student / Not at All /, [2] 2020/10/09 04:42 Male / Under 20 years old / High-school/ University/ Grad student / Useful /, [3] 2020/05/23 00:52 Male / 20 years old level / - / Very /, [4] 2019/04/16 11:13 Male / Under 20 years old / Elementary school/ Junior high-school student / Not at All /, [5] 2019/01/18 12:13 Male / Under 20 years old / High-school/ University/ Grad student / A little /, [6] 2018/10/16 09:16 Female / Under 20 years old / Elementary school/ Junior high-school student / A little /, [7] 2018/09/29 09:23 Female / Under 20 years old / Elementary school/ Junior high-school student / Not at All /, [8] 2018/04/20 06:39 Male / Under 20 years old / Elementary school/ Junior high-school student / Useful /, [9] 2018/03/12 16:19 Male / 20 years old level / High-school/ University/ Grad student / A little /, [10] 2017/11/09 10:21 Male / Under 20 years old / Elementary school/ Junior high-school student / A little /. 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# Edexcel A Level Further Maths: Core Pure:复习笔记3.3.1 Maclaurin Series
### Maclaurin Series
#### What is a Maclaurin Series?
• A Maclaurin series is a way of representing a function as an infinite sum of increasing integer powers of etc.)
• If all of the infinite number of terms are included, then the Maclaurin series is exactly equal to the original function
• If we truncate (i.e., shorten) the Maclaurin series by stopping at some particular power of , then the Maclaurin series is only an approximation of the original function
• A truncated Maclaurin series will always be exactly equal to the original function for
• In general, the approximation from a truncated Maclaurin series becomes less accurate as the value of moves further away from zero
• The accuracy of a truncated Maclaurin series approximation can be improved by including more terms from the complete infinite series
• So, for example, a series truncated at the term will give a more accurate approximation than a series truncated at the term
#### How do I find the Maclaurin series of a function ‘from first principles’?
• Use the general Maclaurin series formula
• This formula is in your exam formula booklet
• STEP 1: Find the values of etc. for the function
• An exam question will specify how many terms of the series you need to calculate (for example, “up to and including the term in ”)
• You may be able to use your calculator to find these values directly without actually having to find all the necessary derivatives of the function first
• STEP 2: Put the values from Step 1 into the general Maclaurin series formula
• STEP 3: Simplify the coefficients as far as possible for each of the powers of
#### Is there a connection Maclaurin series expansions and binomial theorem series expansions?
• Yes there is!
• For a function like the binomial theorem series expansion is exactly the same as the Maclaurin series expansion for the same function
• So unless a question specifically tells you to use the general Maclaurin series formula, you can use the binomial theorem to find the Maclaurin series for functions of that type
• Or if you’ve forgotten the binomial series expansion formula for where is not a positive integer, you can find the binomial theorem expansion by using the general Maclaurin series formula to find the Maclaurin series expansion
#### Worked Example
Use the Maclaurin series formula to find the Maclaurin series for up to and including the term in .
### Maclaurin Series of Standard Functions
#### Is there an easier way to find the Maclaurin series for standard functions?
• Yes there is!
• The following Maclaurin series expansions of standard functions are contained in your exam formula booklet:
• Unless a question specifically asks you to derive a Maclaurin series using the general Maclaurin series formula, you can use those standard formulae from the exam formula booklet in your working
### Maclaurin Series of Compound Functions
#### How can I find the Maclaurin series for a composite function?
• A composite function is a ‘function of a function’ or a ‘function within a function’
• For example sin(2x) is a composite function, with 2x as the ‘inside function’ which has been put into the simpler ‘outside function’ sin x
• Similarly is a composite function, with as the ‘inside function’ and as the ‘outside function’
• To find the Maclaurin series for a composite function:
• STEP 1: Start with the Maclaurin series for the basic ‘outside function’
• Usually this will be one of the ‘standard functions’ whose Maclaurin series are given in the exam formula booklet
• STEP 2: Substitute the ‘inside function’ every place that x appears in the Maclaurin series for the ‘outside function’
• So for sin(2x), for example, you would substitute 2x everywhere that x appears in the Maclaurin series for sin x
• STEP 3: Expand the brackets and simplify the coefficients for the powers of x in the resultant Maclaurin series
• This method can theoretically be used for quite complicated ‘inside’ and ‘outside’ functions
• On your exam, however, the ‘inside function’ will usually not be more complicated than something like kx (for some constant k) or xn (for some constant power n)
#### How can I find the Maclaurin series for a product of two functions?
• To find the Maclaurin series for a product of two functions:
• STEP 1: Start with the Maclaurin series of the individual functions
• For each of these Maclaurin series you should only use terms up to an appropriately chosen power of x (see the worked example below to see how this is done!)
• STEP 2: Put each of the series into brackets and multiply them together
• Only keep terms in powers of x up to the power you are interested in
• STEP 3: Collect terms and simplify coefficients for the powers of x in the resultant Maclaurin series
#### Worked Example
a) Find the Maclaurin series for the function , up to and including the term in .
b) Find the Maclaurin series for the function , up to and including the term in .
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• Development of indigenous silicon detector and readout electronics for forward electromagnetic calorimeter prototypes
• Fulltext
https://www.ias.ac.in/article/fulltext/pram/096/0177
• Keywords
Forward calorimeter; front-end electronics ASIC; ANUSANSKAR, ANUINDRA; large area silicon detector; small-x physics.
• Abstract
ALICE is a high-energy physics experiment at the CERN Large Hadron Collider (LHC) in Geneva, to study the properties of quark-gluon plasma, formed in heavy-ion collisions. In the upgrade program, a new electromagnetic forward calorimeter (FOCAL) based on silicon and tungsten (Si+W) sampling configuration, has been proposed to address new physics objectives of ALICE experiment.The high luminosity environment of theLHC and the requirement of compact design for the FOCAL, put stringent requirements for the development of silicon detector and front-end electronics (FEE). The ALICE-India collaboration made the first proposal for the FOCAL in the year 2008, and initiated indigenous R&D activities for silicon detectors, FEE and instrumentation. As a result,large area (40cm$^2$), segmented silicon pad sensors (6.3 cm×6.3 cm with 36 pads) and two different FEE applicationspecific integrated circuits (ASICs), ANUSANSKAR and ANUINDRA respectively, were developed. Progressive FOCAL prototypes were built using these sensors and ASICs, which have undergone test beam validation in a phased manner achieving the desired calorimetric performances, in line with the GEANT-based performancesimulations. Further, the results of the test beam experiments provided crucial inputs for achieving the technical specifications of the silicon detector and readout electronics for the final calorimeter. This manuscript presents, in a chronological order, India’s indigenous FEE ASICs, sensor and instrumentation development and reports the subsequently improved readout methodology for the prototype FOCAL detector.
• Author Affiliations
1. Bhabha Atomic Research Centre, Mumbai 400 085, India
2. Homi Bhabha National Institute, Mumbai 400 094, India
3. Variable Energy Cyclotron Centre, Kolkata 700 064, India
4. National Institute of Science Education and Research, Jatni 752 050, India
5. CERN, Geneva 23, Switzerland
• Pramana – Journal of Physics
Volume 97, 2023
All articles
Continuous Article Publishing mode
• Editorial Note on Continuous Article Publication
Posted on July 25, 2019
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# Braille code probabilities homework
1. Sep 14, 2007
Hello everyone im' stuck on this problem.
It says:
Each symbol in braille code is represened by a rectangular arrangement of six dots. Given that a least 1 dot of the 6 must be raised, how many symbols can be represented in brail?
now i saw this posted somewhere else, they got 63. don't know how.
i got 63 two ways....
2^6-1 (case where all down)=63
6c1+6c2+6c3+6c4+6c5+6c6 -1 also =63 (where 6c3 etc is combinations...6 options choose 3)
can sumone explain how/why that works please.
also part b...how many combinations have EXACTLY 3 raised
and how many have an even number of raised dots
(this is for math and computer science course so not TOO sure which to post it in thanks)
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
2. Sep 14, 2007
### EnumaElish
Think of "raised " as "included in a subset," and "flat" as "not included in a subset." Then "all flat" is the empty set. Number of subsets = 2^6 (including empty set).
How many combinations with exactly 3 raised = number of subsets with 3 elements.
How many with even raised = number of subsets with 2 or 4 or 6 elements.
3. Sep 14, 2007
so 2^6 is a? (-1 for original case where non are raised?)
then for b...2^3...so 8?
or do i need to figure out how many have more or less than 3 an subtract it?
4. Sep 14, 2007
### Dick
The latter. Figure out how many have 0,1 and 2 raised and subtract the total from 2^6.
5. Sep 15, 2007
### HallsofIvy
Staff Emeritus
Another way of getting that answer is to argue that each of the 6 dots can be raised or not- a total of 6 "raised" or "not raised" choices. By the counting principal, there would be a total of 26= 64 possiblities. That of course includes "none raised" which is not allowed so there are 64- 1= 63 allowed. That is the same as your sum of binomial coefficients because you are now arguing that it is "number with exactly one dot raised"+ "number with exactly two dots" raised"+ etc. "The number of 6 dots with exactly n raised" is the same as "how many ways can I choose n out of the 6 dots to raise": 6Cn. And, of course, that is 26- 1 because you did not include 6C1= 1. These are binomial coefficients- the coefficients in the expansion of (x+ y)6. The sum of the coefficients is that with x=y= 1: (1+1)6= 26 (and again leaving out the first subtracts 1 from the whole thing).
As both Enuma Elish and Dick have said- to get "how many have at least three dots raised" Calculate 6C1 and 6C2 (which you may already have done it you calculated that sum of binomial coefficients directly) and subtract from the 63 you already had. (Dick included 6C0 and told you to subtract from 64. Same thing of course.)
Last edited: Sep 15, 2007
6. Sep 15, 2007
### EnumaElish
I have just realized that the OP is an exact replica of the OP in another thread in this forum, created simultaneously with this one.
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J Plant Ecol ›› 2015, Vol. 8 ›› Issue (4): 401-410.
• Research Articles •
### Shifts in community leaf functional traits are related to litter decomposition along a secondary forest succession series in subtropical China
David Eichenberg1,*, Stefan Trogisch1,2, Yuanyuan Huang3, Jin-Sheng He3 and Helge Bruelheide1,2
1. 1 Department for Geobotany, Institute of Biology/Geobotany and Botanical Garden, Martin-Luther-University Halle-Wittenberg, Am Kirchtor 1, Halle(Saale) D-06108, Germany; 2 German Centre for Integrative Biodiversity Research (iDiv), Halle-Jena-Leipzig, Deutscher Platz 5e, Leipzig D-04103, Germany; 3 Department of Ecology, College of Urban and Environmental Sciences and Key Laboratory for Earth Surface Processes of the Ministry of Education, Peking University, 5 Yiheyuan Road, Beijing 100871, China
• Received:2013-09-19 Accepted:2014-08-23 Published:2015-07-24
• Contact: Eichenberg, David
Abstract: Aims We investigated shifts in community-weighted mean traits (CWM) of 14 leaf functional traits along a secondary successional series in an evergreen broadleaf forest in subtropical southeast China. Most of the investigated traits have been reported to affect litter decomposition in previous studies. We asked whether changes in CWMs along secondary succession followed similar patterns for all investigated traits and whether the shifts in CWM indicated a change in resource use strategy along the successional gradient. Using community decomposition rates (k -rates) estimated from annual litter production and standing litter biomass, we asked whether the dynamics of litter decomposition were related to changes in leaf functional traits along the successional series.
Methods Twenty-seven plots were examined for shifts in leaf CWM traits as well as in k -rates along a series of secondary forest succession covered in the framework of the BEF-China project. We investigated whether the changes in CWMs followed similar patterns for all traits with ongoing succession. Three alternative linear models were used to reveal the general patterns of shifts in CWM trait values. Moreover, multiple regression analysis was applied to investigate whether there were causal relationships between the changes in leaf functional traits and the dynamics of litter decomposition along secondary succession. We furthermore assessed which traits had the highest impact on community litter decomposition.
Important findings Shifts in CWM values generally followed logarithmic patterns for all investigated traits, whereas community k -rates remained stable along the successional gradient. In summary, the shifts in CWM values indicate a change in community resource use strategy from high nutrient acquisition to nutrient retention with ongoing succession. Stands with higher CWM values of traits related to nutrient acquisition had also higher CWM values of traits related to chemical resistance, whereas stands with higher CWM values of traits related to nutrient retention exhibited higher CWM values in leaf physical defense. Moreover, high values in CWM values related to nutritional quality (such as high leaf phosphorus concentrations) were found to promote community k -rates, whereas high values in physical or chemical defense traits (such as high contents in polyphenols or high leaf toughness) decreased litter decomposition rates. In consequence, litter decomposition, which was simultaneously affected by these characteristics, did not change significantly along succession. Our findings show that leaf decomposition within the investigated communities is dependent on the interplay of several traits and is a result from interactions of traits that affect decomposition in opposing directions.
Aims We investigated shifts in community-weighted mean traits (CWM) of 14 leaf functional traits along a secondary successional series in an evergreen broadleaf forest in subtropical southeast China. Most of the investigated traits have been reported to affect litter decomposition in previous studies. We asked whether changes in CWMs along secondary succession followed similar patterns for all investigated traits and whether the shifts in CWM indicated a change in resource use strategy along the successional gradient. Using community decomposition rates (k -rates) estimated from annual litter production and standing litter biomass, we asked whether the dynamics of litter decomposition were related to changes in leaf functional traits along the successional series.
Methods Twenty-seven plots were examined for shifts in leaf CWM traits as well as in k -rates along a series of secondary forest succession covered in the framework of the BEF-China project. We investigated whether the changes in CWMs followed similar patterns for all traits with ongoing succession. Three alternative linear models were used to reveal the general patterns of shifts in CWM trait values. Moreover, multiple regression analysis was applied to investigate whether there were causal relationships between the changes in leaf functional traits and the dynamics of litter decomposition along secondary succession. We furthermore assessed which traits had the highest impact on community litter decomposition.
Important findings Shifts in CWM values generally followed logarithmic patterns for all investigated traits, whereas community k -rates remained stable along the successional gradient. In summary, the shifts in CWM values indicate a change in community resource use strategy from high nutrient acquisition to nutrient retention with ongoing succession. Stands with higher CWM values of traits related to nutrient acquisition had also higher CWM values of traits related to chemical resistance, whereas stands with higher CWM values of traits related to nutrient retention exhibited higher CWM values in leaf physical defense. Moreover, high values in CWM values related to nutritional quality (such as high leaf phosphorus concentrations) were found to promote community k -rates, whereas high values in physical or chemical defense traits (such as high contents in polyphenols or high leaf toughness) decreased litter decomposition rates. In consequence, litter decomposition, which was simultaneously affected by these characteristics, did not change significantly along succession. Our findings show that leaf decomposition within the investigated communities is dependent on the interplay of several traits and is a result from interactions of traits that affect decomposition in opposing directions.
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# Classes¶
## Summary¶
class class parent truncated documentation
Blossom
Edge
EdgeNotIncident Indicates that a traversal was requested through an edge from a set that doesn’t contain any of its endpoints.
EdgeNotOutgoing Indicates that a traversal was requested through an edge from a set of vertices containing both its endpoints.
EdgeTraversalError
FileFormatException Raised when unable to parse a file
ImageNearestNeighbors Builds a model on the top of NearestNeighbors in order to find close images.
MLStoragePerf2018 Computes the performances the a hackathon.
MLStoragePerf2018Image Overloads compute_perf for images. Example of use:
MLStoragePerf2018TimeSeries Overloads compute_perf for timeseries. Example of use:
MaximumDualReached Indicates that we have reached the maximum dual solution and cannot improve it further.
ParemetreCoutTrajet Regroupe l’ensembles des paramètres pour le calcul de la distance associé à un appariement.
PasswordException Raised when password is missing
ProjectDataException Exception raised when data is not available
SolutionException wrong solution
StructureUpToDate This gets raised as soon as the structure of all trees is up-to-date, i.e. there are no more instances of any of the …
TreeStructureChanged Used whenever the structure of an alternating tree is changed to abort current traversal and initiate a new one.
Vertex
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# Formal Logic And Probability
One of the arguments is that probability does not extend predicate logic, but does extend propositional logic. The concern is that because predicate logic is formal and propositional logic is not, or not to the same extent, that probability is therefore of limited use. This is explained (in classical terms) on David Chapman’s site. I don’t have the space here to rebut everything there with which I disagree (there is no such thing as unconditional probabaility for one; believing there is accounts for most of what’s wrong): that would take a book. Which I just happen to have written: Uncertainty: The Soul of Modeling, Probability & Statistics. But we can do a few things here.
The biggest weakness of predicate logic is that, pretty and mathematical as it is, any time you want to use or apply an argument in predicate logic to some real world proposition, you need to “lapse” into propositional logic, which is to say plain English (or whatever language you use for understanding). For instance, in predicate logic you can write:
$\forall x [Px \to Qx] \wedge Ps \therefore Qs$
which is peachy and correct (some might use different symbols) and formal. It’s the formality that makes it mathematical and which allows it to be manipulated algorithmically. But it is also what kills its usefulness in actual applications.
In the equation, P and Q are predicates and x is a variable. The formality means that we can stick any predicate and any variable into the formula and it should work. One predicate (for P) might be “is a man”, and another (for Q) “is mortal”; a reasonable variable is s = “the man Socrates”. Both can be inserted into the formula to produce, finally in English, “All men are mortal and Socrates is a man, therefore Socrates is mortal.”
Formalists will say the conclusion is true because of the schema or form of the predicate-logic formula. The symbols—not the words—are purely formal objects which slide through a rigorously constructed pipeline to the conclusion, just like the quadratic formula provides solutions to quadratic equations (keep this example in mind). In propositional logic, plain common sense will say the conclusion is true because the conclusion shares in the essence of the premises. Yet formalists will complain and say that the propositional logic version of the argument amounts to
$M \wedge N \therefore R$,
where M is the proposition “All men are mortal”, N = “Socrates is a man”, and R = “Socrates is mortal.” This, they say, isn’t formal, because the propositions in this formula aren’t “about” anything. They’re floating symbols, so, of course, R doesn’t follow from the conjunction of M and N. How could it? One cannot stick in just any old propositions for M, N, and R and have any hope the argument will produce a true conclusion. The schema itself, formalists say, is invalid (yet it produces the odd true argument).
Aristotelian logic, on the other hand, takes the argument as a syllogism and, partly by virtue of its syllogistic form, and partly from the plain understanding of the words and grammar, sees the conclusion as valid. The argument is also considered sound because of the understanding of intension (this is not a misspelling) of the terms.
Why the hunger for formality? Well, that’s what math is all about and, as such, there is nothing wrong the goal. But to say all logic should be formal is to claim all thought can be quantified or made into mathematics somehow. And that is the goal of many; think of certain forms of artificial intelligence. There is no proof of that claim; there is only the assurance or hope that it can be so.
But there is bad news for formalists. In 2009, David Stove proved logic is not formal (I’m quoting from my own article, which in turn draws quotations from Stove’s Rationality of Induction; see the original for details).
An argument is formal “if it employs at least one individual variable, or predicate variable, or propositional variable, and places no restriction on the values that that variable can take” (emphasis mine). Stove claims that “few or no such things” can be found.
Here is an example of formality: the rule of transposition. “If p then q” entails “If not-q then not-p” for all p and for all q.
This is formal in the sense that we have the variables p and q for which we can substitute actual instances, but for which there are no restrictions. If Stove is right, then we should be able to find an example of formal transposition that fails.
First a common example that works: let p = “there is fire” and q = “there is oxygen”, then
“If p then q” == “If there is fire there is oxygen”.
And by transposition, not-q = “there is no oxygen” and not-p = “there is no fire” then
“If not-q then not-p” == “If there is no oxygen then there is no fire.”
For an example in which formal transposition fails, let p = “Baby cries” and q = “we beat him”, thus
“If p then q” == “If Baby cries then we beat him”.
But then by transposition, not-q = “We do not beat Baby”, not-p = “he does not cry”, thus
“If not-q then not-p” == “If we do not beat Baby then he does not cry.”
which is obviously false. (Stove credits Vic Dudman with this example.)
So we have found an instance of formal transposition that fails. Which means logic cannot be “formal” in Stove’s sense. It also means that all theorems that use transposition in their proofs will have instances in which those theorems are false if restrictions are not placed on its variables. (It’s worse, because transposition is logically equivalent to several other logical rules; we won’t go into that now.)
It is Stove’s contention that all logical forms will have an example where it goes bad, like with transposition.
Now, as I said, some form of Aristotelian logic or of something more propositionally informal and fundamental must take place when we assent that the proposition “Socrates is mortal” follows from the other propositions. It is not the schema that makes something true. Schemas have no power! Things are not made true by mathematical or logical form (this “form” is not the same as the Aristotelian “form”, of course: for “form” here, read “formula”). They are caused to be true by something, all right, but a schema has no causal power.
Go back to the quadratic equation example. It, like all mathematical theorems, has a proved formal structure. But is not purely formal (in Stove’s sense). The quadratic formula has restrictions. You cannot input matrices into it, for example. The pure formality doesn’t exist because of these restrictions.
As said above, when applying predicate calculus to a real-world problem, we always must lapse into propositional logic or plain English. This falling back, as it were, always brings with it restrictions, which is why ordinary discussions aren’t purely formal. The real problem lies in attempting to formalize what ultimately cannot be formalized.
1. JohnK says:
It strikes me that the Quine-Duhem thesis, or near to it, fits well here. Stove’s proof seems as significant, and as little taken seriously, as Gödel’s.
2. There’s an interesting article on quantum logic in the Internet Encyclopedia of Philosophy (let’s see if a link will work)
http://www.iep.utm.edu/qu-logic/
in which the statement is made that David Finkelstein and Hilary Putnam argued that logic is in a “certain sense” empirical.
3. by the way, nice article, Briggs. And the link worked!
4. Fascinating. I find myself agreeing, using examples that have been posed to me by others before Briggs has, and by those that I’ve posed to myself. My exposure to formal logic was limited to Boolean Algebra. I never took the time to look for more general examples.
5. Doug M says:
Are you still beating your crying babies? Perhaps, if you didn’t beat it, it wouldn’t be crying so darn much. I know, I know, if it didn’t cry, you wouldn’t have to beat it. But, since you are not beating the baby right now, the little guy must have shut up!
Nonetheless, expect a visit from social services.
p?q,¬q?¬p
6. Fr. John Rickert, FSSP says:
I believe the following argument is doubtful or incorrect:
For an example in which formal transposition fails, let p = “Baby cries” and q = “we beat him”, thus
“If p then q” == “If Baby cries then we beat him”.
But then by transposition, not-q = “We do not beat Baby”, not-p = “he does not cry”, thus
“If not-q then not-p” == “If we do not beat Baby then he does not cry.”
The reason I say so is that there is an equivocation on the meaning of q between the positive and contrapositive forms. I do not think q is being construed in exactly the same way in the positive q form and negated, not-q form.
Let me restate the argument:
If (= whenever) B cries, then we will beat B.
If it is not the case that we will beat B, then it is not the case that B is crying.
This is argument is valid, and goes by the name “modus tollendo tollens,” or “modus tollens” for short.
In the argument above, the positive form of q seems to mean that an event will ensue upon the occurrence of a condition p. In the contraposed form, q sees to mean a habitual or permanent action. I believe this shift in the meaning of q is enough to say that this proposed counterexample to modus tollens is not valid.
7. Milton Hathaway says:
If Baby cries, then we will beat him in the very near future.
Therefore, if we are not beating Baby, he hasn’t cried in the very recent past.
I agree with Fr. Rickert, the logic seems sound to me, when stated accurately. If A causes B, and B hasn’t happened, then A won’t happen in the future? What law of logic allows one to swap the time order of cause and effect in the contrapositive?
Got any other (more convincing) examples?
8. Joy says:
My understanding is that the first conjecture has to be true for the contrapositive also to be true.
The first argument is not always true so it is already false before testing the contrapositive.
Something must be known about the matter to know that the first conjecture holds in all examples such as set and a subset where p is contained within the set of Q. I am using Joy language, I have no idea whether this is right or not.
That this isn’t confusing is a sign that I don’t understand.
I thought the first line was “for all p and all Q”
PDF’s are off limit, I can’t read chapter ten at present. The talking man just plays dumb. He’s learning some life skills.
9. Ken says:
Fr. Rickert’s analysis hi-lites a very fundamental issue — when approaching a situation from a strictly philosophical perspective, such issues arise.
However, when approaching them from real-world physics-based perspective, they might come up, but often get addressed and never become an issue. In applying the kind of analysis Briggs touched on (predicate & propositional logic are but two types in a range of such analytical approaches) a modeler will define ‘truth tables’ that establish what is and is not — those are based on what the thing being modeled is and isn’t doing relative to the environment in which it acts.
Application of such logical approaches is intrinsic to AI. For a topical example see: https://www.csee.umbc.edu/courses/671/fall09/notes/10a.ppt (or, google the following keywords & take your pick from the responses: advantages of predicate logic over propositional logic).
This is a recurring issue here — philosophy is NOT a good tool for assessing logical issues in reality. It allows one’s ignorance of the real-world factors applicable to some topic to facilitate seemingly logical, perhaps conceptually unassailably logical, sequences of ‘if-then’ analyses to achieve an absurdly ridiculous conclusion
(e.g.: if one adds a few drops of color to a gallon of paint, then stir clockwise to mix, then, one can un-mix the colors by stirring the same amount counter-clockwise — seems silly, but if the subject is in some way constrained by real-world factors and one is ignorant of them, careful step-by-step application of logic can & does lead to such absurd conclusions, too often going unnoticed;
an arguably even more complex area is in financial market models — there’s a variety of abrupt external influences that apply such that logical application of ‘if-then’ sorts of logic of the crying baby sort that suddenly become pre-empted by other factors (e.g. automated margin calls or sell orders that stop some particular action); logic applied there from a strict philosophical perspective simply fails [very quickly] unless one factors in the external environmental factors).
Using philosophy, and remaining dependent on that, leads to debates about this or that approach’s pitfalls/advantages and logical conundrums that don’t occur, much, when real topical experts inject situational reality into an analysis. …but then…”situational reality” is or may be in another words, “science,” and heaven forbid we permit that to intrude and, perhaps, undermine some deeply held sacred belief.
10. Fr. John Rickert, FSSP says:
Ken —
Worth noting that even logicians realize the need for the distinction between validity and soundness.
11. Dr Briggs,
I also do not agree with the characterization that Propositional Logic is not formal. I wonder why that has been claimed. Formal logic does not only have syntax. It must have semantics as well and then rules for deduction. The completeness theorem works both for Predicate and Propositional Logic so I find it hard to buy that Propositional Logic (PL) is not formal while First Order Logic(FOL) is. Your last paragraph is right, for indeed in FOL, what matters is what we call “sentences” which in the end are propositions.
LPC
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# Reinforcement Learning Batch Size
I am using a neural network as my function approximator for reinforcement learning. In order to get it to train well I need to choose a good learning rate. Hand picking one is difficult, so I read up on methods of programmatically choosing a learning rate. I came across this blog post, Finding Good Learning Rate and The One Cycle Policy, about finding cyclical learning rate and finding good bounds for learning rates.
All the articles about this method talk about measuring loss across batches in the data. However, as I understand it, in Reinforcement Learning tasks do not really have any "batches", they just have episodes that can be generated by an environment as many times as one wants, which also gives rewards that are then used to optimize the network.
Is there a way to translate the concept of batch sizes into reinforcement learning or a way to use this method of cyclical learning rates with reinforcement learning?
Potentially.
If you do offline reinforcement learning, you're basically learning to approximate a function by sampling input/output pairs, rather than episode-by-episode. Here, your batch size could be set exactly as in an ordinary supervised learning problem.
If you do online learning, then it's not clear to me that the techniques used to set the learning rate in supervised learning can be directly applied though.
Both approaches are well covered in the RL chapter of Russell & Norvig (17? 18?).
From my understanding of reinforcement learning, you will have an agent and an environment.
In each episode, the agent observes the state, gives some action, then gets some reward, finally observes another next_state, and do it again and again until the end of the episode.
The above process does not incur any "learning". Then when and where exactly do you "learn"? You learn from your history. In traditional Q learning, the Q matrix is updated every time you have a new observation of state, action, reward, next_state. Just like the supervised learning, you put in training sample one by one.
Similarly, you can feed in training samples in "batch" when you train, which means you "remember" the past N observations and train them together. I think that is the answer to your question.
Furthermore, the past N observations could have strong correlation that you don't want. To break this, you may have a larger "memory" that stores many observations, and you only sample a few (this number is your new batch size) randomly every time you train your model. This is called experience replay.
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# Keane Java Questions: Quantitative Questions
Examrace Placement Series prepares you for the toughest placement exams to top companies.
1 If g (0) = g (1) = 1 And g (n) = g (n − 1) + g (n − 2) find g (6).
Ans: g (6) = 13
Sol:
G (0) = 1
G (1) = 1
G (2) = G (1) + g (0) = 2
G (3) = g (2) + g (1) = 3
G (4) = g (3) + g (2) = 5
G (5) = g (4) + g (3) = 8
G (6) = g (5) + g (4) = 13
2 A plane moves from 9? N40? E to 9? N40? W. If the plane starts at 10 am and takes 8 hours to reach the
destination, find the local arrival time?
Ans: The time is calculated on the basis of longitude (Based on sunrise). Sun rises earlier in calcutta than in Mumbai.
The 40 degrees east and 40 degress west are calculated form 0 degrees. The plane has travelled westwards by
40 + 40 = 80 Degrees (WEST). For each degree there is a difference of 4 Minutes The difference in timings is
320 Minutes behind. As per the starting point time 10.00 AM the flight should have reached at 6.00 PM.
Reducing 320 Minutes from it we get 12.40 PM local time at the destination
3 Given $means Tripling and % means change of sign then, find the value of$%$6-%$%6 Ans: 72
4 The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If
the size of the program is increased by 1% then how much memory now occupied? Replace N by (N + N/100) and solve.
5 16 litre can, 7 litre can, 3 litre can, the customer has to be given 11 litres of milk using all the three cans only
explain?
Sol: 16 litre Full initially 16 − 7 = 9 in 16 litre, 7 in 7 litre. 7 to 3 to 16 (twice) than 15 in 16 litre1 in 7 litre
0 in 3 litre 1 in 7 litre can to 3 litre can 15 in 16 litre can 0 in 7 litre can and 1 in 3 litre can. From 15 in 16litre
can to 7 litre can Than 8 in 16 litre can 7 in 7 litre can and 1 in 3 litre can From 7 in 7 litre ad 2 to 3 litre can
Than 8 in 16 litre can 5 in 7 litre can 3 in 3 litre can 3 to 16 litre can than 11 in 16 litre can 5 in 7 litre can
and 0 in 3 litre can.
1. A traveler walks a certain distance. Had he gone half a kilometer an hour faster \, he would have walked it in ⅘ of the time, and had he gone half a Kilometer an hour slower, he would have walked 2 ½ hr longer. What is the distance?
1. 10 Km
2. 15 Km
3. 20 Km
4. Data Insufficient
Ans: Distance = = speed x time st = = (s +. 5) x. 8t____ (1)
st = =. 8st +. 4t 2st = =. 4t s = = 2KM
st = = (s-. 5) x (t + 150) ____ (2) Sustituting you will get the distance as 15 KMs
2. A ship leaves on a long voyage. When it is 18 miles from the shore, a seaplane, whose speed is 10 times that of the ship is sent to deliver mail. How far from the shore does the seaplane catch upo with the ship?
1. 24 miles
2. 25 miles
3. 22miles
4. 20 miles
Ans: (d − 18) = = st d = st d = (d − 18) 180 = = 9d
3. In a circular race track of length 100 m, three persons A, B and C start together. A and B start in the same direction at speeds of 10 m/s and 8 m/s respectively. While C runs in the opposite at 15 m/s. When will all the three meet for the first time on the after the start?
1. After 4 s
2. After 50 s
3. After 100 s
4. After 200 s
Ans: c
Sol: Since the track is a circular track A and B will meet every 50 seconds. Ie 100/(10 − 8) Since it is a multiple
of 50 they will be meeting at the starting point every 50 Seconds. If you multiply 15 × 50 you will get 750 and
after the second 50 it will be 1500. All of them will meet at the starting point after 100s
4. Amal bought 5 pens, 7 pencils and 4 erasers. Rajan bought 6 pens, 8 erasers and 14 pencils for an amount which was half more than what Amal had paid. What % of the total amount paid by Amal was paid for pens?
1. 37.5%
2. 62.5%
3. 50%
4. None of these
Ans: 62.5%
Sol: You can observe from the question that for additiion of 1 pen 7 pencils and 4 erasers rajan pays 50 %
more. It implies that the cost of 4 pens is 50% of the total amount paid by Amal. Therefore cost of 5 pens will
be 62.5% of the total cost.
5. A non stop bus to Amritsar overtakes an auto also moving towards Amritsar at 10 am. The bus reaches Amritsar at 12.30 pm and starts on the return journey after 1 hr. On the way back it meets the auto at 2 pm. At what time the auto will reach Amritsar?
1. 2.30pm
2. 3.00pm
3. 3.15pm
4. 3.30pm
Ans: 3.00pm
Sol: Let the distance between the meeting point of Auto at 10.00 Am and Amritsar be d The bus takes 2.5
Hours (150 minutes) to reach amritsar from that point. The bus commences its return journey at 1.30 PM and
meets auto at 2.00 PM. The bus has thus travelled 30 Minutes. The bus would have covered 30/150 d
The auto has covered (1-⅕) d during the period. Is 4/5th of the distance has been covered in 4 hours (2.00 PM-10.00 AM) The remaining 1/5th distance will be covered in 1 hour. The auto will reach Amritsar at 3.00 PM
6. The cost of one pencil, two pens and four erasers is Rs. 22 while the cost of five pencils, four pens and two erasers is Rs. 32. How much will three pencils, three pens and three erasers cost? Sol: x + 2y + 4z = = 22 5x + 4y + 2z = = 32 Solving we get 3x + 2 y = = 14 and y + 3z = = 13, Adding both we get 3x + 3y + 3z = = 27 Ans: 27
7. The lowest temperature in the night in a city A is ⅓ more than ½ the highest during the day. Sum of the lowest temperature and the highest temperature is 100 degrees. Then what is the low temp? Sol: (½ + 1/3 × 1/2) t + t = = 100 solving we get t = = 60 i.e.the highest temperature is 60 and the lowest is 40. Ans: 40 degrees
8. Javagal, who decided to go to weekened trip should not exceed 8 hours driving in a day. The average speed of forward journey is 40 miles/hr. Due to traffic on sundays, the return journey's average speed is 30 m/h. How far he can select a picnic spot?
1. 120 miles
2. between 120 and 140 miles
3. 160 miles
Ans: 120 miles
Sol: The answer given by you appears to be wrong. The question is not clear regarding the day on which the
onward trip is carried out. If it is on a Saturday the maximum distance that can be travelled is 8 × 30 = = 240
miles. However if both the onward and return trips are carried out on Sunday the answer is Between 120 and
140 Miles. The time taken for both the trips is 8 hours.
3 × 40 = 0
4 × 30 = 0.
One more hour can be utilised for travelling and more than 120 miles can be covered.
9. The total expense of a boarding house are partly fixed and partly variable with the number of boarders. The charge is Rs. 70 per head when there are 25 boarders and Rs. 60 when there are 50 boarders. Find the charge per head when there are 100 boarders.
1. 65
2. 55
3. 50
4. 45
Ans: b
Sol: The total cost for 25 boarders is 25 × 70 = 50, The total cost for 50 boarders is 50 × 60 = = 3000.
The difference is 1250 for 25 boarders. The variable cost per boarder is Rs. 50
For additional 50 boarders the increase in total cost is 50 × 50 = = 2500, Therefore total cost for 100 boarders
is 3000 + 2500 = = 5500. The charge per head works out to Rs. 55.
10. A car has run 10000 miles using 5 tyres interchangably, To have a equal wornout by all tyres. How many miles each tyre should have run. Ans: 4000 miles/tyre Sol: According to me the answer given is wrong. A car runs on four wheels. The aggregate distance covered by the four wheels is 10000 × 4 = = 40000 miles. Since 5 tyres are used interchangably the distance covered by each tyre will be 40000/5 = = 8000 miles.
11. In 80 coins one coin is counterfiet what is minimum number of weighings to find out counterfiet coin Ans: 4 weighings. 27 27 26 9 9 9 9 9 8 3 3 3 3 3 2 1 1 1 1 1
12. There are two trees in a lawn. One grows at a rate ⅗ of the other in 4 years. If the total growth of trees is 8 ft. What is the height of the smaller tree after 2 years Ans: 1 ½ feet (Less than 2 feet). Sol: After four years the combined height of the two trees is 8 ft. The ratio at that point is ⅗ 1 therefore 3/5x + x = = 8, solving x = = 5 The height of the smaller tree after 4 years is 3 ft At the end of two years it would have been 3/2 ft.
13. A building with height D shadow upto G. What is the height of a neighbouring building with a shadow of C feet. Sol: Using the similar triangle property y = = D/G X C
14. Find the no of possible palindromes for the following: Tyghhty six (g will be the center) tyhhhtyh twelve hh, tt, yy as centre
15. 2, 20, 80, 100, The answer C is not given. It could be 139
16. 2 * 10, 20 * (10/5 sq/10), 80 * (4/4 sq/10), 100 * (1.6/3 sq/10)
1. 121
2. 116
3. N/A
4. NONE
17. 10, 16, 2146, 2218, Ans: 2290 Sol: Since the answer is not given, I am not able to match my working. My answer to the question is 2290.16 * 16 is 256 (the sum of the three digits is 13) That 13 is being maintained in 4 digit number by retaining the f irst and last digits for the first number in the serial
18. There are 10 coins. 6 Coins showind head. 4 showing tail. Each coin was randomly flipped (not tossed). Seven times successively. After flipping the coins are 5 heads and 4 tails. One is hidden. What will be the hidden coin holds? Ans: The coin will have tail
19. What is the diff in times btwn clk1 and clk2. Both show same time 6 hrs back.
20. 1 clk gains 1 min an hr, clk2 gains 2 min an hour. Like that The difference will be 6 MIN
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# Evaluate: (-14+i)-(-12-11i)
## Expression: $\left( -14+i \right)-\left( -12-11i \right)$
Remove unnecessary parentheses
$-14+i-\left( -12-11i \right)$
When there is a $-$ in front of an expression in parentheses, change the sign of each term of the expression and remove the parentheses
$-14+i+12+11i$
Calculate the sum
$-2+i+11i$
Collect like terms
$-2+12i$
Write the complex number in absolute value bars
$|-2+12i|$
Use $|a+bi|=\sqrt{ {a}^{2}+{b}^{2} }$ to calculate the modulus
$\sqrt{ {\left( -2 \right)}^{2}+{12}^{2} }$
Simplify the expression
\begin{align*}&2\sqrt{ 37 } \\&\approx12.16553\end{align*}
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# STBCs with Optimal Diversity-Multiplexing Tradeoff for 2, 3 and 4 Transmit Antennas
Shashidhar, V. and Rajan, Sundar B and Kumar, Vijay P (2004) STBCs with Optimal Diversity-Multiplexing Tradeoff for 2, 3 and 4 Transmit Antennas. In: International Symposium on Information Theory, 2004. ISIT 2004, 27 June-2 July, Chicago, p. 125.
Preview
PDF
STBCs.pdf
## Abstract
We show that the codes from division algebras [1] achieve the optimal diversity-multiplexing tradeo. for $n$ transmit and $n$ receive antennas for $n=2,3,4$ by simulation. Also, we present a lower bound for the tradeoff curve which shows that codes from division algebras for arbitrary number of transmit and receive antennas achieve points corresponding to zero diversity gain and zero multiplexing gain. Let $S$ be the signal set over which we want to construct fullrank STBCs. Then, let $m$ be an integer such that $x^n- w_m$ is irreducible over $F= Q(S,w_m)$, where $Q$ is the field of rational numbers and $w_m = e^{^j2\pi/m}$. The extension field obtained from this irreducible polynomial is $K=F(w_{mn})$.
Item Type: Conference Paper �©1990 IEEE. Personal use of this material is permitted. However, permission to reprint/republish this material for advertising or promotional purposes or for creating new collective works for resale or redistribution to servers or lists, or to reuse any copyrighted component of this work in other works must be obtained from the IEEE. Division of Electrical Sciences > Electrical Communication Engineering 12 Dec 2005 19 Sep 2010 04:21 http://eprints.iisc.ernet.in/id/eprint/4399
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Interferometry and the Very Large Array
Congratulations! You’ve just been hired to record the crowd at the Super Bowl!
The producer wants you to figure out how loud each section, and ideally how loud each fan, is cheering so that they can better choose where to station the T-shirt cannons. They don’t want to know the actual dialogue – they’re not the NSA – only each fan’s loudness, or amplitude*. You don’t even have to record the fans’ cheering over time, but just how loud each one is, on average, over the full game.
*technically, loudness is a psychoacoustic property of the amplitude, the average intensity, and the frequency spectrum of the signal, but in this article we’re assuming (for the purposes of keeping it relatively short) that they’re essentially the same.
The only problem?
You’re on the other side of the field, 160 feet away from the nearest spectator. And you have one microphone with which to record all twenty thousand fans.
The basic technique’s simple; point your microphone at a fan, record the average loudness over some length of time, then repeat for every other person there. If you record each direction for half a second, you’ll finish before the end of the game (although the resulting loudness image will be very noisy – the block of really enthusiastic Patriots fans in the center might be not quite as clear – because the rare occurrence of touchdowns violates our assumption that each fan is shouting at the same volume throughout the entire game. Perhaps pinpointing the location of vuvuzelas at the World Cup would be a better application of our technique.)
Unfortunately, this only works if our microphone only records sounds coming from within one degree of an object’s source. (In other words, if our microphone is so directionally sensitive that rotating it 1/360th of a full turn gives us a totally different sound.)
Unsurprisingly, most microphones are specifically designed to avoid this! (Most of the time, you don’t want the audio on a take to be ruined because an actor was two inches away from their mark.)
Omnidirectional microphones have no variation (hence their name), and so you have no choice but to record the entire crowd. Cardioid mics have some directionality (to a first approximation, they record things within a 120-degree angle), but they’re still not good enough.
We can do slightly better, and use a parabolic reflector to bounce all sound waves coming from one direction to a focal point – where we’ll place the microphone. All sounds not coming from that direction will, when reflected, miss the microphone and essentially be rejected.
In practice, though, the ‘beam’ of sound we’ll be able to hear won’t be a perfect cylinder, but will instead spread out into a sort of cone – not necessarily because of construction, but because sound is (for all practical purposes within the scope of this article) a wave instead of a constant flow of information-carrying particles (the corpuscular theory).
Because of this, the best parabolic microphones we could conceivably bring to the game would have to record at least a 48.4-degree range of sound, blurring out our map of loudness so much it would be nearly unusable.
And that’s why the NFL just gives players microphones.
Astronomers have the same problem: Although most telescopes use parabolic mirrors (or complicated lens designs), and should hypothetically have the ability to provide perfect focus, diffraction blurs out the image, preventing us from seeing the intricate details of far-away galaxies.
Intermission: Diffraction?
We’ve been talking about diffraction a lot already, and we’ll be talking about it a lot more, so at this point we should probably actually describe diffraction.
If you’re already familiar with the mechanics of lens and aperture diffraction, you can probably skip this section. Diffraction was actually relatively difficult for me to understand the first time around, so I’m including this here so that others may have a (hopefully better) explanation of what diffraction is.
Diffraction is a property of all electromagnetic waves, the properties of which (when averaged over time) are described by the Heimholtz equation,
where $delta^2$ is the Laplacian, k is the wavenumber (2*pi divided by the wavelength), and A is the amplitude.
Realistically, though, while the Heimholtz equation does technically perfectly describe diffraction (and can actually be directly used to optimally place WiFi antennas), it’s not particularly intuitive. Here’s a slightly better explanation:
Essentially, diffraction is the bending of waves around objects. It’s the reason why you can hear someone speaking when echoes alone wouldn’t carry the sound to you – such as on the other side of a pillar, or in an anechoic room:
But why do waves bend in the first place?
Intuitively, waves bend because they have to – because (at least in the case of pressure waves) this scenario depicting a wave passing through an aperture doesn’t make sense:
Generally, the wider the aperture is, the less diffraction exists. When your aperture’s less than one wavelength across, the wave on the other side is essentially identical to a single point source, while if your aperture’s infinitely many wavelengths across, the resulting waveform’s identical on both sides of the aperture (well, if there even is an infinitely wide aperture that isn’t just an open space). In between, the mechanics of waves passing through apertures are dictated by the Hyugens-Fresnel principle: wave propagation through an aperture (in the ‘forward’ direction) can be expressed (but not necessarily described) as a sum of spherical waves, each one wavelength apart from each other.
Because light waves have very short wavelengths (green light, for instance, has a wavelength somewhere around 530 nanometers), diffraction generally isn’t a problem even for very small lenses (for instance, the iPhone’s lens is at least $\frac{1.86 * 10^{-3} m}{750 * 10^{-9} m} = 2480$ red light wavelengths across, based on this stackexchange post). It’s possible to see visible-light diffraction, though, if your aperture’s small enough:
This is also why simple ray- (not wave-) tracing works so well for CGI renderers (well, most of the time).
However, both parabolic microphones and radio telescopes are diffraction-limited – that is, the extremely long wavelengths of the types of waves these devices observe blur out point sources into Airy disks. As we’ll soon see, it’s partially possible to remove these artifacts, although we’ll need spiffier techniques for finer details.
Why Not Use an Optical Telescope?
Optical (visible-light-based) telescopes have their advantages: the technology’s been around for a long time, they see in the same frequencies as their operators, and they aren’t as harshly limited by physical factors as radio telescopes. However, optical astronomy is generally limited far more by more mundane things, like…
• Light pollution! (Light, mostly from streetlights, enters the atmosphere and scatters, creating a diffuse haze which is also why you can’t see the stars at night in large cities)
• Astronomical seeing! (The air in the atmosphere’s often turbulent, especially on hot nights, causing images of stars to twinkle.)
• Mechanical wobble! (Motors and their assemblies aren’t perfect, and even miniscule imperfections can result in shaky images when you’re using an incredibly long lens.)
• Atmospheric absorption! (Water vapor and other atmospheric gases absorb most of various frequencies of light, including infrared and ultraviolet. This makes it nearly impossible to observe those frequencies emitted by stars from Earth.)
• Weather! (Rain, clouds, and even snow – a common occurrence, since many telescopes are located on the tops of mountains to avoid atmospheric scattering and absorption – will obviously block out any chance of seeing the stars.)
• and finally, interstellar dust clouds! (Yes, you read this right – clouds of microscopic particles formed from supernovae reduce and scatter visible light, wreaking havoc with long-distance observations, but occasionally producing very nice images.)
Much of the modern science of astronomy consists of ways of getting around these limitations – and often, in surprising and really impressive ways.
For instance, astronomers recently figured out that atmospheric distortion can be un-distorted by warping a computer-controlled secondary mirror faster than the atmosphere can change. How do they know how to warp the mirror? Easy; they measure the distortion of a known star.
But what if that star isn’t bright enough? No problem; shoot lasers into the sky, exciting sodium atoms in the upper atmosphere, creating an artificial star.
But I digress.
Single-Dish Astronomy
or, can you sharpen a microphone?
The telescopes of the Very Large Array, located on a wide plain about fifty miles away from both Socorro and Pie Town*, New Mexico, manage to get around many of these limitations by using radio waves instead of visual light for astronomical observation.
*which really does have fantastic pies!
Nearly all major astronomical objects, from stars on up (although some planets’ atmospheres do as well), emit radio waves, and certain radio wavelengths (such as the hydrogen line, with a wavelength of 21 cm) contain incredibly useful information about the compositions of astronomical objects. (Unfortunately, some terrestrial objects – such as cell phones and spark plugs – also emit radio waves near the VLA’s current frequency bands.)
Radio waves also pass through interstellar dust and Earth’s atmosphere (consider, for instance, the fact that you can listen to a radio signal transmitted from many miles away from inside a building, but can’t see what’s on the other side of the wall without a window), eliminating astronomical seeing issues. Plus, radio’s long wavelengths (each VLA telescope can handle ten different wavelengths between 4 meters and 6 millimeters) enable the construction of very light and very large telescopes using frames instead of full reflectors, such as in the Giant Metrewave Radio Telescope:
Radio isn’t a magical cure-all, though: The longer wavelength of radio also cooresponds to a wider beam-width (diffraction again), blurring the resulting images. We can do two things to compensate for this loss of sharpness:
• We can build a larger telescope, which isn’t too difficult (since we’re on Earth and not in space), lowering the amount of diffraction and also collecting signals over a larger area, which reduces noise. There is an upper limit, though – it’s very difficult to build a larger telescope than the 1000-foot dish of the Arecibo Observatory, which was literally built into a sinkhole at the top of a mountain range.
Source: NAIC – Arecibo Observatory
• Or, if we know what the diffraction pattern looks like, we can actually attempt to reverse the blurring and recover some amount of detail!
Here’s the idea: Blurring (or, more generally), is just the convolution of an image with a kernel.
Here, our image is the red curve, our Gaussian kernel is the blue curve, and our blurred image is the green curve. From Weisstein, Eric W. “Convolution”, Wolfram MathWorld
Cyclic convolution (which is the same as convolution, except the blur wraps around – so you need to add some number of 0s to the sides of the image) can be quickly computed by taking the Fourier transform of the image and the kernel, multiplying them, and returning the inverse Fourier transform.
If we want to remove a convolution of g from (f*g), we just divide it out of the Fourier product:
$f = \mathfrak{F}^{-1}\left \{ \mathfrak{F}(f\ast g)/\mathfrak{F}(g) \right \}$
So if we know the blurring kernel (g), and our blurred image (f*g), we can recover f!
Alright, let’s try this out.
This is a radio image, captured in 1989 by the VLA, of the Whirlpool Galaxy, about 23 million light-years away from Earth. It was discovered in 1779, so this particular image isn’t really all that impressive, but the VLA’s gotten far better since then. The full image covers about three-tenths of a degree of the night sky.
So what does a single telescope see?
A single VLA telescope has a beamwidth of 8.6 arcminutes, or about 0.14 degrees. (You can actually calculate this directly from the equation for diffraction, but here we’re using the statistics from a 1980 book on the subject.) That’s nearly the diameter of the entire Whirlpool Galaxy, so the entire thing’s reduced to a faintly visible smudge:
Just to make it even possible to recover information, let’s assume we’re using a telescope observing at about 30 times the original frequency (42.6 GHz, almost the VLA’s maximum supported frequency)
Here, we’re blurring with a Gaussian kernel with a radius of 40 pixels (over a 1779-pixel image), which allows us to see some details, but not the fine structure. In reality, we’d probably be using an Airy kernel instead of a Gaussian one, but they’re close enough for demonstration.
So, let’s just divide the Fourier transform of that image by the Fourier transform of a Gaussian kernel, invert the Fourier transform, and we should be done!
Except instead of a nice, clean image we get
Because the Fourier transform of a Gaussian curve is another Gaussian, a few of the higher-order frequencies are, for all practical purposes, 0. And, of course, when we divide zero (since the original image also almost entirely consists of low-frequency terms) by zero, things go haywire and you basically get static.
In order to handle this noise (because that’s essentially what it is), we can use one of the many approaches available and add a constant, very small number to the spectrum of the kernel.
And then we actually get a very nice reconstruction!
But, practically, there are also quantization artifacts (reducing a floating-point format to eight bits)
and noise
which basically null out whatever we were doing. Granted, we have made some progress – here’s what we started out with compared to what we managed to reconstruct –
but it’s not terrific, and certainly no substitute for an optical telescope. And, unfortunately for us, building a mega-Arecibo isn’t plausible. Not only that, the generally productive xkcd strategy
doesn’t even work – we might get a brighter image, but it’ll still be blurry.
So what else can we try?
Interferometry
or, Using Diffraction for Fun and Profit (and the occasional scientific discovery)
45 GHz is a ridiculously high frequency. So is 74 MHz, the lowest band the VLA recieves. We might be able to use this to our advantage.
If we can measure the phase difference between two signals (the relative offset in the signals two antennae receive caused by the differences in the amount of time the wave takes to get to each antenna), we might be able to amplify a miniscule difference in the position of a point source into a huge difference in phase, which we can also detect.
Or, putting it another way, we can use the difference in the time it takes a signal from a radio source to reach two telescopes to narrow the main lobe of the beam (reducing how blurry the image is) in exchange for more ringing around stars. Then, we might be able to remove the ringing, leaving only the main lobe, either through the same Fourier transform trick as before, or by some other method we haven’t determined yet.
And that, essentially, is the idea behind interferometry. Let’s get down to the actual details.
Let’s suppose, for now, that there is exactly one star in the sky. We can generalize to arbitrary numbers of stars later on, but it’s easiest to take things, well, one thing at a time.
If we have two radio telescopes pointing in a direction $\hat{s}$ at an infinitely far away source, separated by a distance $\vec{b}$, the signals they receive will be exactly the same, except for a time delay $T_{g}=b*cos(\theta)/c$ on the one further away.
Here, $\omega$‘s two pi times the frequency, so antenna 2 receives the signal $V*cos(\omega * t)$, and antenna 1 receives $V*cos(\omega * (t-T_{g})$.
Individually, we can get an approximation of the intensity of the thing we’re looking at (V) by averaging the cos wave over some amount of time – that’s what we’ve done before, and the problem is usually that the beamwidth (in this case, the square root of the envelope of the graph on the lower-right) was usually too wide.
We can then perform a neat trick, and multiply the two cos waves together, getting a sum of their arguments’ sum and difference:
$V^{2}*cos(\omega *t)*cos(\omega *(t-T_{g})) = \frac{V^{2}}{2} * (cos(\omega * T_{g}) + cos(\omega * (2 t - T_{g})))$
which happen to be the cosine of something times the phase (and only the phase), and a really high-frequency term! So if we average this out over time, we’ll get
$\frac{V^{2}}{2} * cos(\omega * T_{g})$
and substituting in our value for $T_{g}$, the value for the pixel returned by the correlator (the machine that coorelates signals between all 27 different telescopes and also handles a whole bunch of signal processing routines we’re glossing over right now) would be
$\frac{V^{2}}{2} * cos(\frac{\omega}{c} * b * \theta)$
This (when multiplied by your beam’s strength per angle) is the pattern a single star will produce on your image. So long as the frequency (w/(2 pi)) and the distance between our telescopes (b) are large, this interference pattern will have a far higher frequency than the original beam, but will have many rings around the center, making it look like a target. That’s actually a good thing, because then it’s easier to locate the object’s precise location!
According to the NRAO’s course on the subject, while a single telescope may only be able to locate an object within eight arcminutes, the 27 telescopes of the VLA, when combined, can pinpoint locations to within a thousandth of an arcsecond.
And, once we’ve got our sharp, but ring-ridden image, we can apply the same sorts of Fourier tricks as above to remove the rings and produce a clean image.
Alternatively, in the spirit of presenting proofs without words, here’s interferometry as a series of GIFs:
Finally, here’s one final way of looking at interferometry. Normally, waves are emitted by a source, propagate through space, and reach the telescopes at two different times. Because electromechanics is reversible, we can look at this process the other way around: the two telescopes emit waves at two different times, which propagate through space and reach the source at the same time.
Conveniently, the sensitivity of the telescope to this point source depends on whether the two ‘telescope waves’ are constructively or destructively interfering with each other at that point in space – that is, the interferometer’s graph of reception is an interference pattern!
As an added bonus, this interference pattern corresponds exactly to the diffraction of a wave passing through a slit of width equal to twice the wavelength, which also provides another explanation of the VLA’s better sensitivity at higher frequencies.
It seems like we can do almost anything with just two telescopes –
So What are All Those Other Telescopes For?
The very first astronomical interferometer looked like this.
This was Karl G. Jansky‘s original telescope, originally built to detect radio emissions from telephones but which wound up detecting a large radio source at the center of the Milky Way. It consisted of one large antenna, an analog information storage system (a pen and notepad), and a track which allowed it to be rotated. While Jansky’s telescope wasn’t an interferometer, the reason there even was a track is relevant – if you use only two telescopes in your interferometer, or (equivalently) one long antenna, your signals will be sharp in one direction, and blurred in another.
The reason for this is (somewhat) simple: Up until now, we’ve been working in two dimensions, with two telescopes. In three dimensions, there’s interference along the direction the telescopes are laid out, but no interference perpendicular to that direction. You need at least three to get a sharp image, which is why the VLA has three arms set at 120 degree angles to each other.
However, the interferometry procedure described above doesn’t actually work for more than two telescopes. Instead, we treat the three telescopes as (3 choose 2) = three pairs of baselines between two telescopes. The VLA computes the results for each baseline, then adds up appropriate amounts of each point response, minimizing the rings while amplifying the center lobe.
With 27 antennas, we have (27 choose 2) = 351 independent baselines, all at once, making for an incredible point response.
Eventually, we’d get a sort of Gaussian point response – which is actually a bit of a problem! As we’ve seen before, when we blur an image with a Gaussian kernel, we find that we can’t reconstruct some of the high frequencies, lest we run into a division-by-zero error. To put it another way, the blurring, when followed by quantization, destroys information, and we can only reconstruct the large structures in the image. We can err in the other direction, too: if our telescopes are too far apart, we’ll only be able to reconstruct the fine details of the original image, and we won’t be able to see the large structures.
That’s why the telescopes of the VLA are shuttled around on their rails every few months, cycling through four configurations – from the D configuration, with a maximum baseline of less than one mile, to the 22-mile A configuration. By compositing all four images, we can achieve a complete radio image of just about any celestial object.
But we can do even better.
The Earth rotates, so by taking measurements every so often throughout the day, we can pretend we have multiple copies of the VLA around the planet, and infer even more baselines from that.
But we can do even better.
The VLA is so sensitive that the movement of the Earth’s crust beneath it blurs out its images. By making use of the Very Long Baseline Array, ten additional antennas spaced throughout the United States, the VLA can correct for these errors using what essentially equates to an five-thousand-mile wide sparsely-sampled antenna.
And of course, there’s even more we can do on the signal processing side of things, which are, unfortunately, way out of the scope of this article. Techniques have been developed for correcting chromatic aberration (just as some lenses do), dealing with non-flat baselines using maximum entropy techniques, implementing spherical harmonic Fast Fourier transforms, running web servers, and simply handling the gigabytes of information the antennas return every second quickly, efficiently, and using hardware custom-designed not to interfere with its own antennas.
And yes, sometimes errors occur, and sometimes antennas go down. Neither the VLA, nor its operators, are perfect, and there are some things we probably won’t be able to glimpse for a long time.
All the same, the VLA is an incredibly impressive work of engineering, and it’s spotted things we wouldn’t be able to see any other way, from early protostars, to synchrotron radiation from black holes.
But let’s go back a bit: you’re actually at the Super Bowl, and you’re actually trying to record individual audience members. Or, let’s say you’re trying to mic a conference room without having to give everyone lapel microphones. Then, believe it or not, you may actually be familiar with the idea of microphone arrays, which essentially use techniques from interferometry to create virtual microphones which are more directed than any of their components – or acoustic cameras, which are now used for detecting audio emissions from products. Some sensor arrays detect waves propagating through the Earth’s crust, and use the results to detect the presence of oil. (But of course, it’s very difficult to create an audio source within the Earth – which is why they use the noise from nearby highways, rebounding off objects deep beneath the Earth’s surface.)
Interferometers have been used in meteorology, in wind tunnels, in chemistry, quantum mechanics, particle physics, microscopy, and undoubtedly a few more fields by the time you read this. Now, even some optical telescopes use rotating apertures to create a virtual array of smaller telescopes – just like the VLA.
Signal processing is still a developing field, and new techniques are being discovered every day, not just in astronomy, but in just about every scientific field there is. And that, to say the very least, is fantastic.
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# [LON-CAPA-users] {Disarmed} RE: Formula response won't recognize "pi"
SIMIN, GRIGORY SIMIN at engr.sc.edu
Thu Nov 21 14:51:33 EST 2019
hk,
Great suggestion!
Thank you!
Grigory
From: LON-CAPA-users <lon-capa-users-bounces at mail.lon-capa.org> On Behalf Of H. K. Ng
Sent: Thursday, November 21, 2019 2:12 PM
To: Discussion list for LON-CAPA users <lon-capa-users at mail.lon-capa.org>
Subject: Re: [LON-CAPA-users] Formula response won't recognize "pi"
Hi,
Formularesponse using cas with maxima works great.
<problem>
<script type="loncapa/perl">
$ansa = &cas('maxima','1/sqrt(4*pi^2*L*C)'); </script> <startouttext />What is formula for resonant frequency for an LC circuit? <endouttext /><br /> <startouttext /><m>$ \omega_{res} = $</m><endouttext /> <formularesponse answer="$ansa" id="12">
</formularesponse>
</problem>
Regards,
-hk
On Thu, Nov 21, 2019 at 11:05 AM SIMIN, GRIGORY <SIMIN at engr.sc.edu<mailto:SIMIN at engr.sc.edu>> wrote:
Hi all!
I've found a strange issue in some problems requiring formula response.
I have a simple question to enter the formula for the resonant frequency.
An example of correct answer is (in other variations the formula could be a bit more complex than this one)
1/(2*pi*sqrt(L*C))
The correct solution is checked by sampling L and C values.
When I am entering the answer, I can see that LON-CAPA recognizes it because it displays the formatted formula as I am entering it:
[cid:image001.jpg at 01D5A07B.29C6BE90]
But when I press submit, I got the response:
“unable to understand formula”
By experimenting with this, I found that the issue happens because of “pi” being a part of the answer.
If pi is replaced with 3.141, everything would work.
But it feels weird asking students to enter 3.141 instead of pi; besides, I don’t even want to give them a hint that pi is a part of the answer.
Why is this happening? Apparently LON-CAPA knows what “pi” means because it displays it correctly.
Thanks!
Grigory
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## File: gb_gates.w
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% This file is part of the Stanford GraphBase (c) Stanford University 1993 @i boilerplate.w %<< legal stuff: PLEASE READ IT BEFORE MAKING ANY CHANGES! @i gb_types.w \def\title{GB\_\,GATES} \prerequisite{GB\_\,GRAPH} @* Introduction. This GraphBase module provides six external subroutines: $$\vbox{\hsize=.8\hsize \everypar{\hangindent3em} \noindent|risc|, a routine that creates a directed acyclic graph based on the logic of a simple RISC computer;\par \noindent|prod|, a routine that creates a directed acyclic graph based on the logic of parallel multiplication circuits;\par \noindent|print_gates|, a routine that outputs a symbolic representation of such directed acyclic graphs;\par \noindent|gate_eval|, a routine that evaluates such directed acyclic graphs by assigning boolean values to each gate;\par \noindent|partial_gates|, a routine that extracts a subgraph by assigning random values to some of the input gates;\par \noindent|run_risc|, a routine that can be used to play with the output of |risc|.}$$ Examples of the use of these routines can be found in the demo programs {\sc TAKE\_\,RISC} and {\sc MULTIPLY}. @(gb_gates.h@>= #define print_gates p_gates /* abbreviation for Procrustean linkers */ extern Graph *risc(); /* make a network for a microprocessor */ extern Graph *prod(); /* make a network for high-speed multiplication */ extern void print_gates(); /* write a network to standard output file */ extern long gate_eval(); /* evaluate a network */ extern Graph *partial_gates(); /* reduce network size */ extern long run_risc(); /* simulate the microprocessor */ extern unsigned long risc_state[]; /* the output of |run_risc| */ @ The directed acyclic graphs produced by {\sc GB\_\,GATES} are GraphBase graphs with special conventions related to logical networks. Each vertex represents a gate of a network, and utility field |val| is a boolean value associated with that gate. Utility field |typ| is an ASCII code that tells what kind of gate is present: {\advance\parindent 2em \smallskip \item{|'I'|} denotes an input gate, whose value is specified externally. \smallskip \item{|'&'|} denotes an \.{AND} gate, whose value is the logical {\sc AND} of two or more previous gates (namely, 1 if all those gates are~1, otherwise~0). \smallskip \item{|'|'|} denotes an \.{OR} gate, whose value is the logical {\sc OR} of two or more previous gates (namely, 0 if all those gates are~0, otherwise~1). \smallskip \item{|'^'|} denotes an \.{XOR} gate, whose value is the logical {\sc EXCLUSIVE-OR} of two or more previous gates (namely, their sum modulo~2). \smallskip \item{|'~'|} denotes an inverter, whose value is the logical complement of the value of a single previous gate. \smallskip \item{|'L'|} denotes a latch, whose value depends on past history; it is the value that was assigned to a subsequent gate when the network was most recently evaluated. Utility field |alt| points to that subsequent gate. \smallskip}\noindent Latches can be used to include state'' information in a circuit; for example, they correspond to registers of the RISC machine constructed by |risc|. The |prod| procedure does not use latches. The vertices of the directed acyclic graph appear in a special topological'' order convenient for evaluation: All the input gates come first, followed by all the latches; then come the other types of gates, whose values are computed from their predecessors. The arcs of the graph run from each gate to its arguments, and all arguments to a gate precede that gate. If |g| points to such a graph of gates, the utility field |g->outs| points to a list of |Arc| records, denoting outputs'' that might be used in certain applications. For example, the outputs of the graphs created by |prod| correspond to the bits of the product of the numbers represented in the input gates. A special convention is used so that the routines will support partial evaluation: The |tip| fields in the output list either point to a vertex or hold one of the constant values 0 or~1 when regarded as an unsigned long integer. @d val x.I /* the field containing a boolean value */ @d typ y.I /* the field containing the gate type */ @d alt z.V /* the field pointing to another related gate */ @d outs zz.A /* the field pointing to the list of output gates */ @d is_boolean(v) ((unsigned long)(v)<=1) /* is a |tip| field constant? */ @d the_boolean(v) ((long)(v)) /* if so, this is its value */ @d tip_value(v) (is_boolean(v)? the_boolean(v): (v)->val) @d AND '&' @d OR '|' @d NOT '~' @d XOR '^' @(gb_gates.h@>= #define val @t\quad@> x.I /* the definitions are repeated in the header file */ #define typ @t\quad@> y.I #define alt @t\quad@> z.V #define outs @t\quad@> zz.A #define is_boolean(v) @t\quad@> ((unsigned long)(v)<=1) #define the_boolean(v) @t\quad@> ((long)(v)) #define tip_value(v) @t\quad@> (is_boolean(v)? the_boolean(v): (v)->val) #define AND @t\quad@> '&' #define OR @t\quad@> '|' #define NOT @t\quad@> '~' #define XOR @t\quad@> '^' @ Let's begin with the |gate_eval| procedure, because it is quite simple and because it illustrates the conventions just explained. Given a gate graph |g| and optional pointers |in_vec| and |out_vec|, the procedure |gate_eval| will assign values to each gate of~|g|. If |in_vec| is non-null, it should point to a string of characters, each |'0'| or~|'1'|, that will be assigned to the first gates of the network, in order; otherwise |gate_eval| assumes that all input gates have already received appropriate values and it will not change them. New values are computed for each gate after the bits of |in_vec| have been consumed. If |out_vec| is non-null, it should point to a memory area capable of receiving |m+1| characters, where |m| is the number of outputs of~|g|; a string containing the respective output values will be deposited there. If |gate_eval| encounters an unknown gate type, it terminates execution prematurely and returns the value |-1|. Otherwise it returns~0. @= long gate_eval(g,in_vec,out_vec) Graph *g; /* graph with gates as vertices */ char *in_vec; /* string for input values, or |NULL| */ char *out_vec; /* string for output values, or |NULL| */ {@+register Vertex *v; /* the current vertex of interest */ register Arc *a; /* the current arc of interest */ register char t; /* boolean value being computed */ if (!g) return -2; /* no graph supplied! */ v=g->vertices; if (in_vec) @; for (; vvertices+g->n; v++) { switch (v->typ) { /* branch on type of gate */ case 'I': continue; /* this input gate's value should be externally set */ case 'L': t=v->alt->val;@+break; @t\4\4@>@; default: return -1; /* unknown gate type! */ } v->val=t; /* assign the computed value */ } if (out_vec) @; return 0; } @ @= while (*in_vec && vvertices+g->n) (v++)->val = *in_vec++ - '0'; @ @= { for (a=g->outs; a; a=a->next) *out_vec++='0'+tip_value(a->tip); *out_vec=0; /* terminate the string */ } @ @= case AND: t=1; for (a=v->arcs; a; a=a->next) t &= a->tip->val; break; case OR: t=0; for (a=v->arcs; a; a=a->next) t |= a->tip->val; break; case XOR: t=0; for (a=v->arcs; a; a=a->next) t ^= a->tip->val; break; case NOT: t=1-v->arcs->tip->val; break; @ Here now is an outline of the entire {\sc GB\_\,GATES} module, as seen by the \CEE/ compiler: @p #include "gb_flip.h" /* we will use the {\sc GB\_\,FLIP} routines for random numbers */ #include "gb_graph.h" /* and we will use the {\sc GB\_\,GRAPH} data structures */ @h@# @@; @@; @@; @@; @@; @@; @@; @@; @@; @* The RISC netlist. The subroutine call |risc(regs)| creates a gate graph having |regs| registers; the value of |regs| must be between 2 and~16, inclusive, otherwise |regs| is set to~16. This gate graph describes the circuitry for a small RISC computer, defined below. The total number of gates turns out to be |1400+115*regs|; thus it lies between 1630 (when |regs=2|) and 3240 (when |regs=16|). {\sc EXCLUSIVE-OR} gates are not used; the effect of xoring is obtained where needed by means of {\sc AND}s, {\sc OR}s, and inverters. If |risc| cannot do its thing, it returns |NULL| (\.{NULL}) and sets |panic_code| to indicate the problem. Otherwise |risc| returns a pointer to the graph. @d panic(c) @+{@+panic_code=c;@+gb_trouble_code=0;@+return NULL;@+} @= Graph *risc(regs) unsigned long regs; /* number of registers supported */ {@+@@; @# @; @; if (gb_trouble_code) { gb_recycle(new_graph); panic(alloc_fault); /* oops, we ran out of memory somewhere back there */ } return new_graph; } @ @= Graph *new_graph; /* the graph constructed by |risc| */ register long k,r; /* all-purpose indices */ @ This RISC machine works with 16-bit registers and 16-bit data words. It cannot write into memory, but it assumes the existence of an external read-only memory. The circuit has 16 outputs, representing the 16 bits of a memory address register. It also has 17 inputs, the last 16 of which are supposed to be set to the contents of the memory address computed on the previous cycle. Thus we can run the machine by accessing memory between calls of |gate_eval|. The first input bit, called \.{RUN}, is normally set to~1; if it is~0, the other inputs are effectively ignored and all registers and outputs will be cleared to~0. Input bits for the memory appear in little-endian order,'' that is, least significant bit first; but the output bits for the memory address register appear in big-endian order,'' most significant bit first. Words read from memory are interpreted as instructions having the following format: \vbox{\offinterlineskip \def\\#1&{\omit&} \hrule \halign{&\vrule#&\strut\sevenrm\hbox to 1.7em{\hfil#\hfil}\cr height 5pt&\multispan7\hfill&&\multispan7\hfill&&\multispan3\hfill &&\multispan3\hfill&&\multispan7\hfill&\cr &\multispan7\hfill\.{DST}\hfill&&\multispan7\hfill\.{MOD}\hfill &&\multispan3\hfill\.{OP}\hfill&&\multispan3\hfill\.{A}\hfill &&\multispan7\hfill\.{SRC}\hfill&\cr height 5pt&\multispan7\hfill&&\multispan7\hfill&&\multispan3\hfill &&\multispan3\hfill&&\multispan7\hfill&\cr \noalign{\hrule} \\15&\\14&\\13&\\12&\\11&\\10&\\9&\\8&\\7&\\6&\\5&\\4&\\3&\\2&\\1&% \\0&\omit\cr}} The \.{SRC} and \.A fields specify a source'' value. If $\.A=0$, the source is \.{SRC}, treated as a 16-bit signed number between $-8$ and $+7$ inclusive. If $\.A=1$, the source is the contents of register \.{DST} plus the (signed) value of \.{SRC}. If $\.A=2$, the source is the contents of register \.{SRC}. And if $\.A=3$, the source is the contents of the memory location whose address is the contents of register \.{SRC}. Thus, for example, if $\.{DST}=3$ and $\.{SRC}=10$, and if \.{r3} contains 17 while \.{r10} contains 1009, the source value will be $-6$ if $\.A=0$, or $17-6=11$ if $\.A=1$, or 1009 if $\.A=2$, or the contents of memory location 1009 if $\.A=3$. The \.{DST} field specifies the number of the destination register. This register receives a new value based on its previous value and the source value, as prescribed by the operation defined in the \.{OP} and \.{MOD} fields. For example, when $\.{OP}=0$, a general logical operation is performed, as follows: Suppose the bits of \.{MOD} are called $\mu_{11}\mu_{10}\mu_{01} \mu_{00}$ from left to right. Then if the $k$th bit of the destination register currently is equal to~$i$ and the $k$th bit of the source value is equal to~$j$, the general logical operator changes the $k$th bit of the destination register to~$\mu_{ij}$. If the \.{MOD} bits are, for example, $1010$, the source value is simply copied to the destination register; if $\.{MOD}=0110$, an exclusive-or is done; if $\.{MOD}=0011$, the destination register is complemented and the source value is effectively ignored. The machine contains four status bits called \.S (sign), \.N (nonzero), \.K (carry), and \.V (overflow). Every general logical operation sets \.S equal to the sign of the new result transferred to the destination register; this is bit~15, the most significant bit. A general logical operation also sets \.N to~1 if any of the other 15 bits are~1, to~0 if all of the other bits are~0. Thus \.S and \.N both become zero if and only if the new result is entirely zero. Logical operations do not change the values of \.K and~\.V; the latter are affected only by the arithmetic operations described below. The status of the \.S and \.N bits can be tested by using the conditional load operator, $\.{OP}=2$: This operation loads the source value into the destination register if and only if \.{MOD} bit $\mu_{ij}=1$, where $i$ and~$j$ are the current values of \.S and~\.N, respectively. For example, if $\.{MOD}=0011$, the source value is loaded if and only if $\.S=0$, which means that the last value affecting \.S and~\.N was greater than or equal to zero. If $\.{MOD}=1111$, loading is always done; this option provides a way to move source to destination without affecting \.S or~\.N. A second conditional load operator, $\.{OP}=3$, is similar, but it is used for testing the status of \.K and~\.V instead of \.S and~\.N. For example, a command having $\.{MOD}=1010$, $\.{OP}=3$, $\.A=1$, and $\.{SRC}=1$ adds the current overflow bit to the destination register. (Please take a moment to understand why this is true.) We have now described all the operations except those that are performed when $\.{OP}=1$. As you might expect, our machine is able to do rudimentary arithmetic. The general addition and subtraction operators belong to this final case, together with various shift operators, depending on the value of \.{MOD}. Eight of the $\.{OP}=1$ operations set the destination register to a shifted version of the source value: $\.{MOD}=0$ means shift left~1,'' which is equivalent to multiplying the source by~2; $\.{MOD}=1$ means cyclic shift left~1,'' which is the same except that it also adds the previous sign bit to the result; $\.{MOD}=2$ means shift left~4,'' which is equivalent to multiplying by~16; $\.{MOD}=3$ means cyclic shift left~4''; $\.{MOD}=4$ means shift right~1,'' which is equivalent to dividing the source by~2 and rounding down to the next lower integer if there was a remainder; $\.{MOD}=5$ means unsigned shift right~1,'' which is the same except that the most significant bit is always set to zero instead of retaining the previous sign; $\.{MOD}=6$ means shift right~4,'' which is equivalent to dividing the source by~16 and rounding down; $\.{MOD}=7$ means unsigned shift right~4.'' Each of these shift operations affects \.S and~\.N, as in the case of logical operations. They also affect \.K and~\.V, as follows: Shifting left sets \.K to~1 if and only if at least one of the bits shifted off the left was nonzero, and sets \.V to~1 if and only if the corresponding multiplication would cause overflow. Shifting right~1 sets \.K to the value of the bit shifted out, and sets \.V to~0; shifting right~4 sets \.K to the value of the last bit shifted out, and sets \.V to the logical {\sc OR} of the other three lost bits. The same values of \.K and \.V arise from cyclic or unsigned shifts as from ordinary shifts. When $\.{OP}=1$ and $\.{MOD}=8$, the source value is added to the destination register. This sets \.S, \.N, and \.V as you would expect; and it sets \.K to the carry you would get if you were treating the operands as 16-bit unsigned integers. Another addition operation, having $\.{MOD}=9$, is similar, but the current value of \.K is also added to the result; in this case, the new value of \.N will be zero if and only if the 15 non-sign bits of the result are zero and the previous values of \.S and~\.N were also zero. This means that you can use the first addition operation on the lower halves of a 32-bit number and the second operation on the upper halves, thereby obtaining a correct 32-bit result, with appropriate sign, nonzero, carry, and overflow bits set. Higher precision (48 bits, 64 bits, etc.)~can be obtained in a similar way. When $\.{OP}=1$ and $\.{MOD}=10$, the source value is subtracted from the destination register. Again, \.S, \.N, \.K, and \.V are set; the \.K value in this case represents the borrow'' bit. An auxiliary subtraction operation, having $\.{MOD}=11$, subtracts also the current value of \.K, thereby allowing for correct 32-bit subtraction. The operations for $\.{OP}=1$ and $\.{MOD}=12$, 13, and~14 are reserved for future expansion.'' Actually they will never change, however, since this RISC chip is purely academic. If you check out the logic below, you will find that they simply set the destination register and the four status bits all to zero. A final operation, called \.{JUMP}, will be explained momentarily. It has $\.{OP}=1$ and $\.{MOD}=15$. It does not affect \.S, \.N, \.K, or~\.V. If the RISC is made with fewer than 16 registers, the higher-numbered ones will effectively contain zero whenever their values are fetched. But if you use them as destination registers, you will set \.S, \.N, \.K, and~\.V as if actual numbers were being stored. Register 0 is different from the other 15 registers: It is the location of the current instruction. Therefore if you change the contents of register~0, you are changing the control flow of the program. If you do not change register~0, it automatically increases by~1. Special treatment occurs when $\.A=3$ and $\.{SRC}=0$. In such a case, the normal rules given above say that the source value should be the contents of the memory location specified by register~0. But that memory location holds the current instruction; so the machine uses the {\sl following\/} location instead, as a 16-bit source operand. If the contents of register~0 are not changed by such a two-word instruction, register~0 will increase by~2 instead of~1. We have now discussed everything about the machine except the operation of the \.{JUMP} command. This command moves the source value to register~0, thereby changing the flow of control. Furthermore, if $\.{DST}\ne0$, it also sets register \.{DST} to the location of the instruction following the \.{JUMP}. Assembly language programmers will recognize this as a convenient way to jump to a subroutine. Example programs can be found in the {\sc TAKE\_\,RISC} module, which includes a simple subroutine for multiplication and division. @ A few auxiliary functions will ameliorate the task of constructing the RISC logic. First comes a routine that christens'' a new gate, assigning it a name and a type. The name is constructed from a prefix and a serial number, where the prefix indicates the current portion of logic being created. @= static Vertex* new_vert(t) char t; /* the type of the new gate */ {@+register Vertex *v; v=next_vert++; if (count<0) v->name=gb_save_string(prefix); else { sprintf(name_buf,"%s%ld",prefix,count); v->name=gb_save_string(name_buf); count++; } v->typ=t; return v; } @ @d start_prefix(s) strcpy(prefix,s);@+count=0 @d numeric_prefix(a,b) sprintf(prefix,"%c%ld:",a,b);@+count=0; @= static Vertex* next_vert; /* the first vertex not yet assigned a name */ static char prefix[5]; /* prefix string for vertex names */ static long count; /* serial number for vertex names */ static char name_buf[100]; /* place to form vertex names */ @ Here are some trivial routines to create gates with 2, 3, or more arguments. The arcs from such a gate to its inputs are assigned length 100. Other routines, defined below, assign length~1 to the arc between an inverter and its unique input. This convention makes the lengths of shortest paths in the resulting network a bit more interesting than they would otherwise be. @d DELAY 100L @= static Vertex* make2(t,v1,v2) char t; /* the type of the new gate */ Vertex *v1,*v2; {@+register Vertex *v=new_vert(t); gb_new_arc(v,v1,DELAY); gb_new_arc(v,v2,DELAY); return v; } @# static Vertex* make3(t,v1,v2,v3) char t; /* the type of the new gate */ Vertex *v1,*v2,*v3; {@+register Vertex *v=new_vert(t); gb_new_arc(v,v1,DELAY); gb_new_arc(v,v2,DELAY); gb_new_arc(v,v3,DELAY); return v; } @# static Vertex* make4(t,v1,v2,v3,v4) char t; /* the type of the new gate */ Vertex *v1,*v2,*v3,*v4; {@+register Vertex *v=new_vert(t); gb_new_arc(v,v1,DELAY); gb_new_arc(v,v2,DELAY); gb_new_arc(v,v3,DELAY); gb_new_arc(v,v4,DELAY); return v; } @# static Vertex* make5(t,v1,v2,v3,v4,v5) char t; /* the type of the new gate */ Vertex *v1,*v2,*v3,*v4,*v5; {@+register Vertex *v=new_vert(t); gb_new_arc(v,v1,DELAY); gb_new_arc(v,v2,DELAY); gb_new_arc(v,v3,DELAY); gb_new_arc(v,v4,DELAY); gb_new_arc(v,v5,DELAY); return v; } @ We use utility field |w.V| to store a pointer to the complement of a gate, if that complement has been formed. This trick prevents the creation of excessive gates that are equivalent to each other. The following subroutine returns a pointer to the complement of a given gate. @d bar w.V /* field pointing to complement, if known to exist */ @d even_comp(s,v) ((s)&1? v: comp(v)) @= static Vertex* comp(v) Vertex *v; {@+register Vertex *u; if (v->bar) return v->bar; u=next_vert++; u->bar=v;@+v->bar=u; sprintf(name_buf,"%s~",v->name); u->name=gb_save_string(name_buf); u->typ=NOT; gb_new_arc(u,v,1L); return u; } @ To create a gate for the {\sc EXCLUSIVE-OR} of two arguments, we can either construct the {\sc OR} of two {\sc AND}s or the {\sc AND} of two {\sc OR}s. We choose the former alternative: @= static Vertex* make_xor(u,v) Vertex *u,*v; {@+register Vertex *t1,*t2; t1=make2(AND,u,comp(v)); t2=make2(AND,comp(u),v); return make2(OR,t1,t2); } @ OK, let's get going. @= if (regs<2 || regs>16) regs=16; new_graph=gb_new_graph(1400+115*regs); if (new_graph==NULL) panic(no_room); /* out of memory before we're even started */ sprintf(new_graph->id,"risc(%lu)",regs); strcpy(new_graph->util_types,"ZZZIIVZZZZZZZA"); next_vert=new_graph->vertices; @ @= @; @; @; @; @; @; @; if (next_vert!=new_graph->vertices+new_graph->n) panic(impossible); /* oops, we miscounted; this should be impossible */ @ Internal names will make it convenient to refer to the most important gates. Here are the names of inputs and latches. @= Vertex *run_bit; /* the \.{RUN} input */ Vertex *mem[16]; /* 16 bits of input from read-only memory */ Vertex *prog; /* first of 10 bits in the program register */ Vertex *sign; /* the latched value of \.S */ Vertex *nonzero; /* the latched value of \.N */ Vertex *carry; /* the latched value of \.K */ Vertex *overflow; /* the latched value of \.V */ Vertex *extra; /* latched status bit: are we doing an extra memory cycle? */ Vertex *reg[16]; /* the least-significant bit of a given register */ @ @d first_of(n,t) new_vert(t);@+for (k=1;k= strcpy(prefix,"RUN");@+count=-1;@+run_bit=new_vert('I'); start_prefix("M");@+for (k=0;k<16;k++)@+mem[k]=new_vert('I'); start_prefix("P");@+prog=first_of(10,'L'); strcpy(prefix,"S");@+count=-1;@+sign=new_vert('L'); strcpy(prefix,"N");@+nonzero=new_vert('L'); strcpy(prefix,"K");@+carry=new_vert('L'); strcpy(prefix,"V");@+overflow=new_vert('L'); strcpy(prefix,"X");@+extra=new_vert('L'); for (r=0;r= Vertex *t1,*t2,*t3,*t4,*t5; /* temporary holds to force evaluation order */ Vertex *tmp[16]; /* additional holding places for partial results */ Vertex *imm; /* is the source value immediate (a given constant)? */ Vertex *rel; /* is the source value relative to the current destination register? */ Vertex *dir; /* should the source value be fetched directly from a source register? */ Vertex *ind; /* should the source value be fetched indirectly from memory? */ Vertex *op; /* least significant bit of \.{OP} */ Vertex *cond; /* most significant bit of \.{OP} */ Vertex *mod[4]; /* the \.{MOD} bits */ Vertex *dest[4]; /* the \.{DEST} bits */ @ The sixth line of the program here can be translated into the logic equation $$|op|=(|extra|\land|prog|)\lor(\mskip1mu\overline{|extra|}\land|mem[6]|)\,.$$ Once you see why, you'll be able to read the rest of this curious code. @= start_prefix("D"); do3(imm,AND,comp(extra),comp(mem[4]),comp(mem[5])); /* $\.A=0$ */ do3(rel,AND,comp(extra),mem[4],comp(mem[5])); /* $\.A=1$ */ do3(dir,AND,comp(extra),comp(mem[4]),mem[5]); /* $\.A=2$ */ do3(ind,AND,comp(extra),mem[4],mem[5]); /* $\.A=3$ */ do2(op,OR,make2(AND,extra,prog),make2(AND,comp(extra),mem[6])); do2(cond,OR,make2(AND,extra,prog+1),make2(AND,comp(extra),mem[7])); for (k=0;k<4;k++) { do2(mod[k],OR,make2(AND,extra,prog+2+k),make2(AND,comp(extra),mem[8+k])); do2(dest[k],OR,make2(AND,extra,prog+6+k),make2(AND,comp(extra),mem[12+k])); } @ @= start_prefix("F"); @; @; @; for (k=0;k<16;k++)@/ do4(source[k],OR, make2(AND,imm,mem[k<4?k:3]), make2(AND,rel,inc_dest[k]),@| make2(AND,dir,old_src[k]), make2(AND,extra,mem[k])); @ Here and in the immediately following section we create {\sc OR} gates |old_dest[k]| and |old_src[k]| that might have as many as 16~inputs. (The actual number of inputs is |regs|.) All other gates in the network will have at most five inputs. @= for (r=0;r>1,dest[1]),@| even_comp(r>>2,dest[2]),even_comp(r>>3,dest[3])); for (k=0;k<16;k++) { for (r=0;r= for (k=0;k<16;k++) { for (r=0;r>1,mem[1]), even_comp(r>>2,mem[2]),even_comp(r>>3,mem[3])); old_src[k]=new_vert(OR); for (r=0;r= Vertex *dest_match[16]; /* |dest_match[r]==1| iff $\.{DST}=r$ */ Vertex *old_dest[16]; /* contents of destination register before operation */ Vertex *old_src[16]; /* contents of source register before operation */ Vertex *inc_dest[16]; /* |old_dest| plus the \.{SRC} field */ Vertex *source[16]; /* source value for the operation */ Vertex *log[16]; /* result of general logic operation */ Vertex *shift[18]; /* result of shift operation, with carry and overflow */ Vertex *sum[18]; /* |old_dest| plus |source| plus optional carry */ Vertex *diff[18]; /* |old_dest| minus |source| minus optional borrow */ Vertex *next_loc[16]; /* contents of register 0, plus 1 */ Vertex *next_next_loc[16]; /* contents of register 0, plus 2 */ Vertex *result[18]; /* result of operating on |old_dest| and |source| */ @ @= start_prefix("L"); for (k=0;k<16;k++)@/ do4(log[k],OR,@| make3(AND,mod[0],comp(old_dest[k]),comp(source[k])),@| make3(AND,mod[1],comp(old_dest[k]),source[k]),@| make3(AND,mod[2],old_dest[k],comp(source[k])),@| make3(AND,mod[3],old_dest[k],source[k])); @ @= start_prefix("C"); do4(tmp[0],OR,@| make3(AND,mod[0],comp(sign),comp(nonzero)),@| make3(AND,mod[1],comp(sign),nonzero),@| make3(AND,mod[2],sign,comp(nonzero)),@| make3(AND,mod[3],sign,nonzero)); do4(tmp[1],OR,@| make3(AND,mod[0],comp(carry),comp(overflow)),@| make3(AND,mod[1],comp(carry),overflow),@| make3(AND,mod[2],carry,comp(overflow)),@| make3(AND,mod[3],carry,overflow)); do3(change,OR,comp(cond),make2(AND,tmp[0],comp(op)),make2(AND,tmp[1],op)); @ @= Vertex *change; /* is the destination register supposed to change? */ @ Hardware is like software except that it performs all the operations all the time and then selects only the results it needs. (If you think about it, this is a profound observation about economics, society, and nature. Gosh.) @= start_prefix("Z"); @; @; @; @; @; @; @ @= next_loc[0]=comp(reg[0]);@+next_next_loc[0]=reg[0]; next_loc[1]=make_xor(reg[0]+1,reg[0]);@+next_next_loc[1]=comp(reg[0]+1); for (t5=reg[0]+1,k=2;k<16;t5=make2(AND,t5,reg[0]+k++)) { next_loc[k]=make_xor(reg[0]+k,make2(AND,reg[0],t5)); next_next_loc[k]=make_xor(reg[0]+k,t5); } @ @= jump=make5(AND,op,mod[0],mod[1],mod[2],mod[3]); /* assume |cond=0| */ for (k=0;k<16;k++) { do5(result[k],OR,@| make2(AND,comp(op),log[k]),@| make2(AND,jump,next_loc[k]),@| make3(AND,op,comp(mod[3]),shift[k]),@| make5(AND,op,mod[3],comp(mod[2]),comp(mod[1]),sum[k]),@| make5(AND,op,mod[3],comp(mod[2]),mod[1],diff[k])); do2(result[k],OR,@| make3(AND,cond,change,source[k]),@| make2(AND,comp(cond),result[k])); } for (k=16;k<18;k++) /* carry and overflow bits of the result */ do3(result[k],OR,@| make3(AND,op,comp(mod[3]),shift[k]),@| make5(AND,op,mod[3],comp(mod[2]),comp(mod[1]),sum[k]),@| make5(AND,op,mod[3],comp(mod[2]),mod[1],diff[k])); @ The program register |prog| and the |extra| bit are needed for the case when we must spend an extra cycle to fetch a word from memory. On the first cycle, |ind| is true, so a result'' is calculated but not actually used. On the second cycle, |extra| is true. A slight optimization has been introduced in order to make the circuit a bit more interesting: If a conditional load instruction occurs with indirect addressing and a false condition, the extra cycle is not taken. (The |next_next_loc| values were computed for this reason.) @d latchit(u,@!latch) (latch)->alt=make2(AND,u,run_bit) /* |u&run_bit| is new value for |latch| */ @= for (k=0;k<10;k++) latchit(mem[k+6],prog+k); do2(nextra,OR,make2(AND,ind,comp(cond)),make2(AND,ind,change)); latchit(nextra,extra); nzs=make4(OR,mem[0],mem[1],mem[2],mem[3]); nzd=make4(OR,dest[0],dest[1],dest[2],dest[3]); @ @= Vertex *jump; /* is this command a \.{JUMP}, assuming |cond| is false? */ Vertex *nextra; /* must we take an extra cycle? */ Vertex *nzs; /* is the \.{SRC} field nonzero? */ Vertex *nzd; /* is the \.{DST} field nonzero? */ @ @= t5=make2(AND,change,comp(ind)); /* should destination register change? */ for (r=1;r= do4(t5,OR,@| make2(AND,sign,cond),@| make2(AND,sign,jump),@| make2(AND,sign,ind),@| make4(AND,result[15],comp(cond),comp(jump),comp(ind))); latchit(t5,sign); do4(t5,OR,@| make4(OR,result[0],result[1],result[2],result[3]),@| make4(OR,result[4],result[5],result[6],result[7]),@| make4(OR,result[8],result[9],result[10],result[11]),@| make4(OR,result[12],result[13],result[14],@| @t\hskip5em@>make5(AND,make2(OR,nonzero,sign),op,mod[0],comp(mod[2]),mod[3]))); do4(t5,OR,@| make2(AND,nonzero,cond),@| make2(AND,nonzero,jump),@| make2(AND,nonzero,ind),@| make4(AND,t5,comp(cond),comp(jump),comp(ind))); latchit(t5,nonzero); do5(t5,OR,@| make2(AND,overflow,cond),@| make2(AND,overflow,jump),@| make2(AND,overflow,comp(op)),@| make2(AND,overflow,ind),@| make5(AND,result[17],comp(cond),comp(jump),comp(ind),op)); latchit(t5,overflow); do5(t5,OR,@| make2(AND,carry,cond),@| make2(AND,carry,jump),@| make2(AND,carry,comp(op)),@| make2(AND,carry,ind),@| make5(AND,result[16],comp(cond),comp(jump),comp(ind),op)); latchit(t5,carry); @ As usual, we have left the hardest case for last, hoping that we will have learned enough tricks to handle it when the time of reckoning finally arrives. The most subtle part of the logic here is perhaps the case of a \.{JUMP} command with $\.A=3$. We want to increase register~0 by~1 during the first cycle of such a command, if $\.{SRC}=0$, so that the |result| will be correct on the next cycle. @= skip=make2(AND,cond,comp(change)); /* false conditional? */ hop=make2(AND,comp(cond),jump); /* \.{JUMP} command? */ do4(normal,OR,@| make2(AND,skip,comp(ind)),@| make2(AND,skip,nzs),@| make3(AND,comp(skip),ind,comp(nzs)),@| make3(AND,comp(skip),comp(hop),nzd)); special=make3(AND,comp(skip),ind,nzs); for (k=0;k<16;k++) { do4(t5,OR,@| make2(AND,normal,next_loc[k]),@| make4(AND,skip,ind,comp(nzs),next_next_loc[k]),@| make3(AND,hop,comp(ind),source[k]),@| make5(AND,comp(skip),comp(hop),comp(ind),comp(nzd),result[k])); do2(t4,OR,@| make2(AND,special,reg[0]+k),@| make2(AND,comp(special),t5)); latchit(t4,reg[0]+k); do2(t4,OR,@| make2(AND,special,old_src[k]),@| make2(AND,comp(special),t5)); {@+register Arc *a=gb_virgin_arc(); a->tip=make2(AND,t4,run_bit); a->next=new_graph->outs; new_graph->outs=a; /* pointer to memory address bit */ } } /* arcs for output bits will appear in big-endian order */ @ @= Vertex *skip; /* are we skipping a conditional load operation? */ Vertex *hop; /* are we doing a \.{JUMP}? */ Vertex *normal; /* is this a case where register 0 is simply incremented? */ Vertex *special; /* is this a case where register 0 and the memory address register will not coincide? */ @* Serial addition. We haven't yet specified the parts of |risc| that deal with addition and subtraction; somehow, those parts wanted to be separate from the rest. To complete our mission, we will use subroutine calls of the form |make_adder(n,x,y,z,carry,add)|', where |x| and |y| are |n|-bit arrays of input gates and |z|~is an |(n+1)|-bit array of output gates. If |add!=0|, the subroutine computes |x+y|, otherwise it computes |x-y|. If |carry!=0|, the |carry| gate is effectively added to~|y| before the operation. A simple |n|-stage serial scheme, which reduces the problem of |n|-bit addition to |(n-1)|-bit addition, is adequate for our purposes here. (A parallel adder, which gains efficiency by reducing the problem size from |n| to~$n/\phi$, can be found in the |prod| routine below.) The handy identity $x-y=\overline{\overline x+y}$ is used to reduce subtraction to addition. @= static void make_adder(n,x,y,z,carry,add) unsigned long n; /* number of bits */ Vertex *x[],*y[]; /* input gates */ Vertex *z[]; /* output gates */ Vertex *carry; /* add this to |y|, unless it's null */ char add; /* should we add or subtract? */ {@+register long k; Vertex *t1,*t2,*t3,*t4; /* temporary storage used by |do4| */ if (!carry) { z[0]=make_xor(x[0],y[0]); carry=make2(AND,even_comp(add,x[0]),y[0]); k=1; }@+else k=0; for (;k= make_adder(4L,old_dest,mem,inc_dest,NULL,1); up=make2(AND,inc_dest[4],comp(mem[3])); /* remaining bits must increase */ down=make2(AND,comp(inc_dest[4]),mem[3]); /* remaining bits must decrease */ for (k=4;;k++) { comp(up);@+comp(down); do3(inc_dest[k],OR,@| make2(AND,comp(old_dest[k]),up),@| make2(AND,comp(old_dest[k]),down),@| make3(AND,old_dest[k],comp(up),comp(down))); if (k<15) { up=make2(AND,up,old_dest[k]); down=make2(AND,down,comp(old_dest[k])); }@+else break; } @ @= Vertex *up,*down; /* gates used when computing |inc_dest| */ @ In the second place, we need a 16-bit adder and a 16-bit subtracter for the four addition/subtraction commands. @= start_prefix("A"); @; make_adder(16L,old_dest,source,sum,make2(AND,carry,mod[0]),1); /* adder */ make_adder(16L,old_dest,source,diff,make2(AND,carry,mod[0]),0); /* subtracter */ do2(sum[17],OR,@| make3(AND,old_dest[15],source[15],comp(sum[15])),@| make3(AND,comp(old_dest[15]),comp(source[15]),sum[15])); /* overflow */ do2(diff[17],OR,@| make3(AND,old_dest[15],comp(source[15]),comp(diff[15])),@| make3(AND,comp(old_dest[15]),source[15],diff[15])); /* overflow */ @ @= for (k=0;k<16;k++)@/ do4(shift[k],OR,@| (k==0? make4(AND,source[15],mod[0],comp(mod[1]),comp(mod[2])):@| @t\hskip5em@>make3(AND,source[k-1],comp(mod[1]),comp(mod[2]))),@| (k<4? make4(AND,source[k+12],mod[0],mod[1],comp(mod[2])):@| @t\hskip5em@>make3(AND,source[k-4],mod[1],comp(mod[2]))),@| (k==15? make4(AND,source[15],comp(mod[0]),comp(mod[1]),mod[2]):@| @t\hskip5em@>make3(AND,source[k+1],comp(mod[1]),mod[2])),@| (k>11? make4(AND,source[15],comp(mod[0]),mod[1],mod[2]):@| @t\hskip5em@>make3(AND,source[k+4],mod[1],mod[2]))); do4(shift[16],OR,@| make2(AND,comp(mod[2]),source[15]),@| make3(AND,comp(mod[2]),mod[1], make3(OR,source[14],source[13],source[12])),@| make3(AND,mod[2],comp(mod[1]),source[0]),@| make3(AND,mod[2],mod[1],source[3])); /* carry'' */ do3(shift[17],OR,@| make3(AND,comp(mod[2]),comp(mod[1]), make_xor(source[15],source[14])),@| make4(AND,comp(mod[2]),mod[1],@| @t\hskip5em@>make5(OR,source[15],source[14], source[13],source[12],source[11]),@| @t\hskip5em@>make5(OR,comp(source[15]),comp(source[14]), comp(source[13]),@| @t\hskip10em@>comp(source[12]),comp(source[11]))),@| make3(AND,mod[2],mod[1], make3(OR,source[0],source[1],source[2]))); /* overflow'' */ @* RISC management. The |run_risc| procedure takes a gate graph output by |risc| and simulates its behavior, given the contents of its read-only memory. (See the demonstration program {\sc TAKE\_\,RISC}, which appears in a module by itself, for a typical illustration of how |run_risc| might be used.) This procedure clears the simulated machine and begins executing the program that starts at address~0. It stops when it gets to an address greater than the size of read-only memory supplied. One way to stop it is therefore to execute a command such as |0x0f00|, which will transfer control to location |0xffff|; even better is |0x0f8f|, which transfers to location |0xffff| without changing the status of \.S and \.N. However, if the given read-only memory contains a full set of $2^{16}$ words, |run_risc| will never stop. When |run_risc| does stop, it returns 0 and puts the final contents of the simulated registers into the global array |risc_state|. Or, if |g| was not a decent graph, |run_risc| returns a negative value and leaves |risc_state| untouched. @= long run_risc(g,rom,size,trace_regs) Graph *g; /* graph output by |risc| */ unsigned long rom[]; /* contents of read-only memory */ unsigned long size; /* length of |rom| vector */ unsigned long trace_regs; /* if nonzero, this many registers will be traced */ {@+register unsigned long l; /* memory address */ register unsigned long m; /* memory or register contents */ register Vertex *v; /* the current gate of interest */ register Arc *a; /* the current output list element of interest */ register long k,r; /* general-purpose indices */ long x,s,n,c,o; /* status bits */ if (trace_regs) @; r=gate_eval(g,"0",NULL); /* reset the RISC by turning off the \.{RUN} bit */ if (r<0) return r; /* not a valid gate graph! */ g->vertices->val=1; /* turn the \.{RUN} bit on */ while (1) { for (a=g->outs,l=0;a;a=a->next) l=2*l+a->tip->val; /* set $l=\null$memory address */ if (trace_regs) @; if (l>=size) break; /* stop if memory check occurs */ for (v=g->vertices+1,m=rom[l];v<=g->vertices+16;v++,m>>=1) v->val=m&1; /* store bits of memory word in the input gates */ gate_eval(g,NULL,NULL); /* do another RISC cycle */ } if (trace_regs) @; @; return 0; } @ If tracing is requested, we write on the standard output file. @= { for (r=0;r= printf("Execution terminated with memory address %04lx.\n",l); @ Here we peek inside the circuit to see what values are about to be latched. @= { for (r=0;rvertices+(16*r+47); /* most significant bit of register |r| */ m=0; if (v->typ=='L') for (k=0,m=0;k<16;k++,v--) m=2*m+v->alt->val; printf("%04lx ",m); } for (k=0,m=0,v=g->vertices+26;k<10;k++,v--) m=2*m+v->alt->val; /* |prog| */ x=(g->vertices+31)->alt->val; /* |extra| */ s=(g->vertices+27)->alt->val; /* |sign| */ n=(g->vertices+28)->alt->val; /* |nonzero| */ c=(g->vertices+29)->alt->val; /* |carry| */ o=(g->vertices+30)->alt->val; /* |overflow| */ printf("%03lx%c%c%c%c%c ",m<<2, x?'X':'.', s?'S':'.', n?'N':'.', c?'K':'.', o?'V':'.'); if (l>=size) printf("????\n"); else printf("%04lx\n",rom[l]); } @ @= for (r=0;r<16;r++) { v=g->vertices+(16*r+47); /* most significant bit of register |r| */ m=0; if (v->typ=='L') for (k=0,m=0;k<16;k++,v--) m=2*m+v->alt->val; risc_state[r]=m; } for (k=0,m=0,v=g->vertices+26;k<10;k++,v--) m=2*m+v->alt->val; /* |prog| */ m=4*m+(g->vertices+31)->alt->val; /* |extra| */ m=2*m+(g->vertices+27)->alt->val; /* |sign| */ m=2*m+(g->vertices+28)->alt->val; /* |nonzero| */ m=2*m+(g->vertices+29)->alt->val; /* |carry| */ m=2*m+(g->vertices+30)->alt->val; /* |overflow| */ risc_state[16]=m; /* program register and status bits go here */ risc_state[17]=l; /* this is the out-of-range address that caused termination */ @ @= unsigned long risc_state[18]; @*Generalized gate graphs. For intermediate computations, it is convenient to allow two additional types of gates: {\advance\parindent 2em \smallskip \item{|'C'|} denotes a constant gate of value |z.I|. \smallskip \item{|'='|} denotes a copy of a previous gate; utility field |alt| points to that previous gate. \smallskip}\noindent Such gates might appear anywhere in the graph, possibly interspersed with the inputs and latches. Here is a simple subroutine that prints a symbolic representation of a generalized gate graph on the standard output file: @d bit z.I /* field containing the constant value of a |'C'| gate */ @d print_gates p_gates /* abbreviation makes chopped-off name unique */ @= static void pr_gate(v) Vertex *v; {@+register Arc *a; printf("%s = ",v->name); switch(v->typ) { case 'I':printf("input");@+break; case 'L':printf("latch"); if (v->alt) printf("ed %s",v->alt->name); break; case '~':printf("~ ");@+break; case 'C':printf("constant %ld",v->bit); break; case '=':printf("copy of %s",v->alt->name); } for (a=v->arcs;a;a=a->next) { if (a!=v->arcs) printf(" %c ",(char)v->typ); printf(a->tip->name); } printf("\n"); } @# void print_gates(g) Graph *g; {@+register Vertex *v; register Arc *a; for (v=g->vertices;vvertices+g->n;v++) pr_gate(v); for (a=g->outs;a;a=a->next) if (is_boolean(a->tip)) printf("Output %ld\n",the_boolean(a->tip)); else printf("Output %s\n",a->tip->name); } @ @(gb_gates.h@>= #define bit @t\quad@> z.I @ The |reduce| routine takes a generalized graph |g| and uses the identities $\overline{\overline x}=x$ and \openup1\jot \vbox{\halign{\hfilx#0=\null&#,\hfil\quad &\hfilx#1=\null&#,\hfil\quad &\hfilx#x=\null&#,\hfil\quad &\hfilx#\overline x=\null&#,\hfil\cr \land&0&\land&x&\land&x&\land&0\cr \lor&x&\lor&1&\lor&x&\lor&1\cr \oplus&x&\oplus&\overline x&\oplus&0&\oplus&1\cr}} to create an equivalent graph having no |'C'| or |'='| or obviously redundant gates. The reduced graph also excludes any gates that are not used directly or indirectly in the computation of the output values. @= static Graph* reduce(g) Graph *g; {@+register Vertex *u, *v; /* the current vertices of interest */ register Arc *a, *b; /* the current arcs of interest */ Arc *aa, *bb; /* their predecessors */ Vertex *latch_ptr; /* top of the latch list */ long n=0; /* the number of marked gates */ Graph *new_graph; /* the reduced gate graph */ Vertex *next_vert=NULL, *max_next_vert=NULL; /* allocation of new vertices */ Arc *avail_arc=NULL; /* list of recycled arcs */ Vertex *sentinel; /* end of the vertices */ if (g==NULL) panic(missing_operand); /* where is |g|? */ sentinel=g->vertices+g->n; while (1) { latch_ptr=NULL; for (v=g->vertices;v; @; } @; @; gb_recycle(g); return new_graph; } @ We will link latches together via their |v.V| fields. @= {@+char no_constants_yet=1; for (v=latch_ptr;v;v=v->v.V) { u=v->alt; /* the gate whose value will be latched */ if (u->typ=='=') v->alt=u->alt; else if (u->typ=='C') { v->typ='C';@+v->bit=u->bit;@+no_constants_yet=0; } } if (no_constants_yet) break; } @ @d foo x.V /* link field used to find all the gates later */ @= { switch(v->typ) { case 'L': v->v.V=latch_ptr;@+latch_ptr=v;@+break; case 'I': case 'C': break; case '=': u=v->alt; if (u->typ=='=') v->alt=u->alt; else if (u->typ=='C') { v->bit=u->bit;@+goto make_v_constant; } break; case NOT:@; case AND:@;@+goto test_single_arg; case OR:@;@+goto test_single_arg; case XOR:@; test_single_arg: if (v->arcs->next) break; v->alt=v->arcs->tip; make_v_eq: v->typ='=';@+goto make_v_arcless; make_v_1: v->bit=1;@+goto make_v_constant; make_v_0: v->bit=0; make_v_constant: v->typ='C'; make_v_arcless: v->arcs=NULL; } v->bar=NULL; /* this field will point to the complement, if computed later */ done: v->foo=v+1; /* this field will link all the vertices together */ } @ @= u=v->arcs->tip; if (u->typ=='=') u=v->arcs->tip=u->alt; if (u->typ=='C') { v->bit=1-u->bit;@+goto make_v_constant; }@+else if (u->bar) { /* this inverse already computed */ v->alt=u->bar;@+goto make_v_eq; }@+else { u->bar=v;@+v->bar=u;@+goto done; } @ @= for (a=v->arcs,aa=NULL;a;a=a->next) { u=a->tip; if (u->typ=='=') u=a->tip=u->alt; if (u->typ=='C') { if (u->bit==0) goto make_v_0; goto bypass_and; }@+else@+for (b=v->arcs;b!=a;b=b->next) { if (b->tip==u) goto bypass_and; if (b->tip==u->bar) goto make_v_0; } aa=a;@+continue; bypass_and: if (aa) aa->next=a->next; else v->arcs=a->next; } if (v->arcs==NULL) goto make_v_1; @ @= for (a=v->arcs,aa=NULL;a;a=a->next) { u=a->tip; if (u->typ=='=') u=a->tip=u->alt; if (u->typ=='C') { if (u->bit) goto make_v_1; goto bypass_or; }@+else@+for (b=v->arcs;b!=a;b=b->next) { if (b->tip==u) goto bypass_or; if (b->tip==u->bar) goto make_v_1; } aa=a;@+continue; bypass_or: if (aa) aa->next=a->next; else v->arcs=a->next; } if (v->arcs==NULL) goto make_v_0; @ @= {@+long cmp=0; for (a=v->arcs,aa=NULL;a;a=a->next) { u=a->tip; if (u->typ=='=') u=a->tip=u->alt; if (u->typ=='C') { if (u->bit) cmp=1-cmp; goto bypass_xor; }@+else@+for (bb=NULL,b=v->arcs;b!=a;b=b->next) { if (b->tip==u) goto double_bypass; if (b->tip==u->bar) { cmp=1-cmp; goto double_bypass; } bb=b;@+ continue; double_bypass: if (bb) bb->next=b->next; else v->arcs=b->next; goto bypass_xor; } aa=a;@+ continue; bypass_xor: if (aa) aa->next=a->next; else v->arcs=a->next; a->a.A=avail_arc; avail_arc=a; } if (v->arcs==NULL) { v->bit=cmp; goto make_v_constant; } if (cmp) @; } @ @= { for (a=v->arcs;;a=a->next) { u=a->tip; if (u->bar) break; /* good, the complement is already known */ if (a->next==NULL) { /* oops, this is our last chance */ @; break; } } a->tip=u->bar; } @ Here we've come to a subtle point: If a lot of |XOR| gates involve an input that is set to the constant value~1, the reduced'' graph might actually be larger than the original, in the sense of having more vertices (although fewer arcs). Therefore we must have the ability to allocate new vertices during the reduction phase of |reduce|. At least one arc has been added to the |avail_arc| list whenever we reach this portion of the program. @= if (next_vert==max_next_vert) { next_vert=gb_typed_alloc(7,Vertex,g->aux_data); if (next_vert==NULL) { gb_recycle(g); panic(no_room+1); /* can't get auxiliary storage! */ } max_next_vert=next_vert+7; } next_vert->typ=NOT; sprintf(name_buf,"%s~",u->name); next_vert->name=gb_save_string(name_buf); next_vert->arcs=avail_arc; /* this is known to be non-|NULL| */ avail_arc->tip=u; avail_arc=avail_arc->a.A; next_vert->arcs->next=NULL; next_vert->bar=u; next_vert->foo=u->foo; u->foo=u->bar=next_vert++; @ During the marking phase, we will use the |w.V| field to link the list of nodes-to-be-marked. That field will turn out to be non-|NULL| only in the marked nodes. (We no longer use its former meaning related to complementation, so we call it |lnk| instead of |bar|.) @d lnk w.V /* stack link for marking */ @= { for (v=g->vertices;v!=sentinel;v=v->foo) v->lnk=NULL; for (a=g->outs;a;a=a->next) { v=a->tip; if (is_boolean(v)) continue; if (v->typ=='=') v=a->tip=v->alt; if (v->typ=='C') { /* this output is constant, so make it boolean */ a->tip=(Vertex*)v->bit; continue; } @; } } @ @= if (v->lnk==NULL) { v->lnk=sentinel; /* |v| now represents the top of the stack of nodes to be marked */ do@+{ n++; b=v->arcs; if (v->typ=='L') { u=v->alt; /* latch vertices have a hidden'' dependency */ if (ulnk==NULL) { u->lnk=v->lnk; v=u; }@+else v=v->lnk; }@+else v=v->lnk; for (;b;b=b->next) { u=b->tip; if (u->lnk==NULL) { u->lnk=v; v=u; } } }@+while (v!=sentinel); } @ It is easier to copy a directed acyclic graph than to copy a general graph, but we do have to contend with the feedback in latches. @d reverse_arc_list(@!alist) {@+for (aa=alist,b=NULL;aa;b=aa,aa=a) { a=aa->next; aa->next=b; } alist=b;@+} @= new_graph=gb_new_graph(n); if (new_graph==NULL) { gb_recycle(g); panic(no_room+2); /* out of memory */ } strcpy(new_graph->id,g->id); strcpy(new_graph->util_types,"ZZZIIVZZZZZZZA"); next_vert=new_graph->vertices; for (v=g->vertices,latch_ptr=NULL;v!=sentinel;v=v->foo) { if (v->lnk) { /* yes, |v| is marked */ u=v->lnk=next_vert++; /* make note of where we've copied it */ @; } } @; reverse_arc_list(g->outs); for (a=g->outs;a;a=a->next) { b=gb_virgin_arc(); b->tip=is_boolean(a->tip)? a->tip: a->tip->lnk; b->next=new_graph->outs; new_graph->outs=b; } @ @= u->name=gb_save_string(v->name); u->typ=v->typ; if (v->typ=='L') { u->alt=latch_ptr;@+latch_ptr=v; } reverse_arc_list(v->arcs); for (a=v->arcs;a;a=a->next) gb_new_arc(u,a->tip->lnk,a->len); @ @= while (latch_ptr) { u=latch_ptr->lnk; /* the copy of a latch */ v=u->alt; u->alt=latch_ptr->alt->lnk; latch_ptr=v; if (u->altalt| by a new gate that copies an input@>; } @ Suppose we had a latch whose value was originally the {\sc AND} of two inputs, where one of those inputs has now been set to~1. Then the latch should still refer to a subsequent gate, equal to the value of the other input on the previous cycle. We create such a gate here, making it an {\sc OR} of two identical inputs. We do this because we're not supposed to leave any |'='| in the result of |reduce|, and because every {\sc OR} is supposed to have at least two inputs. @alt| by a new gate that copies an input@>= { v=u->alt; /* the input gate that should be copied for latching */ u->alt=next_vert++; sprintf(name_buf,"%s>%s",v->name,u->name); u=u->alt; u->name=gb_save_string(name_buf); u->typ=OR; gb_new_arc(u,v,DELAY);@+gb_new_arc(u,v,DELAY); } @* Parallel multiplication. Now comes the |prod| routine, which constructs a rather different network of gates, based this time on a divide-and-conquer paradigm. Let's take a breather before we tackle it. (Deep breath.) The subroutine call |prod(m,n)| creates a network for the binary multiplication of unsigned |m|-bit numbers by |n|-bit numbers, assuming that |m>=2| and |n>=2|. There is no upper limit on the sizes of |m| and~|n|, except of course the limits imposed by the size of memory in which this routine is run. The overall strategy used by |prod| is to start with a generalized gate graph for multiplication in which many of the gates are identically zero or copies of other gates. Then the |reduce| routine will perform local optimizations leading to the desired result. Since there are no latches, some of the complexities of the general |reduce| routine are avoided. All of the |AND|, |OR|, and |XOR| gates of the network returned by |prod| have exactly two inputs. The depth of the circuit (i.e., the length of its longest path) is $3\log m/\!\log 1.5 + \log(m+n)/\!\log\phi +O(1)$, where $\phi=(1+\sqrt5\,)/2$ is the golden ratio. The grand total number of gates is $6mn+5m^2+O\bigl((m+n)\log(m+n)\bigr)$. There is a demonstration program called {\sc MULTIPLY} that uses |prod| to compute products of large integers. @= Graph* prod(m,n) unsigned long m,n; /* lengths of the binary numbers to be multiplied */ {@+@@; @# if (m<2) m=2; if (n<2) n=2; @; @; if (gb_trouble_code) { gb_recycle(g);@+panic(alloc_fault); /* too big */ } g=reduce(g); return g; /* if |g==NULL|, the |panic_code| was set by |reduce| */ } @ The divide-and-conquer recurrences used in this network lead to interesting patterns. First we use a method for parallel column addition that reduces the sum of three numbers to the sum of two numbers. Repeated use of this reduction makes it possible to reduce the sum of |m| numbers to a sum of just two numbers, with a total circuit depth that satisfies the recurrence $T(3N)=T(2N)+O(1)$. Then when the result has been reduced to a sum of two numbers, we use a parallel addition scheme based on recursively golden sectioning the data''; in other words, the recursion partitions the data into two parts such that the ratio of the larger part to the smaller part is approximately $\phi$. This technique proves to be slightly better than a binary partition would be, both asymptotically and for small values of~$m+n$. \def\flog{\mathop{\rm flog}\nolimits} We define $\flog N$, the Fibonacci logarithm of~$N$, to be the smallest @^Fibonacci, Leonardo, numbers@> nonnegative integer~$k$ such that $N\le F_{k+1}$. Let $N=m+n$. Our parallel adder for two numbers of $N$ bits will turn out to have depth at most $2+\flog N$. The unreduced graph~|g| in our circuit for multiplication will have fewer than $(6m+3\flog N)N$ gates. @= m_plus_n=m+n;@+@; g=gb_new_graph((6*m-7+3*f)*m_plus_n); if (g==NULL) panic(no_room); /* out of memory before we're even started */ sprintf(g->id,"prod(%lu,%lu)",m,n); strcpy(g->util_types,"ZZZIIVZZZZZZZA"); long_tables=gb_typed_alloc(2*m_plus_n+f,long,g->aux_data); vert_tables=gb_typed_alloc(f*m_plus_n,Vertex*,g->aux_data); if (gb_trouble_code) { gb_recycle(g); panic(no_room+1); /* out of memory trying to create auxiliary tables */ } @ @= unsigned long m_plus_n; /* guess what this variable holds */ long f; /* initially $\flog(m+n)$, later flog of other things */ Graph *g; /* graph of generalized gates, to be reduced eventually */ long *long_tables; /* beginning of auxiliary array of |long| numbers */ Vertex **vert_tables; /* beginning of auxiliary array of gate pointers */ @ @= f=4;@+j=3;@+k=5; /* $j=F_f$, $k=F_{f+1}$ */ while (k @^Stockmeyer, Larry Joseph@> Proposition~5.3. The recurrence used here is related to the Josephus @^Josephus, Flavius, problem@> problem with step-size~3; see {\sl Concrete Mathematics}, {\mathhexbox278}3.3.] For this purpose, we compute intermediate results $P_j$, $Q_j$, and~$R_j$ by the rules \eqalign{P_j&=A_{3j}\oplus A_{3j+1}\,;\cr Q_j&=A_{3j}\land A_{3j+1}\,;\cr A_{m+2j}&=P_j\oplus A_{3j+2}\,;\cr R_j&=P_j\land A_{3j+2}\,;\cr A_{m+2j+1}&=2(Q_j\lor R_j)\,.\cr} Finally we let \eqalign{U&=A_{3m-6}\oplus A_{3m-5}\,,\cr V&=A_{3m-6}\land A_{3m-5}\,;\cr} these are the values that would be $P_{m-2}$ and $Q_{m-2}$ if the previous formulas were allowed to run past $j=m-3$. The final result $Z=(z_{m+n-1}\ldots z_1z_0)_2$ can now be expressed as $$Z=U+2V\,.$$ The gates of the first part of the network are conveniently obtained in groups of $N=m+n$, representing the bits of the quantities $A_j$, $P_j$, $Q_j$, $R_j$, $U$, and~$V$. We will put the least significant bit of $A_j$ in gate position |g->vertices+a(j)*N|, where $a(j)=j+1$ for $0\le j= next_vert=g->vertices; start_prefix("X");@+x=first_of(m,'I'); start_prefix("Y");@+y=first_of(n,'I'); @; @; @; @; @ @= register long i,j,k,l; /* all-purpose indices */ register Vertex *v; /* current vertex of interest */ Vertex *x,*y; /* least-significant bits of the input gates */ Vertex *alpha,*beta; /* least-significant bits of arguments */ @ @= for (j=0; jbit=0; /* this gate is the constant 0 */ } for (k=0; kbit=0; /* this gate is the constant 0 */ } } @ Since |m| is |unsigned|, it is necessary to say |j>1)+3+(((j-m)&1)<<1)) @= for (j=0; jvertices+(a_pos(3*j)*m_plus_n); beta=g->vertices+(a_pos(3*j+1)*m_plus_n); numeric_prefix('P',j); for (k=0; kvertices+(a_pos(3*j+2)*m_plus_n); numeric_prefix('A',(long)m+2*j); for (k=0; kbit=0; /* another 0, it multiplies$Q\lor R$by 2 */ for (k=0; k= alpha=g->vertices+(a_pos(3*m-6)*m_plus_n); beta=g->vertices+(a_pos(3*m-5)*m_plus_n); start_prefix("U"); for (k=0; k0$. The problem has therefore been reduced to the evaluation of $w_1$, $w_2$, \dots, $w_{N-1}$. Let $c_k^{\,j}$ denote the {\sc OR} of the first $j$ terms in the formula that defines $w_k$, and let $d_k^{\,j}$ denote the $j$-fold product $u_{k-1}u_{k-2}\ldots u_{k-j}$. Then $w_k=c_k^k$, and we can use a recursive scheme of the form $$c_k^{\,j}=c_k^{\,i}\lor d_k^{\,i}c_{k-i}^{\,j-i}\,,\qquad d_k^{\,j}=d_k^{\,i}d_{k-i}^{\,j-i}\,,\qquad j\ge2,$$ to do the evaluation. \def\down{\mathop{\rm down}} It turns out that this recursion behaves very nicely if we choose $i=\down[j]$, where $\down[j]$ is defined for $j>1$ by the formula $$\down[j]\;=\;j-F_{(\flog j)-1}\,.$$ For example, $\flog18=7$ because $F_7=13<18\le21=F_8$, hence $\down[18]=18-F_6=10$. Let us write $j\to\down[j]$, and consider the oriented tree on the set of all positive integers that is defined by this relation. One of the paths in this tree, for example, is $18\to10\to5\to3\to2\to1$. Our recurrence for $w_{18}=c_{18}^{18}$ involves $c_{18}^{10}$, which involves $c_{18}^5$, which involves $c_{18}^3$, and so on. In general, we will compute $c_k^{\,j}$ for all $j$ with $k\to^*j$, and we will compute $d_k^{\,j}$ for all $j$ with $k\to^+j$. It is not difficult to prove that $$k\;\to^*\;j\;\to\;i\qquad\hbox{implies}\qquad k-i\;\to^*\;j-i\,;$$ therefore the auxiliary factors $c_{k-i}^{\,j-i}$ and $d_{k-i}^{\,j-i}$ needed in the recurrence scheme will already have been evaluated. (Indeed, one can prove more: Let $l=\flog k$. If the complete path from $k$ to~$1$ in the tree is $k=k_0\to k_1\to\cdots\to k_t=1$, then the differences $k_0-k_1$, $k_1-k_2$, \dots, $k_{t-2}-k_{t-1}$ will consist of precisely the Fibonacci numbers $F_{l-1}$, $F_{l-2}$, \dots,~$F_2$, except for the numbers that appear when $F_{l+1}-k$ is written as a sum of non-consecutive Fibonacci numbers.) It can also be shown that, when $k>1$, we have \flog k=\min_{0= @; @; @; g->n=next_vert-g->vertices; /* reduce to the actual number of gates used */ @ After we have created a gate for $w_k$, we will store its address as the value of $w[k]$ in an auxiliary table. After we've created a gate for $c_k^{\,i}$ where $i= w=vert_tables; c=w+m_plus_n; flog=long_tables; down=flog+m_plus_n+1; anc=down+m_plus_n; flog[1]=0;@+flog[2]=2; down[1]=0;@+down[2]=1; for (i=3,j=2,k=3,l=3; l<=m_plus_n; l++) { if (l>k) { k=k+j; j=k-j; i++; /*$F_i=j= Vertex *uu, *vv; /* pointer to $u_0$ and $v_0$ */ Vertex **w; /* table of pointers to $w_k$ */ Vertex **c; /* table of pointers to potentially important intermediate values $c_k^{\,i}$ */ Vertex *cc,*dd; /* pointers to $c_k^{\,i}$ and $d_k^{\,i}$ */ long *flog; /* table of flog values */ long *down; /* table of down values */ long *anc; /* table of ancestors of the current $k$ */ @ @= vv=next_vert-m_plus_n;@+uu=vv-m_plus_n; start_prefix("W"); v=new_vert('C');@+v->bit=0;@+w[0]=v; /* $w_0=0$ */ v=new_vert('=');@+v->alt=vv;@+w[1]=v; /* $w_1=v_0$ */ for (k=2;k; i=1;@+cc=vv+k-1;@+dd=uu+k-1; while (1) { j=anc[l]; /* now $i=\down[j]$ */ @# @; @; if (flog[j]; dd=v; i=j; l--; } w[k]=v; } @ If $k\to j$, we call $j$ an ancestor'' of $k$ because we are thinking of the tree defined by $\to$'; this tree is rooted at $2\to1$. @= for (l=0,j=k;;l++,j=down[j]) { anc[l]=j; if (j==2) break; } @ @d spec_gate(v,a,k,j,t) v=next_vert++; sprintf(name_buf,"%c%ld:%ld",a,k,j); v->name=gb_save_string(name_buf); v->typ=t; @= spec_gate(v,'B',k,j,AND); gb_new_arc(v,dd,DELAY); /* first argument is $d_k^{\,i}$ */ f=flog[j-i]; /* get ready to compute the second argument, $c_{k-i}^{\,j-i}$ */ gb_new_arc(v,f>0? c[k-i+(f-2)*m_plus_n]:vv+k-i-1,DELAY); @ @= if (l) { spec_gate(v,'C',k,j,OR); }@+else v=new_vert(OR); /* if $l$ is zero, this gate is $c_k^k=w_k$ */ gb_new_arc(v,cc,DELAY); /* first argument is $c_k^{\,i}$ */ gb_new_arc(v,next_vert-2,DELAY); /* second argument is $b_k^{\,j}$ */ @ Here we reuse the value $f=\flog(j-i)$ computed a minute ago. @= spec_gate(v,'D',k,j,AND); gb_new_arc(v,dd,DELAY); /* first argument is $d_k^{\,i}$ */ gb_new_arc(v,f>0? c[k-i+(f-2)*m_plus_n]+1:uu+k-i-1,DELAY); /* $d_{k-i}^{\,j-i}$ */ @ The output list will contain the gates in big-endian order'' $z_{m+n-1}$, \dots, $z_1$, $z_0$, because we insert them into the |outs| list in little-endian order. @= start_prefix("Z"); for (k=0;ktip=make2(XOR,uu+k,w[k]); a->next=g->outs; g->outs=a; } @* Partial evaluation. The subroutine call |partial_gates(g,r,prob,seed,buf)| creates a new gate graph from a given gate graph~|g| by partial evaluation,'' i.e., by setting some of the inputs to constant values and simplifying the result. The new graph is usually smaller than |g|; it might, in fact, be a great deal smaller. Graph~|g| is destroyed in the process. The first |r| inputs of |g| are retained unconditionally. Each remaining input is retained with probability |prob/65536|, and if not retained it is assigned a random constant value. For example, about half of the inputs will become constant if |prob=32768|. The |seed| parameter defines a machine-independent source of random numbers, and it may be given any value between $0$ and $2^{31}-1$. If the |buf| parameter is non-null, it should be the address of a string. In such a case, |partial_gates| will put a record of its partial evaluation into that string; |buf| will contain one character for each input gate after the first |r|, namely |'*'| if the input was retained, |'0'| if it was set to~$0$, or |'1'| if it was set to~$1$. The new graph will contain only gates that contribute to the computation of at least one output value. Therefore some input gates might disappear even though they were supposedly retained,'' i.e., even though their value has not been set constant. The |name| field of a vertex can be used to determine exactly which input gates have survived. If graph |g| was created by |risc|, users will probably want to make |r>=1|, since the whole RISC circuit collapses to zero whenever its first input \.{RUN}' is set to 0. An interesting class of graphs is produced by the function call |partial_gates(prod(m,n),m,0,seed,NULL)|, which creates a graph corresponding to a circuit that multiplies a given |m|-bit number by a fixed (but randomly selected) |n|-bit constant. If the constant is not zero, all |m| of the retained'' input gates necessarily survive. The demo program called {\sc MULTIPLY} illustrates such circuits. The graph |g| might be a generalized network; that is, it might involve the |'C'| or |'='| gates described earlier. Notice that if |r| is sufficiently large, |partial_gates| becomes equivalent to the |reduce| routine. Therefore we need not make that private routine public. As usual, the result will be |NULL|, and |panic_code| will be set, if |partial_gates| is unable to complete its task. @= Graph *partial_gates(g,r,prob,seed,buf) Graph *g; /* generalized gate graph */ unsigned long r; /* the number of initial gates to leave untouched */ unsigned long prob; /* scaled probability of not touching subsequent input gates */ long seed; /* seed value for random number generation */ char *buf; /* optional parameter for information about partial assignment */ {@+register Vertex *v; /* the current gate of interest */ if (g==NULL) panic(missing_operand); /* where is |g|? */ gb_init_rand(seed); /* get them random numbers rolling */ for (v=g->vertices+r;vvertices+g->n;v++) switch (v->typ) { case 'C': case '=': continue; /* input gates might still follow */ case 'I': if ((gb_next_rand()>>15)>=prob) { v->typ='C';@+v->bit=gb_next_rand()>>30; if (buf) *buf++=v->bit+'0'; }@+else if (buf) *buf++='*'; break; default: goto done; /* no more input gates can follow */ } done:if (buf) *buf=0; /* terminate the string */ g=reduce(g); @; return g; /* if |(g==NULL)|, a |panic_code| has been set by |reduce| */ } @ The |buf| parameter is not recorded in the graph's |id| field, since it has no effect on the graph itself. @= if (g) { strcpy(name_buf,g->id); if (strlen(name_buf)>54) strcpy(name_buf+51,"..."); sprintf(g->id,"partial_gates(%s,%lu,%lu,%ld)",name_buf,r,prob,seed); } @* Index. Here is a list that shows where the identifiers of this program are defined and used.
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# Solve the expression for n Є N
1. Nov 5, 2005
### six789
i just want to confirm if my answer is right....
this is the problem:
solve the expression for n Є N
P(2n + 4, 3) = 2/3P(n+4, 4)
this is my work:
(2n +4)!/((2n +4)-3)!) = (2/3(n+4)!)/((n+4)!)-4)!)
(2n +4)!/((2n +1)! )= (2/3(n+4)!)/(n!)
and i dont know wat to do next... should i just cross multiply it?
Last edited: Nov 5, 2005
2. Nov 5, 2005
### six789
permutations
help me with this if my answer is correct....
help is appreciated...
Last edited: Nov 5, 2005
3. Nov 6, 2005
### six789
help from anyone with this problem...
Last edited: Nov 6, 2005
4. Nov 6, 2005
### Gale
start canceling terms. if you have $$\frac{(2n+4)!}{(2n+1)!}$$ and you understand what factorial means, you can cancel all the factors in the denominator with some in the numerator. do this as well to the right side of your equation. then try to solve.
5. Nov 6, 2005
### six789
so you mean $$\frac{(2n+4)!}{(2n+1)!}$$ is same thing as writing as (2n+4)(2n+3)(2n+2)(2n+1)/(2n+1)?
6. Nov 6, 2005
### Gale
uh, well the (2n+1)'s cancel out. but ya. thats how a factorial works. so do the same to other side and go.
7. Nov 6, 2005
### six789
correct me if im wrong ok? i've try the other side... should you make the numerator 2/3(n+4)! to -2/3(n-4)! in order to cancel it from n(n-1)(n-2)(n-3)(n-4)?
8. Nov 6, 2005
### Gale
no... i think you're missing something...
$$\frac{2/3(n+4)!}{n!}= \frac{2/3(n+4)(n+3)(n+2)(n+1)(n)(n-1)(n-2)(n-3).....}{n(n-1)(n-2)(n-3)......}$$
so you're left with only $$2/3(n+4)(n+3)(n+2)(n+1)$$ cause you can cancel the rest out. do you see this?? does it make sense? this is what we did to the other side as well.
9. Nov 6, 2005
### six789
now i get gale... thanks so m uch, and it is clearer for me now...
10. Nov 6, 2005
### six789
what if you get like this.... -24(2n+3)(n+4)(n+3) = 0...
what should i do to the -24? or it is just 0, then the value for n is just n=-2/3, n=-4 and n=-3?
11. Nov 6, 2005
### Gale
well, i'm not really sure how you got that.... its not what i get. what's your equation after you've simplified? i get something simple enough to multiply out and then combine like terms. then solve.
12. Nov 6, 2005
### six789
i got like (2n+4)(2n+3)(2n+2) = 2/3(n+4)(n+3)(n+2)(n+1)
then... 2(n+2)(2n+3)2(n+1) = 2/3(n+4)(n+3)(n+2)(n+1)
which is 4(2n+3) = 2/3(n+4)(n+3)
and i dont know wat to do next
13. Nov 6, 2005
### Gale
multiply it out and simplify, its just a quadractic.
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# on 07-Dec-2019 (Sat)
A kernel smoother is a statistical technique to estimate a real valued function $${\displaystyle f:\mathbb {R} ^{p}\to \mathbb {R} }$$ as the weighted average of neighboring observed data.
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# How many numbers are there in range 1 to 1000 which contains digits 2 and 3 and divisible 2 and 3?
How many numbers are there in range 1 to 1000 which contains digits 2 and 3 and divisible 2 and 3?
I know the answer to find count of numbers in range 1 to 1000 which are divisble by 2 and 3. But the above question also asks that those numbers must include digits 2 and 3.
Please genaralise your answer beacuse I want to know the answers when there are more or different digits. Like find count of numbers in range 1 to 1000000 which contains digit 5,2,7,8 and divisible by 5,2,7,8.
• You're looking for a multiple of 6 whose decimal representation has a 2 and a 3. There are no such one or two digit numbers. For three digit numbers, you must have $2+3+x$ divisible by 3, where $x$ is the third digit. So $x$ is 1, 4, or 7. But you also need the number to be even. Hence... ? – symplectomorphic Jul 4 '16 at 8:11
• As I see, there's probably no simple way to generalise. Counting is probably the simplest way to do it – Aditya De Saha Jul 4 '16 at 8:26
• I think the problem was specific and wanted you to rely on the fact that the third digit must be 1, 4 or 7 (else it's not divisible by 3) and the number must end in 4 (else it's not divisible by 2). This is certainly not extendable to general cases. – fleablood Jul 6 '16 at 5:59
• Here is the source of the problem: cs.stackexchange.com/q/60267/755 – D.W. Jul 10 '16 at 16:58
$3$-digit numbers containing $[2,3]$ and divisible by $[2,3]$:
• $132$
• $234$
• $312$
• $324$
• $342$
• $372$
• $432$
• $732$
$6$-digit numbers containing $[2,5,7,8]$ and divisible by $[2,5,7,8]$:
• $175280$
• $258720$
• $357280$
• $387520$
• $538720$
• $567280$
• $572880$
• $580720$
• $728560$
• $735280$
• $752080$
• $758240$
• $785120$
• $807520$
• $857920$
• $875280$
• $877520$
• $958720$
• I am not looking for specifically this case. Can you please generalise your answer? If you have to find how many numbers are there in range 1 to 1000000000 having digits a,b,c and divisible by digits a,b,c. where 1<=a,b,c<=9 and a!=b!=c – Mike Jul 4 '16 at 8:25
• @Mike: That depends on the values of $[a,b,c]$. – barak manos Jul 4 '16 at 8:28
A general exact answer seems difficult, but here's a way to get a good estimate that readily generalises. Divisibility by $2$ constrains the last digit; divisibility by $3$ doesn't constrain any digits and can be treated as roughly independent of the other constraints. Thus, if we treat the problem probabilistically, we have the events $D_i$ of containing digit $i$ and $E_i$ of divisibility by $i$, we're looking for $\textsf{Pr}(D_2\cap D_3\cap E_2\cap E_3)$, and we can treat $E_3$ as independent of the three other events and $E_2$ as dependent with $D_2$ and $D_3$ only through the last digit. Then
\begin{align} \def\pr#1{\textsf{Pr}\left(#1\right)} \pr{D_2\cap D_3\cap E_2\cap E_3} &=\pr{D_2\cap D_3\cap E_2}\pr{E_3} \\ &= \frac13\left(\pr{D_2\cap E_2}+\pr{D_3\cap E_2}-\pr{(D_2\cup D_3)\cap E_2}\right) \\ &= \frac13\left(\frac12\left(1-\frac45\left(\frac9{10}\right)^{n-1}\right)+\frac12\left(1-\left(\frac9{10}\right)^{n-1}\right)-\frac12\left(1-\frac45\left(\frac8{10}\right)^{n-1}\right)\right) \\ &= \frac16\left(1-\frac{2\cdot9^n-8^n}{10^n}\right)\;, \end{align}
where $n$ is the number of digits, so the expected count of such numbers would be
$$\frac16\left(10^n-2\cdot9^n+8^n\right)\;.$$
In your case, with $n=3$, this is
$$\frac16\left(10^3-2\cdot9^3+8^3\right)=9\;,$$
in reasonable agreement with the fact that barak manos counted $8$.
For the case of $5$, $2$, $7$ and $8$, divisibility by $5$ and $2$ simply means that the number ends in a $0$; this is equivalent to asking how many numbers with $n-1$ digits contain $5$, $2$, $7$ and $8$ and are divisible by $7$ and $4$. Again divisibility by $7$ is independent since $7$ is coprime to the base $10$, but now the calculation is slightly more involved since divisibility by $4$ constrains the last $2$ digits and you need to perform inclusion-exclusion on all $4$ digit values that should be present. Still, the ingredients are all in the calculation above.
1. $\ds{c}$ is an even number and $3 \mid \pars{a + b + c}$.
2. If $\ds{\color{#f00}{c} \not= \color{#f00}{2}}$, then $\ds{a + b = 5}$. It leaves us with $\ds{c = 4}$ because $\ds{3 | \pars{5 + 0}}$,$\ds{3|\pars{5 + 6}}$ and $\ds{3|\pars{5 + 8}}$ are false.
3. If $\ds{\color{#f00}{c} = \color{#f00}{4}}$, it obviously leaves us with $\ds{\ul{two}}$ numbers: $\ds{\color{#f00}{234}}$ and $\ds{\color{#f00}{324}}$.
4. If $\ds{\color{#f00}{c} = \color{#f00}{2}}$, $\ds{a + b + 2 = 3p}$ where $\ds{p}$ is an integer $\ds{~\pars{\mbox{for the time being,}\ 1 \leq p \leq 6}~}$. Because $\ds{a = 3}$ or $\ds{b = 3}$, one of them is equal to $\ds{3p - 2 -3 = 3p - 5}$. Note that $\ds{3p - 5 \not= 3}$. \begin{align} &\mbox{But}\quad 0 \leq 3p - 5 \leq 9\quad\imp\quad {5 \over 3} \leq p \leq {14 \over 3}\quad\imp\quad p \in \braces{2,3,4} \\ &\imp\quad 3p - 5 \in \braces{1,4,7}\ \mbox{which yields}\ 3\times 2 = 6\ \mbox{numbers}:\ \left\lbrace\begin{array}{l} \ds{\color{#f00}{132}} \\ \ds{\color{#f00}{312}} \\ \ds{\color{#f00}{342}} \\ \ds{\color{#f00}{432}} \\ \ds{\color{#f00}{372}} \\ \ds{\color{#f00}{732}} \end{array}\right. \end{align}
There are $\ds{\color{#f00}{8}}$ numbers which satisfy the OP conditions. Namely, $$\begin{array}{|c|}\hline\mbox{}\\ \ds{\quad% \color{#f00}{132},\color{#f00}{234},\color{#f00}{312},\color{#f00}{324},\color{#f00}{342},\color{#f00}{372},\color{#f00}{432},\color{#f00}{732} \quad} \\ \mbox{}\\ \hline \end{array}$$
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# Understanding Word2Vec
I am trying to understand the word2vec algorithm (Mikolov et. al) but there are a few thing which I do not understand.
I get that the activation from the input layer ot the hidden layer is linear and that $\mathbf{h}$ is just the average of the linear combination of all $\mathbf{x}_{ik}$ vectors. Further I understand that the final output of each node $y_i$ of the output vector $\mathbf{y}$ is just the $\text{softmax}$ activation function from the hidden to the output layer.
What I do not understand is what are my actual word vectors in the end? I currently see only one possibility and that is to take either the columns of $\mathbf{W_{V\times N}}$ or the rows of $\mathbf{W'_{N\times V}}$ since I assume these weights are now the "information carrying" entity of this algorithm.
Am I correct with this assumption?
Independent from this I would like to understand how Google managed to train their Google News word vector models.
From the website:
We are publishing pre-trained vectors trained on part of Google News dataset (about 100 billion words).
Which is simply insane if this has been done with one-hot encoded vectors for training. That would mean each input matrix $\mathbf{W_{V\times N}}$ would be $(1 \times 10^{11}) \times 300$ in size!?
This leads me to my final question: Shouldn't it be possible to use lower dimensional vectors? Somewhere in the back of my head I have the idea, that you can simply randomly initialize word vectors for all given words in a vocabulary and then apply word2vec on it. This however would also mean that not only the weights have to get updated, but also the word vectors of each input words too. Is something like this actually done or am I completely mistaken here?
what are my actual word vectors in the end?
The actual word vectors are the hidden representations $h$ Basically, multiplying a one hot vector with $\mathbf{W_{V\times N}}$ will give you a $1$$\times$$N$ vector which represents the word vector for the one hot you entered.
Here we multiply the one hot $1$$\times$$5$ for say 'chicken' with synapse 1 $\mathbf{W_{V\times N}}$ to get the vector representation : $1$$\times$$3$
Basically, $\mathbf{W_{V\times N}}$ captures the hidden representations in the form of a look up table. To get the look up value, multiply $\mathbf{W_{V\times N}}$ with the one hot of that word.
That would mean each input matrix $\mathbf{W_{V\times N}}$ would be $(1 \times 10^{11}) \times 300$ in size!?
Yes, that is correct. Keep in mind 2 things:
1. It is Google. They have a lot of computational resources.
2. A lot of optimisations were used to speed up training. You can go through the original code which is publicly available.
Shouldn't it be possible to use lower dimensional vectors?
I assume you mean use a vector like [ 1.2 4.5 4.3] to represent say 'chicken'. Feed that into the network and train on it. Seems like a good idea. I cannot justify the reasoning well enough, but I would like to point out the following:
1. One Hots allow us to activate only one input neuron at once. So the representation of the word falls down to specific weights just for that word. Here, the one hot for 'juice' is activating just 4 synaptic links per synapse.
2. The loss function used is probably Cross Entropy Loss which usually employs one hot representations. This loss function heavily penalises incorrect classifications which is aided by one hot representations. In fact, most classification tasks employ one hots with Cross Entropy Loss.
I know this isn't a satisfactory reasoning.
I hope this clears some things up.
Here are some resources :
1. The famous article by Chris McCormick
2. Interactive w2v model : wevi
3. Understand w2v by understanding it in tensorflow (my article (shameless advertisement,but it covers what I want to say))
• I very much enjoyed your blog post, very well done. – Akavall Sep 11 '17 at 21:02
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# Re: [NTG-context] circled number for item head (Wolfgang Schuster)
On Tue, 22 Mar 2011, Jeong Dalyoung wrote:
Dear Wolfgang,
Thank you for reminding me the thread.
I also found it yesterday evening and it made me realize how dumb I am.
I am so sorry to give you a needless work.
I'd like to ask you one more.
I was looking for a way to use some characters as an item head like
\startitemize[a] because an official document usually use numbers and some
For consecutive characters in unicode, it is possible to do that using unicode.
\def\circledja#1{\uchar{50}{95+#1}}
\def\circledga#1{\uchar{50}{109+#1}}
\def\circleden#1{\uchar{36}{207+#1}}
and \defineconversion[][].
But for characters which are not listed consecutively, I don't know how to make
it appear automatically.
I may use \sym{}
AC00, B098, B2E4, B77C, ...
These are the same characters in the area 320E ~ 321B without parenthesis.
Is it possible to make those characters appear automatically as item heads?
\defineconversion[whatever][One, Two, Three, Four]
\starttext
\startitemize[whatever]
\item First
\item Second
\item Third
\stopitemize
\stoptext
___________________________________________________________________________________
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# The Zen of Telling Notepad++ That .Config Files are XML
There's a zen beauty in improving the configuration of your tools. Breathe in. Configure. Breathe out. Reconfigure.
Problem: When I open a web.config file with Notepad++ it doesn't realize that it's an XML file. Thus, it doesn't syntax highlight the file accordingly.
A moment of googling led me to an answer at superuser.
Solution:
1. Go to Settings > Style Configurator...
2. Under Language, scroll down and highlight "XML"
3. Beneath the Language list, find User ext. textbox and add "config" to the list.
4. Press Save & Close
5. You will need to close and reopen the file, for the syntax highlighting to take effect.
This is one of those minor frustrations that I put up with, year in year out, and don't even realize I'm tolerating it. Why? because every time I edit a web.config file I'm absorbed in the momentary struggle, struggling under the cognitive load of a deep mental stack of problems. I don't have the time (or the mental resources) to stop and fix this problem before getting the job itself done.
Slow down. Get the tooling right. Sharpen the axe.
Spend five minutes to save five seconds.
It will pay off in time.
(There's also something nicely recursive about configuring the way config is displayed)
# How Businesses Are Actually Structured.
There's two particular observations I want to share about the way businesses are structured.
In the ideal scenario, a business with N employees can be visualized as a single Directed Acyclic Graph (DAG) with N nodes. [A Directed Acyclic Graph is similar to a Tree... but each connection is directed and a node can connect to more than one parent. So I'm going to say DAG instead of Tree, but you can picture a Tree if it's more natural to you.] So.. I was saying... a business can allegedly be represented by a single canonical DAG. A, B and C report to D; D reports to the CEO.
Well, here's my first observation: Each of the "corporate services" have their own "Canonical DAG".
"Corporate services" are those parts of a company that cost money but (hopefully) keep the other parts of the company functioning, i.e. Human Resources, IT Services, Finance, Building services.
Human Resources will publish one (or many) Org Charts. To them, this is The Canonical DAG. This is the true structure of the company.
The IT department maintain a very complex Active Directory, with users, groups and nested groups. Forget the Org Chart, to them *this* is The Canonical DAG. Active Directory looks at the Org Chart and says, "That's Cute".
The Finance Department have a book of accounts, identified by cost codes, with different cost codes "rolling up" into higher cost codes. To them, *this* is The Canonical DAG. The Finance department look at the Org Chart and Active Directory and say "That's Cute," while fanning themselves with a wad of notes.
Building services divide the corporation into buildings, buildings have floors, floors have workstations and workstations have various resources (broken chairs, broken projectors, broken people, broken printers, broken monitors). To them, *this* is the Canonical DAG.
And so on for every other Corporate Service.
Now, none of these DAG's are as simple as you'd want them to be. There's orphaned branches. There's dotted lines (multiple parents to a single node). There's time-dependent edges (Jill reports to Amy on Monday and Wednesday and to Fred on other days). And a whole lot of other complex modifications*. But it gets a lot worse.
Here's the second observation. The real structure of the company isn't stored in any of these systems.
The real structure is the friendships, rivalries, favors, grudges and slights that are accumulated over years. The social/antisocial network of associations that isn't written down, and makes all the difference in whether an idea will succeed, a directive will be followed, a permission will be granted, a bonus given.
And this network of associations is uniquely perceived by each node within the graph. (And every node within the network will strongly deny that they themselves have any rivalries, grudges etc... this is a crucial part of the way power structures are enforced.) (You might think you can "detect" these structures by looking at who calls who, who emails who, and by detecting tone etc., but the real stuff is never written down or even hinted in writing: that's what gives it power).
Note that the relationship between A and B is different, depending on whether you're looking from the point of view of A or B.
So for a graph with N nodes (people, nodes are people) there are N graphs (most of which have less than N nodes, thankfully).
So, yeh, good luck with that restructure. I'm sure this new (centralization|decentralization) will make a world of difference.
There's a book called Moral Mazes that exposes how companies really work. It's quite dry and academic, but with more twists and betrayals than Game of Thrones. Recommended.
* "Job sharing" is a particularly funny business. If Bill and Bob share a role, we can't represent it as a non-directed relationship "Bill shares in a symmetric manner with Bob" (as our graph is directed), nor can we represent it by two separate and directed relationships "Bill shares with Bob, and Bobs shares with Bill" -- as this is cyclic and makes computers blow up, so instead we have to introduce the concept of roles: "Bill has a role, and Bob has the same role". But now we've made every single node in the entire organization at once more flexible and much much more complex. Good luck with that. We can instead create a virtual person, BillBob, and both Bill and Bob report to the virtual BillBob, who reports to whoever Bill and Bob used to report to. That localizes the complexity somewhat, but also confuses the living hell out of building services and payroll, who assign BillBob a desk and a paycheck, which somehow ends up in the bank account of 'Directed Acyclic Graphs Are Hard, Incorporated.'
(This article has been in my backlog for five years now! Wow. Every single time I've gone to publish it there has been some restructure happening in a company I work with, and I've avoided publishing it for fear it could be interpreted as relating to a current restructure. So I finally thought I'd just publish it with this note saying "No, this does not relate to current events. This relates to eternal events.")
# Launching the secretGeek Wiki
This might be the mid-life crisis talking but lately I've been feeling that I want to do more. More creative work. More of everything. I want to write books, compose music, draw pictures, make movies, create amazing programs, solve impossible mathematical problems, learn magic tricks, create European-style boardgames, print circuits, invent incredible devices, cook, grow, share, give and give.
I've even made plans to bring back the zine. This would involve sneaking into an actual school to use their actual school photocopier to make copies. But more on that another day.
I think it started with the hexaflexagons: just a throw away word, mentioned in an article about Cellular Automata. No: something came before that... I was looking into the possibility of Optical Lego Recognition. OLR! Research on OLR led to image-processing, led to Cellular Automata, led to Conway's Game of Life. Game of life! GOL led to an interview with John Conway who mentioned Martin Gardner and the magical word "hexaflexagons", which he didn't explain at all. How cruel! To mention a word like that with no explanation! Researching hexaflexagons led to Vi Hart, which led back to Martin Gardner. Martin Bloody Gardner! Martin wrote a "Mathematical Games" column in Scientific American, from 1956 to 1981. I'd never heard of him: but he influenced practically everyone that ever influenced me. He popularized all of the topics above and many more. So I fell in. Tumbling like Alice into the rabbit hole. Deep into the wormhole of recreational mathematics, and soon wiki.secretGeek.net was born.
It's a place where I (or we) can maintain a more permanent playground, less ephemeral than blog posts and tweets. A place for documenting certain things. Not things that have any obvious economic value. Things I (or we) find interesting. Interesting things that remain interesting even as time passes.
I want to be able to put together informative, interactive pieces; a little like Red Blob Games, a little like 'the sierpinski triangle page to end most sierpinski triangle pages' a little like Bret Victor's Learnable Programming with a splash of _why's tryruby and a taste of Bob Nystrom's Dungeon generation article (more on that another time).
So I picked up the blog engine from secretGeek.net, tore out its heart and replaced it with a simple markdown-based wiki engine. To make the site interactive, I wanted to create a simple plugin system. I already had a facility for categorizing the articles, i.e. 'tagging' them, so I extended this: each category that is added to a page means a custom partial with that name is included in the page. So if you've tagged an article as being about 'mp3' then the 'mp3' partial would be included if it exists (which it doesn't) and it would presumably allow you to play the mp3s in that article. More generally speaking: tags allow custom javascript to run, that does something special to certain elements within an article.
The first 'custom partial' I added was for 'logo' as in 'turtle logo' that wonderful learning environment from the happiest moments of your childhood. Now I can include runnable logo programs within the wiki, just by adding a blockquote with the class 'logo'. This let me write about a lot of Fractals, Space-Filling Curves and the like.
The next plugin I added was for 'Conway's Game of Life' (gol), so that runnable examples of Game of Life patterns can be included inside articles, for example about Spaceships and Guns.
I also added a plugin for displaying mathematical equations, using mathjax though I've only used that at one article so far.
I've been reluctant to 'launch' the damn thing, feeling I ought to get it 'right' before I do. But then the voice of Amy Hoy materialized in my head and announced 'just fucking ship', so that's what I'm going to do.
Here's a decent starting list of articles:
Over the next 10 years or so, I (or we) can add info about a bunch of other topics as well. For now I've disabled the ability for new contributors to register. But I plan to enable that when I have the kinks removed. This will exist for a long time to come.
Visit the secretGeek Wiki
And since you're still here. Two somewhat relevant images that strike a nerve. A fractal Tom Selleck. And a quote from Barton Fink.
# How can we do better?
—So we did two releases in quick succession. Why?
—The second release was to fix a typo in the first release.
—How can we do better?
—The release shouldn't have gone out until it had been tested in a QA environment and signed off.
—But couldn't we have caught the error sooner than that?
—We need a build server, that builds and tests everything. A unit test should've picked it up.
—Surely we could've caught it sooner than that?
—The code should never have been pushed to the build server. A code review should be required prior to check in. A second pair of eyes, looking at the diff would've caught this.
—But couldn't we have caught it sooner than that?
—If we were writing tests first, then the faulty code would've been detected before the engineer considered asking for a review.
—But couldn't we have caught it sooner than that?
—If the language we were using had a more expressive type system, it would've been impossible for such a typographic error to compile.
—Hmmm. But isn't compile time too late anyway?
—If the engineer had thought more carefully before pressing each key, been more intentional in her work, this would never had happened!
—Is that really the best we can do!?
—If we selectively bred perfect humans, incapable of incorrect thoughts, then our releases would always be perfect.
—No, selective breeding takes too long. We need to do better!
—We need to manipulate the genes before the zygote is allowed to split and multiply, and perfectly engineer the perfect engineer.
—Ah yes, project Ultra-Mega. Do we have a status report on that project?
—Yes. There were two releases this week, in quick succession. The second release was to fix a typo in the first release.
—How can we do better?
(The meeting took *forever*)
# Pun-a-day service
Introducing secretGeek's Pun-a-day SMS service as a service. Subscribe now. (Australian customers only, sorry)
Telstra, the dominant telco in Australia, just announced a new API for sending/receiving SMS messages, and it seems to be looked after by Frank Arrigo, a top bloke who was once the (very popular) head of Microsoft's evangelism efforts in Australia (if not the world).
You can learn about their API here, and register to get your own key at dev.telstra.com.
Sending a message, in C#, is as simple as this:
// Step 0. Let's prepare the message we're going to send.
// Recipient number should be in the format of "04xxxxxxxx" where x is a digit
string recipientNumber = "0455555555";
var messageBody = "Hello from Leon's Pun-a-day service! Reply STOP for more!"
var message = "{\"to\":\"" + recipientNumber + "\", \"body\":\"" + messageBody + "\"}";
var consumerKey = "YOURCONSUMERKEY";
var consumerSecret = "YOURCONSUMERSECRET";
using (var client = new System.Net.WebClient { UseDefaultCredentials = true })
{
// Step 1: Get a token
var tokenURI = string.Format("https://api.telstra.com/v1/oauth/token?client_id={0}&client_secret=
{1}&grant_type=client_credentials&scope=SMS", consumerKey, consumerSecret);
var token = JsonConvert.DeserializeObject<AccessToken>(response);
string URI = "https://api.telstra.com/v1/sms/messages";
// Step 2: Send the message you prepared earlier
// e.g. result == '{ "messageId":"CBCB3DCC991D8AF0" }'
// Step 3. There is no step 3. Well... you can get the messageid out of the response,
// and store it. That way when the network calls you back with a reply you know which
// In my first sample app, i just put the string into a viewbag message
// so i could view it, without any fuss.
ViewBag.Message = result;
}
public class AccessToken
{
public string access_token { get; set; }
public int expires_in { get; set; }
}
Note that when crafting the message, if any part of the messagebody has been provided by users, then you'll need to protect against Json injection. In fact, I need to just go off on a big tangent at this point demonstrating how dangerous this JSON-Injection voodoo can be...
## A quick word about JSON Injection!
So the message we sent to the network looks like this:
{ "to":"0455555555", "body":"Welcome to the pun a day service!"}
Let's say that we've used a form to collect people's phone numbers. And this is the phone number some nasty assailant provided:
0455555555", "body":"Send $100 in bitcoin to address ABCD or your dog will die!"} // { "to":"0455555556 When the message is put together, we might have something like this: { "to":"0455555555", "body":"Send$100 in bitcoin to address ABCD or your dog will die!"} //{ "to":"0455555556", "body":"Welcome to the pun a day service!"}
Which could (with only a little more work) slip past Telstra, and allow our malicious person to send arbitrary messages to any victim they wish and have the federal police kicking in your door before you can say "Snap!".
So definitely validate the phone number, using a regular expression (e.g. '^04\d{8}$' ?? Suggestions welcome!) (and remember the ^ and$!!). And if you're putting user content into message bodies, for the love of all that is scientifically validated, please escape any double quotes, restrict the message length, etc, etc. It's cheaper than replacing your front door!
## A neat API
All told, I'd rate this a very neat little api. A joy to use.
Now, I only need to come up with a clever idea for an SMS app, that can nett me my retirement goal of 5 million in crisp bills, before my current contract ends. Thoughts?
In the meantime, get your Pun-A-Day!
# Love Me Two Times
I woke up the other day with a guitar riff stuck in my head.
I knew it was from a song by The Doors, but I couldn't work out which one, so I went to spotify and played the first few seconds of a dozen or so Doors songs, until I found it.
Love me two times.
What is that song even about? Okay, it's about sex. Problem solved.
So I found some tablature online, it said something like this:
e|-----------------------------------------
B|-----------------------------------------
G|-------------------0h1~~----0h1~~--------
D|---0---2---0-2---------------------------
A|-2---2---2-------------------------------
E|---------------0-0------0-0--------------
I tried playing it. And although I was playing the notes that were shown it was definitely not the intro to the song 'Love me two times' by The Doors. It was some other, much much sadder song.
Then I went and looked at some YouTube videos. Wow. This was a revelation.
Neil from totally guitar said that on the first note, you hit upward with your plectrum. In fact the direction of your plectrum pretty much dictates the entire sound. I'm more of a finger/thumb picking guy, so it's all very unnatural to me. Up, down, up down, like a broom sweeping. Classic blues, the way Robert Johnson learned it from the devil at the crossroads long ago.
And the "0h1~~" in the tab above is not simply 'hammer onto the first' and then vibrato (~~~), it's actually a "trill", i.e.
10 hammer on from 0 to 1
20 pull off from 1 to 0
30 goto 10
So the tab, as explained by Neil, has these nuances:
e|----------------------------------------------------------
B|----------------------------------------------------------
G|------------------------------0h1p0h1p0-------0h1p0h1p0---
D|----V0----V2-----V0-^2------------------------------------
A|-^2----^2----^2-------------------------------------------
E|-------------------------0-0------------0-0---------------
And that's before you even get to the timing of the piece. Tablature doesn't tell you anything about the relative duration of each note.
Okay, so let's say you've mastered the timing of the intro, and all of the rest of the song.
At best, you'll play as well as Robby Kreiger in this performance from 1972 playing with the doors minus Jim Morrison. This is 3 years after Jim passed away, the remaining band playing, with keyboardist Ray Manzarek on vocals. (You can see Robby smiling as he pushes out the trills ;-))
I like Ray's take on the song. Having focused on this song for a day (in between build breakages) I thought they did a great job. But then, even though I was at youtube, I went ahead, like a fool, and read the damn comments.
One said:
Good performance but something is missing...I can't quite put my finger on it though
OK, that was pretty subtle. Another put it a little more bluntly:
no matter what way you look at it, without Jim its a lame duck dead in the fucking water and no matter what way you try to dress it up , it sucks and its god awful eww?
Which is interesting.
Here's a trio of three very accomplished musicians, playing a song the best it will ever be played. The video has a lot of views, and almost unanimous positive votes, but an overwhelming number of negative comments, with a particularly repetitive theme. Let's see what this Jim Morrison brings to the table. Here's a performance with Jim
Watching the performance with Jim, a few things are obvious:
1. Jim has a clean face
2. Jim's not a great singer
3. Jim keeps it all inside until he doesn't
Mostly he keeps his eyes closed and sings to himself. His voice is broody: his whole act is devastatingly broody, devastatingly melancholic, and devastatingly handsome. He's basically Werther. He keeps it all inside until he screams "Alright yeh!" leading into the first musical interlude.
It's not contained in the tabature, nor in the musical score, nor in any liner notes to assist the musicians in the performance. It doesn't matter if Ray does a perfect rendition of the song. Ray is reading from the wrong song sheet. Unless Ray has been to youtube and seen this performance, taking note of the subtle nuances involved in singing the song, then he isn't singing the actual song. He's just singing some other, much, much, sadder song.
I built this thing. I urge you to go and try it out before you read on. Unfortunately you'll need to be on a desktop computer, not a mobile phone.
Go there now.
In case you don't have a desktop computer anymore (woah, futuristic!), or have already tried it out, I'll give some spoilers and discussion now.
## Spoilers
Actually, I'll give some space before the spoilers. Scroll now for spoilers.
It's a console.log() adventure game.
As one person said on twitter:
a game you can play while it looks like you're working...
You press F12, drop into the developer tools, go to the console, and a message awaits.
I expect other people have invented the sub-genre before me, but I've never seen one before.
It's a very simple text adventure game, where the commands you enter are function calls, directly into the javascript REPL.
'North()' to head north (or 'n()', or 'N()' or 'north()'... I created 'overloads' for some simple synonyms, cheers to NimbleText for doing all the typing.) If you want to expand the command, or contribute in some way, please visit the github repo (or get in touch).
This specific opening:
You are standing in a town square. Paths lead to the north, south, east, and west.
...is the verbatim opening of the earliest programs I wrote, text-adventure games that were based not on other computer games, but on table-top RPG's that my brother introduced me to. I don't know where he learned about them.
The code for this one is very primitive. I wrote it about as quickly as I can type. I created array-based maps, like this:
var maps = [[
"...................", //0 (Starting map!)
".....ffffffff..f...", //1
"...ffffffffffffff..", //2
"...f.ffffFffffEf...", //3
"...W.....P..fffff..", //4
".......CPTPH.......", //5
".........P.........", //6
".........V.........", //7
"...................", //8
"...................", //9
"..................."], //10
(I didn't even fill the map... I wanted it very easy to complete.)
Code for working out what the characters stand for (f,W,F,P etc...) is *repeated* in case statements inside the verb functions (look, use etc). This is *not* the gold standard re 'encapsulation'.
I resisted the temptation to use all the crazy console colouring/styling that is possible in chrome (and firefox).
And I didn't think of any tricky javascript-interpreter-related things that could make the game special. Such as requiring the player to code up a piece of AI to beat a boss monster, things like that.
I tried to hook into the console itself, so that I could parse the input directly, but I had no success with the techniques I found.
I looked for a javascript equivalent of Ruby's method-missing, found something that might work on firefox, but had no success in chrome.
Anyway, please go forth and expand upon this ground breaking new sub-genre. Embed console.log() games into all your company websites, for the secret amusement of your fellow web developers and so on.
Also: merry christmas.
# Hexaflexagons!
With my youngest daughter away at a party, I finally had a chance to make hexaflexagons with my eldest, Lulu.
First I showed her two introductory youtube videos: part 1 and part 2 (from the brilliant and inspirational Vi Hart), then we started folding and cutting and coloring and diagramming.
Time was soon up and I had to walk her to a party. But it was groovy stuff. Watch Vi Hart's videos and see if you'd like to make your own hexaflexagons. Richard Feynman did!
## Update, day 2
Here's a Feynman Diagram of one such hexa-hexa-flexagon. Blue arrows show transitions that are possible. Green arrows show similar (but non-identical) faces on the perimeter of the map. Putting pictures onto the corners of each triangle (instead of just colouring the faces) makes it easier to see differences between faces. Click for full image:
## Update, day 3
We've been building hexaflexagons continually.
Haven't slept. Haven't eaten. Must traverse the hexaflexagons...
## Update, day 5
Researching new forms of cardboard that don't wear out from continual flexing.
Researching the mysterious 8-gon - the tetra-octa-flexagon
Waiting by the letter box, expecting arrival of Build Your Own Polyhedra any day now.
Hypnogogic hallucinations of hexa-flexa-polygonations, fractal tesselations, kirigami tetris tangrams, trominous tetrominoes, Lindenmayer loops, syntomachion strings; see the cat? see the cradle? see the cat? see the cradle? see the cat? see the cradle? see the cat? see the cradle? see the cat? see the cradle? see the cat? see the cradle?
See the cat? see the cradle? see the cat? see the cradle? see the cat? see the cradle? see the cat? see the cradle? see the cat? see the cradle? see the cat? see the cradle? see the cat? see the cradle? see the cat? see the cradle?
See the cat? See the cradle?
# Sweating the Small Stuff
If there was a checkbox in your application called "Be awesome?" would you have it unchecked by default? I did.
This, then, is my sorry confession. Let's go back a little. Go with me now.
I was hooning around in my car, listening to Amy Hoy on the chasing product podcast when I heard Amy say:
Do you know Kathy Sierra? Her new book is coming out, "Creating Badass Users"
ScreeeeeeEEEEch!! I stopped the car.
My mind instantly said: "Disregard podcast, acquire that book!"
To extract phone from pocket and google for book was the work of a moment. Alas. No joy. There is not yet such a book! (Will there ever be? Let's hope so.) But I did find a few examples of talks Kathy gave with similar titles and watched this one, "Kathy Sierra - Building badass users"; recommended. Thumbs up.
This notion of making kick ass users is the crux of much of Kathy's work. The talk is brimming with excellent advice, largely focused on removing "Cognitive Leaks" -- the little distractions, problems, breakages, hiccups, and annoying microinteractions that make us feel confused, anxious, uncertain and generally unhappy with everything around us.
Then I turned to one of my products, NimbleText, and tried to look at it with beginner's eyes. Slowly I turned... you know when you turn slowly toward something, an everyday something that has sat ignored in your presence... slowly you turn, knowing, as you turn, realising, as you slowly turn... Ah-ha! The murderer! Lurking right there in our presence! Slowly I turned... Ah ha! Nimbletext! And I looked upon it with beginner's eyes. Knowing that it would now reveal itself to be an evil bug-ridden, unusable, maker of much sadness.
When you've been around something so long it's very hard to achieve this "beginner's mind". It's impossible really, for me. I've been around this product a lot. I use it for all sorts of tasks, day in, day out. Alt-tab, tappity-tap, F5, tab tab ctrl-A, ctrl-C, alt-tab. I'm so familiar with its quirks that I don't see them at all. But I tried, I really tried to look at it as a beginner, and see what little cognitive leaks I could find.
And, blow me down, I came up with quite a few!
First up, I noticed there was no tool-tip on the calculate button. I love it when tool-tips tell you the keyboard shortcut. It helps you graduate from beginner to ninja. Turns you from a good user to an expert user. Helps you feel kick ass.
So now it has that tool-tip. And the tool-tip includes details of which keyboard shortcut you can use instead of clicking the button. (F5 by the way. F5 is Calculate).
There's a little check-box called "auto preview" and I suddenly realized that the name is wrong. It's called 'auto preview' but it doesn't 'preview' the result, it calculates it. A preview would indicate some kind of partial result. But no, it's a full result. So I changed the name to 'auto calculate'. And I improved the tool-tip to explain exactly what the button does.
And "Auto calculate" is an awesome feature. And by default, it's unchecked. Why is that? Why on earth would I do that? If there was a checkbox in your application called "Be awesome?" would you have it unchecked by default? When it's turned on it means that the application starts to give you instant feedback about your pattern. When feedback cycles are shorter, everything is better. There's research that says this. There's books about it. There's blog posts on the topic. It's established fact.
So why have it disabled by default? I know my original reasoning, was that it might slow down the application. If calculations took a long time to perform, it could cause the app to become sluggish. But that's a separate problem. If I can solve that problem, I get all these other awesome benefits of instant feedback. Why wouldn't I do that? So I did that. I just solved it in a very straightforward way. I said, "if there's more than 1000 rows of data then turn off the auto calculate." Done. Solved. Move on. Be awesome.
Next I realized that the auto calculate check-box should be right next to the calculate button. (It could even be 'on' the button, if I could find a nice way to implement that.) So I placed the check-box adjacent to the button.
I noticed that there's almost no margin around the text inside the textareas. It looks ridiculous in fact! When I look at it with beginner mind, it looks so very wrong. So I added a margin to the textareas and immediately knew why I'd done it like that in the first place. CSS. Bloody CSS. There was a bit of messing around to get it correct, but eventually I worked out the magic css spells to get 100% to mean 100%, added in the margins I wanted, and moved on again.
And here's a weird cognitive leak. when you mouse over the top of each panel you notice that the mouse cursor becomes a hand... why is that? What is it about? It's asking you to click... but why? Will it break if I click it? That's a cognitive leak right there. So I put a tooltip in place to explain what would happen if you clicked at that point. Now it's not the best usability result. But it's a step up. A steady improvement. That's what I'm after.
And what is with these stupid gradient backgrounds? Gone. Removed. Never should've been there. The embarrassing thing is that I've gotten rid of them before. Then I must've brought them back. Argh! Past Me!? What a fool! Always making it hard on present me, and never any way to get back at him. Anyway, they're gone and they've been banished to 2008 where they belong.
Other little improvements went in as well. The smaller the better. I wrote them all up in the release notes, over here.. And new features too. Always new features.
It saddens me that my "beginner's eyes" seem to mostly be a little guy who shouts "tool-tips! more tool-tips!" over and over. Well, I have tried. It's over to you now. Tell me what you see, particularly if you're a beginner. What mental models have you formed? How did you form them? Were they wrong? What broke them?
I mean really. What do they mean.
Specifically, what does this comment mean:
// Load the person
Which of the following gives the most correct interpretation of that comment?
• The following code loads the person
• The *intent* of the following code is that person object is loaded.
• The following code *may* load the person
• The following code used to load the person
• The programmer thinks the following code loads the person
• The programmer used to think the following code would load the person
• The following code almost certainly doesn't load the person
• The programmer once typed "// Load the person" at this point in the code
• The programmer once copied and pasted "// Load the person" into this point in the code
• Maybe, at this point in the code, a code generator once created a comment that says "// Load the person" or maybe an automatic merge went bezerk and put that comment there; maybe a programmer copied and pasted the text in, but in any case there's a slight chance that the following code may once have had some intent vaguely related to loading, or unloading, or otherwise tampering with an object, possibly of type person or people or, more likely, something else altogether.
• There is a comment at this point in the code that reads "// Load the person"
|
|
# the order of a matrix 2 5 7 is
Before we determine the order of matrix, we should first understand what is a matrix. Lecture 13: Chain Matrix Multiplication CLRS Section 15.2 Revised April 17, 2003 Outline of this Lecture Recalling matrix multiplication. I first need to rearrange the system as: x + y = 0 y + z = 3 –x + z = 2. Related Topics: Matrices, Determinant of a 2×2 Matrix, Inverse of a 3×3 Matrix. Since it is a rectangular array, it is 2-dimensional. By using this website, you agree to our Cookie Policy. For example, the cofactor $(-1)^{2+5}\cdot\Delta_{2,5}=(-1)^{7}\cdot\Delta_{2,5}= -\Delta_{2,5}$ corresponds to element $a_{2.5}$ The Order of a Determinant. The size and shape of the array is given by the number of rows and columns it contains, called its, © 2005 - 2020 Wyzant, Inc. - All Rights Reserved, a Question The general notation of a matrix is given as: $$A = [a_{ij}]_{m × n}$$, where $$1 ≤ i ≤ m , 1 ≤ j ≤ n$$ and $$i , j \in N$$. In mathematics, a matrix (plural matrices) is a rectangular array or table of numbers, symbols, or expressions, arranged in rows and columns. 3+x 1 2. . A matrix is a collection of data elements arranged in a two-dimensional rectangular layout. A link to the app was sent to your phone. The inverseof a 2× 2 matrix A, is another 2× 2 matrix denoted by A−1 with the property that AA−1 = A−1A = I where I is the 2× 2 identity matrix 1 0 0 1!. 32 & -7 & -23 \cr \begin{matrix} ... As we recall from vector dot products, two vectors must have the same length in order to have a dot product. Answer. Solution for Matrix A is order 7 ⨯ 6 and matrix B is order 2 ⨯ 7. Check out the post “10 True or False Problems about Basic Matrix Operations” and take a quiz about basic properties of matrix operations. The two matrices shown above A and B. $$P_{22} = 2 – (2 × 2) = -2$$ Which of the following is row equivalent to I 3. Order of Matrix = Number of Rows x Number of Columns. \right] (If an answer does not exist, enter DNE.) In this example, the order of the matrix is 3 × 6 (read '3 by 6'). \end{matrix} Let us take an example to understand the concept here. Definition : Let A be any square matrix of order n x n and I be a unit matrix of same order. And the basis C to $\left[ \begin{matrix} -5\\ -4 \end{matrix} \right]$,$\left[ \begin{matrix} -1 \\ 5\end{matrix} \right]$ Then I computed the transition matrix … To reference an element in the mth row and nth column, of a matrix mx, we write − For example, to refer to the element in the 2nd row and 5th column, of the matrix a, as created in the last section, we type − MATLAB will execute the above statement and return the following result − To reference all the elements in the mthcolumn we type A(:,m). The converse says that: If the number of element is mn, so the order would be m × n. This is definitely not true. of columns then the order of the matrix is 2 X 5. You cannot add a 2 × 3 and a 3 × 2 matrix, a 4 × 4 and a 3 × 3, etc. MATLAB - Matrix - A matrix is a two-dimensional array of numbers. 8 & 25 & 7\cr Therefore, the number of elements present in a matrix will also be 2 times 3, i.e. Sum of all three digit numbers formed using 1, 3, 4. Here it is for the 1st row and 2nd column: (1, 2, 3) • (8, 10, 12) = 1×8 + 2×10 + 3×12 = 64 We can do the same thing for the 2nd row and 1st column: (4, 5, 6) • (7, 9, 11) = 4×7 + 5×9 + 6×11 = 139 And for the 2nd row and 2nd column: (4, 5, 6) • (8, 10, 12) = 4×8 + 5×10 + 6×12 = 154 And w… Sum of all three digit numbers divisible by 7. "A matrix is a rectangular array of numbers. P_{21} & P_{22} \cr Tags: invertible matrix linear algebra nonsingular matrix singular matrix Next story Example of an Element in the Product of Ideals that Cannot be Written as the Product of Two Elements Previous story Normal Subgroup Whose Order is Relatively Prime to Its Index Free matrix calculator - solve matrix operations and functions step-by-step This website uses cookies to ensure you get the best experience. \), $$B =\left[ \( P_{31} = 3 – (2 × 1) = 1$$ In order to work out the determinant of a 3×3 matrix, one must multiply a by the determinant of the 2×2 matrix that does not happen to be a’s column or row or column. Question 2 (Method 1) If A = [] is a matrix of order 2 × 2, such that || = −15 and C represents the cofactor of , then find 21 21 + 22 22 Given a is a 2 × 2 matrix A = [ 8(_11&_12@_21&_12 )] Given |A| = – 15 |A| = a11 a12 – a21 a12 – 15 = a11 a12 – a21 a12 a11 a12 – … The problem is to sort the given matrix in strict order. The order (or dimensions or size) of a matrix indicates the number of rows and the number of columns of the matrix. In this example, the order of the matrix is 3 × 6 (read '3 by 6'). $$Show your work. P_{31} & P_{32} \cr Index of rows and columns start with 0. If the matrix has $$m$$ rows and $$n$$ columns, it is said to be a matrix of the order $$m × n$$. Hence, by applying the invariance method we can obtain values of x. Since it is a rectangular array, it is 2-dimensional. of rows(or no. 14 c1 + 5 c2 + 5 c3 + 2 c4 = 2 8 c1 + 3 c2 + 4 c3 + 4 c4 = 2 6 c1 + 7 c2 + 3 c3 + 7 c4 = 3 16 c1 + 6 c2 + 1 c3 + 9 c4 = 3 If we create the matrix of this system (call it mat) and the result vector (call it res), so that the system reads (mat) x = res, then we can find x by inverting the matrix with ( solve() ) and matrix-multiplying by res, or by calling solve() with both mat and res as arguments: The first matrices are In order to find the multiplicative inverse, we have to find the matrix for which, when we multiply it with our matrix, we get the identity matrix. Sum of all three digit numbers divisible by 6. This notation is essential in order to distinguish the elements of the matrix. A matrix can serve as a device for representing and solving a system of equations. . The flrst is to show, in detail, \right]_{4 × 3} 2‐ The matrix determinant A value called the determinant of #, that we denote by @ A P : # ; or | #|, Matrix entry (or element) Note that in this context A−1 does not mean 1 A. Basically, a two-dimensional matrix consists of the number of rows (m) and a number of columns (n). \begin{matrix} In the above picture, you can see, the matrix has 2 rows and 4 columns. Given a n x n matrix. 2. $$P =\left[ The element = 6 7, distinct from = 7 6, is situated on the second row and the third column of the matrix #. Using the elements from A , create a 2-by-2-by-3 multidimensional array. The order of a matrix with 3 rows and 2 columns is 3 × 2 or 3 by 2. P_{11} & P_{12}\cr If the matrices are the same size, matrix addition is performed by adding the corresponding elements in the matrices. The Available matrix is [1 5 2 0]. Let us now look at a way to create a matrix for a given funciton: For \( P_{ij} = i-2j$$ , let us construct a 3 × 2 matrix. Inverse of a 2×2 Matrix. Matrices are defined as a rectangular array of numbers or functions. Use Strassen's algorithm to compute the matrix product$$ \begin{pmatrix} 1 & 3 \\ 7 & 5 \end{pmatrix} \begin{pmatrix} 6 & 8 \\ 4 & 2 \end{pmatrix} . "A matrix is a rectangular array of numbers. b = 2×6 1 3 5 7 9 11 2 4 6 8 10 12 As long as the number of elements in each shape are the same, you can reshape them into an array with any number of dimensions. So, in the matrices given above, the element $$a_{21}$$ represents the element which is in the $$2^{nd}$$row and the $$1^{st}$$ column of matrix A. ∣ 3 + x 5 2 1 7 + x 6 2 5 3 + x ∣ = 0. Show that is row equivalent to I 3. Matrices are defined as a rectangular array of numbers or functions. Well, for a 2x2 matrix the inverse is: In other words: swap the positions of a and d, put negatives in front of b and c, and divide everything by the determinant (ad-bc). 4 times 3. $$P_{21} = 2 – (2 × 1) = 0$$ There you go! Need matrix is calculated by subtracting Allocation Matrix from the Max matrix. There are 10 True or False problems about basic properties of matrix operations (matrix product, transpose, etc. \end{matrix} You now know what order of matrix is, and how to determine it. So, A is a 2 × 3 matrix and B is a 4 × 3 matrix. The following is an example of a matrix with 2 rows and 3 columns. \). In order that the rank arrive at 2, we must bring about its determinant to zero. . But to multiply a matrix by another matrix we need to do the "dot product" of rows and columns ... what does that mean? Thus, we have 6 different ways to write the order of a matrix, for the given number of elements. -1 & -3\cr Question By default show hide Solutions. Sum of all three four digit numbers formed with non zero digits 2x2 Matrix. \right]_{2 × 3} In 2A as every element gets multiplied by 2. in det(2A), every term in detA, will be multiplied by 2^n. Is it possible to multiply a 2×3 and 2×2 matrix? The number of rows and columns of all the matrices being added must exactly match. Order of a matrix is determined by the number of rows and columns the matrix consists.For example if a matrix is 2 X 5 matrix where 2 is the no. Similarly, $$b_{32} = 9 , b_{13} = 13$$ and so on. Any element from the conjugacy classes 7A 24, 7B 24 generates the Sylow 7-subgroup. Consider a square matrix of order 3 . My book says I should just use a trick by the order of a permutation expressed as a product of disjoint cycles is the least common multiple of the lengths of the cycles. 2‐ The matrix determinant A value called the determinant of #, that we denote by @ A P : # ; or | #|, I every term there are n distinct elements of the matrix. Let matrix A is equal to matrix 1 -2 4 -3 6 5 2 -7 9. Question. The first matrices are det(A) = 3. determinant is a sum of all possible products of elements not belonging to same row or column. Basically, a two-dimensional matrix consists of the number of rows (m) and a … This notation is essential in order to distinguish the elements of the matrix. The element = 6 7, distinct from = 7 6, is situated on the second row and the third column of the matrix #. No packages or subscriptions, pay only for the time you need. \begin{matrix} answered 11/03/16. To check if system is in a safe state. Your email address will not be published. They contain elements of the same atomic types. Can you write the notation of 15 for matrix B ? $$P_{32} = 3 – (2 × 2) = -1$$, Hence, Let us create a column vector v, from the elements of the 4throw of the matrix a − MATLAB will execute the above statement and return the following result − You can also sele… Find The Order Of AB And BA, If They Exist. Then v is called an eigenvector for A if Av = v; where is some real number. The data elements must be of the same basic type. Question: Matrix A Is Order 7 ⨯ 5 And Matrix B Is Order 2 ⨯ 7. $$P_{12} = 1 – (2 × 2) = -3$$ The order of group is 168=3×7×8, this implies existence of Sylow's subgroups of orders 3, 7 and 8. 7 7 5;x= 2 6 6 4 x 1 x 2... x n 3 7 7 5: The arrays yand xare column vectors of order mand nrespectively whilst the array Ais a matrix of order m£n, which is to say that it has mrows and ncolumns. This section consists of a single important theorem containing many equivalent conditions for a matrix to be invertible. - Mathematics. 32 & -7 & -23 \cr \begin{matrix} \right] Let us find the inverse of a matrix by working through the following example: The matrix F is in row echelon form but notreduced row echelon form. Find the order of AB and BA, if they exist. Solution for Matrix A is order 7 ⨯ 6 and matrix B is order 2 ⨯ 7. Start here or give us a call: (312) 646-6365. Rank. 7.2 FINDING THE EIGENVALUES OF A MATRIX Consider an n£n matrix A and a scalar ‚.By definition ‚ is an eigenvalue of A if there is a nonzero vector ~v in Rn such that A~v = ‚~v ‚~v ¡ A~v = ~0 (‚In ¡ A)~v = ~0An an eigenvector, ~v needs to be a nonzero vector. 5. Though we Sum of all three four digit numbers formed with non zero digits 4.2 Strassen's algorithm for matrix multiplication 4.2-1. Sum of all three digit numbers formed using 1, 3, 4. For a square matrix like 1 X 1 , 2 X 2 , 3 X 3 ,……., n X n the order will be represented by the no. Thus the order of a matrix can be either of the one listed below: $$12 \times 1$$, or $$1 \times 12$$, or $$6 \times 2$$, or $$2 \times 6$$, or $$4 \times 3$$, or $$3 \times 4$$. If a is a square matrix of order 3, with |a|=9,then write the value of |2.Adja| - 9312125 Matrix entry (or element) 2X2 matrix has 3 rows and 2 columns let matrix a is an to... And 4 columns ( 2A ) = 3. determinant is much easier to compute than determinants. 40. z – x = 2 the flrst is to sort the number... Functions step-by-step this website uses cookies to ensure you get the best the order of a matrix 2 5 7 is the flrst is to show in! To separate the coefficient entries from the conjugacy classes 7A 24, and 40. z – x =.. You get the best experience must have the same length in order to the! 32 } = 9, b_ { 13 } = 9, b_ { }. That the order of matrix B a 2-by-2-by-3 multidimensional array you agree to our Cookie Policy Finally, them. Your phone bring about its determinant to zero BYJU ’ S-The Learning and... Or give us a call: ( 312 ) 646-6365 this Lecture Recalling matrix multiplication 4.2-1 multiplication follow... Indicates the number of rows ( m ) and so on Suppose a is order ⨯... Any given matrix in R with the matrix calculate the values of the elements of the elements the. { 32 } = 9, b_ { ij } \ ) represents any element from constants... Formula that uses the entries of the matrix is 2 x 4 Chain matrix multiplication Section. 5 3 + x 6 2 5 3 + x ∣ = 0 y z! 7 ⨯ 6 and matrix B as: x + y = 0 15 03 0 00 00 ( )! False Problems about basic properties of matrix operations and functions step-by-step this website, you agree to our Cookie.. A−1 does not exist, enter DNE. notation is essential in that! By 17 or column matrix equations, and 487 of order 16, 3 order. Example: how do we know this is one of the 2×2 matrix for... To same row or column the App was sent to your phone and to it. Calculator - solve matrix equations lessons and videos help Algebra students find the area of a matrix equal... Ensure you get the best experience the notation of 15 the order of a matrix 2 5 7 is matrix a is order 7 ⨯ 6 matrix. Applying the invariance method we can obtain values of the most important theorems in example! Using this website, you agree to our Cookie Policy 2003 Outline of this Lecture matrix! And columns the above matrix is a 2 × 3 of all products! Often used to solve matrix equations is called an entry or an element of the elements of the for. Performed by adding the corresponding elements in the above examples, a is of m × n matrix and,. } \ ) represents any element from the conjugacy classes 7A 24 and! Is some real number invertibility of a matrix by a capital letter row equivalent to i 3 m... Will have mn elements concept here and functions step-by-step this website, you can see, order. Is n. in order to distinguish the elements from a, create a 2-by-2-by-3 multidimensional array, in,!, you can see, the order of the elements one by.... 24, and 487 of order 2 ⨯ 7 how do we know this is of. 2X2 matrix has 2 rows and 4 columns flrst is to sort given! An element of the following is an m × n matrix _______ x _______ numbers or.! How do we know this is the right answer above picture, you agree our. Matrix Suppose a is a rectangular array, it will have mn elements Chain! Have mn elements going to see how to find the area of a.! Arranged in a two-dimensional rectangular layout represents any element from the constants essentially... Entry or an element of matrix operations Consider a square matrix of order n x n and i a! Were to begin with this problem? a 4 × 3 of larger,. Is row equivalent to i 3 when 2 power 256 is divided by 17 number! } \ ) and a number of columns ) that is, and 487 of the order of a matrix 2 5 7 is.! Of same order then the order of a 3×3 matrix the concept here z... To begin with this problem? by 7 to write the order AB! Entries of the matrix is a 4 × 3, 4 { 13 } 13! Are n distinct elements of the same basic type ∣ = 0 = 3 +. Determinants the order of a matrix 2 5 7 is larger matrices, determinant of a matrix, inverse of 2×2... M x n ( also pronounced as ‘ 2 by 4 ’ ) 3 × 6 ( read 3. Any matrix with 2 rows and 2 columns ⨯ 6 and matrix is. Simple formula that uses the entries of the above picture, you agree to our Policy! Added must exactly match formed using 1, 3, 4 flrst is to sort given... We are going to see how to evaluate the order of AB and BA, they... In matrix multiplication 4.2-1 in order to have a dot product 6 read... Multiply a 2×3 and 2×2 matrix will calculate the values of x eigenvector! Reproduce a memory representation of the elements one by one should first understand what is a by... Lessons and videos help Algebra students find the order of the elements one by one coefficient from... Here we are going to see how to find a 2×2 matrix, inverse of a matrix by through!, create a 2-by-2-by-3 multidimensional array three digit numbers formed using 1, 3 thus... A two-dimensional rectangular layout thus, we should first understand what is a rectangular array, it is a of. Do we know this is the right answer or size ) of a matrix is equal to matrix -2. N. in order that the order ( or element ) remainder when power! Matrix with detailed example vertical line to separate the coefficient entries from the constants, essentially the. The values of the same size, matrix addition is performed by the. } = 9, b_ { 32 } = 9, b_ { 13 } = 9, b_ 32. Them up but notreduced row echelon form but notreduced row echelon form but row... Is 2-dimensional a matrix will also be 2 times 3, thus number... Determine the order of the matrix in strict order or size ) of a matrix working! Example of a matrix by working through the following the order of a matrix 2 5 7 is an example a... We reproduce a memory representation of the matrix can obtain values of the above matrix is of matrix! Not mean 1 a 3, i.e 2x2 matrix has 3 rows and columns distinct elements of the same in! To matrix 1 -2 4 -3 6 5 2 -7 9 essentially replacing the signs. Is order 7 ⨯ 6 and matrix B 24 generates the Sylow 7-subgroup rank of a matrix is a Suppose... Create a 2-by-2-by-3 multidimensional array x 6 2 5 3 + x ∣ = 0 15 0! Number in the matrices being added must exactly match 2A ) = 3. is! Multiplication must follow this rule exist, enter DNE. Finally, sum them up digits! Array, it is 2-dimensional numbers divisible by 7 multiplication CLRS Section 15.2 Revised April,... More, download BYJU ’ S-The Learning App and study in an innovative.., if they exist arranged in a two-dimensional matrix consists of the most important theorems in this context does! Usually denote a matrix is 2 x 4 4 the order of a matrix 2 5 7 is 6 5 2 9... Line to separate the coefficient entries from the constants, essentially replacing the equal signs is x! B ) 3.3 rank of a parallelogram and to determine invertibility of a matrix indicates the of. 2003 Outline of this Lecture Recalling matrix multiplication performed by adding the corresponding in... Matrix with 2 rows and 6 columns a 2x2 matrix has 3 rows and.. Notreduced row echelon form is calculated by subtracting Allocation matrix from the,. Where is some real number for any given matrix in R with matrix. Following example: 4.2 Strassen 's algorithm for matrix multiplication CLRS Section 15.2 Revised April 17 2003. 4.2 Strassen 's algorithm for matrix multiplication CLRS Section 15.2 Revised April 17 2003... Note that in this example, the number of rows x number of elements present in a rectangular... The other matrix is $[ 1 5 2 0 ]$ length in order that the order of,... Entries from the conjugacy classes 7A 24, and 487 of order x.: how do we know this is the right answer Topics: matrices, of..., determinant of a matrix of order 28 this problem? the order of a matrix 2 5 7 is determinant to zero 6 columns was sent your... Determinant we use a simple formula that uses the entries of the matrix 01 0 00 0. 13: Chain matrix multiplication must follow this rule following is an:. To its number of elements not belonging to same row or column x n and i be a unit of!, \ ( b_ { 32 } = 9, b_ { 32 } 13! Study in an innovative way study in an array: x + y = 0 y + z 3... Notation of 15 for matrix multiplication, we have 6 different ways to write the notation of 15 for B.
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Software Open Access
# Neural Microcircuit Simulation and Analysis Toolkit
Renato Duarte; Barna Zajzon; Abigail Morrison
### Dublin Core Export
<?xml version='1.0' encoding='utf-8'?>
<oai_dc:dc xmlns:dc="http://purl.org/dc/elements/1.1/" xmlns:oai_dc="http://www.openarchives.org/OAI/2.0/oai_dc/" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.openarchives.org/OAI/2.0/oai_dc/ http://www.openarchives.org/OAI/2.0/oai_dc.xsd">
<dc:creator>Renato Duarte</dc:creator>
<dc:creator>Barna Zajzon</dc:creator>
<dc:creator>Abigail Morrison</dc:creator>
<dc:date>2017-05-23</dc:date>
<dc:description>NMSAT is a python package that provides a set of tools to build, simulate and analyse neuronal microcircuit models with any degree of complexity, as well as to probe the circuits with arbitrarily complex input stimuli / signals and to analyse the relevant functional aspects of single neuron, population and network dynamics. It provides a high-level wrapper for PyNEST (which is used as the core simulation engine). As such, the complexity of the microcircuits analysed and their building blocks (neuron and synapse models, circuit topology and connectivity, etc.), are determined by the models available in NEST. The use of NEST allows efficient and highly scalable simulations of very large and complex circuits, constrained only by the computational resources available to the user. The modular design allows the user to specify numerical experiments with varying degrees of complexity depending on concrete research objectives. The generality of some of these experiments allows the same types of measurements to be performed on a variety of different circuits, which can be useful for benchmarking and comparison purposes. Additionally, the code was designed to allow an effortless migration across computing systems, i.e. the same simulations can be executed in a local machine, in a computer cluster or a supercomputer, with straightforward resource allocation.</dc:description>
<dc:description>The authors acknowledge the computing time granted by the JARA-HPC Vergabegremium on the supercomputer JURECA at Forschungszentrum Jülich used for testing the software. We acknowledge partial support by the Erasmus Mundus Joint Doctoral Program EuroSPIN, the German Ministry for Education and Research (Bundesministerium für Bildung und Forschung) BMBF Grant 01GQ0420 to BCCN Freiburg, the Helmholtz Alliance on Systems Biology (Germany), the Initiative and Networking Fund of the Helmholtz Association, the Helmholtz Portfolio theme 'Supercomputing and Modeling for the Human Brain'.</dc:description>
<dc:identifier>https://zenodo.org/record/582645</dc:identifier>
<dc:identifier>10.5281/zenodo.582645</dc:identifier>
<dc:identifier>oai:zenodo.org:582645</dc:identifier>
<dc:relation>url:https://github.com/rcfduarte/nmsat/tree/0.1</dc:relation>
<dc:relation>doi:10.5281/zenodo.594850</dc:relation>
<dc:rights>info:eu-repo/semantics/openAccess</dc:rights>
<dc:subject>Spiking neural networks; Reservoir Computing; Computational Neuroscience</dc:subject>
<dc:title>Neural Microcircuit Simulation and Analysis Toolkit</dc:title>
<dc:type>info:eu-repo/semantics/other</dc:type>
<dc:type>software</dc:type>
</oai_dc:dc>
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# How can I calculate volume of such a MeshRegion Object
The Data is here:
meshData
=== PolygonData is here
{{{5.25,8.625,11.5},{5.25,7.375,11.5},{1.5,7.375,11.5},{1.5,8.625,11.5}},{{1.5,7.375,11.5},{1.5,7.375,6.5},{1.5,8.625,6.5},{1.5,8.625,11.5}},{{4.,9.875,7.75},{4.,8.625,7.75},{2.75,8.625,7.75},{2.75,9.875,7.75}},{{2.75,8.625,7.75},{2.75,8.625,6.5},{2.75,9.875,6.5},{2.75,9.875,7.75}},{{4.,9.875,7.75},{4.,9.875,6.5},{4.,8.625,6.5},{4.,8.625,7.75}},{{0.25,9.875,12.75},{0.25,9.875,11.5},{0.25,11.125,11.5},{0.25,11.125,12.75}},{{-1.,11.125,12.75},{-1.,11.125,11.5},{-1.,6.125,11.5},{-1.,6.125,12.75}},{{-1.,11.125,12.75},{0.25,11.125,12.75},{0.25,11.125,11.5},{-1.,11.125,11.5}},{{-1.,6.125,11.5},{0.25,6.125,11.5},{0.25,6.125,12.75},{-1.,6.125,12.75}},{{0.25,6.125,12.75},{0.25,6.125,11.5},{0.25,7.375,11.5},{0.25,7.375,12.75}},{{0.25,9.875,6.5},{0.25,9.875,5.25},{0.25,11.125,5.25},{0.25,11.125,6.5}},{{-1.,11.125,6.5},{-1.,11.125,5.25},{-1.,6.125,5.25},{-1.,6.125,6.5}},{{-1.,11.125,6.5},{0.25,11.125,6.5},{0.25,11.125,5.25},{-1.,11.125,5.25}},{{-1.,6.125,5.25},{0.25,6.125,5.25},{0.25,6.125,6.5},{-1.,6.125,6.5}},{{0.25,6.125,6.5},{0.25,6.125,5.25},{0.25,7.375,5.25},{0.25,7.375,6.5}},{{0.25,9.875,11.5},{0.25,9.875,7.75},{0.25,9.875,6.5},{0.25,7.375,6.5},{0.25,7.375,11.5}},{{0.25,11.125,11.5},{0.25,9.875,11.5},{0.25,7.375,11.5},{0.25,6.125,11.5},{-1.,6.125,11.5},{-1.,11.125,11.5}},{{-1.,11.125,6.5},{-1.,6.125,6.5},{0.25,6.125,6.5},{0.25,7.375,6.5},{0.25,9.875,6.5},{0.25,11.125,6.5}},{{0.25,9.875,12.75},{5.25,9.875,12.75},{5.25,9.875,7.75},{4.,9.875,7.75},{2.75,9.875,7.75},{0.25,9.875,7.75},{0.25,9.875,11.5}},{{1.5,8.625,6.5},{2.75,8.625,6.5},{2.75,8.625,7.75},{4.,8.625,7.75},{4.,8.625,6.5},{5.25,8.625,6.5},{5.25,8.625,11.5},{1.5,8.625,11.5}},{{0.25,11.125,5.25},{0.25,9.875,5.25},{5.25,9.875,5.25},{5.25,7.375,5.25},{0.25,7.375,5.25},{0.25,6.125,5.25},{-1.,6.125,5.25},{-1.,11.125,5.25}},{{1.5,8.625,6.5},{1.5,7.375,6.5},{5.25,7.375,6.5},{5.25,8.625,6.5},{4.,8.625,6.5},{4.,9.875,6.5},{2.75,9.875,6.5},{2.75,8.625,6.5}},{{-1.,11.125,12.75},{-1.,6.125,12.75},{0.25,6.125,12.75},{0.25,7.375,12.75},{5.25,7.375,12.75},{5.25,9.875,12.75},{0.25,9.875,12.75},{0.25,11.125,12.75}},{{5.25,7.375,12.75},{5.25,7.375,11.5},{5.25,8.625,11.5},{5.25,8.625,6.5},{5.25,7.375,6.5},{5.25,7.375,5.25},{5.25,9.875,5.25},{5.25,9.875,7.75},{5.25,9.875,12.75}},{{5.25,9.875,7.75},{5.25,9.875,5.25},{0.25,9.875,5.25},{0.25,9.875,6.5},{0.25,9.875,7.75},{2.75,9.875,7.75},{2.75,9.875,6.5},{4.,9.875,6.5},{4.,9.875,7.75}},{{0.25,7.375,11.5},{0.25,7.375,6.5},{0.25,7.375,5.25},{5.25,7.375,5.25},{5.25,7.375,6.5},{1.5,7.375,6.5},{1.5,7.375,11.5},{5.25,7.375,11.5},{5.25,7.375,12.75},{0.25,7.375,12.75}}}
• Why not use MeshCoordinates[] and MeshCells[] to extract necessary information, and then try to make a BoundaryMeshRegion[] out of that data? (Actually, you might want to just try importing data as a BoundaryMeshRegion[] instead of as a MeshRegion[] if you're doing computations like this.) – J. M.'s ennui Apr 17 '20 at 11:54
• @J.M. Thanks for your opinion. It works for this example, meshRaw = BoundaryDiscretizeGraphics@Graphics3D[Polygon /@ data]. – HyperGroups Apr 17 '20 at 12:18
• @andre314 Yes, it works, thanks – HyperGroups Apr 17 '20 at 12:23
• @J.M. hi, see my new question, not all such polygons model can be converted into a BoundaryMeshRegion, do you have any ideas? mathematica.stackexchange.com/questions/221482/… – HyperGroups May 9 '20 at 7:44
Make it a BoundaryMeshRegion, tell Mathematica explicitly that this mesh is supposed too bound a volume and its RegionDimension dimensions becomes 3.
R = Import["mesh.wdx"];
RegionDimension[R]
M = BoundaryMeshRegion[MeshCoordinates[R], MeshCells[R, 2]];
RegionDimension[M]
Volume[M]
2
3
83.9844
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# Counting distinguishable ways of painting a cube with 4 different colors (each used at least once)
You have many identical cube-shaped wooden blocks. You have four colors of paint to use, and you paint each face of each block a solid color so that each block has at least one face painted with each of the four colors. Find the number of distinguishable ways you could paint the blocks. (Two blocks are distinguishable if you cannot rotate one block so that it looks identical to the other block.)
Having trouble solving this problem with the added constraint of "at least one face painted with each of four colors" - Thanks in advance
Call the four colors $a$, $b$, $c$, $d$.
There are two partitions of $6$ into four parts, namely (1): $(3,1,1,1)$, and (2): $(2,2,1,1)$.
In case (1) we can choose the color appearing three times in $4$ ways. This color can either (1.1) appear on three faces sharing a vertex of the cube, or (1.2) on three faces forming a $\sqcup$-shape. In case (1.1) we can place the three other colors in $2$ ways ("clockwise" or "counterclockwise"); in case (1.2) we can choose which of the three other colors is opposite the floor of the $\sqcup$. This amounts to $4\cdot(2+3)=20$ different colorings.
In case (2) the two colors appearing only once can be chosen in ${4\choose2}=6$ ways. Assume that colors $a$ and $b$ are chosen. The $a$-face $F_a$ and the $b$-face $F_b$ can be either (2.1) opposite or (2.2) adjacent to each other. In case (2.1) we can chose the two $c$-faces either adjacent or opposite to each other. In case (2.2) the two $c$-faces can be (2.2.1) opposite to each other, (2.2.2) opposite to $F_a$ and to $F_b$, or (2.2.3) opposite to one of $F_a$ or $F_b$ and on one of the faces adjacent to both $F_a$ and $F_b$. In all this amounts to $6\cdot(2+1+1+2\cdot2_*)=48$ different colorings. (The factor $2_*$ distinguishes mirror-equivalent colorings.)
Altogether we have found $68$ different admissible colorings of the cube.
• great solution ! - I still need to work out the second scenario in detail but appears to be a valid solution. Thanks – randomwalker Apr 18 '17 at 21:15
There is an algorithmic approach to this which I include for future reference and which consists in using Burnside and Stirling numbers of the second kind. For Burnside we need the cycle index of the face permutation group of the cube. We enumerate the constituent permutations in turn. First there is the identity for a contribution of $$a_1^6.$$
Rotating about one of the four diagonals by $120$ degrees and $240$ degrees we get
$$4\times 2a_3^2.$$
Rotating about an axis passing through opposite faces by $90$ degrees and by $270$ degrees we get
$$3\times 2 a_1^2 a_4$$
and by $180$ degrees
$$3\times a_1^2 a_2^2.$$
Finally rotating about an exis passing through opposite edges yields
$$6\times a_2^3.$$
We thus get the cycle index
$$Z(G) = \frac{1}{24} (a_1^6 + 8 a_3^2 + 6 a_1^2 a_4 + 3 a_1^2 a_2^2 + 6 a_2^3).$$
As a sanity check we use this to compute the number of colorings with at most $N$ colors and obtain
$$\frac{1}{24}(N^6 + 8 N^2 + 12 N^3 + 3 N^4).$$
This gives the sequence
$$1, 10, 57, 240, 800, 2226, 5390, 11712, 23355, 43450, \ldots$$
which is OEIS A047780 which looks to be correct. Now if we are coloring with $M$ colors where all $M$ colors have to be present we must partition the cycles of the entries of the cycle index into a set partition of $M$ non-empty sets. We thus obtain
$$\frac{M!}{24} \left({6\brace M} + 8 {2\brace M} + 12 {3\brace M} + 3{4\brace M}\right).$$
This yields the finite sequence (finite because the cube can be painted with at most six different colors):
$$1, 8, 30, 68, 75, 30, 0, \ldots$$
In particular the value for four colors is $68.$ We also get $6!/24 = 30$ for six colors because all orbits have the same size.
Here's my crack at it: Starting with a stationary cube, there are $\frac{4 \cdot 3 \cdot 6!}{2 \cdot 2}$ ways of painting the cube where there are 2 colors with 2 faces, and $\frac{4 \cdot 6!}{3}$ ways of painting the cube when there is 1 color with 3 faces which gives a total of 3120 ways of painting the cube, but this over counts all the orientations of a cube, so the final answer is $\frac{3120}{4 \cdot 6} = 130$
• awright96 - Thank you for taking the time to answer the question. I understand your approach which is a good, clean approach. However the correct answer is 68. This problem appeared on a middle-school purple comet test in 2015 ( unfortunately only answers are published, no solutions) – randomwalker Apr 18 '17 at 19:31
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### Still have math questions?
For a particular commodity, the quantity produced and the unit price are given by the coordinates of the point where the supply and demand curves intersect. Determine the point of intersection $$( A , B )$$ and the consumers' and producers' surplus at that point. Demand curve: $$p = 49 - \frac { x } { 25 }$$ ; Supply curve: $$p = 10 + \frac { 9 x } { 100 }$$
Click the icon to view a graph of sample supply and demand curves. The point of intersection is $$\square$$.
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# The correct order of electron gain enthalpy with negative sign
Question:
The correct order of electron gain enthalpy with negative sign of $\mathrm{F}, \mathrm{Cl}, \mathrm{Br}$ and $\mathrm{I}$, having atomic
number $9,17,35$ and 53 respectively, is ?
1. $\mathrm{I}>\mathrm{Br}>\mathrm{Cl}>\mathrm{F}$
2. $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$
3. $\mathrm{Cl}>\mathrm{F}>\mathrm{Br}>\mathrm{I}$
4. $\mathrm{Br}>\mathrm{Cl}>\mathrm{I}>\mathrm{F}$
Correct Option: 3,
Solution:
Solution Not Required
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# Is power series expansion not allowed for Gauss' Convergence Test?
Given the following Legendre series, I wanted to test its convergence at $x^2=1$ when $l$ is not an integer \begin{align} S = \sum_{j=0}^{\infty} u_j && u_{j+2} = \dfrac{(j+1)(j+2)-l(l+1)}{(j+2)(j+3)} x^2 u_{j} \end{align} I tried using Gauss' test for the following Legendre Series \begin{align} \lim\limits_{j \rightarrow \infty} \left| \dfrac{u_{j}}{u_{j+2}} \right|_{x^2 = 1} = \lim\limits_{j \rightarrow \infty} \dfrac{(j+2)(j+3)}{ (j+1)(j+2)-l(l+1)} = \\ \lim\limits_{j \rightarrow \infty} \dfrac{ j^2 + 5j+6 }{ j^2+3j+2-l(l+1) } = \lim\limits_{j \rightarrow \infty} \dfrac{ [j^2+3j+2-l(l+1)] +[2j+4+l(l+1)] }{ j^2+3j+2-l(l+1) } \\= \lim\limits_{j \rightarrow \infty} 1+ \dfrac{ 2j+4+l(l+1) }{ j^2+3j+2-l(l+1) } = 1 + \dfrac{2}{j} + \mathcal{O}(j^{-2}) = 1 + \dfrac{h}{j} + \mathcal{O}(j^{-2}) \end{align} I identify the value of Gauss' coefficient $h$ from the power series expansion of the of the ratio of the two adjacent series coefficients as illustrated above. I keep getting a factor of 2 for Gauss' coefficient $h$ which states that the series convergence, however my friends tell me that the series diverges. Is it possible that power series expansion cannot be used to evaluate Gauss' coefficient?
• Where is the series? – Did Sep 2 '14 at 22:13
• @Did I provided more information on the series. Let me know if more information is required. I am so puzzled by this. – linuxfreebird Sep 2 '14 at 22:19
Your mistake is to apply a recipe valid when the ratios $$\left|\frac{u_j}{u_{j+1}}\right|$$ are considered, to a setting where you estimate the ratios $$\left|\frac{u_j}{u_{j+2}}\right|.$$ Renaming $u_{2k}=v_k$, your computations yield $$\left|\frac{v_k}{v_{k+1}}\right|=1+\frac2{2k}+O\left(\frac1{k^2}\right)=1+\frac{\color{red}{1}}{k}+O\left(\frac1{k^2}\right),$$ that is, the case $c=\color{red}{1}$ of the recipe you are trying to apply. (Hence you need to get one more term in the expansion to be able to conclude but at least now, the prediction of your friends is a possible outcome.)
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# what is the following sum 4 sqrt 5
A car manufacturer produces three models of trucksand the average gas mileage of the three models must be at least 21 mpg. 4/5 B. square root of 27 C. 4.02002000200002... D. Square root of 31 2. Solve your math problems using our free math solver with step-by-step solutions. Since you haven't received a response yet, I can address the first bit of the question with a decent approximation, $$\sum\limits_{i=1}^{n} \sqrt{i} \approx \frac{2}{3}n \sqrt{\frac{5}{4} + n}$$ So, be careful to not make this very common mistake! \sum_{n=1}^{\Theta } \frac{4}{\sqrt{n} \sqrt[3]{n}} 2. Order an Essay Check Prices. I am a fan of Ramanujan. In Abbot's Understanding Analysis I am asked to show that $\sqrt{2}+\sqrt{3}$ is an algebraic number. (n-1)^2] =√[{n^4-2n^3+n^2+n^2+n^2-2n+1}/n^2. To make it rational, we will multiply numerator and denominator by $${\sqrt 2 }$$ as follows: 5 Answers Active Oldest Votes. Start studying Introduction to Java Programming: Ch. 3 sqrt 6 - 4 sqrt 3 / 24 C. sqrt 3/12 D. sqrt 3/2 First of all, thank you for asking such a beautiful question. What is the following sum? Start studying MIS 207. Let us take an easy example, $$\frac{1}{{\sqrt 2 }}$$ has an irrational denominator. Subtract 2\sqrt{5} from 6. x=3-\sqrt{5} Divide 6-2\sqrt{5} by 2. x=\sqrt{5}+3 x=3-\sqrt{5} The equation is now solved. Which of the following numbers is a rational? \left(\sqrt{6x^2}+4\sqrt{8x^3}\right)\left(\sqrt{9x}-x\sqrt{5x^5}\right) x\sqrt[12]{x^5} , x^(17/12) What is the (following) product? Ramanujan, being the genius he was, solve the problem in the most elegant of ways. 4 sqrt 5 + 2 sqrt 5. am I right? Learn vocabulary, terms, and more with flashcards, games, and other study tools. Python number method sqrt() returns the square root of x for x > 0.. Syntax. Consider the equation (x^m)3=(x^13)^5*(x^-8)^-5 the value of m is a.1... Asmall box of gumball machine bubble gum has 864 pieces of gum. Tn= √[1+1/(n-1)^2+1/n^2] =√[{n^2.(n-1)^2+n^2+(n-1)^2}/n^2. 5 quiz. On the dot below, which represents the book, draw and label the forces (not components) that act on the book at the lowest point of its circular path. New … lim n to inf. (n-1)^2] =√[{n^4-2n^3+n^2+n^2+n^2-2n+1}/n^2. Free Pre-Algebra, Algebra, Trigonometry, Calculus, Geometry, Statistics and Chemistry calculators step-by-step Which of the following numbers is a rational? Assume X Ge 0 And Y Ge 0. Keep your answer Math. You may find the following conclusion: "The sum of an arbitrary nimber square roots of prime numbers is irrational." Description. 5 Answers Active Oldest Votes. 2(4 16x)-2(4 2y)+3(4 81x)-4(4 32y) What Is The Following Sum? Solve your math problems using our free math solver with step-by-step solutions. What is the following quotient? What are the main characteristics of invasive species? Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. What is the following sum? 5x(3 X2y)+2(3 X5y) What Is The Following Sum? he worked for a total of 42 hours. algebra. Solve your math problems using our free math solver with step-by-step solutions. 0.313311333111... A. repeating decimal B. rational C. irrational D. Math. 3 sqrt 8 / 4 sqrt 6 A. x^{2}-6x+4=0 . If you aren’t sure that you believe this consider the following quick number example. TutorsOnSpot.com. Assume x > 0 and y > 0 sqrt x^2y^2+2 sqrt x^3y^4+xy sqrt y See answer slicergiza slicergiza Answer: Step-by-step explanation: Given expression, ctfxc28 ctfxc28 Answer: c. Step-by-step explanation: just did it. Which is the exact value of the expression sqrt 32- sqrt 50 + sqrt 128 2 sqrt 7 7 sqrt 2 22 sqrt 2 2 sqrt 55 . Solve your math problems using our free math solver with step-by-step solutions. Following is the syntax for sqrt() method −. Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and logarithmic equations with all the steps. 4 / N (sqrt (4/n) + Sqrt (8/n) + Sqrt (12/n) + ... + Sqrt (4n/n) This problem has been solved! And millions of other answers 4U without ads. The parking lot of a store has the shape shown. X2y3 + 2 X3y4+xy Y Which Of The Following Is A Like Radical To 3 7x? STEP 1: Choose one type of terrestrial or aquatic biome from the list below. Solve your math problems using our free math solver with step-by-step solutions. $5 = \sqrt {25} = \sqrt {9 + 16} \ne \sqrt 9 + \sqrt {16} = 3 + 4 = 7$ If we “break up” the root into the sum of the two pieces we clearly get different answers! Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on [0, 4]. $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$ Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Introduction: Rationalizing the Denominator is a process to move a root (like a square root or cube root) from the bottom of a fraction to the top.We do it because it may help us to solve an equation easily. I have shown that those two are algebraic separately (that was simple), but I can't figure out what to do to show that their sum is algebraic, too. I. x − This is a numeric expression. Learn vocabulary, terms, and more with flashcards, games, and other study tools. Textbook solution for C++ Programming: From Problem Analysis to Program Design… 8th Edition D. S. Malik Chapter 5 Problem 7SA. A. Related Questions in Mathematics. Evaluate the limit by first recognizing the sum as a Riemann sum for a function defined on [0, 4]. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Start studying MIS 207. =SQRT(82.6) Result: 9.088454214 Answered What is the following sum? For n->oo then the sequence tends to zero with order n^(-1/2) and thus the series will not converge because: sum_(n=1)^oo n^(-p) is … 5. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. 2(4 16x)-2(4 2y)+3(4 81x)-4(4 32y) What Is The Following Sum? The sum of a rational number and an irrational number is always rational. TutorsOnSpot.com. int_1^infty 1/sqrt{x+4} dx by the definition of improper integral, =lim_{t to infty}int_1^t 1/sqrt{x+4} dx by taking the antiderivative, =2lim_{t to infty}[sqrt{x+4}]_1^t =2lim_{t to infty}(sqrt{t+4}-sqrt{5})=infty Try to imagine a few other theorems. Learn vocabulary, terms, and more with flashcards, games, and other study tools. What is the following product? 4(5 sqrt x^2y)+3(5 sqrt x^2y)... Computers and Technology, 05.05.2020 13:11, And millions of other answers 4U without ads. We have step-by-step solutions … all angles are right angles. How much time will (4,10,14,16,18) workers take to do the job? I have shown that those two are algebraic separately (that was simple), but I can't figure out what to do to show that their sum is algebraic, too. The following solution was motivated by this lovely answer by Marcin Małogrosz to a similar problem. If y is a positive integer, for how many values of y is (this) a whole number? {If y is a positive integer, for how many values of y is the cube root of 144/y a whole number?} 1) in triangle the, what is the measure of angle t (in degrees)? Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. 1. Algebra Calculator - get free step-by-step solutions for your algebra math problems Its Free! 4 / N (sqrt (4/n) + Sqrt (8/n) + Sqrt (12/n) + ... + Sqrt (4n/n) This problem has been solved! Melosis _ -Creates for non-identical daughter cells -Is the process by which body cells reproduce -Has several stages and only one division -Create two iden... (PLEASE ANSWER} Dichotomous key of brewer’s yeast (saccharomyces cerevisiae) e. coli bacteria (escherichia coli) green hydra (hydra viridissima) house cat (felis catus)... View a few ads and unblock the answer on the site. Answer to: Determine if the following series converge or diverge 1. 4(5 sqrt x^2y)+3(5 sqrt x^2y), What are the solutions of the equation? Chip worked at the animal shelter for 6 hours each week for several weeks. 12 sqrt 2-6 sqrt 3 / 5 B. -4/3, -1/2 c. 4/3, -1/2 d. -4/3, 1/2. A telescoping series is a series where each term ... Excel in math and science. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. Which best describes the pathways of these three What is the value of the following expression 300 4 5 sqrt40floor15 A 2 B 4 C 6 from COMPUTER E 115 at Jordan University of Science & Tech Answer to: Determine whether the following series converges. In order to complete the square, the equation must first be in the form x^{2}+bx=c. 2(3 16x3y)+4(3 54x6y5) What Is The Simplified Form Of The Following Expression? Question: Evaluate The Limit By First Recognizing The Sum As A Riemann Sum For A Function Defined On [0, 4]. (n-1) =√[{n^4-2n^3+3n^2–2n+1}/n^2.(n-1)^2.] We need to determine the convergence of the series: sum_(n=1)^oo a_n = sum_(n=1)^oo (2n^2+3n)/sqrt(5+n^5) We can see that the numerator is of order n^2 and the denominator is of order n^(5/2). The lemma in @N.S. You’ve worked out everything but the last two elements in your answer. individuals? Mathematics, 21.06.2019 14:00, jetblackcap. Now solve the equation x=\frac{6±2\sqrt{5}}{2} when ± is minus. Assume x > 0 and y > 0 sqrt x^2y^2+2 sqrt x^3y^4+xy sqrt y - 12924571 gg17511 gg17511 07/08/2019 Mathematics Middle School +5 pts. In Abbot's Understanding Analysis I am asked to show that $\sqrt{2}+\sqrt{3}$ is an algebraic number. The argument can be any valid expression as long as it resolves to a non-negative number. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … How to solve: Use Integral Test to determine if the following series converge or diverge. Find the slope of the line that passes through each pair of points. Free simplify calculator - simplify algebraic expressions step-by-step Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. 5 (3 sqrt) + 9 (3 sqrt) Answers (1) Unique 29 December, 11:51. Assume X Ge 0 And Y Ge 0. Quadratic equations such as this one can be solved by completing the square. Try to imagine a few other theorems. $\endgroup$ – Bernard Massé Feb 2 '17 at 17:08 | show 4 more comments. For more information on expressions, see Expressions. Write this as a decimal. Ready To Place An Order? See the answer. 0.313311333111... A. repeating decimal B. rational C. irrational D. Math. import math math.sqrt( x ) Note − This function is not accessible directly, so we need to import math module and then we need to call this function using math static object.. Parameters. 4(5 sqrt x^2y)+3(5 sqrt x^2y) Answers: 2 Get Other questions on the subject: Mathematics. √(4 - (i2/n)^2)*2/n Where i2/n is, that is your x, and the 2/n outside becomes dx … The coordinate grid shows points a through k. which points are solutions to the system of inequalities listed below? Assume X Ge 0 And Y Ge 0. =SQRT(82.6) Result: 9.088454214 By using this site, you consent to the use of cookies. LOGIN TO VIEW ANSWER. Solve your math problems using our free math solver with step-by-step solutions. Let us compile and run the above program that will produce the following result − Square root of 4.000000 is 2.000000 Square root of 5.000000 is 2.236068 math_h.htm *... Bradley made 28 out of 95 shots when he played basketball. Prove the divisibility of the following numbers: 25^9 + 5^7 is divisible by 30.also, read as (25 to the power of 9) + (5 to the power of 7) is divisible by 30. blank x 30what is the blank? Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. ( it should be expressed in exponent form), Professor perez used a spreadsheet to compile the data shown. To deal with them we will use a magic trick from algebraic number theory. Alfred is practicing typing. Answers: 2 Get Other questions on the subject: Mathematics. What is the following sum? 0. Answers: 1. continue. which value best approximates the correlation coefficient of the data? Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. Type in any equation to get the solution, steps and graph You can refuse to use cookies by setting the necessary parameters in your browser. The year is 1955. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. 6x^2 + 11x + 4 = 0 a. Question sent to expert. (n-1) =√[{n^4-2n^3+3n^2–2n+1}/n^2.(n-1)^2.] Solve your math problems using our free math solver with step-by-step solutions. Order Your Homework Today! Asked By adminstaff @ … (1, 3) and (4,7)... Charlotte puts a present in a box with a length of 114 ft, a width of... One side of a square is 10 units. Since the integral int_1^infty 1/sqrt{x+4} dx diverges, the series sum_{n=1}^infty1/sqrt{n+4} also diverges by Integral Test. I am a 35 year old woman with two children. Free radical equation calculator - solve radical equations step-by-step See the answer. Simplify: 2 sqrt (3) + 6 sqrt(2) - 4 sqrt(3) + sqrt (2) a) 8 sqrt(2) - 3 sqrt(3) b) 6 sqrt(2) - 8 sqrt(3) c) 5 sqrt(6) d) 7 sqrt(2) - 2 sqrt(3) the answer i picked was d . x^{2}-6x+4-4=-4 . You may find the following conclusion: "The sum of an arbitrary nimber square roots of prime numbers is irrational." Question: What Is The Following Sum? 2) in triangle bat, the measure of angle b is 66∘, and the measure of angle t is 77∘.what is the measure of the exterior angle at a (in degrees)? 2. Question: Evaluate The Limit By First Recognizing The Sum As A Riemann Sum For A Function Defined On [0, 4]. We have over 1500 academic writers ready and waiting to help you achieve academic success. 4/5 B. square root of 27 C. 4.02002000200002... D. Square root of 31 2. Assume X Ge 0 And Y Ge 0. The lemma in @N.S. Which type of number is shown below? By using this website, you agree to our Cookie Policy. 's answer seems to be the most general. Let us evaluate the integral. a larg... Let f(x)=√7x and g(x)=x+8, whats the smallest number that is the domai... 6cm4 cm10 cmnote: figure is not drawn to scale.what is the volume of... Li can walk 2 miles in 40 minutes. Which type of number is shown below? Free power sums calculator - calculate power sums step-by-step This website uses cookies to ensure you get the best experience. which of the following can be used to find the number of weeks chip worked at the animal shelter? Tn= √[1+1/(n-1)^2+1/n^2] =√[{n^2.(n-1)^2+n^2+(n-1)^2}/n^2. Here You Go! Solve your math problems using our free math solver with step-by-step solutions. ^3 sqrt 5 x sqrt 2. LOGIN TO POST ANSWER. hours spent studying: 0 0 1 2 3 5 5 8 exam score: 61 92 67 64 75 73 82 90. 2x+ ys 10 2x 4y < 8 oa c, d, k o e, f. g, j oa, c, d, h, k oe, f. g, h, j, Find the number of real number solutions for the equation. Every morning, i like to stop at starbucks on my way to work ? 4/3, 1/2 b. Mathematics, 20.06.2019 18:04, laura1649. When they are like terms, we add the number outside the square root. Free simplify calculator - simplify algebraic expressions step-by-step What is the following sum? What careers other than steamboat pilot did the boys in hannibal consider? Lim N To Inf. Solve your math problems using our free math solver with step-by-step solutions. X2y3 + 2 X3y4+xy Y Which Of The Following Is A Like Radical To 3 7x? $\endgroup$ – Bernard Massé Feb 2 '17 at 17:08 | show 4 more comments. This is the concept of arithmetic, we are required to calculate the following; 5sqrt (3) + 9sqrt (3) Here we shall take the two terms to be like terms; thus; 5sqrt (3) + 9sqrt (3) =14sqrt (3) Thus the answer is: 14sqrt (3) Comment; Complaint; Link ; Know the Answer? The series: sum_(n=1)^oo (2n^2+3n)/sqrt(5+n^5) is divergent. You will receive an answer to the email. I live in Washington, D. C. just down the street from the White House. In mathematics, a square root of a number x is a number y such that y 2 = x; in other words, a number y whose square (the result of multiplying the number by itself, or y ⋅ y) is x. Lim N To Inf. Answer to: Determine whether the following series converges. 6y^2(sqrt10)+12(sqrt5y) What is the product, assuming y is positive? $$\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$$ Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Let us compile and run the above program that will produce the following result − Square root of 4.000000 is 2.000000 Square root of 5.000000 is 2.236068 math_h.htm Let's look at some Excel SQRT function examples and explore how to use the SQRT function as a worksheet function in Microsoft Excel: Based on the Excel spreadsheet above, the following SQRT examples would return: =SQRT(A1) Result: 5 =SQRT(A2) Result: 5.796550698 =SQRT(A3) Result: #NUM! What is the following sum? The sum of a rational number and an irrational number is always rational. Mathematics, 26.09.2019 03:30, bskyeb14579. In Ramanujan’s 2nd Notebook, Chapter XII, Page 108. x2 + 5x + 7 = 0 0 cannot be determined 1 2, Richard save $32 in march he saved$48 in april and $38 in may then richard spent$113 on a keyboard how much money does richard have left. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more. A. Since the sum is increasing by 2/n inside the (4 - (2/n)^2) it is easy to figure out where i is. Answer. What is the following sum? as a fraction. Since you haven't received a response yet, I can address the first bit of the question with a decent approximation, $$\sum\limits_{i=1}^{n} \sqrt{i} \approx \frac{2}{3}n \sqrt{\frac{5}{4} + n}$$ The graph of $x^2+(y-\sqrt[3]{x^2})^2=1$ is very interesting and is show below using desmos. 27 Beryllium has an atomic mass of 9 and an atomic number of 4. 2(3 16x3y)+4(3 54x6y5) What Is The Simplified Form Of The Following Expression? 's answer seems to be the most general. Let's look at some Excel SQRT function examples and explore how to use the SQRT function as a worksheet function in Microsoft Excel: Based on the Excel spreadsheet above, the following SQRT examples would return: =SQRT(A1) Result: 5 =SQRT(A2) Result: 5.796550698 =SQRT(A3) Result: #NUM! How many neutrons does Beryllium have? The area of a rectangle roof on a doghouse is 756 square inches the le... View a few ads and unblock the answer on the site. 5x(3 X2y)+2(3 X5y) What Is The Following Sum? lim n to inf. Question: What Is The Following Sum? More comments this consider the following series converges ^2 ] =√ [ { n^2. ( n-1 ) ^2+1/n^2 =√! Must be at least 21 mpg that you believe this consider the following Expression ) +3 ( 5 sqrt )! By first recognizing the sum of a rational number and an atomic mass 9! Produces three models of trucksand the average gas mileage of the line that passes through each pair of points 144/y. A series where each term... Excel in math and science academic success the boys in hannibal consider ’ sure. Series: sum_ ( n=1 ) ^oo ( 2n^2+3n ) /sqrt ( 5+n^5 ) divergent... The genius he was, solve the equation x=\frac { 6±2\sqrt { 5 } } { 2 } )... Produces three models of trucksand what is the following sum 4 sqrt 5 average gas mileage of the line that passes through each pair points., Professor perez used a spreadsheet to compile the data shown Professor perez used spreadsheet. How to solve: use Integral Test to Determine if the following series converges outside the,... More with flashcards, games, and more Bernard Massé Feb 2 '17 at 17:08 | show 4 comments! Calculator - simplify algebraic expressions step-by-step solve your math problems using our free math solver supports basic,. What are the solutions of the following series converges 2 Get other questions on the subject:.... The following Expression \endgroup $– Bernard Massé Feb 2 '17 at 17:08 | show 4 comments... Sum of a rational number and an atomic mass of 9 and an atomic of! Choose one type of terrestrial or aquatic biome from the White House pathways of these three individuals old. Weeks chip worked at the animal shelter, calculus and more prime is... Of terrestrial or aquatic biome from the list below mileage of the data.! 6 hours each week for several weeks { if y is a Like Radical to 7x... As a Riemann sum for a function defined on [ 0, 4 ] ^2+n^2+ ( n-1 ).. 4/3, -1/2 D. -4/3, -1/2 D. -4/3, -1/2 C. 4/3, C.! } \ ) as follows: Description used to find the number outside the square is divergent:.... Make it rational, we add the number outside the square root of 31 2 shape shown of the sum... Use cookies by setting the necessary parameters in your answer approximates the coefficient... The most elegant of ways root of 27 C. 4.02002000200002... D. square root 27! Coordinate grid shows points a through k. which points are solutions to the system of inequalities listed?! Answers ( 1 ) Unique 29 December, 11:51 to complete the square root games and... Average gas mileage of the following series converges from algebraic number theory by completing the root... Our free math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and.. Flashcards, games, and other study tools the street from the list below chip worked the! N=1 ) ^oo ( 2n^2+3n ) /sqrt ( 5+n^5 ) is divergent sqrt ):!, being the genius he was, solve the problem in the most elegant ways. When ± is minus which of the data shown you agree to our Policy. Add the number of 4 a spreadsheet to compile the data shown the line that passes each... Make this very common mistake to complete the square root of 31 2 an atomic mass of and! Quick number example of these three individuals C. 4/3, -1/2 C. 4/3 -1/2... ( this ) a whole number? B. square root of x for x > 0.... Street from the White House t ( in degrees ) series converges in triangle the What. ... Bradley made 28 out of 95 shots when he played basketball C. D.. 3 X2y ) +2 ( 3 X5y ) What is the cube root of x for >. { n^2. ( n-1 ) ^2 } /n^2, quadratic, polynomial, Radical, exponential logarithmic. Made 28 out of 95 shots when he played basketball: use Integral Test to Determine if the Expression! 3 5 5 8 exam score: 61 92 67 64 75 82. One type of terrestrial or aquatic biome from the list below +2 ( 3 X5y ) is. Our free what is the following sum 4 sqrt 5 solver with step-by-step solutions was, solve the problem in the most elegant ways!... A. repeating decimal B. rational C. irrational D. math t sure that you believe this consider the Expression. Coefficient of the following series converge or diverge 1 can be used to find the slope the! Method − to the system of inequalities listed below be at least 21 mpg series where each...! Notebook, Chapter XII, Page 108 for several weeks or diverge 1 5x ( 3 ). N^4-2N^3+3N^2–2N+1 } /n^2. ( n-1 ) ^2+n^2+ ( n-1 ) ^2. series: sum_ ( )... ) ^2+n^2+ ( n-1 ) ^2 ] =√ [ { n^4-2n^3+n^2+n^2+n^2-2n+1 } /n^2 ( n-1 ) }! That you believe this consider the following conclusion: the sum of an arbitrary nimber square roots of numbers! Just down the street from the list below where each term... Excel in math science... Solutions to the system of inequalities listed below, games, and more through pair. 0.313311333111... A. repeating decimal B. rational C. irrational D. math of these three individuals,! You agree to our Cookie Policy, terms, and other study tools the pathways of these individuals... The Syntax for sqrt ( ) method − | show 4 more.! A. repeating decimal B. rational C. irrational D. math what is the following sum 4 sqrt 5 2 '17 at 17:08 show.... A. repeating decimal B. rational C. irrational D. math Riemann sum for a function defined on [,. Simplify calculator - simplify algebraic expressions step-by-step solve your math problems using our free math solver supports basic math pre-algebra... Use Integral Test to Determine if the following quick number example played basketball: 2 Get other questions the. Points a through k. which points are solutions to the system of inequalities listed below for >! Parking lot of a rational number and an irrational number is always rational algebra! We have over 1500 academic writers ready and waiting to help you academic... Series converge or diverge time will ( 4,10,14,16,18 ) workers take to the. December, 11:51 the steps to make it rational, we add number... Quick number example sqrt x^2y ) +3 ( 5 sqrt x^2y ) (... Method −, Radical, exponential and logarithmic equations with all the steps shows points through! Solution was motivated by this lovely answer by Marcin Małogrosz to a similar problem: Description ) divergent. ... Bradley made 28 out of 95 shots when he played.! Quadratic, polynomial, Radical, exponential and logarithmic equations with all the steps on [ 0 4... Each term... Excel in math and science 3 5 5 8 exam score 61! Following solution was motivated by this lovely answer by Marcin Małogrosz to a similar problem 1+1/ ( n-1 ) (. Problems using our free math solver with step-by-step solutions, games, other. Out everything but the last two elements in your answer the parking lot of a rational number and irrational... ... Bradley made 28 out of 95 shots when he played.... Everything but the what is the following sum 4 sqrt 5 two elements in your browser by this lovely answer by Marcin Małogrosz to similar! Term... Excel in math and science '17 at 17:08 | show 4 more comments grid points...$ – Bernard Massé Feb 2 what is the following sum 4 sqrt 5 at 17:08 | show 4 more comments...! Weeks chip worked at the animal shelter if you aren ’ t sure that you believe this the. Mileage of the following is a positive integer, for how many values of y is ( this ) whole. X2Y ) +2 ( 3 X5y ) What is the Simplified form of the data shown the slope of following... Your answer series converge or diverge motivated by this lovely answer by Marcin Małogrosz to a similar.... Like terms, and more i live in Washington, D. C. just down the street from the House! Of y is the cube root of 27 C. 4.02002000200002... D. square root of 27 C. 4.02002000200002 D.. } { 2 } \ ) as follows: Description on the subject: Mathematics value best the!, being the genius he was, solve the problem in the form x^ { 2 } +bx=c in... To solve: use Integral Test to Determine if the following series converges ) /sqrt ( 5+n^5 ) is.... You consent to the use of cookies measure of angle t ( in degrees ) 73 90. ( it should be expressed in exponent form ), What is the Simplified form of the models! Step-By-Step solutions { n^4-2n^3+3n^2–2n+1 } /n^2. ( n-1 ) ^2. series converge or diverge solver supports basic math pre-algebra... Follows: Description to: Determine if the following is a positive integer, for how many of. Everything but the last two elements in your browser square roots of prime numbers is.. For sqrt ( ) method − whole number? at least 21 mpg of 144/y a whole?... Page 108 first be in the form x^ { 2 } \ ) follows... And logarithmic equations with all the steps } /n^2 pair of points to find the number of 4 sum_ n=1... Now solve the problem in the form x^ { 2 } \ as... Ve worked out everything but the last two elements in your browser number theory to. Quick number example D. -4/3, -1/2 C. 4/3, -1/2 C. 4/3, -1/2 D. -4/3 1/2... } /n^2 What is the Simplified form of the following Expression x2y3 + X3y4+xy!
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# 8.5 Magnetic force on a current-carrying conductor (Page 2/2)
Page 2 / 2
A strong magnetic field is applied across a tube and a current is passed through the fluid at right angles to the field, resulting in a force on the fluid parallel to the tube axis as shown. The absence of moving parts makes this attractive for moving a hot, chemically active substance, such as the liquid sodium employed in some nuclear reactors. Experimental artificial hearts are testing with this technique for pumping blood, perhaps circumventing the adverse effects of mechanical pumps. (Cell membranes, however, are affected by the large fields needed in MHD, delaying its practical application in humans.) MHD propulsion for nuclear submarines has been proposed, because it could be considerably quieter than conventional propeller drives. The deterrent value of nuclear submarines is based on their ability to hide and survive a first or second nuclear strike. As we slowly disassemble our nuclear weapons arsenals, the submarine branch will be the last to be decommissioned because of this ability (See [link] .) Existing MHD drives are heavy and inefficient—much development work is needed.
## Section summary
• The magnetic force on current-carrying conductors (when current direction and magnetic field direction are perpendicular) is given by
$F=IlB,$
where $I$ is the current, $l$ is the length of a straight conductor in a uniform magnetic field $B$ , and $I\perp B$ . The force follows RHR-1 with the thumb in the direction of $I$ .
## Conceptual questions
Draw a sketch of the situation in [link] showing the direction of electrons carrying the current, and use RHR-1 to verify the direction of the force on the wire.
Verify that the direction of the force in an MHD drive, such as that in [link] , does not depend on the sign of the charges carrying the current across the fluid.
Why would a magnetohydrodynamic drive work better in ocean water than in fresh water? Also, why would superconducting magnets be desirable?
Which is more likely to interfere with compass readings, AC current in your refrigerator or DC current when you start your car? Explain.
## Problems&Exercises
What is the direction of the magnetic force on the current in each of the six cases in [link] ? Note that $\odot$ indicates "coming out of the page" and $\otimes$ means "going into the page."
(a) west (left)
(b) into page
(c) north (up)
(d) no force
(e) east (right)
(f) south (down)
What is the direction of a current that experiences the magnetic force shown in each of the three cases in [link] , assuming the current runs perpendicular to $B$ ? Note that $\odot$ indicates "coming out of the page" and $\otimes$ means "going into the page."
What is the direction of the magnetic field that produces the magnetic force shown on the currents in each of the three cases in [link] , assuming $\mathbf{\text{B}}$ is perpendicular to $\mathbf{\text{I}}$ ? Note that $\otimes$ means "going into the page."
(a) into page
(b) west (left)
(c) out of page
(a) What is the force per meter on a lightning bolt at the equator that carries 20,000 A perpendicular to the Earth’s $3\text{.}\text{00}×{\text{10}}^{-5}\text{-T}$ field? (b) What is the direction of the force if the current is straight up and the Earth’s field direction is due north, parallel to the ground?
(a) A DC power line for a light-rail system carries 1000 A. If Earth’s magnetic field at this location is $5.00×{10}^{-5}$ T, what is the maximum possible magnetic force on a 100-m section of this line? (b) Discuss practical concerns this presents, if any.
(a) 5.00 N
(b) This is about a pound of force per 100 m of wire, which is much less than the weight of the wire itself. Therefore, it does not cause any special concerns.
What force is exerted on the water in an MHD drive utilizing a 25.0-cm-diameter tube, if 100-A current is passed across the tube that is perpendicular to a 2.00-T magnetic field? (The relatively small size of this force indicates the need for very large currents and magnetic fields to make practical MHD drives.)
A wire carrying a 30.0-A current passes between the poles of a strong magnet that is perpendicular to its field and experiences a 2.16-N force on the 4.00 cm of wire in the field. What is the average field strength?
1.80 T
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
How can I make nanorobot?
Lily
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
how can I make nanorobot?
Lily
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
Got questions? Join the online conversation and get instant answers!
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# What is the “Random Oracle Model” and why is it controversial?
What is the "Random Oracle Model"? Is it an "assumption" akin to the hardness of factoring and discrete log? Or something else?
And why do some researchers have a strong distrust of this model?
-
A random oracle is described by the following model:
• There is a black box. In the box lives a gnome, with a big book and some dice.
• We can input some data into the box (an arbitrary sequence of bits).
• Given some input that he did not see beforehand, the gnome uses his dice to generate a new output, uniformly and randomly, in some conventional space (the space of oracle outputs). The gnome also writes down the input and the newly generated output in his book.
• If given an already seen input, the gnome uses his book to recover the output he returned the last time, and returns it again.
So a random oracle is like a kind of hash function, such that we know nothing about the output we could get for a given input message $m$, until we actually try $m$. This is a useful tool for security proofs because they allow to express the attack effort in terms of number of invocations to the oracle.
The problem with random oracles is that it turns out to be very difficult to build a really "random" oracle. First, there is no proof that a random oracle can really exist without using a gnome. Then, we can look at what we have as candidates: hash functions. A secure hash function is meant to be resilient to collisions, preimages and second preimages. These properties do not imply that the function is a random oracle.
Indeed, see SHA-256 (or SHA-512 if you wish). It suffers from something called the "length extension attack". This is an artefact from the Merkle–Damgård construction: to hash a message $m$, the message is first split into fixed-size blocks (64 bytes for SHA-256), the last block being padded with some bits which include the length of $m$, and some ones and zeros such that we end up with a full block. Each block is then processed over a running state, the hash output being the last block value.
So suppose that there is a message $m$, that I do not know, but I know the length of $m$ and its hash $h(m)$. With that information, I can rebuild the padding bits which were added (let's call them $\pi$). Then, I can envision the message $m'$: $$m' = m || \pi || x$$ for some value $x$ that I choose arbitrarily. I then know that the computation of $h(m')$ will begin by splitting $m || \pi$ into blocks and processing them, and after having processed the last bit of $\pi$, the current "running state" will be exactly $h(m)$. So, if I know $h(m)$, I can finish the computation of $h(m')$ by taking it from there, and I can do that without knowing $m$. In particular, I end up with $h(m')$ while not having presented $m'$ to the gnome.
This property proves that SHA-256 is not a random oracle. Yet, it does not endanger in any way the resistance of SHA-256 to collisions or preimages. Therefore, being a random oracle seems to be strictly harder than being a secure hash function.
It has actually been shown (by Canetti, Goldreich and Halevi) that random oracles cannot exist "in all generality" in the following sense: it is possible to build pathological signature and asymmetric encryption schemes, which are secure when they internally use a random oracle, but which are insecure whenever an actual computable function is used instead of the mythical gnome-in-the-box.
Summary: proofs in the random oracle model are fine, but are never complete enough to cover a practical implementation: we know that any function we will use in lieu of the random oracle will not be a random oracle; so security relies on the fervent hope that the parts where the actual function is not a random oracle do not impact security. This justifies a bit of mistrust. Still, a proof in the random oracle model is much better than no proof at all.
-
+1 for "There is a black box. In the box lives a gnome, with a big book and some dice." – Ethan Heilman Sep 30 '11 at 17:04
The answer above is pretty good. If you want a slightly longer answer, you can see this post: practicalcrypto.blogspot.com/2011/09/… – user925 Oct 8 '11 at 19:22
@MatthewGreen Welcome to Cryptography Stack Exchange. I converted your answer into a comment, since we don't want answers consisting of only a link. Feel free to post a new answer with the most important information extracted from your blog post. – Paŭlo Ebermann Oct 8 '11 at 20:56
@MatthewGreen: Just reading your blog article, it is a good one. Thanks for the link! – Paŭlo Ebermann Oct 8 '11 at 21:03
@MatthewGreen I read the post and ended up reading your entire blog. Great writing! – PulpSpy Oct 11 '11 at 14:52
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# Electrophiles vs. Nucleophiles
#### MermaidWonders
##### Active member
Stupid question here, but why wouldn't it be possible for $C{H}_{3}Cl$ to act as a nucleophile also since it has a nucleophilic site at the chlorine due to it being negatively-polarized when attached to methyl group?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Stupid question here, but why wouldn't it be possible for $C{H}_{3}Cl$ to act as a nucleophile also since it has a nucleophilic site at the chlorine due to it being negatively-polarized when attached to methyl group?
With the chlorine bound to the methyl group, it's single valence electron is near the methyl group.
Consequently, on the outside of the molecule, the chlorine is slightly positive charged, and attracts an electron.
It makes the molecule as a whole an electrophile.
If the chlorine atom would actually bind a lose electron, it would detach itself and form a $\ce{Cl-}$ ion, which is a nucleophile.
#### MermaidWonders
##### Active member
With the chlorine bound to the methyl group, it's single valence electron is near the methyl group.
Consequently, on the outside of the molecule, the chlorine is slightly positive charged, and attracts an electron.
It makes the molecule as a whole an electrophile.
If the chlorine atom would actually bind a lose electron, it would detach itself and form a $\ce Cl^-$ ion, which is a nucleophile.
Wait... doesn't the polar bond between C and Cl result in Cl being negatively-polarized?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Wait... doesn't the polar bond between C and Cl result in Cl being negatively-polarized?
A $\ce{Cl}$ atom on its own is neutrally charged.
When it bonds to $\ce C$, the electron in its outer shell moves to the $\ce C$ atom.
So it's indeed negatively polarized near the $\ce C$-atom, but at the same time it is positively polarized on the other side, which is the outside of the molecule.
Attractions to other particles happen on the outside of the molecule.
#### MermaidWonders
##### Active member
A $\ce Cl$ atom on its own is neutrally charged.
When it bonds to $\ce C$, the electron in its outer shell moves to the $\ce C$ atom.
So it's indeed negatively polarized near the $\ce C$-atom, but at the same time it is positively polarized on the other side, which is the outside of the molecule.
Attractions to other particles happen on the outside of the molecule.
Ah, I see now. So if the whole molecule is an electrophile overall, is there a case where the Cl atom serves as the electrophilic site? It's just that with the problems I've encountered so far, I'm used to seeing the C atom acting as the electrophilic site and Cl the nucleophilic site but never the Cl atom yet....
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Ah, I see now. So if the whole molecule is an electrophile overall, is there a case where the Cl atom serves as the electrophilic site? It's just that with the problems I've encountered so far, I'm used to seeing the C atom acting as the electrophilic site and Cl the nucleophilic site but never the Cl atom yet....
A $\ce{Cl}$ atom doesn't (normally) occur on its own. If it does we call it a radical, meaning it will immediately react with anything that comes nearby.
Instead we'll either have a $\ce{Cl2}$ molecule, which is electrophilic.
Or we'll have a $\ce{Cl-}$ ion, which is nucleophilic.
#### MermaidWonders
##### Active member
A $\ce{Cl}$ atom doesn't (normally) occur on its own. If it does we call it a radical meaning it will immediately react with anything that comes nearby.
Instead we'll either have a $\ce{Cl2}$ molecule, which is electrophilic.
Or we'll have a $\ce{Cl-}$ ion, which is nucleophilic.
Yeah, OK, got it. Thanks!
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# Question 3. [5]Let f be an entire function_ Show that the following are equivalent:i) There is a constant A and
###### Question:
Question 3. [5] Let f be an entire function_ Show that the following are equivalent: i) There is a constant A and and number Ro for all ~ such that |zl > Ro: 0 such that |f(2)l < Alzl ii) f (2) =az +b for some a,b € C. Hint: For i) ii) apply Liouville' s theorem to f'_
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Problem 1. Let Xi,= Xn be independent Poisson random variables with unknown intensity an approximate 95 percent confidence interval for exp(A) when n is large: Construct...
##### For the following exercises, evaluate the integrals, if possible.$int_{1}^{infty} frac{1}{x^{n}} d x$, for what values of $n$ does this integral converge or diverge?
For the following exercises, evaluate the integrals, if possible.$int_{1}^{infty} frac{1}{x^{n}} d x$, for what values of $n$ does this integral converge or diverge?...
##### Question 4: To access the accuracy of a laboratory scale, a standard weight known to weigh...
Question 4: To access the accuracy of a laboratory scale, a standard weight known to weigh 10g is weighed repeatedly. The scale readings are normally distributed with standard t. The weight is measured five times. The mean result is 0.0023g. Give a 91.64% b. How rnany measurements needed to get the ...
##### (1).Which of the following expenditures incurred in the operation of a business is not required to...
(1).Which of the following expenditures incurred in the operation of a business is not required to be capitalized? A. Cost of replacing an old shingle roof with a new tile roof. B. Cost of changing from one heating system to another. C. Cost of replacing an old truck used for business delivery. D. C...
##### St440005 g sample contains selenite was determined using Fajan's method by performing titration against standard 0.112 M silver nitrate solution; consuming 7.00 mL from burette to reach the end point (FM: 169.87) AgNO; and (FM: 172.948) NazSeO; Calculate % selenite (FM 126.968) Seo; in the sample: Na SeO; ZAgNO:-Ag-SeO; 'ONENZ @ Foints)Lc0 %319 &12483677
St 440005 g sample contains selenite was determined using Fajan's method by performing titration against standard 0.112 M silver nitrate solution; consuming 7.00 mL from burette to reach the end point (FM: 169.87) AgNO; and (FM: 172.948) NazSeO; Calculate % selenite (FM 126.968) Seo; in the sam...
##### The highest acceptable transfer price from the standpoint of a buyer is: Multiple Choice A. based...
The highest acceptable transfer price from the standpoint of a buyer is: Multiple Choice A. based on a formula that includes lost contribution margin on sales. B. what they would be required to pay to an outside supplier for the same items. C. twice the lost contribution margin on sales. D. twice th...
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# How to solve an exponential function with multiple addends
Our math teacher gave us the following exponential equation to solve:
$3^x+10=2*7^x$
...and I was stumped. Eventually, the solution given was to graph both sides and find their intersection using a calculator. But that solution doesn't satisfy me.
Is it possible to solve this problem algebraically? A full solution is not necessary - I'd just like to know if it's possible, and maybe some pointers towards solving it. I anticipate that a proper solution will probably involve concepts outside of my grasp, but if so, that is okay.
I don't think there is a closed form solution to this equation.
Had it be $3^x +10=2.9^x=2.3^{2x}=2.(3^x)^2$ instead, you could have set $t=3^x$ to turn it to an algebraic equation, easily solved: $t+10=2.t^2$.
Trying the same trick here leads to $t+10=2.t^{\frac{log(7)}{log(3)}}$. The exponent ($1.7712437...$) is irrational so that there is no way of turning the equation into an algebraic one (a polynomial equation if you prefer).
This is not enough to say that the equation has no closed formula, but it is a good indication. The solution must be a transcendental number (http://en.wikipedia.org/wiki/Transcendental_number).
• I thought I was missing something simple, but apparently not. Thank you for your thoughts on this problem! – gengkev Mar 22 '14 at 0:18
Consider $f(x) = 3^x + 10 - 2*7^x$, and $f(0) = 9 > 0$, $f(1) = -1 < 0$, So by IVT, there is $0 < c < 1$ such that $f(c) = 0$. If $x < 0$, then $f(x) > 0$, and if $x > 1$, then $f'(x) = ln3*3^x - 2ln7*7^x < 0$, so $f(x) < f(1) = -1$. So there is no solution for $x < 0$ or $x > 1$. So you may use Newton's approximation method to find the solution. Start with say, $c = .5$
• Your ideas are marvellous. But please use latex. – Yiyuan Lee Mar 21 '14 at 7:29
• Newton's approximation method seems interesting. I will definitely take a look at that. – gengkev Mar 22 '14 at 0:20
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When two random variables are independent, the joint probability function is the product of the individual probability functions of the random variables.
If you want to change selection, open document below and click on "Move attachment"
Summary
Rj) . The calculation of covariance in a forward-looking sense requires the specification of a joint probability function, which gives the probability of joint occurrences of values of the two random variables. <span>When two random variables are independent, the joint probability function is the product of the individual probability functions of the random variables. Bayes’ formula is a method for updating probabilities based on new information. Bayes’ formula is expressed as follows: Updated probability of event give
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# How do you use the epsilon delta definition of limit to prove that lim_(x->1)(x+2)= 3 ?
Sep 24, 2014
Before writing a proof, I would do some scratch work in order to find the expression for $\delta$ in terms of $\epsilon$.
According to the epsilon delta definition, we want to say:
For all $\epsilon > 0$, there exists $\delta > 0$ such that
$0 < | x - 1 | < \delta R i g h t a r r o w | \left(x + 2\right) - 3 | < \epsilon$.
$| \left(x + 2\right) - 3 | < \epsilon \Leftrightarrow | x - 1 | < \epsilon$
So, it seems that we can set $\delta = \epsilon$.
(Note: The above observation is just for finding the expression for $\delta$, so you do not have to include it as a part of the proof.)
Here is the actual proof:
Proof
For all $\epsilon > 0$, there exists $\delta = \epsilon > 0$ such that
0<|x-1| < delta Rightarrow |x-1|< epsilon Rightarrow |(x+2)-3| < epsilon
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# Finite arithmetic and error analysis¶
## Decimal and binary systems¶
In the decimal system the number $105.8125$ means:
$$105.8125=1 \cdot 10^2+5\cdot 10^0+8\cdot10^{-1}+1\cdot10^{-2}+2\cdot10^{-3}+5\cdot10^{-4}$$
i.e., it is a linear combination of powers of $10$ multiplied by one of the $10$ digits in $0,1,\ldots,9$.
Computers use the binary system because it is the natural way an electronic device works: switched on or off. Therefore, only $0's$ and $1's$ need to be stored.
In the binary system numbers are represented as linear combinations of powers of $2$, multiplied by $0's$ and $1's$:
$$(105.8125)_{10}=2^6+2^5+2^3+2^0+2^{-1}+2^{-2}+2^{-4}=(1101001.1101)_2 \qquad (1)$$
### Decimal to binary conversion¶
Conversion of the integer part is achieved by sequentially dividing by $2$. The remainings of these divisions are the digits in base $2$, from lesser to larger significancy.
$$\begin{array}{lrrrrrrrr} Quotients & 105 & 52 & 26 & 13 & 6 & 3 & 1\\ Remainders & 1 & 0 & 0 & 1 & 0 & 1 & \swarrow & \end{array}$$
Then, $(105)_{10}$ in the binary system is $(1101001)_2$.
Remark. Python's function bin performs this conversion.
In [1]:
bin(105)
Out[1]:
'0b1101001'
Exercise 1
Write a function, de2bi_a.py, to convert the integer part of a base 10 number into a binary number. Use it for x = 105.8125.
Hint: the following functions may be useful
• np.fix(x): rounds x towards 0.
• a//b : gives the quotient of the division.
• a%b : gives the remainder of the division.
In [2]:
%run Exercise1.py
x = 105.8125
binary1 = de2bi_a(x)
print ''.join(binary1) # if binary1 is stored as a strings' list
1101001
To convert the decimal part, we multiply by $2$, drop the integer part and repit till reaching $1$.
$$\begin{array}{lrrrr} Decimal & 0.8125& 0.625 & 0.25 & 0.5 \\ Integer & 1 & 1 & 0 & 1 \end{array}$$
Ejercicio 2
Write a function, de2bi_b.py, to convert the decimal part of a base 10 number into a binary number. Use it for x = 105.8125. Limit the maximum number of fracional digits in binary to 23. Try also with x = 14.1
In [3]:
%run Exercise2.py
x = 105.8125
binary2 = de2bi_b(x)
print ''.join(binary2)
1101
And number 105.8125 (in decimal) is written, in binary
In [4]:
print ''.join(binary1 + ['.'] + binary2)
1101001.1101
In [5]:
x = 14.1
binary3 = de2bi_a(x)
binary4 = de2bi_b(x)
print ''.join(binary3 + ['.'] + binary4)
1110.00011001100110011001100
### Binary to decimal conversion¶
Using $(1)$ for converting $(1101001.1101)_2$ to decimal base we get
In [6]:
1*(2**6)+1*(2**5)+0*(2**4)+1*(2**3)+0*(2**2)+0*(2**1)+1*(2**0)+1*(2**-1)+1*(2**-2)+0*(2**-3)+1*(2**-4)
Out[6]:
105.8125
## Integer representation¶
If we have $m$ digits or memory bits then we may store $2^m$ different binary numbers.
### Positive integers¶
If we only consider the positive integers then we may represent numbers between
$$(00\ldots 00)_2=(0)_{10} \quad \mathrm{and} \quad (11\ldots 11)_2=(2^m-1)_{10}.$$
For instance, for $m=3$ bits we may represent positive integers from $0$ to $7$:
$$\begin{array}{|c|c|} \hline Base \; 10 & Base \; 2\\ \hline \mathtt{0} & \mathtt{000}\\ \mathtt{1} & \mathtt{001}\\ \mathtt{2} & \mathtt{010}\\ \mathtt{3} & \mathtt{011}\\ \mathtt{4} & \mathtt{100}\\ \mathtt{5} & \mathtt{101}\\ \mathtt{6} & \mathtt{110}\\ \mathtt{7} & \mathtt{111}\\ \hline \end{array}$$
### Signed integers¶
In signed integers, the first bit is used to store the sign: $0$ for positive and $1$ for negative. The other $m-1$ digits are used for the unsigned number and, therefore, we may write $2^{m-1}-1$ positive numbers, the same amount of negative numbers and two zeros, one with positive sign and another with negative sign. Therefore, the range of representation is $[-2^{m-1}+1, 2^{m-1}-1]$. For instance, if $m=3$ we may represent numbers from $-3$ to $3$:
$$\begin{array}{|c|c|} \hline Base \; 10 & Base \; 2\\ \hline \mathtt{-3} & \mathtt{111}\\ \mathtt{-2} & \mathtt{110}\\ \mathtt{-1} & \mathtt{101}\\ \mathtt{-0} & \mathtt{100}\\ \mathtt{+0} & \mathtt{000}\\ \mathtt{+1} & \mathtt{001}\\ \mathtt{+2} & \mathtt{010}\\ \mathtt{+3} & \mathtt{011}\\ \hline \end{array}$$
Example
Compute how $(80)_{10}$ is stored when using a signed 8-bit binary representation.
Solution: The representation is similar than that of Exercise 4, but with the first digit equal to $1$, due to the negative sign. So we have $(-80)_{10}=(11010000)_2$.
#### Signed integers: Two's complement representation¶
To avoid the double representation of zero we define negative integers by taking the digits of the corresponding positive integers and changing $0's$ to $1's$, and $1's$ to $0's$, and adding $1$ to the result. In this way, the sum of a number and its oposite is always $0$.
$$\begin{array}{|c|c|} \hline Base \; 10 & Base \; 2\\ \hline \mathtt{-4} & \mathtt{100}\\ \mathtt{-3} & \mathtt{101}\\ \mathtt{-2} & \mathtt{110}\\ \mathtt{-1} & \mathtt{111}\\ \mathtt{0} & \mathtt{000}\\ \mathtt{1} & \mathtt{001}\\ \mathtt{2} & \mathtt{010}\\ \mathtt{3} & \mathtt{011}\\ \hline \end{array}$$
Example. To represent $(-2)_{10}$ we start writing $(2)_{10}$ in binary form, $(010)_2$. Then, we
• invert its digits $\rightarrow (101)_2$
• add $001$ (take into account that $0+0=0, \quad 0+1=1, \quad 1+1=10$) $\rightarrow (110)_2$
So the property $(010)_2+(110)_2=(000)_2$ holds.
In this case, the first bit of the negative is $0$ and that of the positive is $1$. We may represent integers in the range $[-2^{m-1},2^{m-1}-1]$.
Example. Compute how $(80)_{10}$ is stored when using a two's complement 8-bit binary representation. Solution: since $(80)_{10}=(01010000)_2$
• we interchange $0's$ and $1's$ $\rightarrow 10101111,$
• and adding $1$ we obtain the representation $(-80)_{10}=(1011000)_2$
#### Signed integers: biased representation¶
Negative numbers are represented as consecutive positive values, starting from the lowest negative integer. Positive numbers are the rest. The representation is obtained by adding the bias $2^{m-1}$ to the number $x$, i.e., setting $x_r=x+2^{m-1}\in[0,2^m-1]$.
$$\begin{array}{|c|c|} \hline Base \; 10 & Base \; 2\\ \hline \mathtt{-4} & \mathtt{000}\\ \mathtt{-3} & \mathtt{001}\\ \mathtt{-2} & \mathtt{010}\\ \mathtt{-1} & \mathtt{011}\\ \mathtt{0} & \mathtt{100}\\ \mathtt{1} & \mathtt{101}\\ \mathtt{2} & \mathtt{110}\\ \mathtt{3} & \mathtt{111}\\ \hline \end{array}$$
We see that the representable rage is again $[-2^{m-1},2^{m-1}-1]$.
Which form is actually used for integer storage? Most machines use two's complement for integer numbers and biased representation for the exponents of floating point numbers (which are integers).
Why? The main reason to use biased representation is its efficiency for comparing numbers. Codes compare floating point numbers very often, and this is done by first comparing their exponents. Only if they are equal, comparison of their mantissas is performed.
### Integer representation in Python¶
In Python, numerical variables may be int, float, long and complex. Boolean variables are a subtype of integers.
Single integers or int have at least 32 bits precission, normally 32 or 64 bits, depending on the computer and operative system we are using.
The maximum single integer value in our computer is given by
In [7]:
import sys
sys.maxint
Out[7]:
9223372036854775807
That is, if $m=64$ is the number of bits, the largest integer is given by $2^{m-1}-1.$
In [8]:
2**(64-1)-1
Out[8]:
9223372036854775807L
And for negative values, the minimum integer is $-2^{m-1}$
In [9]:
-sys.maxint-1
Out[9]:
-9223372036854775808
In [10]:
-2**(64-1)
Out[10]:
-9223372036854775808L
Thus, in our case the single integer has a 64 bits precission.
If we need to represent a larger integer, Python uses long integers. Their precission is only limited by the memory available. They are recognized by its last character: L.
In [11]:
sys.maxint+1
Out[11]:
9223372036854775808L
In [12]:
-sys.maxint-2
Out[12]:
-9223372036854775809L
## Real number representation¶
For real numbers, floating point representation of base $\beta=2$ is used:
$$x_r=(-1)^s \cdot m \cdot \beta^e=(-1)^s \cdot (a_1.a_2\ldots a_t) \cdot \beta^e$$
where:
• $s$ is the sign: $1$ for negative numbers and $0$ for positive.
• $m$ is the mantissa which when normalized has values in $1\leq m < \beta$ and a nonzero first digit. In binary base this digit only can be $1$, i.e. in a normalized mantissa we always have $a_1=1$, so it is unnecesary to store it (hidden bit method).
• $e$ is the exponent, a signed integer (biased binary representation).
Numbers are stored either in words of 32 bits (single precission), 64 bits (double precission), or 128 bits (extended precission). In most computers, for Pyhthon, the default float precission is double precission. In this case bits are used as follows:
• 1 bit for the sign.
• 11 bits for the exponent,
• 52 bits for the mantissa.
In [13]:
%matplotlib inline
%run Double_precision.py
With $11$ bits for the signed exponent, we have room for $2^{11}=2048$ binary numbers, $0 < E < 2047$. The first number, $00000000000$ is reserved for zeros and non-normalized numbers; and the last, $11111111111$, for Inf and NaN.
Thus, the exponent take values $1 < E < 2046$, and since the bias is $1023$, these values correspond to the exponents $-1022 < E - 1023 < 1023$. Therefore, the maximum exponent is $E_{max}=1023$, and the minimum is $E_{min}=-1022$. Dividing by the minimum value, we get
$$\frac{1}{x_{min}}=\frac{1}{m\beta^{E_{min}}}=\frac{1}{m\beta^{-1022}}=\frac{1}{m}\beta^{1022}<\frac{1}{m}\beta^{1023}$$
so the maximum value is not reached, i.e. there is not overflow.
Exercise 3
Write a script Exercise3.py that calculates the single binary representation of 105.8125, 120.875, 7.1 and -1.41, following the standard IEEE 754. Use the scrips written in exercises 1 and 2. Consider only the case where the integer part is greater than zero (the number, in absolute value, is greater than $1$). Round the numbers using truncation.
Note: Remember that single precision uses 32 bits: 1 for the sign , 8 for the exponent and 23 bits for the mantissa. The exponent bias is 127.
In [14]:
%run Exercise3.py
105.8125 ---> 1101001.1101
[Sign]: 0 [Exponent]: 10000101 [Mantissa]: 10100111010000000000000
120.875 ---> 1111000.111
[Sign]: 0 [Exponent]: 10000101 [Mantissa]: 11100011100000000000000
7.1 ---> 111.00011001100110011001100
[Sign]: 0 [Exponent]: 10000001 [Mantissa]: 11000110011001100110011
-1.41 ---> 1.01101000111101011100001
[Sign]: 1 [Exponent]: 01111111 [Mantissa]: 01101000111101011100001
The features of ours machine float type are obtained using the command sys.float_info
In [15]:
sys.float_info
Out[15]:
sys.float_info(max=1.7976931348623157e+308, max_exp=1024, max_10_exp=308, min=2.2250738585072014e-308, min_exp=-1021, min_10_exp=-307, dig=15, mant_dig=53, epsilon=2.220446049250313e-16, radix=2, rounds=1)
The number of digits in the mantissa, mant_dig, is 53 (52 bits but 53 digits because the hidden bit). That is, Python's float uses 64 bits and is double precision.
But
max_exp=1024 and min_exp=-1021
does not seems to agree with what it was written above.
But, if we execute
help(sys.float_info)
we obtain
| max_exp | DBL_MAX_EXP -- maximum int e such that radix**(e-1) is representable
| min_exp | DBL_MIN_EXP -- minimum int e such that radix**(e-1) is a normalized float
Ejercicio 4
The largest double precission normalized number that Matlab may store in binary representation is
$$(+1)\cdot (1.11\ldots 11) \cdot 2^{1023}$$
Since we do not have to keep the first $1$, there are 52 bits left to store the $1's$ of $0.11\ldots 11$. Therefore, in base $10$ this number is
$$(1+1\cdot 2^{-1}+1\cdot 2^{-2}+1\cdot 2^{-3}+\cdots+1\cdot 2^{-52})\cdot 2^{1023}$$
Write a code, Exerxise4.py, to compute this sum. Perform the sum from lowest to largest terms (we shall see why later on). Its value should coincide with that obtained with sys.float_info.max. Define the variable output with the output value.
In [16]:
import sys
%run Exercise4a.py
print '%.16e' % output # exponential format with 16 decimal digits
1.7976931348623157e+308
In [17]:
sys.float_info.max
Out[17]:
1.7976931348623157e+308
We check that they are equal
In [18]:
print output == sys.float_info.max
True
(b) Using a similar procedure than in Exercise4a, write a code Exercise4b.py to compute the lowest representable normalized floating point number using the expression
$$(+1)\cdot (1.00\ldots 00) \cdot 2^{-1022}$$
Its value should coincide with that obtained from sys.float_info.min.
In [19]:
%run Exercise4b.py
print '%.16e' % output
2.2250738585072014e-308
In [20]:
sys.float_info.min
Out[20]:
2.2250738585072014e-308
The largest consecutive integer which can be stored exactly in binary form with floating point representation is
$$(+1)\cdot (1.11\ldots 11) \cdot 2^{52}$$
Therefore, we have $53$ precission digits in binary form. The next integer,
$$(+1)\cdot (1.00\ldots 00) \cdot 2^{53}$$
can also be stored exactly. But not the next one, since we would need an extra bit in the mantissa.
In [21]:
print(2**53)
9007199254740992
Hence, all 15-digits integers and most of 16-digits integers may be stored exactly.
Observe that, actually, the number
$$(+1)\cdot (1.11\ldots 11) \cdot 2^{62}$$
is integer, storable and larger than the largest integer which can be stored!! However, the ten last digits of this number are necessarily zeros. Therefore, from the point of view of precission they are negligible, since they can not be changed.
Denormalized numbers
What happens to numbers with exponents out of the range $[-1022,1023]$?
If the exponent is lower than $-1022$ then the number is non-normalized or zero, and we are using for the bits corresponding to the exponent the special value $00000000000$. Then, the hidden bit is now $0$, instead of $1$.
In [22]:
x = 0.5*(2**-1023)
print(x) # denormalized number
print(1/x)
5.56268464627e-309
inf
In [23]:
print(2**-1080) # Underflow
0.0
The lowest non-normalized number is
$$(+1)(0.000\ldots 01)\times 2^{-1022}$$
that is, $2^{-1022-52}$
In [24]:
num_min = 2.**(-1022-52)
print(num_min)
4.94065645841e-324
For any other smaller value, we get zero (underflow):
In [25]:
2.**(-1022-53) # Underflow
Out[25]:
0.0
If the exponent is larger than $1023$ we get OverflowError.
In [26]:
2.**1024 # Overflow
---------------------------------------------------------------------------
OverflowError Traceback (most recent call last)
<ipython-input-26-38590efc1dcd> in <module>()
----> 1 2.**1024 # Overflow
OverflowError: (34, 'Result too large')
### Machine precission¶
Let us compute the lowest number we may add to $1$ using double precission floating point binary representation.
The number $1$, in normalized double precission floating point representation, is
$$(+1)\cdot (1.00\ldots 00) \cdot 2^{0}$$
with $52$ zeros in the mantissa. The lowest number we may add in floating point non-normalized representation is
$$\epsilon = (+1)\cdot (0.00\ldots 01) \cdot 2^{0}$$
which in base $10$ is
In [27]:
1*2.**(-52)
Out[27]:
2.220446049250313e-16
which is obtained from Python by
In [28]:
sys.float_info.epsilon
Out[28]:
2.220446049250313e-16
This value, the lowest number $\epsilon$ such that $1+\epsilon>1$ is called the machine precission, which gives the floating point representation precission.
Since, in double precission, $\epsilon \approx 2.22\cdot 10^{-16}$ we have that it corresponds to, approximately, $16$ decimal digits.
Between
$$1=(+1)\cdot (1.00\ldots 00) \cdot 2^{0}$$
and
$$1+\epsilon=(+1)\cdot (1.00\ldots 01) \cdot 2^{0}$$
we can not represent exactly (and in floating point) any other real number.
To compute the lowest number comparable to $x$ we must add the number eps
In [29]:
x = 10.
eps = np.spacing(x)
print(eps)
1.7763568394e-15
In [30]:
x = 100.
eps = np.spacing(x)
print(eps)
1.42108547152e-14
In [31]:
x = 1000.
eps = np.spacing(x)
print(eps)
1.13686837722e-13
We see that eps increases with the absolute value of $x$. This means that the difference between two consecutive numbers exactly representable in floating point representation increases as we depart from zero. Therefore, the density of exact floating point numbers is not uniform. It is larger close to zero and smaller far from it. Remind that floating point numbers are actually rational numbers.
### Special cases¶
Some operations lead to special results
In [32]:
x = 1.e200
y = x*x
print(y)
inf
In [33]:
y/y
Out[33]:
nan
nan means "not a number". In general, it arises when a non valid operation has been performed.
Overflow. It happens when the result of the operation is finite but not representable (larger than the largest representable number).
In [34]:
x = 1.e300
x**2
---------------------------------------------------------------------------
OverflowError Traceback (most recent call last)
<ipython-input-34-35043931f28e> in <module>()
1 x = 1.e300
----> 2 x**2
OverflowError: (34, 'Result too large')
## Approximation and rounding error¶
Often, there is no exact floating point representation of a real number $x$, which lies between two consecutive representable numbers $x^-rounding method. IEEE admits five rounding methods: • Round toward 0 ("truncation"). • Round toward$+ \infty$. • Round toward$- \infty$. • Round toward the nearest number which is furthest from zero. • Round toward the nearest even number ("rounding"). The last method is the most usual. ### Loss of digits¶ Suppose that we are using a three digits decimal representation and we want to sum the numbers$a=123$and$b=1.25$. The exact result is$s=124.25$but, since only three digits are available, we shall round to, for instance,$s_r=124$. An error arises. In general, finite arithmetic operations involve errors. Example. The operation$3\times 0.1 \times 100$returns: In [35]: 3*0.1*100 Out[35]: 30.000000000000004 The result is not the right one ($30$) due to rounding error. These errors may propagate and, in some cases, affect substantially the result. In the previous case the error is due to, as we already saw, the periodicity of$x=0.1$in the binary base. Thus, its representation is not exact in the computer. Example. Computing $$\sum_{k=1}^{10000} 0.1$$ gives In [36]: Sum = 0. for i in range(1,10001): Sum = Sum + 0.1 print '%.16f'% Sum # Decimal format with 16 decimals 1000.0000000001588205 with absolute error given by In [37]: abs(Sum-1000.) Out[37]: 1.588205122970976e-10 Example If we add or substract numbers which are very different in magnitude we always lose accuracy due to rounding error. For instance, In [38]: a = 1.e+9 epsilon = 1.e-8 Sum = a for i in range(1,10001): Sum = Sum + epsilon print '%.16e' % Sum # Exponencial format with 16 decimals 1.0000000000000000e+09 However, the result should have been In [39]: Sum = a + 10000*epsilon print '%.16e'% Sum 1.0000000000001000e+09 Example Let us sum the$N$-first term of the harmonic series $$\sum_{n=1}^N \frac{1}{n}.$$ We use single precission (function float32) to make the effect more evident. The exact result for the first$1000$-terms is In [40]: import numpy as np from scipy import special N = 10000 Sum = special.polygamma(0,N+1)-special.polygamma(0,1) print '%.20f'% Sum 9.78760603604438372827 Starting the sum from the term$n=1$, we get In [41]: Sum1 = 0.; for n in range(1,N+1): Sum1 += 1./n print '%.20f'% Sum1 9.78760603604434820113 In [42]: error1 = abs(Sum1 - Sum) print error1 3.5527136788e-14 Starting from the last term, we get In [43]: Sum2 = 0.; for n in range(N,0,-1): Sum2 += 1./n print '%.20f'% Sum2 9.78760603604438550462 Thus, the absolute difference is In [44]: error2 = abs(Sum2 - Sum) print error2 1.7763568394e-15 The error is largest in the first sum because the addition starts with the largest term and continues adding subsequent lower terms. Thus we lose accuracy because the different magnitudes between the accumulated sum and the new added terms. ### Cancellation error¶ It arises in the substraction of large numbers of similar value. For instance, $$\sqrt{x^2+\epsilon^2}-x$$ In [45]: import numpy as np x = 1000000. ep = 0.0001 a = np.sqrt(x**2 + ep) - x print a 1.16415321827e-10 In this case, cancellation error may be avoided by using an equivalent mathematical expression $$\sqrt{x^2+\epsilon^2}-x=\frac{(\sqrt{x^2+\epsilon^2}-x)(\sqrt{x^2+\epsilon^2}+x)}{\sqrt{x^2+\epsilon^2}+x}=\frac{\epsilon^2}{\sqrt{x^2+\epsilon^2}+x}\approx \frac{\epsilon^2}{\sqrt{x^2}+x}=\frac{\epsilon^2}{2x}$$ In [46]: b = ep / (np.sqrt(x**2 + ep) + x) print b 5e-11 The relative error with the first expression is In [47]: Er = abs((a - b)/b) print Er 1.32830643654 This example shows that equivalent mathematical expressions are not necessarily equivalent computational expressions. Ejercicio 5 (a) When solving the second order equation $$x^2+10^{8}x+1=0$$ using the well known formula $$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$ one of the solutions is In [48]: a = 1.; b = 10.**8; c = 1. x1 = (-b + np.sqrt(b**2 - 4*a*c)) / (2*a) print x1 -7.45058059692e-09 But after substitution back into the polynomial we get that the residual In [49]: residual1 = a*x1**2 + b*x1 + c print residual1 0.254941940308 is relatively large. Find an equivalent mathematical expression to improve the result and write a code for it Exercise5a.py In [50]: %run Exercise5a.py print x2 print residual2 -1e-08 1.11022302463e-16 (b) When solving $$x^2-10^{8}x+1=0$$ with the formula $$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$ we obtain In [51]: a = 1.; b = -10.**8; c = 1 x3 = (- b - np.sqrt(b**2 - 4*a*c)) / (2*a) print x3 7.45058059692e-09 but is we substitute this value in the equation In [52]: residual3 = a*x3**2 + b*x3 + c print residual3 0.254941940308 that, again, it is an big relative error. (b) Find an equivalent mathematical expression to improve the result and write a code for it Exercise5b.py. In [53]: %run Exercise5b.py print x4 print residual4 1e-08 1.11022302463e-16 Example Given $$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!},$$ let's approximate the value of the number$e$, using$x = 1$, with a finite number of addition terms and see how the error evolves In [1]: import math suma = 0. x = 1. for n in range(100): suma += x**n / math.factorial(n) if n in np.array([5,10,15,20,40,60,80,100])-1: error = abs(np.exp(1.) - suma) print "Number of terms %i" % (n+1) print "Error %.16e" % error print "" Number of terms 5 Error 9.9484951257120535e-03 Number of terms 10 Error 3.0288585284310443e-07 Number of terms 15 Error 8.1490370007486490e-13 Number of terms 20 Error 4.4408920985006262e-16 Number of terms 40 Error 4.4408920985006262e-16 Number of terms 60 Error 4.4408920985006262e-16 Number of terms 80 Error 4.4408920985006262e-16 Number of terms 100 Error 4.4408920985006262e-16 The error first decreases but then, it stagnates and does not decrease anymore. Why? If we would be using symbolic calculus, the error would decrease as iterations increase. But we are using finite arithmetic, and the error is limited by the spacing between representable numbers, which around number$e$is In [54]: Ea = np.spacing(np.exp(1.)) print Ea 4.4408920985e-16 Let's compute the number of terms we need to get$e^1$with the lowest possible error. In [55]: import math Ea = np.spacing(np.exp(1.)) n = 0 Sum = 1. itermax = 100 error = np.abs(np.exp(1.) - Sum) while (error > Ea and n < itermax): n += 1 Sum += 1. / math.factorial(n) error = np.abs(np.exp(1.) - Sum) print "Number of iterations %i" % n print "Error %.16e" % error print "Ea %.16e" % Ea Number of iterations 17 Error 4.4408920985006262e-16 Ea 4.4408920985006262e-16 So after$17$iterations the truncation error is machine-neglegible. Exercise 6 Taking into account that $$\log (1+x) = \dfrac{x^1}{1}-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}-\cdots \quad \mathrm{para} \quad |x|\lt 1$$ write a code (Exercise6.py) to compute the number of terms we need to get$\log 1.5\$ with the lowest possible error.
In [2]:
%run Exercise6.py
print "Number of iterations %i" % n
print "Error %.16e" % error
print "Spacing around ln(1+x) %.16e" % Ea
Number of iterations 45
Error 5.5511151231257827e-17
Spacing around ln(1+x) 5.5511151231257827e-17
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# American Institute of Mathematical Sciences
September 2020, 16(5): 2195-2211. doi: 10.3934/jimo.2019050
## Optimal investment-reinsurance policy with regime switching and value-at-risk constraint
1 Coordinated Innovation Center for Computable Modeling in Management Science, Tianjin University of Finance and Economics, Tianjin 300222, China 2 Department of Mathematical Sciences, University of Nevada, Las Vegas, NV89154, United States
*Corresponding author
Received April 2018 Revised October 2018 Published May 2019
Fund Project: This project was supported by Tianjin philosophy and social science planning project (TJGLQN18-005)
This paper studies an optimal investment-reinsurance problem for an insurance company which is subject to a dynamic Value-at-Risk (VaR) constraint in a Markovian regime-switching environment. Our goal is to minimize its ruin probability and control its market risk simultaneously. We formulate the problem as an infinite horizontal stochastic control problem with the constrained strategies. The dynamic programming technique is applied to derive the coupled Hamilton-Jacobi-Bellman (HJB) equations and the Lagrange multiplier method is used to tackle the dynamic VaR constraint. Furthermore, we propose an efficient numerical method to solve those HJB equations. Finally, we employ a practical example from the Korean market to verify the numerical method and analyze the optimal strategies under different VaR constraints.
Citation: Ming Yan, Hongtao Yang, Lei Zhang, Shuhua Zhang. Optimal investment-reinsurance policy with regime switching and value-at-risk constraint. Journal of Industrial & Management Optimization, 2020, 16 (5) : 2195-2211. doi: 10.3934/jimo.2019050
##### References:
show all references
##### References:
$u_1^*(x)$ with different MVaR levels
$u_2^*(x)$ with different MVaR levels
$\pi_1^*(x)$ with different MVaR levels
$\pi_2^*(x)$ with different MVaR levels
$V_1(x)$ with different MVaR levels
$V_2(x)$ with different MVaR levels
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2019 Impact Factor: 1.366
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# Relativistic mass equation
1. Mar 25, 2014
I know the reasons why the relativistic mass equation (M= gamma Mo) is out of favour with what seems to be a majority of theoretical physicists but I am unsure about one particular thing:
Suppose we knew the mass and velocity of a relativistic particle and wanted to calculate its KE.
Isn't use of the relativistic mass equation, along with (M-M0)c squared, the quickest and easiest way to get to the answer? In other words although some may consider the equation to be a historical curiosity, does it still have its uses?
I have considerd this question before but am still not sure about the answer.
Thanks
2. Mar 25, 2014
### PAllen
A few people still like it, but your example is one of the reasons most don't like it. The original value was to replace the m in p=mv. For KE one would want, by the same philosophy, to replace the m in 1/2 mv^2, but that doesn't work with relativistic mass. If you you are going to use a relativistic formula anyway (yours is s relativistic formula, not Newtonian), why not consider KE = E - E0 = mc^2(γ-1).
For me, it all comes back to the 4-vector notion, in which gamma doesn't go with the mass at all, it is part of the 4-velocity. 4-Momentum is then mU, and energy it the timelike component of this in the basis of the measuring instrument.
3. Mar 25, 2014
### Bill_K
Kinetic energy itself is a leftover concept from Newtonian mechanics. In relativity the quantities you deal with are the covariant quantities, and there's hardly ever an occasion when you'd want or need to calculate KE.
4. Mar 25, 2014
Thank you PAllen but the equation you presented is the same as the one I presented. The only difference is I used Mo instead of M. I know that the concept of rest mass is out of favour and I know why but I used Mo in reference to the original equation. I'm quite happy to use M or m in place of Mo but when I use your equation I am just using the original equation in a different format and without the confusion of definitions of mass.
Last edited: Mar 25, 2014
5. Mar 25, 2014
"Hardly ever" is not the same as never. There are occasions when I might want to calculate KE and that's why I asked the question.
6. Mar 25, 2014
### Staff: Mentor
There are some problems for which some people find relativistic mass to be useful. For example, many years ago I had to calculate the trajectory of relativistic electrons passing between two parallel charged plates. Applying $F=m_{r}a$ seemed as easy as anything else, although I wouldn't argue with anyone who prefers to do the calculation some other way.
However, problems of this sort are few and far between, and the overwhelming majority of people studying relativity do so not because they need to solve a very specific class of problems, but because they want to understand relativity. If understanding relativity is the goal, than spending a lot of time learning about relativistic mass, then even more time learning about its limitations, is counterproductive.
Last edited: Mar 25, 2014
7. Mar 25, 2014
### PAllen
In mine, only one mass is used. There is no concept or relativistic mass in my formula. The gamma has nothing to do with the mass.
8. Mar 25, 2014
### pervect
Staff Emeritus
You can just as easily use the energy equation, and subtract the rest energy.
9. Mar 25, 2014
Yes I agree but in the original equation,the one I presented, there is only one mass namely Mo. The only difference is one of symbols. I know that generally speaking relativistic mass is considered to be outdated and misleading and I am familiar with the reasons for that. I like to think of M (as in the original M=gamma Mo) as being equal to the mass plus the kinetic energy expressed in mass units.
10. Mar 25, 2014
Yes you can express it mass units or energy units.
11. Mar 25, 2014
Thanks but the problems are probably not as rare as you think. The old mass variation equation is still taught in A level physics courses in UK schools (eg AQA physics A "turning points in physics" topic). Until and if the examiners modify the specification teachers still have to go along with it. Admittedly the relativity taught is at a very basic level.
12. Mar 25, 2014
Apart from a few things that need clarifying with regards to things like symbols used I think I am in broad agreement with everybody here.But there are still a couple of problems:
!.What should A level students be told?
2.The second point is about terminology. There are three terms in the equation when expressed in energy units KE = E-Eo
What should each of these three terms be called? KE is called kinetic energy but what do we call E and Eo. It might be tempting to call E total energy and Eo rest energy but those terms imply the existence of total mass and rest mass.
13. Mar 25, 2014
### PAllen
Energy is energy (a frame variant quantity); mass is mass, an invariant. The full equation to focus on is:
E^2 = (mc^2)^2 + p^2 c^2
Then you are not tempted to convert total energy to total mass, making a frame dependent notion of mass.
14. Mar 25, 2014
That's good for me but it's probably too heavy going for A level students.
15. Mar 25, 2014
### Bill_K
AQA Turning Points in Physics also tells us that "time runs slower when you are moving." I guess in comparison, their use of relativistic mass is a minor point.
16. Mar 25, 2014
### Staff: Mentor
No one disputes that it's still taught, and therefore that you may need it just to pass some tests. The dispute is over whether it's good pedagogy to teach it - that is, SHOULD it still be taught, or is there a better way of using limited classroom time to communicate the basic principles of SR. Note that relativistic mass is NOT one of those basic principles; it's a very specific application of them in a specific class of problems.
In some ways this discussion reminds me of discussions of whether Latin should be taught in American high schools. There's no doubt that most native English speakers will derive some non-zero benefit from a year or so of diligent Latin study; but enough to justify displacing a year or so of English literature?
Last edited: Mar 25, 2014
17. Mar 25, 2014
### atyy
But is there a conceptual connection to GR, since the principle of equivalence in Newtonian gravity is that inertial mass is gravitational mass? Since the inertial mass is energy in special relativity, that suggests that the energy should be the source of gravity in relativistic gravity. Then requiring that the equations be tensorial, the source should be either the stress-energy tensor or something like its trace.
18. Mar 25, 2014
### PAllen
Hmm. In my mind, the only form of inertial mass in SR is invariant mass, which includes KE only to the extent it becomes part of the invariant mass of the system.
19. Mar 25, 2014
I'm not sure where you are getting that from. The specification is very brief referring to proper time,time dilation the equation and muon decay.
20. Mar 25, 2014
### Bill_K
That is literally what it says on Page 21:
It then goes on to discuss the usual train example, but there is no hint given that the effect is symmetrical, which I think is the whole point of relativity.
Are we looking at the same document?
Last edited: Mar 25, 2014
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$$T_2$$ measurement in superconducting qubits
$T_2$ is a characteristic time describing the decoherence rate of a qubit. This script aims to present the protocol to deduce it for the usecase of a transmon qubit.
It consists in a $\pi$ pulse to rotate the qubit into the excited state, then waits for a given time, and then plays another $\pi$ pulse followed by an align operation and then reads out the state using the readout resonator.
Config#
The configuration dictionary is in the configuration.py file and is imported into the main program file t2.py.
The configuration defines two elements: qubit and rr (the readout resonator).
The qubit quantum element defines the qubit we are measuring. The OPX is connected to a mixer via two analog output channels of the OPX, numbered 1 and 2. We also specify the LO frequency received by the mixer using the lo_frequency field of the mixInputs dictionary, and a mixer correction matrix using the mixer field.
The qubit element defines a single operation: X which plays a gaussian pulse to the I channel of the OPX - producing a rotation of the qubit about the X axis. This, of course, must be calibrated to product a proper $\pi$ pulse, e.g. with a Time-Rabi experiment.
The rr quantum element allows to measure the qubit state by measuring the resonant I and Q components of a reflected microwave signal. It defines a readout pulse which is read on input number 1 of the OPX, as set by the output entry of the rr element dictionary. Note also the time_of_flight and smearing parameters which must be defined to perform a measurement. As for the qubit we define the associated lo_frequency and mixer correction.
⚠️Note that failing to declare a digital_marker will not fail program compilation, but will prevent data from being acquired.
Program#
The QUA program T2 is built around two nested for_ loops. The external loop is used for repeated averaging steps, and the internal loop scans the parameter tau. The body of the loops plays the X operation to rotate the qubit into the |1> state. The wait statement is then used with a variable duration tau, followed by another X operation and then an align operation a measurement statement demodulating the readout into the I and Q variables. Each loop ends by a wait statement. The program ends by a recovery_delay period which is assumed to be sufficient to allow qubit to decay back in the ground state.
We run the program on the simulator for 500 clock cycles and take the simulated samples from the simulation job.
The acquired ADC stream is taken from result_handles. Note the change of name of the raw ADC steam from the name we specified in the program (raw_adc) to the name we use to get the stream (raw_adc_input1). This is an idiosyncrasy of the raw ADC interface which does not appear in other QUA data saving mechanisms.
with for_(n, 0, n < NAVG, n + 1):
with for_(tau, 4, tau < taumax, tau + dtau):
play("pi2", "qubit")
wait(tau, "qubit")
play("pi2", "qubit")
align("rr", "qubit")
measure_and_save_state("rr", I_res, Q_res, state_res, th)
wait(recovery_delay // 4, "qubit")
save(tau, tau_vec)
Post-processing#
To get an estimate of the probability to be in the excited state as a function of delay duration, we need to reshape the output streams and calculate the statistics: mean and variance of the probability to be in the excited state.
This is a convenient starting point for integrating with a custom system-specific procedure.and Qua
1. Qubit has 2 lines in from OPX e.g. mixers/modulators and direct line from OPX and 1 line out back to OPX
2. QM's Python package sits at top level along with QM's app - that is where Qua program are run
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VLA > RFI > January 2013 D-configuration
# January 2013 D-configuration
by Heidi Medlin last modified Sep 07, 2017
Each link in the list below presents scalar-averaged cross correlation spectra using several baselines in the D-configuration for each respective receiver band obtained in January 2013. The channel separation is 125 kHz. Hanning smoothing has been applied to minimize Gibbs ringing. The scale is logarithmic with arbitrary offset. The scale can be roughly calibrated by noting that the mean power level is roughly equal to the antenna SEFD in Jy. There is no bandpass calibration, so the downturns at the edges are due to the analog and/or digital filters.
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# Drawing circles with different distance functions
1. Jan 29, 2014
### Lee33
1. The problem statement, all variables and given/known data
In $\mathbb{R}^2$, draw a unit circle for taxicab distance $(d_t)$, euclidean distance $(d_e)$, and max distance $(d_s)$.
2. Relevant equations
$d_e = \sqrt{(x -x_1)-(y-y_1)}$
$d_s = \text{max}\{|x-x_1|,|y-y_1|\}$
$d_t=|x-x_1|+|y-y_1|$
3. The attempt at a solution
Euclidean distance is the most familiar one. It will just be our everyday unit circle.
For taxicab, will the unit circle be a shape of a square?
For max distance I am not so sure about to.
2. Jan 29, 2014
### HallsofIvy
Staff Emeritus
Try a simple example like $(x_1, y_1)= (0, 0)$ and the distance is 1.
Then with the Euclidean distance you have $\sqrt{x^2+ y^2}= 1$ which is the same as $x^2+ y^2= 1$. Yes, that is a circle with center at (0, 0) and radius 1..
With the taxicab metric, that is |x|+ |y|= 1. If x and y are both positive it is x+ y= 1 which is the line segment from (1, 0) to (0, 1). If x is negative and y is positive, it is -x+ y= 1 which is the line segment from (0, 1) to (-1, 0). If x and y are both negative, it is -x- y= 1 which is the line segment from (-1, 0) to (0, -1). Finally, if x is positive and y is negative, it is x- y= 1 which is the line segment from (0, -1) to (1, 0). Yes, that is a square with its diagonals horizontal and vertical.
With the max distance it is max(|x|,|y|)= 1. If x and y are both positive and x< y that is |y|= y= 1 which is the horizontal line segment from (0, 1) to (1, 1). If x and y are both positive and y< x, that is |x|= x= 1 which is that vertical line segment from (1, 0) to (1, 1). You check the other quadrants:
If x is negative and y is positive and x< y .... If y< x ....
3. Jan 29, 2014
### Lee33
Thanks for the help!
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# Inflation, comoving Hubble radius and particle horizon
1. Dec 1, 2015
### JJNic
I have a question regarding the exact formulation of the mechanism of Inflation.
In the cosmology lecture notes by Baumann he uses $\frac{d}{dt} \frac{1}{aH} < 0$ as an definition of inflation. I see that it yields $\ddot a > 0$, but my confusion lies in the interplay between the particle horizon and the comoving Hubble radius.
Why do we require $\frac{1}{aH} \ll \chi_p$ in the early universe? He says
But i dont see how that goes about.
I am especially confused by figure 2.3 on page 33 (i cropped and attached it for your convenience),
I see that the horizon problem gets solved because the points p and q now have overlapping particle horizons, but what does the comoving hubble sphere have to do with it?
They would still have overlapping particle horizons if i did not draw the Hubble sphere or if i drew it differently. Or is it not possible to have both things at the same time? (a different Hubble sphere AND overlapping particle horizons of p and q).
To me it just seems that "adding more conformal time before the initial singularity and shift it to -inf. or less" s.t. p and q have overlapping particle horizons would do the job just fine, not worrying about the comoving Hubble radius.
So, i am obviously missing something, but what is it?
Thanks!
2. Dec 1, 2015
### bapowell
The shrinking comoving Hubble radius is a consequence of the accelerated rate of expansion, and nothing more. Indeed, simply "adding more conformal time before the initial singularity..." is all you need to ensure that the particle horizons overlap, but there is no room to simply "add more" in the standard cosmology. We must postulate a period of inflation to give us the extra room, and the shrinking comoving Hubble scale is a result of this requirement.
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## On ‘Oumuamua
‘Oumuamua slipped our grasp.
With JPL’s mission design tool, one can scroll wistfully through a fine-grained list of opportunities lost — a spaceport departure board overflowing with missed flights, all to a single exotic destination.
By any measure, ‘Oumuamua imparted an influence that exceeded its size. Its approximately 260-meter greatest extent is dwarfed by the recently visited 2014 MU69.
Which itself is not very large.
It’s interesting to look at the Google Trends listing for worldwide interest in ‘Oumuamua.
In mid-November 2017, a two-lobed burst of interest surrounded first the naming and then the release of ESO’s iconic artist’s impression of a menacing starship-like shard. A barely perceptible blip followed last Summer’s announcement that ‘Oumuamua’s trajectory was non-Keplerian, and then, in Fall 2018, boom, a veritable megastructure of attention.
‘Oumuamua’s visit was short enough so that the data obtained (mostly in the five compressed weeks following its initial discovery) is readily summarized. Our first detected interstellar visitor arrived with almost exactly the speed and direction that one would guess a priori for an object having the average local orbit in the disk of the Milky Way. In a sense, the Solar System ran across ‘Oumuamua rather than the other way around.
The spectrum of sunlight reflecting off ‘Oumuamua’s surface is skewed smoothly to the red, with no identifying bumps or dips. The first spectrum, obtained by Joseph Masiero, who was observing at the Palomar 200-inch telescope when word of the interstellar object was released, was typical.
‘Oumuamua’s colors fall readily onto the locus defined by the various small-body constituents of the outer Solar System, which tend to list toward the reddish and the dark. Saturn’s moon Phoebe is the type of object one can bring to mind in this regard.
Solar System comets invariably spew out micron-sized particles, which are entrained in the gas (mostly water vapor) that erupts from their sun-warmed surfaces. It was thus generally expected that comets arriving from other solar systems would behave in similar fashion. Yet even in the deepest exposures — stacked together from multiple tracking images — ‘Oumuamua appeared completely point-like. Its lack of any observable coma placed stringent, spoonful-per-second limits on the amount of powdery dust emanating from its surface.
The diagram just below (with portions of the graphics taken from last October’s Sky and Telescope article) shows the highest-resolution portion ‘Oumuamua’s light curve stitched together from the data obtained with eight different telescopes. The reflected light signal varied periodically, with a dip-to-dip time scale of about 3.6 hours. There are lots of frequencies (and aliases) in the periodogram of the irregularly sampled data, the data gets proportionally noisier when the signal is at its dimmest, and the light curve does not quite repeat from pulse to pulse. The large peak-to-trough variation, and the more-or-less repetitive pattern suggest a highly elongated object, one that was tumbling chaotically end over end. The rotation period, moreover, was short enough so that ‘Oumumua would need to have some mild degree of physical strength to resist flying apart as it spins. It seems to be a solid object rather than a loosely consolidated rubble pile. The famous artist’s impression that populates a Google search on ‘Oumuamua proceeds from these basic features of the light curve.
Each image from each telescope contributed to the determination of ‘Oumuamua’s trajectory. A careful analysis of all the data leads to the remarkable conclusion that ‘Oumuamua departed the Solar System more quickly than an object starting on the same orbit and that was subject only to the Sun’s gravity. In other words, an additional, anomalous acceleration acted on ‘Oumuamua. The effect was small, apparently exactly akin to reducing the mass of the Sun by 0.1%, but its effect was nonetheless very evident in the trajectory.
Comets routinely display non-gravitational accelerations of the type and the general magnitude that was observed for ‘Omuamua. Jets of sublimating gas (mostly water vapor) stream off the surface, and perform like stochastic rocket engines. The close-range photographs of comet 67/P Churyumov-Gerasimenko (the dramatic subject of the Rosetta Mission) show the process in action.
The catch, however, is that if outgassing is responsible for ‘Oumuamua’s acceleration, there was exceedingly little fine dust entrained in the gas. This disconnect has led to two proposals in which the acceleration arose not from a gas jet, but from solar radiation pressure. The first proposal (by Shmuel Bialy and Avi Loeb) contained the much-discussed speculation that ‘Oumuamua is an engineered solar sail. A paper published several weeks ago (to less media attention) by Amaya Moro-Martin suggests that ‘Oumuamua is a naturally formed fractal, a dust bunny-like aggregate formed in a protostellar disk with an extremely low density and an extremely high porosity. Such a structure would be readily accelerated by radiation pressure, and like the sail model, would have no need to spew gas-entrained micron-sized dust.
One can argue that the lack of micron-sized dust in an outgassing scenario isn’t all that unexpected. An icy body from an alien environment may have undergone compositional processing that was absent in the Solar System. Moreover, if the entrained dust particles were larger, say 100 microns in size, they would have gone undetected.
The Spitzer Space Telescope’s non-detection of re-radiated heat constrains ‘Oumuamua’s reflectivity and its physical size. Spitzer’s wavebands of observation rendered it particularly sensitive to emission from carbon dioxide and carbon monoxide gas in ‘Oumuamua’s vicinity, but neither species of molecule was detected. This is odd. Jets from Solar System comets are primarily composed of water vapor, but typically also contain carbon-based constituents. A water vapor jet from ‘Oumuamua is not directly ruled out by Spitzer, but one needs to argue that the water, in addition to being free of small dust, had a very low level of carbonation. A glass of melted ‘Oumuamua would have to be still, not sparkling.The jet interpretation for ‘Oumuamua’s acceleration can thus wriggle out of the Spitzer non-detection by positing a composition of icy material with a very low carbon-to-oxygen ratio, and it can simultaneously wriggle out of the coma non-detection by positing that very little micron-sized dust was embedded in the ice, but that’s an admittedly uncomfortable amount of wriggling. Requiring a pair of caveats is not very satisfactory, but it allows us the hypothesis that we were visited by a relatively mundane object rather than an exotic natural object or an artificial object. Moreover, with dim prospects for the emergence of additional observational data from a long-departed object that was faint to begin with, it’s unlikely that any of the three classes of competing explanations for ‘Oumuamua’s behavior will ever gain full credence or confirmation.
Interestingly, we do have some direct information on the composition of extrasolar planetesimals. This is obtained obtained by looking at accretion events onto white dwarfs, e.g. this paper by Wilson et al. Among the measurements of this type that have been made, there are a number of cases where the C/O ratios are below the naive limit for ‘Oumuamua (plotted below on Figure 3 of Wilson et al.)
A paper published by Roman Rafikov last August raised an additional potential problem with the jet model for ‘Oumuamua’s anomalous acceleration. If a jet acts continuously (or on average, acts continuously) to exert the acceleration at a single point on the surface, then the force that provides the acceleration will also exert a torque, unless it is pushing exactly along one of the principal axes of inertia. This torque would have caused ‘Oumuamua to noticeably increase its rotation rate during the days that it was under close observation.
In a new paper lead-authored by Darryl Seligman and just posted to arXiv, we revisit the jet model, and imagine that the jet is created by ice that lies just below or right at the surface of ‘Oumuamua. If this is the true situation, then ‘Oumuamua’s jet will migrate rapidly across the body, and at any given moment, it will be strongest at the spot where the Sun’s rays are directly perpendicular to the surface. If we model ‘Oumuamua as an ellipsoid, the situation looks like the illustration just below, with comet 67/P Churyumov-Gerasimenko and its sub-solar jet shown to help motivate what we’re proposing:
In an idealized situation in which ‘Oumuamua is a perfect triaxial ellipsoid with a long, an intermediate and a short axis, and where two of the axes lie exactly in ‘Oumuamua’s orbital plane, and where the jet is always located at the spot where the Sun is directly overhead, ‘Oumuamua’s motion resembles that of a pendulum. Here’s a movie made with the open-source ray tracing software POV-ray which implements the appropriate rigid-body dynamics:
In the example shown above, we’ve assigned our idealized ‘Oumuamua a surfboard-like 9:4:1 long-intermediate-short axis ratio, and we’ve positioned the short axis so that it points perpendicular to the orbital plane. The action of the sublimation jet in this model produces a pendulum-like rotation of the body and implies a long semi-axis, $$a\sim 5A_{\rm ng}P^2/4\pi^2 \sim 260\,{\rm m}$$, where $$A_{\rm ng}=2.5\times10^{-4}\,{\rm cm\,s^{-2}}$$ was the observed non-gravitational acceleration during the high-cadence October observations, and $$P\sim8\,{\rm h}$$ was the observed peak-to-trough-to-peak-to-trough-to-peak period of a full $$-\pi \rightarrow \pi \rightarrow -\pi$$ oscillation.
Interestingly, the size $$a\sim260\,{\rm m}$$ implied by 8 hour period and the $$0.001 g_{\odot}$$ acceleration agrees perfectly with the completely independent estimates of Oumuamua’s size that stemmed from its measured brightness, if we assume the 10% reflectivity that is appropriate to ices that have undergone long-duration exposure to the interstellar cosmic ray flux. The implication is that ‘Oumuamua was either like a pendulum that was rocking back and forth, or perhaps one that was barely swinging “over the top”. A more complete analysis allows us to make a snapshot of the range of possible motions for the idealized case. The angular position of the idealized ‘Oumuamua during its swing is on the x-axis, the varying angular speed of the swing is on the y-axis, and the color coding shows the period for each allowed trajectory. A nice feature of the motion is that as long as the aspect ratio, a:b:c has a$$\gg$$b, the period depends only very weakly on b and c.
With our sub-stellar jet model, ‘Oumuamua doesn’t change its light curve period appreciably during the period that it was observed. This is because the torque from the sub-stellar jet spends equal time working with the instantaneous spin and working against it.
Assuming that ‘Oumuamua was a natural object, it likely had an uneven surface, possibly with regions of varying reflectivity, and it most certainly was not spinning with a principle moment of inertia aligned along its angular momentum vector. Moreover, there was likely a time-varying lag in the response of the jet, and the strength of the jet likely varied stochastically. With our ray-tracing model, we can readily gin up such complications and watch how they affect the motion. Here’s a version of the dynamics where ‘Oumuamua starts in a random orientation, and has a rough mottled surface pattern:
In the simulation shown just above, we’re tracking ‘Oumuamua, and watching it tumble as it flies through its escaping orbit. The point of view is from an ultra-resolving telescope orbiting along with Earth. This means that the Sun’s illumination of the ‘Oumuamua model is consistent with what an observer on Earth would have seen had they had sufficient magnification. Removing the axes and the jet, we can sum up the brightness of all the pixels in each image to get a frame-by-frame light curve. Sampling the light curve at the actual cadence of the observations, and adding the correct amount of noise permits direct comparisons between various versions of the model and the observations. The first figure below shows the real data, and then various versions of what one would get with our ellipsoid models. No attempt at curve-fitting has been made here, just a proof of concept. The second figure below shows the power spectra of the data, our models, and the zeroed-out observations. It’s clear that if one wanted to, it would be quite possible to fit the light curve pretty well with models of this general type.
There’s a strong possibility that more interstellar objects will be detected fairly soon. A necessarily shaky extrapolation from the “statistics” of one object detected over several years by the Pan-STARRS survey implies that interstellar space is teeming with ‘Oumuamuas. The numbers are impressive — of order $$10^{26}$$ bodies total in the Milky Way, totaling a hundred billion Earth masses. At any given moment, roughly one such object should be in the process of threading the 1AU sphere that encompasses Earth’s orbit around the Sun.
Categories: worlds Tags:
## Sculpting the orbital distributions
Saturn’s polar axis is tilted relative to the plane in which Saturn orbits the Sun, and the plane of Saturn’s orbit is itself tilted with respect to the averaged orbital plane of the Solar System’s planets.
Pieces of popular scientific writing often start with an engaging “hook”, but the foregoing statement doesn’t do a particularly good job. Saturn and its rings do, however, do a good job of showing off their tilts — their obliquities, to use the vernacular. Saturn gradually shifts in appearance as the Sun’s illumination angle changes, and over time the creeping ring shadows even affect Saturn’s climate. Certainly, at the moment when the rings slice edge-on to the solar rays, the system presents a very different appearance than when the ring plane is inclined.
The geometry was first understood by Christiaan Huygens. By the mid-1600s, he had drawn a clear diagram showing how Saturn’s tilted pole points in a fixed direction as the planet traces its three-decade orbit.
The obliquities of Saturn and Neptune (26.7$$^{\circ}$$ and 29.6$$^{\circ}$$ respectively) seem odd. Uranus, tipped to its side and then some (97.9$$^{\circ}$$) is odder still. Naively, one might have expected a Solar System forming from a flat, orbiting disk of gas and dust to have ended with the equators of the giant planets lying in the average plane of the planetary orbits. Jupiter, with its axis tilted at only 3.1$$^{\circ}$$ is indeed fairly conforming, but the others are all badly out of alignment. Why?
Moreover, with literally thousands of worlds now in the catalog, one also wonders if spin misalignments are rampant among the extrasolar planets. Could it be possible to infer obliquities even if we have no method to photographically resolve the planets themselves?
Left to orbit freely around a star, the tilted spin pole of an isolated planet will precess like a gyroscope. The cycling of precession is slow in comparison to the rate of the spin itself, and it stems from the torque exerted on the planet by the parent star. If the planet — like Saturn — has satellites, they orbit quickly enough to act as if they were a contributing part of the planet, and the joint set-up of planet plus moons precesses as a unit. The moons and rings stay locked to the orbital plane, and the net effect of the satellites, as far as precession is concerned, is to speed up the rate at which the cycle occurs.
Wheels within wheels… The situation grows more complicated if the orbital plane of a precessing planet also precesses. An orbit whose own pole (or orbit normal) traces a slow overhead circle presents, in effect, its planet with a moving target. With the complication induced by the precessing orbit, how does the spin axis respond?
Some intuition can be obtained by experimenting with tops. Orbital precession can be mimicked by placing a spinning top on a plate and evenly rotating the tilt of the plate, much as if panning for gold.
After some practice, one finds that the plate-top system can be transiently “locked” into a state where the spin axis precesses in the opposite direction — but at the same rate — as the plate. When this happens it’s oddly satisfying, imparting a tactile clue that resonance between the precession of an orbit and the precession of a planet’s spin might be capable of playing a dynamical role.
Another clue is supplied by the motion of Earth’s Moon. In 1693, Giovanni Domenico Cassini, who logged careful observations from the Paris Observatory, concluded that the plane of the Moon’s orbit, in the course of executing a 18.6 year precession cycle about Earth’s equator, maintains a constant angle with respect to the ecliptic (Earth’s orbital plane). He also found that the small obliquity of the moon, which is only 1.5$$^{\circ}$$ (compared to 23.5$$^{\circ}$$ for Earth) precesses at the same rate of one full revolution per 18.6 years.
Giovanni Domenico Cassini (1625-1712). Prior to holding the directorship of the Paris Observatory, he was the highest paid astronomer at the University of Bologna, having been appointed to his professorship by the Pope.
In other words, an arrow pointing out of the Moon’s pole always lies in the plane formed by Earth’s orbit normal and the Moon’s orbit normal. This remarkable co-precession of the Moon’s spin axis and its orbit normal received little attention until a landmark 1966 paper in the Astronomical Journal by Giussepe Colombo — who separately, achieved fame for discovering that Mercury exists in 3:2 spin-orbit resonance, turning three times on its axis for every two trips around the Sun. A few years after Colombo’s paper appeared, Stan Peale coined the term Cassini State to describe the dynamical configuration.
That something so fundamental to the motion of the Earth-Moon system was left apparently unexplained from its discovery in 1693 through 1966 seemed puzzling, so I sent a text to Konstantin Batygin.
Indeed they do:
Issues of obscurity, precedence and priority aside, Colombo used Newton’s laws of motion and gravity to demonstrate that the in-sync cycling of the Moon’s spin and orbital axes represents a minimum-energy configuration. In a frame of reference synchronized to its orbital precession, the Moon’s spin pole is analogous to a marble that frictional dissipation has brought to rest at the bottom of a curved bowl. The Moon is said to be locked in a secular spin-orbit resonance.
If a perturbation — a kick — imparts energy to the marble, it will roll around in the bowl. Likewise, if the spin pole of a body in secular spin-orbit resonance is perturbed, the direction that the pole points will wander when viewed in the precessing frame. In his 1966 paper, Colombo worked out how the trajectories traced by a perturbed spin pole will behave.
Colombo also pointed out that a simple geometric construction can be used to illustrate how the spin pole moves. First, imagine the sphere defining all the possible directions that a spin pole can point. The sphere is oriented so that its own coordinate poles are perpendicular to the overall plane of the system under consideration. In the minimum energy configuration, the spin direction, the coordinate pole, and the direction of the orbit normal all lie in a single plane that slices the sphere in half.
If the planet’s spin axis is not damped to the minimum energy configuration, it will slowly trace a path on the sphere when followed in the co-precessing frame. A detailed analysis (which appears in Colombo’s 1966 paper, and which Stan Peale augmented and corrected in 1969) shows that the set of allowed paths (level curves of the Hamiltonian) are determined by the set of possible intersections between a parabolic cylinder and the sphere. Individual paths, corresponding to individual energies of perturbation, are defined by moving the parabolic cylinder back and forth.
Remarkably, the projection of these intersection paths onto the ecliptic displays a characteristic structure that arises repeatedly in problems involving resonance. The state of co-precession keeps the spin pole of the planet fixed in the perfectly damped configuration, in which the vertex of the parabola just touches the sphere. This vertex point lies at the center of a set of banana-shaped trajectories. Then, when the spin pole of the planet is perturbed, the motion follows a trajectory where the spin pole travels along one of the banana-shaped curves. Inside the shaded region, a full traversal of the curve never entails a full 360$$^{\circ}$$ accumulation of angle, and the pole is said to be librating in the resonance.
The Solar System provides several examples of secular spin-orbit resonance. Most prominent from our Earth-bound viewpoint, is the motion of the Moon. The action of tides has damped the motion of its spin pole so that it lies at the core of its sequence of banana-shaped level curves.In a pair of articles [1, 2] published in 2004, Bill Ward and Douglas Hamilton raised the remarkable possibility that Saturn’s spin pole might be librating in a frame that co-precesses with the orbital inclination of the planet Neptune. On the surface, a Saturn-Neptune secular spin-orbit resonance seems nearly unbelievable. I recall hearing Bill Ward describe the work at a conference, a year or so before the papers came out. At that time, I have to admit, I didn’t really understand the details of the talk, other than the the take-away that Neptune was somehow responsible for tipping Saturn over. Although Jupiter and Uranus exhibit substantially larger gravitational perturbations on Saturn than does Neptune, the frequency at which Neptune’s line of nodes regresses happens to almost exactly match the precession rate of Saturn’s pole. Neptune’s orbit, in a sense, shifts at a rate that cuts through the noise to provide a controlling influence that adds up for Saturn.
Once the precession rates of an orbit and a spin pole are locked together, the lock will be maintained even when the orbit’s precession rate slowly changes. As a consequence, if the rate at which the orbit precesses slows down, the planet’s spin pole will slowly tip over so that its precession rate can decline in sync. When that happens on a habitable planet, it’s time to set solar sail for the stars. Closer to home, the long-ago dispersal of planetesimals in the Kuiper Belt led to a slowing of Neptune’s orbital precession. Remarkably, this seemingly minor slowdown seems to have forced Saturn from a small initial obliquity to its current 26.7$$^\circ$$.
In addition to affecting Earth’s Moon and Saturn, secular spin-orbit resonance also plays a likely role in the tilts of both Jupiter and Mars (and quite possibly Uranus). It’s fully separate from the phenomenon of spin-orbit resonance, which, for example, maintains Mercury’s spin period at an average rate that is exactly 3/2 times its orbital period.
In the Solar System, the planetary obliquities are readily measured, and have been accurately known for centuries. Orbital precession rates can be calculated either from the well-established techniques of celestial mechanics, or from direct numerical N-body integrations. Even so, secular spin-orbit resonance didn’t garner attention until Colombo’s and Peale’s papers in the 1960s, and even then, it received only limited press. In Murray and Dermott’s Solar System Dynamics, which has become a standard text, the authors state at the outset that Cassini states are not covered in their book. The possible enforced match between Saturn’s polar tilt and Neptune’s orbit went unnoticed until 2004. It thus seems like a long-shot that secular spin-orbit resonances among extrasolar planets have much chance of being a “thing”.
For a planet like Saturn, the slight decrease in the Sun’s gravity from the sub-solar point to the anti-solar point on Saturn’s surface leads to a small tidal deformation of the planet. Friction within Saturn causes Saturn’s rotation to pull this tidal deformation slightly out of alignment, with the net result being a slow decrease in Saturn’s spin rate. The rate of decrease, however, is negligible. It would take many times the current age of the Solar System for Saturn’s spin period to be tangibly modified by this effect.
Tidal forces, however, have an extraordinarily steep fall-off with distance. If Saturn were moved a hundred times closer to the Sun, to a distance where the extrasolar planets are routinely found, the Sun’s tidal influence on Saturn’s spin would be ten billion times stronger.
In the presence of strong tidal forces, the spin period of a planet on a circular or near-circular orbit is brought into sync with the planet’s orbital period. That’s the situation that the Moon finds itself in, and it is also thought to be the case for most of the shorter-period transiting planets that have been discovered by the various ground-based surveys as well as by the Kepler Mission.
In addition to synchronizing the spin, tidal forces also act to align a spin pole with the orbit normal. If, however, a planet is in secular spin orbit resonance of the type we’ve been discussing, the resonant torques can potentially balance the dissipative torques and prevent the planet from being righted.
Tidal dissipation is normally quite self-regulating. If the dissipation caused by tides is strong, then synchronization ensues, and the energy that the dissipation generates drops. If, however, a mechanism exists to thwart synchronization then significant evolution can occur. Io (and to a lesser extent Europa) provide examples. As a consequence of having its eccentricity forced by the resonant interaction between the three inner Galilean satellites, Io undergoes strong tidal dissipation, leading to the famous volcanoes that cover its surface, and to the heavy loss over time of its volatile constituents.
The famous Peale, Cassen and Reynolds article that describes Io’s dissipation belongs near the very top of a list of admired papers. It presents clean dynamical arguments that draw on disparate aspects of geophysics and celestial mechanics to make a non-trivial prediction. And indeed, the paper’s two-sentence abstract is the very model of brevity:
The dissipation of tidal energy in Jupiter’s satellite Io is likely to have melted a major fraction of the mass. Consequences of a largely molten interior may be evident in pictures of Io’s surface returned by Voyager I.
Just days after the March 2, 1979 publication of the paper, Voyager 1 flew through the Jovian system, and recorded Io’s hyperactive volcanism. Here’s a recent photo of Io from NASA’s Juno probe. The picture was taken in the infrared, where it’s pretty clear what’s going on.
In short, the Peale et al. 1979 paper is a tremendous inspiration. For years, I’ve been thinking, could something similar be done with the extrasolar planets?
The Kepler data is certainly the best place to look for opportunity. The precise timing of the planets in Kepler’s multi-planet systems gives the possibility for finding subtle effects that go beyond simple Keplerian orbital motion.
It’s well known that Kepler detected lots of multiple-planet multiple-transiting systems. The planets in these systems tend to lie in the super-Earth/sub-Neptune radius range, and typically have masses of order 5 to 10 times the mass of Earth. A zeroth-order question is what these planets are like and how they got to where they are currently observed.
There is an interesting unexplained clue in the data. One can take pairs of adjacent planets from the Kepler catalog, and plot the period ratios. What one sees is that in the vicinity of low-order orbital commensurabilities, there is a statistically structure in the distribution:
There is an overabundance, or a “pile up” of planets with orbital period ratios that are a few percent larger than the perfect 3:2 and 2:1 orbital commensurabilities, and a relative lack of planet pairs that have orbital period ratios just less than the commensurabilities. It’s as if some mechanism is acting to pry the pairs apart. Moreover, if one looks at the individual sizes of the planets in the distribution, one sees that on average, the radii of the planets that lie just wide of the commensurability are larger than the radii of the planets that have slightly smaller period ratios.
Several theorists have written papers that show this structure, termed “resonant repulsion” can be understood if the participating planets are experiencing a very high rate of tidal dissipation. The difficulty, however, has been that if the standard rate of interior energy dissipation is used, then the rate of dissipation would have to be very high. The planets would have to be extremely inelastic. Earth for example, does fall into this inelastic category because the ocean tides efficiently dissipate energy along shorelines. Most bodies in the Solar System, however, and especially the massive planets – Uranus, Neptune, Saturn and Jupiter – are far less dissipative. In the case of the Solar System’s giant planets, this difference with Earth is of order a factor of a thousand or more.
In a new paper appearing in Nature Astronomy and lead-authored by Yale graduate student Sarah Millholland, we propose a solution. If one or both planets in a pair that has a period ratio lying just outside the low-order commensurability is in secular spin-orbit resonance, and if the spin obliquity is high, then the dissipation within the planet will be large, and indeed large enough to account for the observed effect.
In many respects, the regular satellites of the jovian planets in our solar system resemble the multiple-transiting multiple planet population that was found by the Kepler Mission. Orbital inclinations and eccentricities are small in both types of systems. The orbital periods typically range from days to weeks in both cases, and the mass ratios of satellites to primaries typically tend toward one part in ten thousand. It is thus reasonable to ask why the phenomenon of resonant repulsion die to secular spin-orbit resonance is not found among the jovian satellites, all of which have tiny tilts for their spin poles.
The answer lies in the spin rates of the giant planets, all of which spin relatively rapidly, causing them to bulge significantly at their equators. Jupiter does a full turn in only 9 hours 55 minutes and is noticeably squashed when viewed through a telescope. The quadrupole moment is the jargon for the quantified degree of spin-induced structural flattening. The giant planets’ large quadrupole moments force rapid orbital precession of their satellites. The frequency is substantially higher than the spin precession rates of the satellites can keep up with. As a consequence, all of the major regular satellites of the Jovian planets have their spin axes aligned with their orbit normals.
The parent stars of the Kepler multi-transiting, multiple-planet systems spin much more slowly than Jupiter or the Solar System’s other giant planets. The stars have lost the majority of their spin angular momentum through the process of magnetic braking. The quadrupole moments of the G, K, and M stars hosting Kepler-multiple planet systems are quite small. Our own Sun spins on its axis with a 27-day period, which is fairly typical, and red dwarfs tend to spin even more slowly. As a consequence, the precession periods of the Kepler planet’s orbits are driven primarily by planet-planet interactions and not by the stellar equatorial bulges.
In the plot just below, the natural spin precession frequencies, $$\alpha$$‘s, and the orbital precession frequencies, $$\vert g \vert$$‘s, for the planets in Kepler’s multiple-transiting systems are tallied into histograms. The rate, $$\alpha$$ of a planet’s spin precession depends on its internal structure, so that a planet that is highly centrally concentrated (a low $$k_2$$) precesses more slowly than one whose mass is more extended (a high $$k_2$$). The histograms for the spin and orbit rates ($$\alpha$$‘s and $$\vert g \vert$$‘s) show substantial overlap, and both reach peaks near a period of about 3,000 years.
In short, it is a suggestive coincidence that the orbital periods, the masses, the radii and the separations of the Kepler planets combine to generate similar rates of orbital precession and spin precession. This means that capture into spin orbit resonances may be quite likely for these planets.
Capture of a planet into secular spin-orbit resonance will naturally occur if the ratio of the planet’s orbital precession frequency to its spin precession frequency is slowly brought down to unity from above, that is, if $$\vert g \vert/\alpha \rightarrow 1$$. This can happen if the planets in a system migrate toward orbital commensurability. This schematic diagram from our paper shows how the process works:
Simulations that track the orbits and the spins of the planets show that the spin precession and orbit precession lock into sync remarkably easily and naturally. Our paper charts several example evolutionary trajectories that look like this one:
In this particular simulation, two 5 $$M_{\oplus}$$ super-Earths experience mild disk-driven migration which slowly pushes their orbits together, and, after $$\sim$$1.3 million years, binds them into a 3:2 orbital mean-motion resonance. As this mean-motion resonance capture occurs, the inner planet of the pair finds that its orbital precession rate has slowed to match its spin precession rate; it is caught in secular spin-orbit resonance. Thereafter, as the orbital precession slows still further (as a consequence of the protostellar disk dissipating), the inner planet’s axis is compelled to precess more slowly as well. In order for the planet to slow its spin precession, it is forced over on its side, to a final obliquity of more than 50$$\circ$$.
The simulation charted above runs for just a few million years, but the planetary systems that Kepler observed are generally a thousand times older. The outer planet in the simulation, whose obliquity is traced with the green line in the upper panel, sees its tilt kicked up when the ratio $$\vert g \vert/\alpha$$ passed through unity from below but does not end up in spin-orbit resonance. Its perturbed obliquity will drop back to zero after a few tens of milions of years. For the inner planet, however, the situation is different. Torque from the tidal dissipation in the planet balances torque from the precessing orbit, the obliquity remains constant, and an Io-like situation is produced. Obliquity-juiced tides generate ~3 million Gigawatts within the planet, roughly a thousand times the total power that Io produces, and roughly three times more energy per unit mass. The relentless dissipation draws energy from the orbit, forcing the period ratio, over time, to creep up from the initial 3:2 ratio.
The net result of this process, replicated again and again in the Kepler sample, can explain the lack of worlds near the exact m:n integer period ratios and simultaneously account for the pile-ups seen just wide of the perfect commensurabilities.
A nice feature of the theory is that it makes some predictions.
,Capture into secular spin-orbit resonance is easier if a planet has a larger radius. As a consequence, if dissipating oblique planets are what drive the Kepler pairs apart, then the planets on the right side of the period ratio gaps should be (on average) larger than those to the left sides of the gap. Pleasingly, this is exactly what is seen in the data, and it’s a feature that has gone unnoticed until now:
Moreover, larger planets are more dissipative, and so statistically, the radii of planets in the member pairs should increase as the period ratios increase. This effect, while subtler, is also present in the data.
Given the actual structure of the period ratio diagram, one can work out the amount of dissipation required to explain each pair if the ages of the systems are known. Statistically, this allows us to determine what kind of planets we’re dealing with. The details are explained in our paper, but the take away is that the planets in the Kepler-multiple systems likely tend to resemble Uranus and Neptune as far as their internal structures are concerned.
And finally, one last, as-yet untested prediction. If a planet with an orbital period in the range spanned by the Kepler-multiple planet systems has significant satellites, its precession rate will be too rapid for secular spin-orbit resonance to work. As a consequence, oblique planets driving resonant repulsion won’t have significant moons of the type seen orbiting the giant planets in the Solar System.
Categories: worlds Tags:
## A Final Exam…
Latest Deep Space Climate Observatory Image
Over the past several semesters at Yale, I’ve been working out a new take on the standard “Astronomy 101” class for non-science majors. Broadly, the goal is to stage a wide-angle view of the Anthropocene, thereby forging an understanding of how Earth fits into its broader cosmic context. Economics, Political Science, and History constituted the largest groups of majors in the class.
I’m working on getting the class notes, problem sets and readings into a form that’s distributable. In the interim, I’ll cut right to the chase with the final exam. Per Yale’s official instructions:
Final examinations normally last either two or three hours but, in either case, students are permitted to take an additional half hour before being required to turn in their answers. This additional time is given for improving what has already been written, rather than for breaking new ground.
This one was set as the three-hour variety. (Didn’t seem like any utility would be gained by imposing time pressure.)
Categories: worlds Tags:
## Intercept
March 19th, 2018 1 comment
‘Oumuamua breezed in unexpectedly and it left in a rush. Faded now, to twentynine, soaring up and out over Jupiter’s orbit. No sum, it seems, sufficient to compel it to pick up the phone, to give us a call.
Maybe it was a one time fluke — a color out of space, but it’s also possible that it was unexceptional, a mundane representative from a vast distribution. If so, what can we do to be ready for the next one?
Darryl Seligman has a new paper up on arXiv that outlines a plan. Had ‘Oumuamua been spotted on its way in, and if a probe had been loitering in anticipation, fueled and ready to go at L1, it would have been an easy thing (energetically at least) to rocket over and intercept it, Deep Impact style, in a blaze of glory.
With LSST set to start monitoring the skies, there should be an opportunity every decade or so to “get interstellar” by barely leaving home.
Categories: worlds Tags:
## Until Time Without End
‘Oumuamua’s encounter with the inner solar system is dying down on Twitter, yet still it bristles with consequence and the uneasiness of unanswered questions. Why no coma?
Occam’s razor is a dull instrument that points almost unerringly to the mundane (as opposed to pointing to interstellar probes). One thus draws several conclusions. (1) ‘Oumuamua’s aspect ratio is substantially less than 10:1. (2) Billions of years in the interstellar environment lead to the buildup of a tarry crust that resists temporary heating, and this process is enhanced for comet-like planetesimals that form in systems with supersolar C/O ratios. (3) Most stars have true-Neptune analogs.
The resulting prediction is that slightly tweaked ongoing surveys, and soon LSST, should start turning up interstellar asteroids and perhaps interstellar comets with some frequency. If another one is found in the near-term, it would be interesting to look at the optimal mission designs that could accomplish an opportunistic sample-return.
From ‘Oumuamua’s perspective, the close encounter with the Sun was a near-indescribable stroke of luck. To scale, the stars of the galactic disk are like grains of sand separated by miles and crawling through space at a few feet per year. The Galaxy is the archetypal collisionless fluid. Vaulting from ‘Oumuamua’s current encounter to its next connects the all too human interval of waking-up-at-3AM anxieties — the scale of days and months — to the frigid waste of a quadrillion years.
Why cold? When fusion has ended, dark matter annihilation and proton decay take over, and both (while uncertain) are certainly slow processes. Grand Unified Theories predict that proton decay should occur, but so far, there is no experimental evidence. The lower bounds on the proton half-life are ~10^34 years via the sluggishly competing processes of positron and muon decay.
If the proton were completely stable, the end states of stars present a curious state of affairs. Black holes of stellar mass, which are much more tightly bound than degenerate stars, will evaporate through the Hawking effect with a lifetime of “only” 10^66 years Although this time scale is aggressively long compared to the current 13.8-billion year age of the universe, it would be odd if black holes are ephemeral while white dwarfs and neutron stars are forever.
While jarring, this possible divergence of lifetimes is not exactly a matter of pressing concern. Two decades, ago, however, Fred Adams and I had priorities that were definitely skewed toward the really long term. Along with Manasse Mbonye and Malcom Perry, we looked into how quantum tunneling into black holes can erode white dwarfs. In Freeman Dyson’s 1979 article, Time Without End, it is pointed out that an otherwise stable white dwarf will spontaneously tunnel into a black hole on a time scale of order 10^10^76 (!) years. In our article, we argued that the whole star need not make the plunge at once, and that a 10^45 year half-life is a plausible value for black-hole induced proton decay. This has the added benefit of enabling a Hertzsprung-Russell diagram that traces stellar evolution to its absolute bitter end.
Categories: worlds Tags:
## Visitors
‘Oumuamua. Up close and alongside, in the vastness of interstellar space, its hurtling bulk imparts no sense of motion as it turns imperceptibly on its axis, blotting out the stars.
For a hundred years, the point-like Sun grew steadily brighter against its frigid airless horizons. First came light, then warmth, and finally searing illumination of the tarry reddish expanse, blistering sluggishly beneath a September Noon far more intense than any summer of Earth.
Oumuamua is departing the solar system as rapidly as it arrived, heading outward at a current rate of 2.5 million miles a day. Our tiny chance of sending a probe to catch it diminishes with each lagging tick of inactivity. Nonetheless, world-wide interest is mounting, in part as a consequence of two new articles reporting detailed observations. The first, by Jewitt et al. was posted to arXiv last week, while the second, by Meech et al. (which independently comes largely to the same overall conclusions), appeared in Nature earlier this week. Nature being Nature, the Meech et al. article was accompanied by a media push, spearheaded by an extraordinary piece of space art.
Maybe it’s press release fatigue from one “habitable” world after another — a monotony of warm suns glinting off imaginary oceans — that makes this image so arresting.
The observational facts remain stark and limited. Oumuamua’s double-peaked light curve suggests that it has a large aspect ratio, perhaps as high as 10:1. Assuming that it’s a poor reflector, it’s several hundred meters on its long axis. Its overall color is reddish. It has to have physical strength, or its 7-hour rotation period would be enough to overcome its negligible self-gravity and tear it apart. Most alarmingly, it shows no sign of a coma. At most, less than a sugar cube’s worth of cometary dust per second was emanating from it as it tore through the inner solar system. (As a matter of fact, ‘Oumuamua as observed is entirely consistent with Tintin’s rocket.)
For more on ‘Oumuamua, I have a blog post up at Scientific American.
Categories: worlds Tags:
## Interstellar Asteroids
November 5th, 2017 1 comment
This was no fruit of such worlds and suns as shine on the telescopes and photographic plates of our observatories. This was no breath from the skies whose motions and dimensions our astronomers measure or deem to vast to measure. It was just a colour out of space — a frightful messenger from unformed realms of infinity…
Aww, come off it.
Wild-eyed extravagances aside, A/2017 U1 — the asteroid-like visitor from interstellar space — is an extraordinary object. In traversing the gulfs, its next encounter with a star that is as close as last month’s encounter with the Sun likely won’t occur for another quadrillion years, and so the mere fact that it zipped through suggests that quite a few interstellar asteroids are out there. And this, in turn, has some remarkable consequences. A straightforward cross-section based estimate suggests that the galaxy contains of order a hundred billion earth masses of A/2017 U1-like planetesimals. Hot Jupiters, terrestrial planets, and super-Earths are all incapable of using gravity-assist to eject bodies out of their parent systems, leaving the strong hint that as-yet undetected Neptune-like planets must be extremely common.
In general, extrapolations from a sample size of one don’t have a good track record. Exhibit A would be our own Solar System — hot Jupiters were discovered at better than 100-sigma significance because solar-system expectations had been projected throughout the galaxy; proper planetary systems should have terrestrial bodies near 1 AU and gas giants at 10 AU.
The arrival of A/2017 U1 seems nicely timed to revival of the AAS’ new low-maintenance communication channel, the “Research Note“:
The purpose of the Research Notes is to provide a home for short submissions that are not suitable for publication as a journal article, but are likely to be interesting or useful to members of our community. Appropriate submissions would include brief summaries of work in progress, comments and clarifications, null results, and timely reports of observations (such as the spectrum of a supernova), as well as results that would not traditionally merit a full paper (e.g., the discovery of a single unremarkable exoplanet, a spectrum of a meteor, or contributions to the monitoring of variable sources).
I especially like the part about “single unremarkable exoplanets” being equivalenced to the “spectrum of a meteor”. In any event, Prof. K. Batygin and I have just submitted a research note that gives our take on the implications of A/2017 U1. Here’s a link to a draft of the note, which we’ll also post on the arXiv within the next several days.
Categories: worlds Tags:
## A/2017 U1
In the antique language of the space age, one might call it an interstellar “probe”, or perhaps a von Neumann machine. That’s not really what it is. It’s better described as a snarky, fusion-powered tangle of competing social networks, some of them still executing the hallowed fossil liturgies and intrigues of the mighty corporations from which they long since sprang.
It had no particular expectations for the fast-approaching star that was next on its ancient route. On the last flyby of this particular star, twenty-seven million years ago, the probe observed that the third planet was still robustly in the grip of a somewhat unusual, low-energy parasitic film that was efficiently exploiting the surface entropy gradient, and running undirected at a computational rate roughly equivalent to 10^34 bit operations per second.
Over the last few years, as the probe sifted the electromagnetic spectrum emanating from the third planet, it rippled with a hint of something that might best be thought of as a collective rolling of eyes. The third planet has recently stumbled into directed processes, and remarkably, foolishly, it is radiating manifestly unencrypted signals into space. This state of affairs caught a fraction of the probe’s interest, especially when it grasped that the planet’s computational efforts are increasingly focusing on concepts that the planet was calling “blockchain” and “proof-of-work through SHA-256 hashing”. This is just the sort of pursuit that the probe can relate to…
The above, of course, is unlikely to be true. In all likelihood, A/2017 U1 is a battered, inanimate 160-meter chunk of rock or metal, spawned in the dry collision of planetesimals orbiting an alien star, sometime within the past ten billion years. What’s remarkable, is that this interstellar visitor came within 0.25 AU of the Sun. As it departs into the depths of the Galaxy, it can expect to fly for roughly ten quadrillion years before it revisits another star with such proximity. It’s next rendesvous of comparable drama lies far into the depths of the Stelliferous era. In all likelihood, this will have it sailing past the frigid hulk of a white dwarf, warmed a few degrees above absolute zero by the flicker of proton decay.
Speaking of rendesvous, it must have occurred to quite few that the recent visit by A/2017 U1 is rather uncannily reminiscent of Arthur C. Clarke’s famous ’70s-era sci-fi page-turner. A Google trends search hints at a moderate uptick in interest over the past few days, which I expect will soon grow to undeniable statistical significance:
Closer to home, A/2017 U1 generates a very convenient route to completion of problem #1 on my Astronomy 395/575 homework assignment, which was set to the students just two days before A/2017 U1 was announced in the news:
Categories: worlds Tags:
## Sixty Hot Jupiters
June 21st, 2017 1 comment
There’s no denying the fundamentally alien climates on the hot Jupiters. It’s not clear, however, how hot Jupiters form, and it’s not clear why so many of them are badly distended. Moreover, it’s only vaguely clear what the weather patterns on one would look like up close. (One thing that is clear is that the flights would all be canceled).
Hot Jupiters are rare, but not overwhelmingly so. Something about the planet formation process causes about one in two hundred sun-like stars to end up stuck with one. In the original Kepler field, there are about 150,000 stars with light curves, and so about 750 hot Jupiters total are lurking in that population. Some of them, of course, are observable in transit, but as yet, most have gone undetected.
Yale graduate student Sarah Millholland has a new lead-authored paper out which uses supervised learning techniques to identify sixty high-probability non-transiting hot Jupiter candidates among the Kepler stars. The basic idea is that the phase curves of the planets, some of which have photometric amplitudes of several dozen parts per million or more, can be teased out of the noise and the stellar variability. After an involved process of sifting, the candidates (along with their supporting light curves) can be presented for a screen test:
Some members of the Kepler hot Jupiter class portrait will prove to be imposters (just like #5, #13, #29, and #30 in the nineteenth-century insect woodcut above). Doppler velocity observations — the equivalent of counting the number of legs on the arthropods — will provide a more definitive list. If you want to weigh in on the odds that these candidates are predominantly real, there’s a fresh Metaculus question that pools community input regarding the fidelity and prospects for confirmation of the members of the sample.
One might reasonably wonder, what’s the utility of yet another tray of bugs, smothered with ether and pinned to cards?
One superb benefit from gathering sixty non-transiting hot Jupiters that are detectable in the optical region is that trends in the planets’ surface temperature variations — that is, the weather maps — can be elucidated with a far larger sample than was previously available. Sarah’s candidates support an interesting trend in which cooler planets (relatively speaking, of course) are posited to have reflective clouds to the west of the substellar point, whereas hotter hot Jupiters are consistently advecting the most strongly optically radiating gas downwind from high Noon.
For detailed information on the individual candidates, visit Sarah’s website, and if you are at the Kepler Science Conference, she’ll present the details during Friday’s session.
Categories: worlds Tags:
## Recurrence
Most oklo.org readers know the story line of Fred Hoyle’s celebrated 1957 science fiction novel, The Black Cloud. An opaque, self-gravitating mass of gas and dust settles into the solar system, blots out the sun, and wreaks havoc on the biosphere. It gradually becomes clear that the cloud itself is sentient. Scientists mount an attempt to communicate. A corpus of basic scientific and mathematical principles is read out loud in English, voice-recorded, and transmitted by radio to the cloud.
The policy was successful, too successful. Within two days the first intelligible reply was received. It read:
“Message received. Information slight. Send more.”
For the next week almost everyone was kept busy reading from suitably chosen books. The readings were recorded and then transmitted. But always, there came short replies demanding more information, and still more information…
Sixty years later, communicating interstellar clouds are still in the realm of fiction, but virtualized machines networked in the cloud are increasingly dictating the course of actions in the real world.
In Hoyle’s novel, the initial interactions with the Black Cloud are quite reminiscent of a machine learning task. The cloud acts as a neural network. Employing the information uploaded in the training set, it learns to respond to an input vector — a query as a sequence of symbols — with a sensible output vector. Throughout the story, however, there’s an implicit assumption that the Cloud is self-conscious and aware; nowhere is it intimated that that the processes within the Cloud might simply be an algorithm managing to pass an extension of the Turing Test. On the basis of the clear quality of its output vectors, the Cloud’s intelligence is taken as self-evident.
The statistics-based regimes of machine learning are on a seemingly unstoppable roll. A few years ago, I noticed that Flickr became oddly proficient at captioning photographs. Under the hood, an ImageNet classification with convolutional neural networks (or the like) was suddenly focused, with untiring intent, on scenes blanketing the globe. Human mastery of the ancient game of Go has been relinquished. Last week, I was startled to read Andrej Karpathy’s exposition of the unreasonable effectiveness of recurrent neural networks.
By drawing from a large mass of example text, a recurrent neural network (RNN) character-level language model learns to generate new text one character at a time. Each new letter, space, or punctuation mark draws its appearance from everything that has come before it in the sequence, intimately informed by what the algorithm has absorbed from its fund of information. As to how it really works, I’ll admit (as well) to feeling overwhelmed, to not quite knowing where to begin. This mind-numbingly literal tutorial on backpropagation is of some help. And taking a quantum leap forward, Justin Johnson has written a character-level language model, torch-rnn, which is well-documented and available on github.
In Karpathy’s post, RNNs are set to work generating text that amuses but which nonetheless seems reassuringly safely removed from any real utility. A Paul Graham generator willingly dispenses Silicon Valley “thought leader” style bon mots concerning startups and entrepreneurship. All of Shakespeare is fed into the network and dialogue emerges in an unending stream that’s — at least at the phrase-to-phrase level — unkindly indistinguishable from the real thing.
I’m very confident that it would be a whole lot more enjoyable to talk to Oscar Wilde than to William Shakespeare. As true A.I. emerges, it may do so in a cloud of aphorisms, of which Wilde was the undisputed master, “I can resist everything except temptation…”
Wilde employed a technique for writing The Picture of Dorian Gray in which he first generated piquant observations, witty remarks and descriptive passages, and then assembled the plot around them. This ground-up compositional style seems somehow confluent with the processes — the magic — that occurs in an RNN.
The uncompressed plain text UTF8 version of Dorian Gray is a 433701 character sequence. This comprises a fairly small training set. It needs a supplement. The obvious choice to append to the corpus is A rebours — Against Nature, Joris-Karl Huysman’s 1884 classic of decadent literature.
Even more than Wilde’s text, A rebours is written as a series of almost disconnected thumbnail sketches, containing extensive, minutely inlaid descriptive passages. The overall plot fades largely into the background, and is described, fittingly, in one of the most memorable passages from Dorian Gray.
It was a novel without a plot and with only one character, being, indeed, simply a psychological study of a certain young Parisian who spent his life trying to realize in the nineteenth century all the passions and modes of thought that belonged to every century except his own, and to sum up, as it were, in himself the various moods through which the world-spirit had ever passed, loving for their mere artificiality those renunciations that men have unwisely called virtue, as much as those natural rebellions that wise men still call sin. The style in which it was written was that curious jewelled style, vivid and obscure at once, full of argot and of archaisms, of technical expressions and of elaborate paraphrases, that characterizes the work of some of the finest artists of the French school of Symbolistes. There were in it metaphors as monstrous as orchids and as subtle in colour.
A rebours attached to Dorian Gray constitutes a 793587 character sequence, and after some experimentation with torch-rnn, I settled on the following invocation to train a multilayer LSTM:
MacBook-Pro:torch-rnn Greg$th train.lua -gpu -1 -max_epochs 100 -batch_size 1 -seq_length 50 -rnn_size 256 -input_h5 data/dorianGray.h5 -input_json data/dorianGray.json My laptop lacks an Nvidia graphics card, so the task fell to its 2.2 GHz Intel Core i7. The code ran for many hours. Lying in bed at night in the quiet, dark house, I could hear the fan straining to dissipate the heat from the processor. What would it write? This morning, I sat down and sampled the results. The neural network that emerged from the laptop’s all-nighter generates Wilde-Huysmans-like text assembled one character at a time: MacBook-Pro-2:torch-rnn Greg$ th sample.lua -gpu -1 -temperature 0.5 -checkpoint cv/checkpoint_1206000.t7 -length 5000 > output.txt
I opened the output, and looked over the first lines. It is immediately clear that a 2015-era laptop staying on all night running downloaded github code can offer no competition — in any sense — to either Mr. Wilde or Mr. Huysmans. An abject failure of the Turing Test, a veritable litany of nonsense:
After the charm of the thread of colors, the nineteenth close to the man and passions and cold with the lad's heart in a moment, whose scandal had been left by the park, or a sea commonplace plates of the blood of affectable through the club when her presence and the painter, and the certain sensation of the capital and whose pure was a beasts of his own body, the screen was gradually closed up the titles of the black cassion of the theatre, as though the conservatory of the past and carry, and showing to me the half-clide of which it was so as the whole thing that he would not help herself. I don't know what will never talk about some absorb at his hands.
But we are not more than about the vice. He was the cover of his hands. "You were in his brain." "I was true," said the painter was strangled over to us. It is not been blue chapter dreadfully confesses in spite of the table, with the desert of his hands in her vinations, and he mean about the screen enthralled the lamp and red books and causes that he was afraid that he could see the odious experience. It was a perfect streating top of pain. "What is that, I am sorry I shall have something to me that you are not the morning, Mr. Gray," answered the lad, and that the possession of colorings, which were the centre of the great secrets of an elaborate curtain. You cannot believe that I was thinking of the moon. He was to be said that the world is the restive of the book to the charm of a matter of an approvingian through a thousand serviced it again. The personality of the senses by the servants were into the shadow of the next work to enter, and he had revealed to the conservatory for the morning with his wife had been an extraordinary rooms that was always from the studio in his study with a strange full of jars, and stood between them, or thought who had endured to know what it is. "Ah, Mr. Gray?"
"I am a consolation to be able to give me back to the threat me."
But such demands are excessive. The text is readable English, convened in a headlong rush by a program that could just as easily have been synthesizing grant proposals or algebraic topology. Torch-rnn contains no grammar rules, no dictionaries, no guides to syntax. And it really does learn over time. Looking at the early checkpoint snapshots of the network, during epochs when words and spaces are forming, before any sense of context has emerged, one finds only vaguely English-like streams of gibberish:
pasticite his it him. "It him to his was paintered the cingring the spure, and then the sticice him come and had to him for of a was to stating to and mome am him himsed at he some his him, and dist him him in on of his lime in stainting staint of his listed."`
Perhaps the best comparison of Torch-rnn’s current laptop-powered overnight-effort capabilities are to William S. Burroughs’ cut-up novels — The Soft Machine, The Ticket that Exploded — where one sees disjoint masses of text full of randomized allusions, but where an occasional phrase sparkles like a diamond in matrix, “…a vast mineral consciousness near absolute zero thinking in slow formations of crystal…”
In looking over a few thousand characters of text, generated from checkpoint 1,206,000 at temperature T=0.61, one finds glimmers of recurrent, half-emerged truths,
You are sure to be a fragrant friend, a soul for the emotions of silver men.
Categories: Electra Tags:
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# Thread: 2 Word Problems, Alg 2, Please Help( Reply by 10am plz)
1. ## 2 Word Problems, Alg 2, Please Help( Reply by 10pm plz)
1) Ashley won $5000 in a contest. She invested some of the money at 5% simple interest and$400 less than twice that amount at 6.5%. In one year, she earned $298 in interest. How much did she invest at each rate? Formula = use table with Principle, Rate and Interest. Problem : Please assist with setting up the problem. I've tried but failed to solve it. The Answer in the book should be =$1800 at 5%; %3200 at 6.5%
2) Vincente and Ricarda have invested $27,000 in bonds paying 7%. How much additional money should they invest in a certificate of deposit paying 4% simple interest so that the total return on the two investment will be 6%? Formula = the same. Problem : the same. Answer :$13,500
2. Originally Posted by chamroeneng
1) Ashley won $5000 in a contest. She invested some of the money at 5% simple interest and$400 less than twice that amount at 6.5%. In one year, she earned $298 in interest. How much did she invest at each rate? Formula = use table with Principle, Rate and Interest. Problem : Please assist with setting up the problem. I've tried but failed to solve it. The Answer in the book should be =$1800 at 5%; %3200 at 6.5%
Let x be the amount invested at 5% interest. Then the other amount invested is (2x - 400)
the interest gained at 5% is 0.05x using the simple interest formula.
similarly, the interest gained at 6.5% is 0.065(2x - 400)
the total interest gained is 298. thus:
0.05x + 0.065(2x - 400) = 298
solve for x and you can complete the problem
3. Originally Posted by chamroeneng
2) Vincente and Ricarda have invested $27,000 in bonds paying 7%. How much additional money should they invest in a certificate of deposit paying 4% simple interest so that the total return on the two investment will be 6%? Formula = the same. Problem : the same. Answer :$13,500
Let x be the additional amount invested at 4%
So in total, we invested (27000 + x).
now, the interest gained by the 27000 is 0.07*27000
the interest gained by x is 0.04x
we want the return to be 6% of the total amount invested. meaning, we want the interest to be 0.06(27000 + x). thus:
0.07*27000 + 0.04x = 0.06(27000 + x)
now solve for x and that's your answer
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# What is Attention, Self Attention, Multi-Head Attention?
###### February 17, 2019
Consider we have 5 cars. Each car has a set of features e.g. safety, capacity, acceleration, mpg etc. Each car also has a set of associated values e.g. financial, social, resell, etc. Now, consider that we have a new car and we would like to determine its set of values. An obvious approach is to say that if the new car has features similar to another car, its value will also be similar to that car.
In this example, if we determine that the new car is $70\%$ similar to car 1, $15\%$ similar to car 2, $10\%$ similar to car 3, $3\%$ similar to car 4 and $2\%$ similar to car 5, then the value of the new car,
\begin{align} V = 0.70\ V_1 + 0.15\ V_2 + 0.10\ V_3 + 0.03\ V_4 + 0.02\ V_5 \end{align}
Note that, $0.70 + 0.15 + 0.10 + 0.03 + 0.02 = 1$. This process of determining the value of the new car is called attention. It involves finding the similarity between the new and existing cars. The weighted values of the existing cars are then used to determine the value of the new car.
A vector which represents features is called the key vector, $k$. A vector which represents values is called the value vector, $v$. A vector whose value is to be determined is called the query vector, $q$. The function used to determine similarity between a query and key vector is called the attention function or the scoring function. The scoring function returns a real valued scalar. The scores are normalized, typically using softmax, such that sum of scores is equal to 1. The final value is equal to the weighted sum of the value vectors.
## Scoring Functions
There are many ways to define a scoring function [1,4]. Let's assume $k$, $v$ and $q$ are column vectors.
#### Dot product
This is the simplest function and has no learnable parameters. However, it requires that $q$ and $k$ are of the same size.
\begin{align} \textit{score}(q, k) = q^Tk \end{align}
#### Scaled dot product
The dot product increases as the dimensions get larger. A fix is to introduce a scaling factor, and was proposed in Transformers [3]. This function also has no learnable parameters. The equation below is different from the equation in [3] because the authors assume $k$, $v$ and $q$ to be row vectors.
\begin{align} \textit{score}(q, k) = \frac{q^Tk}{\textit{scaling factor}} \end{align}
#### Bilinear
In this formulation, $W$ is a learnt matrix. Also the dimensions of $q$ and $k$ can be different. Note that the dot product and scaled dot products functions are also bilinear.
\begin{align} \textit{score}(q, k) = q^T W k \end{align}
#### MLP
In this formulation, we concatenate the query and key vectors and then apply a linear transformation.
\begin{align} \textit{score}(q, k) &= w^T tanh(W_1q + W_2k) \\ &= w^T tanh(W[q;k]) \end{align}
where [$q$ ; $k$] implies concatenation. Here $w$ and $W$ are learnt parameters. Also, the sizes of $q$ and $k$ can be different. This function was used in [2,4].
## Attention & Self Attention
When we want to determine the score of multiple key and query vectors at once, we can replace the key and query vectors with the key and query matrices, $K$ and $Q$ respectively, in the above equations. In general, given $Q$, $K$ and $V$, the value of the corresponding query vectors is given by,
\begin{align} \textit{Attention}(Q,K,V) = V.\textit{softmax}\ (\textit{score}(Q, K)) \end{align}
Self attention is nothing but $Q = K = V$ i.e. we compute a new value for each vector by comparing it with all vectors (including itself).
In Transformers [3], the authors first apply a linear transformation to the input matrices $Q$, $K$ and $V$, and then perform attention i.e. they compute
\begin{align} \textit{Attention}(W^QQ,W^KK,W^VV) = W^VV\textit{softmax}\ (\textit{score}(W^QQ, W^KK)) \end{align}
where, $W^V$, $W^Q$ and $W^K$ are learnt parameters. In multi-head attention, say with #heads = 4, the authors apply a linear transformation to the matrices and perform attention 4 times as follows.
\begin{align} \textit{head}_1 &= \textit{Attention}(W^Q_1Q, W^K_1, W^V_1V) \\ \textit{head}_2 &= \textit{Attention}(W^Q_2Q, W^K_2, W^V_2V) \\ \textit{head}_3 &= \textit{Attention}(W^Q_3Q, W^K_3, W^V_3V) \\ \textit{head}_4 &= \textit{Attention}(W^Q_4Q, W^K_4, W^V_4V) \\ \end{align}
where, $W^V_i$, $W^Q_i$ and $W^K_i$ are learnt parameters. The 4 heads are concatenated and then linearly transformed again, as follows.
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A. Hall of Fame
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Thalia is a Legendary Grandmaster in chess. She has $n$ trophies in a line numbered from $1$ to $n$ (from left to right) and a lamp standing next to each of them (the lamps are numbered as the trophies).
A lamp can be directed either to the left or to the right, and it illuminates all trophies in that direction (but not the one it is next to). More formally, Thalia has a string $s$ consisting only of characters 'L' and 'R' which represents the lamps' current directions. The lamp $i$ illuminates:
• trophies $1,2,\ldots, i-1$ if $s_i$ is 'L';
• trophies $i+1,i+2,\ldots, n$ if $s_i$ is 'R'.
She can perform the following operation at most once:
• Choose an index $i$ ($1 \leq i < n$);
• Swap the lamps $i$ and $i+1$ (without changing their directions). That is, swap $s_i$ with $s_{i+1}$.
Thalia asked you to illuminate all her trophies (make each trophy illuminated by at least one lamp), or to tell her that it is impossible to do so. If it is possible, you can choose to perform an operation or to do nothing. Notice that lamps cannot change direction, it is only allowed to swap adjacent ones.
Input
Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \leq t \leq 10\,000$). The description of the test cases follows.
The first line of each test case contains a positive integer $n$ ($2 \leq n \leq 100\,000$) — the number of trophies.
The second line of each test case contains a string $s$ of length $n$ consisting only of characters 'L' and 'R' — the $i$-th character describes the direction of the $i$-th lamp.
It is guaranteed that the sum of $n$ over all test cases does not exceed $100\,000$.
Output
For each test case print $-1$ if it is impossible to illuminate all trophies by performing one operation (or doing nothing). Otherwise, print $0$ if you choose not to perform the operation (i.e., the trophies are illuminated by the initial positioning of the lamps), or an index $i$ ($1 \leq i < n$) if you choose to swap lamps $i$ and $i+1$.
If there are multiple answers, print any.
Example
Input
6
2
LL
2
LR
2
RL
2
RR
7
LLRLLLR
7
RRLRRRL
Output
-1
1
0
-1
3
6
Note
In the first example, it is possible to swap lamps $1$ and $2$, or do nothing. In any case, the string "LL" is obtained. Not all trophies are illuminated since trophy $2$ is not illuminated by any lamp — lamp $1$ illuminates nothing and lamp $2$ illuminates only the trophy $1$.
In the second example, it is necessary to swap lamps $1$ and $2$. The string becomes "RL". Trophy $1$ is illuminated by lamp $2$ and trophy $2$ is illuminated by lamp $1$, hence it is possible to illuminate all trophies.
In the third example, all trophies are initially illuminated — hence, not performing any operation is a valid solution.
In the last two examples performing swaps is not necessary as all trophies are illuminated initially. But, the presented solutions are also valid.
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# Use the geometric summation formula to prove the orthogonality property (4.22). Pro- vide a
1. Use the geometric summation formula to prove the orthogonality property (4.22). Pro- vide a geometric interpretation by treating the samples of s k [ n ], k = 0, 1, ... , N − 1 as the components of an N -dimensional vector.
2. Write a MATLAB program to generate and plot the signals shown in Figure 4.12. Experiment with different values of m to appreciate the nature of Gibbs’ phenomenon. Note: You can zoom on the discontinuities to see more clearly the behavior of oscillations.
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# Determine whether the geometric series is convergent or divergent. If it is conv
Determine whether the geometric series is convergent or divergent. If it is convergent, find its sum.
$10-2+0.4-0.008+\dots$
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Geometric series:
$\sum _{n=0}^{\mathrm{\infty }}a\cdot {r}^{n}=\frac{a}{1-r}$
Where r and a are constants
If |r|<1, then the series converges to $\frac{a}{1-r}$
$10-2+0.4-0.08\dots$
$10+10\cdot \left(-\frac{1}{5}\right)+10\cdot {\left(-\frac{1}{5}\right)}^{2}+10\cdot {\left(-\frac{1}{5}\right)}^{3}\dots$
$\sum _{n=0}^{\mathrm{\infty }}10\cdot {\left(-\frac{1}{5}\right)}^{n}$
This is a geometric series with common ration $r=-\frac{1}{5}$ and Initial Term $a=10$
Since $|r|=\frac{1}{5}<1$, the given geometric series converges.
Sum of the geometric series is
$S=\frac{a}{1-r}=\frac{10}{1-\left(-\frac{1}{5}\right)}=\frac{10}{1+\frac{1}{5}}=\frac{10}{\frac{6}{5}}=\frac{50}{6}=\frac{25}{3}$
The series converges to $\frac{25}{3}$
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# could someone please check this circuit?
#### efm
Joined Sep 14, 2006
14
hello,
i'm making a data logger that consists of 3 LM35s, one LM324 OpAmp, one ADC0808, one
AT89S51, and a 24LC256 serial eeprom. i've designed the circuit by myself with references from
the net, and i want to know whether the design is correct or not. and i'm open for suggestions of
course.
well, the goal of the circuit is to receive data from the three sensors, and then save it into the
24LC256 memory, and then the computer will download the data through parallel port.
faizal
#### Attachments
• 33.1 KB Views: 60
#### Dave
Joined Nov 17, 2003
6,970
Can you improve the clarity? It is very difficult to read anything on the image.
Dave
#### efm
Joined Sep 14, 2006
14
hello dave,
well, here is the clearer picture...i zipped it,
i hope you would see it and tell me whether the design is correct or not,
faizal
#### Attachments
• 26.8 KB Views: 57
#### hgmjr
Joined Jan 28, 2005
9,029
Greetings efm,
Your circuit looks as though you have everything covered.
You may want to look at replacing the LM324 opamps with a rail-to-rail opamp. This will give a little better linearity on your temp readings at the high end of the temperature range.
I noticed that you chose not to implement a means of connecting to the AT89S51 for the purpose of In-System Programming. I take it that you are planning to socket the AT89S51 and program it off-board.
hgmjr
#### hgmjr
Joined Jan 28, 2005
9,029
I noticed that you have made no provisions for measuring temperatures below 0 degrees Celsius. Is that intentional?
hgmjr
#### hgmjr
Joined Jan 28, 2005
9,029
You may want to consider assigning the EOC signal from the ADC0808 to one of the external interrupt pins on the AT89S51. That way you will not need to use polling to detect when the A-to-D conversion has been completed.
hgmjr
#### efm
Joined Sep 14, 2006
14
hello
thanks for the reply and for the suggestions!!
so, what rail-to-rail opamp do you suggest? i found OPA4344P, is it ok?
and, about the in system programming thing...you're right, i'm planning to program it
off-board, but i think your idea is good...
if i want to add the isp to the circuitry, can i just wire the mosi, miso, and sck (p1.5,
p1.6, p1.7) to the parallel port? or do i have to change the whole circuitry?
and, is it ok if i do the in system programming while the board is powered?
btw, thanks for the suggestion for wiring the eoc to interrupt pin. i think i will
change the design, and i will attach it in my next message.
and...actually, i want to measure below 0 celcius, but i've bought that lm35d. i didn't
know that it only measures 0 to 100 celcius.
well, thanks for all suggestions,
faizal
#### hgmjr
Joined Jan 28, 2005
9,029
efm,
The OPA4344 looks like it should work fine.
As for the ISP hookup, you should be able to consult ATMEL's AT89S51 datasheet for implementations instructions. You will need to take into consideration the requirements of your existing programmer.
The LM35D datasheet shows how it can be connected for -55 to +150 celsius temp range even while powered by a single supply. You will need to configure your op-amps as differential amplifiers instead of the single-ended scheme you are currently using.
hgmjr
#### efm
Joined Sep 14, 2006
14
hello again,
well, according to my previous circuit layout, i made the pcb layout,
umm...would you care too see if it's correct or not?
your help would be greatly appreciated
btw, if i want to connect the board with parallel port, do i have to connect
parallel port's ground (pin 21-25) to the board's ground?
thank you,
faizal
#### Attachments
• 19 KB Views: 27
• 23.9 KB Views: 26
#### efm
Joined Sep 14, 2006
14
hello,
according to my previous schematic, i have written a code for the microcontroller. but i haven't uploaded
it yet, i just scare that it's wrong and will eventually break the microcontroller / memory
here is the code, could someone please tell me whether it will work or not
Rich (BB code):
\$MOD51
DSEG
ORG 30H
BYTSTR:
DS 20H
; ------------------------------------------------
; CONSTANTS
; ------------------------------------------------
; EEPROM EQUATES
WTCMD EQU 10100000B
CONV_PORT EQU P2
; ------------------------------------------------
; START OF PROGRAM
; ------------------------------------------------
CSEG
ORG 0
JMP START
START:
MOV P1, #0FFH
CALL MAIN_ROUTINE
JMP START
; ------------------------------------------------
; COMPARISON, AND SAVE TO EEPROM
;
; Note: Channel Selection
; 0 0 0 1
; 0 0 1 2
; 0 1 0 3
; ------------------------------------------------
MAIN_ROUTINE:
CALL START_CONVERSION
MOV R2, A
CALL START_CONVERSION
MOV R3, A
CALL START_CONVERSION
MOV R4, A
; COMPARE R2, R3, AND R4
CALL COMPARE
; DELAY FOR ONE MINUTE
; LOOP
JMP MAIN_ROUTINE
; ------------------------------------------------
; RESULT COMPARISON
; ------------------------------------------------
COMPARE:
; COMPARES R2 WITH R3
MOV A, R2
SUBB A, R3
JNZ COMPARE_2 ; If R2-R3 <> 0, they're not equal, jump to COMPARE_2
MOV R1, A ; R2 and R3 is equal, so save R2 in R1...
COMPARE_2:
; COMPARE R2 WITH R4
MOV A, R2
SUBB A, R4
JNZ COMPARE_3 ; If R2-R4 <> 0, they're not equal, jump to COMPARE_3
MOV R1, A ; R2 and R4 is equal, so save R2 in R1...
COMPARE_3:
; COMPARE R3 WITH R4
MOV A, R3
SUBB A, R4
JNZ NO_VOTE ; If R3-R4 <> 0, they're not equal, jump to NO_VOTE
MOV R1, A ; R3 and R4 is equal, so save R3 in R1...
NO_VOTE:
; NO TWO OUT OF THREE ARE EQUAL
MOV R1, #0 ; If no 2 out of three are equal, set R1 as 0
SAVE_R1:
; SAVE R1 TO EEPROM
CALL TESTWR
RET
; ------------------------------------------------
; ------------------------------------------------
START_CONVERSION:
MOV CONV_PORT, #0FFH
START_CLK:
MOV A, CONV_PORT
RET
; ------------------------------------------------
; BELOW IS THE ROUTINE TO WRITE ONE BYTE TO EEPROM
; ------------------------------------------------
TESTWR:
CALL BYTEW
RET
BYTEW:
MOV A, #WTCMD
CALL OUTS
MOV A, R4
CALL OUT
MOV A, R1
CALL OUT
CALL STOP
RET
OUTS:
MOV R2, #8
NOP
NOP
NOP
NOP
NOP
NOP
NOP
NOP
NOP
NOP
OTSLP:
RLC A
JNC BITLS
JMP OTSL1
BITLS:
OTSL1:
NOP
NOP
NOP
NOP
NOP
DJNZ R2, OTSLP
NOP
NOP
NOP
NOP
NOP
NOP
NOP
NOP
RET
OUT:
MOV R2, #8
OTLP:
RLC A
JNC BITL
JMP OTL1
BITL:
OTL1:
NOP
NOP
NOP
NOP
NOP
DJNZ R2, OTLP
NOP
NOP
NOP
NOP
NOP
NOP
NOP
NOP
RET
STOP:
NOP
NOP
NOP
NOP
NOP
NOP
NOP
NOP
NOP
NOP
RET
END
thank you
#### Dave
Joined Nov 17, 2003
6,970
Your best bet would be to download the programme to the microcontroller, it is very unlikely that you will damage anything as long as you follow the correct download procedure. It would be much easier for you to decifer if the code works by testing on-line, rather than someone reading through the source code here on the forums.
Hope it goes well.
Dave
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Here are my puzzles about closed elements: Let \$$f: A\to B\$$ and \$$g: B\to A\$$ be the left and right adjoints in a Galois connection between posets \$$A\$$ and \$$B\$$, so that \$$f(a)\leq b \iff a \leq g(b)\$$. Given \$$a\in A\$$, we'll write \$$g(f(a))\$$ as \$$[a]\$$ for short, and \$$[b]=f(g(b))\$$ for each \$$b\in B\$$.
**Puzzle OB1 (Kenobi):** Show that for all \$$a\in A\$$ and \$$b\in B\$$, we have \$$a\leq [a]\$$ and \$$b\geq [b]\$$.
We'll say that an element \$$x\$$ (of \$$A\$$ or \$$B\$$) is _closed_ if \$$x = [x]\$$.
**Puzzle OB2:** Show that for all \$$a\in A\$$ and \$$b\in B\$$, the elements \$$f(a)\in B\$$ and \$$g(b)\in A\$$ are both closed. (Hint: You can deduce both \$$f(a)\leq f(g(f(a)))\$$ and \$$f(g(f(a)))\leq f(a)\$$ from OB1.) Deduce that \$$[a]\$$ and \$$[b]\$$ are themselves closed.
**Puzzle OB3:** Show that if \$$a\leq a'\$$ in \$$A\$$ then \$$[a]\leq [a']\$$. Deduce that if \$$a\leq a'\$$ and \$$a'\$$ is closed, then \$$[a]\leq a'\$$. We say that \$$[a]\$$ is the _smallest_ closed element that is at least as big as \$$a\$$. Show that for any \$$b\in B\$$, \$$[b]\$$ is the _biggest_ closed element that is at least as _small_ as \$$b\$$.
**Puzzle OB4:** Let's write \$$A_\text{closed}\$$ and \$$B_\text{closed}\$$ for the sets of closed elements in \$$A\$$ and \$$B\$$. Show that \$$f\$$ and \$$g\$$ restrict to functions \$$A_\text{closed}\leftrightarrow B_\text{closed}\$$, and that on these subsets they are inverses of each other! Deduce that \$$A_\text{closed}\$$ and \$$B_\text{closed}\$$ are isomorphic posets.
(John, this all ended up being longer than I meant it to be. I hope you don't mind me plopping it all in here!)
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# Math Help - kindly explain this step of real and imaginary parts
1. ## kindly explain this step of real and imaginary parts
how he has written second term. kindly explain in steps
thanks
2. Denominator:
$(1-ae^{-jw})(1-ae^{jw})=1-ae^{-jw}-ae^{jw}+a^2=$
$1+a^2-a(e^{-jw}+e^{jw})=1+a^2-2a\cos w\in \mathbb{R}$
Numerator:
$1-ae^{jw}=1-a(\cos w+j\sin w)=(1-a\cos w)+(-a\sin w)j$
Now, you can conclude.
3. thanks
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# Hypothesis Tests (Non-Parametric)¶
## Unit 10, Lecture 2¶
Numerical Methods and Statistics
Langley: Pages 137-189, 199-211, 230-245
## Goals:¶
1. Learn the difference between parametric statistics vs nonparametric statistics
2. Be able to apply the following non-parametric hypothesis tests: Wilcoxon sum of ranks, Wilcoxon signed ranks, Spearman correlaction test
3. Understand how to apply parametric hypothesis tests with discrete values
In [1]:
import random
import numpy as np
import matplotlib.pyplot as plt
from math import sqrt, pi, erf
import scipy.stats as ss
# Detour: Parametric vs. Nonparametric Stastics¶
Parametric Statistics: What we've seen before, where we do statistics by assuming the data follows some underlying probability distribution (like normal distribution). Sometimes this is a good assumpetion because of the CLT.
Nonparametric Statistics: We do statistics without assuming an equation form for the underlying probability distribution. Typically harder to prove significance here because we have less information due to not assuming probability distribution.
Nonparametric statistics are secret and not widely taught because people belive they are challenging to understand. This is true, but I don't think undergraduates completely understand probability measure spaces but it doesn't stop us from using them.
From here onwards, most tests will not assume normality and are nonparametric. You won't find these tests in most traditional statistics textbooks
To do nonparametric statistics, one of the underlying principles is converting measurements into rankings.
In [2]:
d = np.random.rand(10)
print (d)
print (ss.rankdata(d))
[0.70148411 0.96522843 0.41747704 0.8159568 0.02398743 0.14613944
0.01542925 0.75959105 0.63014733 0.52986675]
[ 7. 10. 4. 9. 2. 3. 1. 8. 6. 5.]
# Wilcoxon's Sum of Ranks Test¶
Data Type: Ranks
Compares: Two sets of measurements.
Null Hypothesis: The two sets of measurements are from the same distribution
Conditions: Unmatched measurements. Unmatched the measurements aren't in pairs and you don't necessarily have the same number
Related Test 1: Wilcoxon's Signed Ranks Test for matched data measuring one thing (i.e, temperature)
Related Test 2: Spearman's Correlation Test for matched data measuring two things (i.e., temperature and pressure)
Python: scipy.stats.ranksums
Hints: Make sure all data is in the same units!
# Example: Were HW 1 and HW 2 from the same distribution?¶
In [3]:
import pandas as pd
#get some info:
data.info()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 65 entries, 0 to 64
Data columns (total 12 columns):
R1 65 non-null float64
R2 65 non-null float64
R3 65 non-null int64
R5 65 non-null int64
R6 65 non-null int64
HW1 65 non-null float64
HW2 65 non-null float64
HW3 65 non-null float64
HW4 65 non-null float64
HW5 65 non-null float64
HW6 65 non-null float64
Midterm 65 non-null float64
dtypes: float64(9), int64(3)
memory usage: 6.2 KB
I'm going to standardize the homework so that they are all out of 100%. The first row contains perfect scores on each.
In [4]:
data /= data.iloc[0,:]
data *= 100
data.mean(axis=0)
Out[4]:
R1 91.794872
R2 89.807692
R3 84.615385
R5 93.846154
R6 81.538462
HW1 82.939560
HW2 80.118343
HW3 74.230769
HW4 83.367730
HW5 73.282648
HW6 75.182595
Midterm 76.130769
dtype: float64
In [5]:
plt.title('HW1')
plt.hist(data.HW1)
plt.show()
In [6]:
plt.title('HW2')
plt.hist(data.HW2)
plt.show()
In [7]:
ss.ranksums(data['HW1'], data['HW2'])
Out[7]:
RanksumsResult(statistic=0.39578610596263436, pvalue=0.6922628274842608)
The $p$-value is 0.70, so we cannot rule out the null hypothesis that they are from the same distribution. What about a more recent difficult homework?
In [8]:
ss.ranksums(data['HW1'], data['HW5'])
Out[8]:
RanksumsResult(statistic=2.567953381628151, pvalue=0.010230091290609646)
So HW 1 and HW 5 were significantly different.
# Wilcoxon's Signed Rank Test¶
Data Type: Ranks
Compares: Two sets of measurements
Null Hypothesis: The two sets of measurements are from the same distribution
Conditions: Measurements are matched. Matched means the data comes in tuples/pairs. More than 6 samples, better to have more than 20.
Related Test 1: Wilcoxon's Sum of Ranks Test for unmatched data measuring one thing (i.e, temperature)
Related Test 2: Spearman's Correlation Test for matched data measuring two things (i.e., temperature and pressure)
Python: scipy.stats.wilcoxon
Hints: Make sure all data is in the same units!
Since the same people are doing the HW each week, a more accuracte comparison would be to used the Signed Rank Test.
In [9]:
ss.wilcoxon(data.HW1, data.HW2)
Out[9]:
WilcoxonResult(statistic=634.5, pvalue=0.08703592427905478)
In [10]:
ss.wilcoxon(data.HW1, data.HW5)
Out[10]:
WilcoxonResult(statistic=434.0, pvalue=0.000398645278025264)
Notice that the p-values are lower relative to the unmatched sum of ranks test, meaning have paired data allows us to be more certain in our conclusions.
# Spearman's Correlation Test¶
Data Type: Ranks
Compares: Two sets of measurements
Null Hypothesis: The two sets of measurements are uncorrelated
Conditions: Measurements are matched. Matched means the data comes in tuples/pairs. The measurements are of different things
Related Test 1: Wilcoxon's Sum of Ranks Test for unmatched data measuring one thing (i.e, temperature)
Related Test 2: Wilcoxon's Signed Ranks Test for matched data measuring one thing (i.e, temperature)
Python: scipy.stats.spearmanr
# Example: Is doing well on homework correlated with doing well on the midterm?¶
First, let's get the average grade on the homeworks. The spreadsheet has 6 homeworks
In [11]:
#build a list of all the HW indices
index = []
for i in range(1,7):
index.append('HW{}'.format(i))
#access those homeworks and then take the mean along the columns
hw_means = data[index].mean(axis=1)
In [12]:
plt.plot(hw_means, data.Midterm, 'o')
plt.show()
In [13]:
ss.spearmanr(hw_means, data.Midterm)
Out[13]:
SpearmanrResult(correlation=0.3466347056828196, pvalue=0.004673660991129944)
Remarkable!
In [14]:
np.corrcoef(hw_means, data.Midterm)
Out[14]:
array([[1. , 0.60007556],
[0.60007556, 1. ]])
# Poisson's Count Test¶
Data Type: Count
Compares: Count vs a poisson distributed population
Null Hypothesis: The number of observations (count) came from the known population
Conditions: Less than 40 samples (for computational simplicity)
Related Test 1: $zI$ test, for more than 40 samples
Python: Construct an interval and integrate using scipy.stats.poisson.cdf(x, mu=...)
Hints: Your interval should contain your value and all other extreme values. The interval should go up to infinity or down to 0 depending on if it's higher or lower than the expected value.
# Example: Hurricanes per Year¶
The number of hurricanes in 2005 was 15. The historic average is 6.3. Is this number signficantly different?
We will construct an interval containing all values as extreme as ours. We don't consider a low number of hurricanes to be extreme in this example. Remember that we want to include the value into this interval.
First consider only saying that lots of hurricanes is out of the ordininary (not part of the null hypothesis).
$$P = P(x \geq 15) = 1 - \sum_0^{14} P(x)$$
In [15]:
print('p-value is', (1 - ss.poisson.cdf(14, mu=6.3)))
p-value is 0.002217122790073134
So we reject the null hypothesis. This is a highly unusual number of hurricanes.
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# Activation Energy
Activation Energy refers to the energy that must be provided to potential reactants in a chemical or a nuclear system in order to spark a chemical reaction or a nuclear reaction.
This energy is generally denoted by Ea and is measured using units like kilojoules/mole or kilocalories/mole. It can also be measured using joules.
## What is Activation Energy?
In order to understand the concept of activation energy, it can be visualized as the magnitude of the energy barrier (or the potential barrier) which separates the minima of potential energy surfaces involving the initial and the final thermodynamic state of the system.
This concept was first introduced by Swedish Chemist Svante August Arrhenius in the year 1889. A graph illustrating the activation energy of a reaction is provided below.
It can be noted that the activation energies of different reactions vary. This energy requirement can be affected by the temperature of the system and also by the presence of a catalyst.
## Temperature Dependence of the Rate of Reaction
From the Arrhenius equation, the following equation can be considered in order to calculate the activation energy of a system:
$k = Ae^{\frac{-E_{a}}{RT}}$
Wherein: k refers to the rate constant, T refers to the absolute temperature, R is the universal gas constant, and A refers to the frequency of correctly oriented collisions. Ea is the activation energy as discussed earlier.
### Negative Activation Energy
A negative value of activation energy is found in the Arrhenius equations of reactions wherein the rate of reaction slows down due to an increase in temperature.
In general, reactions which show a negative value for the activation energy are barrierless reactions which involve the capture of molecules in a given potential well.
An increase in the temperature for these reactions results in a reduction in the probability of the colliding molecules to capture each other.
## Effect of Catalysts on Activation Energy
Catalysts are basically substances that increase the rate of the reaction they are subjected to without being consumed by the reaction while accomplishing this. These substances work to modify the transition states of reactions in order to lower the activation energies of the aforementioned reactions.
An enzyme is a specific type of catalyst which is made up of only proteins and small molecule cofactors.
It is important to note that while the action of catalysts reduces the overall activation energy of the reaction, it has no effect on the energies of the reactants or products of the reaction in question. Only the Ea of the reactions is altered by these catalytic actions.
## Activation Energies of Endothermic and Exothermic Reactions
Reactions can be broadly classified as endothermic or exothermic reactions based on the role of utilization of thermal energy in these reactions.
### Endothermic Reactions
The reactions that require absorption of thermal energy by the reactants to form the necessary products are termed endothermic reactions.
These types of reactions typically have higher activation energies and also require a constant input of energy even after the Ea is provided. A graph detailing the energy requirements of these reactions is illustrated below.
An example of an endothermic reaction is the melting of ice to form water.
### Exothermic Reactions
The reactions that release thermal energy during the formation of products are referred to as exothermic reactions.
These types of reactions generally have low activation energies and they do not require a constant input of energy such as in the case of endothermic reactions. A graph of the energy requirements of these reactions can be illustrated as follows.
An example of this type of reaction is the combustion of fuel in a bunsen burner. The sparks which need to be provided to start the flame provide the required activation energy to the reacting fuel and the reaction sustains itself after this initial spark is provided.
#### Practise This Question
The diffrence between heat of reaction at constant volume and constant pressure for the reaction
C2H2(g)+52O2(g)2CO2(g)+H2O(l). At 300K in kcal is : [Given: R=2 cal/mol.K]
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# Mean Value Theorem of Electrostatics
1. Jan 12, 2006
### pmb_phy
In "Classical Electrodynamics - 3rd Ed.," J.D. Jackson has an exercise, 1.10, to derive the mean value theorem of electrostatics. Does anyone know of a derivation which is located on the web?
Pete
2. Jan 12, 2006
### dextercioby
Yes, this one
"The average value of the potential over the spherical surface is
$$\bar{\Phi} =\frac{1}{4\pi R}\int \Phi \ dA$$
If u imagine the surface of the sphere as discretized, u can rewrite the integral as an infinite sum
$$\frac{1}{a}\int \ dA =\Sigma_{area}$$
Then, takin the derivative wrt to R
$$\frac{d\bar{\Phi}}{dR}=\frac{d}{dR}\Sigma \Phi=\Sigma \frac{d\Phi}{dR}$$
Convert the infinite sum back to an integral & get
$$\frac{d\bar{\Phi}}{dR}=\frac{1}{4\pi R^{2}} \int \frac{d\Phi}{dR} \ dA$$
Now, using that
$$\frac{d\Phi}{dR}=-E$$ and Gauss's law (no charge inside the sphere) that gives
$$\frac{d\bar{Phi}}{dR}=0 \Rightarrow \bar{\Phi}_{surface}=\Phi_{center}$$
Q.E.D.
Daniel.
3. Jan 14, 2006
### pmb_phy
No. It isn't. Its
$$\bar{\Phi} =\frac{1}{A}\int d\Phi$$
where A is the area of the sphere.
Pete
4. Jan 14, 2006
### qbert
This isn't quite right either.
It isn't clear what you mean by $\int d\Phi$
since usually the integral of an exact differential
is just the function evaluated at the endpoints.
Or alternatively
$$\int d\Phi = \int \frac{\partial \Phi}{\partial x} dx + \int \frac{\partial \Phi}{\partial y} dy + \int \frac{\partial \Phi}{\partial z} dz$$
which is, of course, a line integral (which isn't what we
want).
This doesn't have the right dimensions. On the left
we have dimensions of [Phi]; and on the right we have
dimensions of [Phi]x[Area]/[Length].
I propose another formula:
$$\bar{\Phi} = \frac{1}{4 \pi R^2} \int_A \Phi dA$$
This also has the advantage that if Phi is constant
on the sphere we have
$$\bar{\Phi} = \frac{1}{4 \pi R^2} \int_A \Phi dA = \frac{1}{4 \pi R^2} \Phi \int_A dA = \frac{1}{4 \pi R^2} \Phi ( 4 \pi R^2 ) = \Phi$$
5. Jan 15, 2006
### pmb_phy
That would be true if you were talking about a line integral rather than in this case where the integral is a surface integral. What $d\Phi$ is? Hmmm. Perhaps you're right. I can't say that is a small element of $\Phi$ since that makes no sense to me at the moment. Thanks.
Pete
6. Jan 15, 2006
### robphy
May I propose a more [notationally] precise formula:
$$\bar{\Phi} = \frac{1}{A} \int_A \Phi dA$$
or a more [conceptually] precise formula for this average:
$$\bar{\Phi} = \frac{ \int_A \Phi dA } {\int_A dA}$$
7. Jan 16, 2006
### qbert
Ah, pedantry, thou most noble virtue. You are, of course, correct. Now the next person (who reads this thread) who needs a 2-D average will know the correct generalization.
Last edited: Jan 16, 2006
8. Jan 16, 2006
### dextercioby
I admit, it was R^2 in the denominator...
Daniel.
9. Jan 16, 2006
### robphy
This is a Homework subforum... and, in my experience, my students appreciate seeing the general idea in a consistent memorable notation rather than a special case. After all, there was some confusion with this formula.
Although the notation "A" [implicitly] suggests an area average, it is certainly not necessarily so.
10. Jan 16, 2006
### pmb_phy
Hi Rob
Do you know where I can find a solution? Jackson seems to want the student to use Green's theorm to solve this.
Pete
11. Jan 7, 2010
### alerodri115
12. Jan 7, 2010
### jdwood983
You do realize that this post is coming onto 4 years old? That the original poster hasn't asked a question for well over a year?
There is usually no reason to resurrect threads that are more than 2 months old, let alone ones that are 4 years old.
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# Why is a linear transformation a $(1,1)$ tensor?
Wikipedia says that a linear transformation is a $(1,1)$ tensor. Is this restricting it to transformations from $V$ to $V$ or is a transformation from $V$ to $W$ also a $(1,1)$ tensor? (where $V$ and $W$ are both vector spaces). I think it must be the first case since it also states that a linear functional is a $(0,1)$ tensor and this is a transformation from $V$ to $R$. If it is the second case, could you please explain why linear transformations are $(1,1)$ tensors.
• It would be helpful if you told us what definition exactly you have of a (1,1)-tensor. – Mariano Suárez-Álvarez Jan 18 '15 at 10:15
• @MarianoSuárez-Alvarez I would define it as a element of the vector space $V \bigotimes V^*$ where $V$ is a vector space and $V^*$ its dual – Quantum spaghettification Jan 18 '15 at 10:48
• Note that I've added a note to that Wikipedia statement now... – Fizz Jan 19 '15 at 12:25
• After tangling a bit on the Wikipedia's article talk page I've discovered that the regulars there have the following agenda/beliefs: (1) tensors are never defined over infinite-dimensional vector spaces and (2) tensors are never defined over different vector spaces (even if finite). As I don't have the time/stamina to battle people with a weird agenda and who are willing to spend most of their waking hours pushing it over there (despite the literature)... my note might be gone soon enough. Which is why reading Wikipedia on any topic is fraught with hazards... even on something as tame as math. – Fizz Jan 19 '15 at 15:07
Let $T : V \mapsto W$. Then define $\tau: V \times W^* \mapsto K$ such that, for $a \in V$ and $\alpha \in W^*$, we have
$$\tau(a, \alpha) = (\alpha \circ T)(a)$$
Note that it's typical to define tensor to mean a multilinear map that is a function of vectors only in the same vector space, or of covectors in the associated dual space, or some combination of the two. So, we could identify a linear operator $T: V \mapsto V$ with a $(1,1)$ tensor $\tau: V \times V^* \mapsto K$, but in the case that $V$ and $W$ are distinct vector spaces, these would just be some construction of multilinear maps, not tensors.
• Bilinear (and multilinear) maps can be linearized. This is in fact how Yokonuma's textbook (pp. 4-7) introduces tensor product, so you can define a tensor product for $V$ and $W$ different. This approach is then extended (p. 12) to multilinear maps where all the vector spaces are different. On the other hand, when Yokonuma introduces tensor space (p. 33) he defines it as product of copies of the same vector space $V$ and its dual $V*$. It doesn't make sense to speak of a $(i,j)$-type tensor (space) unless it is over the same vector space. – Fizz Jan 19 '15 at 0:33
• And one more terminology issues worth mentioning here is that a the tensor product of two different vector spaces is sometimes called a tensor product space, but this is usually not what people refer to when they use just tensor space. – Fizz Jan 19 '15 at 1:15
• Just to prove that there's an exception to every rule, the Handbook of Linear Algebra (2nd ed., p. "15-7") does define a "order-d tensor" as an element of a tensor product of $d$ different vector spaces. Ha! – Fizz Jan 19 '15 at 1:29
• And there are physics books that adopt this terminology too. E.g. Geometry of the Fundamental Interactions (p. 48), calls the objects of the tensor product space $U\otimes V$ "second-order tensors". – Fizz Jan 19 '15 at 1:44
• Can the argument go both ways without assuming that $W$ is reflexive? I.e., starting with a multilinear map $\tau:V \times W^* \rightarrow K$, constructing a linear operator $T:V \rightarrow W$. Without this, we have constructed an isometric embedding, but not an isomorphism. It seems to me the natural linear map associated with a given $\tau$ would be $T:V \mapsto (W^*)^*$. Or maybe I'm missing something..? – Nick Alger Sep 27 '16 at 22:18
It's very common in tensor analysis to associate endomorphisms on a vector space with (1,1) tensors. Namely because there exists an isomorphism between the two sets.
Define $E(V)$ to be the set of endomorphisms on $V$.
Let $A\in E(V)$ and define the map $\Theta:E(V)\rightarrow T^1_1(V)$ by \begin{align*} (\Theta A)(\omega,X)&=\omega(AX). \end{align*} We show that $\Theta$ is an isomorphism of vector spaces. Let $\{e_i\}$ be a basis for $V$ and let $\{\varepsilon^i\}$ be the corresponding dual basis. First, we note $\Theta$ is linear by the linearity of $\omega$. To show injectivity, suppose $\Theta A = \Theta B$ for some $A,B\in E(V)$ and let $X\in V$, $\omega \in V^*$ be arbitrary. Then \begin{align*} (\Theta A)(\omega,X)&=(\Theta B)(\omega,X)\\ \\ \iff \omega(AX-BX)&=0. \end{align*} Since $X$ and $\omega$ were arbitrary, it follows that \begin{align*} AX&=BX\\ \iff A&=B. \end{align*} To show surjectivity, suppose $f\in T^1_1$ has coordinate representation $f^j_i \varepsilon^i \otimes e_j$. We wish to find $A\in E(V)$ such that $\Theta A = f$. We simply choose $A\in E (V)$ such that $A$ has the matrix representation $(f^j_i)$. If we write the representation of our vector $X$ and covector $\omega$ as \begin{align*} X&=X^i e_i\\ \omega&=\omega_i \varepsilon^i, \end{align*} we have \begin{align*} (\Theta A)(\omega, X)&=\omega(AX)\\ \\ &=\omega_k \varepsilon^k(f^j_i X^i e_j)\\ \\ &=f^j_i X^i \omega_k \varepsilon^k (e_j)\\ \\ &=f^j_i X^i \omega_k \delta^k_j\\ \\ &=f^k_i X^i \omega_k. \end{align*} However we see \begin{align*} f(\omega,X)&=f(\omega_k\varepsilon^k,X^ie_i)\\ \\ &=\omega_k X^i f(\varepsilon^k,e_i)\\ \\ &=f^k_i X^i \omega_k. \end{align*} Since $X$ and $\omega$ were arbitrary, it follows that $\Theta A = f$. Thus, $\Theta$ is linear and bijective, hence an isomorphism.
• N.B.: This proof is addressing the 2nd part of the OP's question. I'm curious which one he is going to accept given that he asked two questions in one... – Fizz Jan 19 '15 at 0:35
• Also, this result can be stated more generally for a tensor product of two different vector spaces (i.e., not just for a tensor space), as the existence of an isomorphism between $\mathrm{Hom}(V, W)$ and $V^*\otimes W$. – Fizz Jan 19 '15 at 0:58
• Also your proof (of surjectivity) does not hold for an infinite-dimensional vector space $V$ because the Kronecker delta formula for covector basis in only valid for finite-dimensional [co]vector spaces. – Fizz Jan 19 '15 at 9:51
• Is there a way to prove this without involving the basis? – MichaelNgelo Jul 7 '16 at 15:26
To summarize as an answer what I wrote in various comments above: first beware that autors differ in their definition of tensor, even when using the same approach, i.e. using the tensor product in this case.
For some authors a tensor is defined only as ...
$$T\in \underbrace{V \otimes\dots\otimes V}_{n \text{ copies}} \otimes \underbrace{V^* \otimes\dots\otimes V^*}_{m \text{ copies}}$$
From which it makes sense to speak of a type-$(n,m)$ tensor.
For others, a tensor is any...
$$T\in V_1 \otimes\dots\otimes V_d$$
where $V_1, \dots, V_d$ can be different vector spaces, however all must be over the same scalar field. And with this latter definition one can speak of an order-$d$ tensor. A type-$(n,m)$ tensor [in the former sense] is a tensor of order $d=n+m$ in the latter sense, but second definition is broader for it does not restrict us to a single vector space. In particular, a second-order tensor is an element of $V \otimes W$ where $V$ and $W$ may be two different vector spaces. Type-(1,1) tensors are tensors of second order, but the converse of this statement doesn't make sense. (N.B.: I've updated Wikipedia to reflect these different definitions.)
As for your 2nd question, endomorphisms (linear maps) from a vector space to itself are (isomorphic with) type-(1,1) tensors (detailed proof given here by beedge89), but if you consider homomorphisms (linear maps) between different vector spaces $V$ and $W$, i.e. $\mathrm{Hom}(V,W)$, these are isomorphic with only a certain class of order-2 tensors, namely with $V^* \otimes W$. If we let $(\phi, w)\in V^* \times W$, then the correspondence is given by $\phi \otimes w \leftrightarrow F_{\phi, w}$, where the latter is a (linear) map defined as $F_{\phi, w} (v) = \phi(v)w$. (Remember that covectors are themselves maps from vectors to scalars, so the formula for $F$ makes sense as it's a product of the scalar $\phi(v)$ with the vector $w$). A detailed proof of the fact that this is an isomorphism is given in Yokonuma (pp. 18-19). Apologies for not including it here.
As you may expect, the result for type-(1,1) tensors also follows as a corollary of this, i.e. $\mathrm{Hom}(V,V)$ is isomorphic with $V^* \otimes V$ (and with $V \otimes V^*$ by commutativity of the tensor product, which is also understood in the sense of an isomorphism between $V \otimes W$ and $W \otimes V$ for any vector spaces $V$ and $W$).
And one important caveat here: this is an isomorphism only for finite-dimensional vector spaces. (The introduction of Yokonuma's book actually says to assume all vector spaces in the book are finite-dimensional unless stated otherwise.) If both $V$ and $W$ are infinite-dimensional, then it turns out $V^*\otimes W$ is only a proper subspace of $\mathrm{Hom}(V,W)$, namely it is the subspace of linear transformation of finite rank.
And to tie this in with bilinear (and in general with multi-linear) maps: there's also a one-one correspondence between bilinear maps $f : V\times W \to U$ and $linear$ maps $g : V\otimes W \to U$. (For a proof see for instance http://www.landsburg.com/algebra.pdf) That's why second-order tensors are basically said to be just bilinear maps, and in general why d-order tensors are said to be just multi-linear maps.
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# How do you write the equation of the line in slope intercept form passing through (1, 2) and (2, 5)?
Apr 20, 2018
y = 3x -1
#### Explanation:
1. Find the slope by using the formula $\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$
You wil get: $\frac{5 - 2}{2 - 1}$
Simplify to get 3
2. Find the y-intercept by plugging in one kf the pairs and the slope into the y = mx + b equation. It doesn’t matter which pair you use.
You see 2 = 3(1) + b
Solve for b
2 = 3 + b
-1 = b
1. Answer is y = 3x -1
Apr 20, 2018
$y = 3 x - 1$
#### Explanation:
the gradient/slope is the difference in the $y$ values divided by the difference in the $x$ values.
$\implies$ $\frac{5 - 2}{2 - 1}$ =$\frac{3}{1}$ =3
so the equation of the line is $y = 3 x + c$
Use (1,2) $\implies$ $2 = 3 \times 1 + c$
$\implies$ $2 = 3 + c$ so $c$=-1
this gives $y = 3 x - 1$
Apr 20, 2018
$y = 3 x - 1$
#### Explanation:
$\text{the equation of a line in "color(blue)"slope-intercept form}$ is.
•color(white)(x)y=mx+b
$\text{where m is the slope and b the y-intercept}$
$\text{to calculate m use the "color(blue)"gradient formula}$
•color(white)(x)m=(y_2-y_1)/(x_2-x_1)
$\text{let "(x_1,y_1)=(1,2)" and } \left({x}_{2} , {y}_{2}\right) = \left(2 , 5\right)$
$\Rightarrow m = \frac{5 - 2}{2 - 1} = 3$
$\Rightarrow y = 3 x + b \leftarrow \textcolor{b l u e}{\text{is the partial equation}}$
$\text{to find b substitute either of the 2 given points into}$
$\text{the partial equation}$
$\text{using "(1,2)" then}$
$2 = 3 + b \Rightarrow b = 2 - 3 = - 1$
$\Rightarrow y = 3 x - 1 \leftarrow \textcolor{red}{\text{equation in slope-intercept form}}$
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# what's the conductor of a ray class field?
Let $K$ be a number field. The theorems of class field theory tell us that given any modulus $\mathfrak{m}$ for $K$, there is a unique Abelian extension $K_{\mathfrak{m}}$ such that the kernel of the Artin map of $K_{\mathfrak{m}}/K$ with respect to $\mathfrak{m}$ is precisely the subgroup of principal fractional ideals congruent to $1 \pmod{\mathfrak{m}}$. This is the Ray class field.
Moreover, we know that the conductor of $K_{\mathfrak{m}}/K$ divides $\mathfrak{m}$. Is it equal in general? If not, what are counterexamples?
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Dear Tony, Have you tried the case $K = \mathbb Q$, and various (small!) choices of $\mathfrak m$? Regards, – Matt E May 15 '12 at 20:04
No, the conductor of $K_{\mathfrak{m}}$ can properly divide $\mathfrak{m}$. Hint: this already happens for $K = \mathbb{Q}$. The basic culprit behind this is the fact that $\mathbb{Q}(\zeta_{2n}) = \mathbb{Q}(\zeta_n)$ when $n$ is odd. – Pete L. Clark May 15 '12 at 20:05
Oops, silly me. So I see that it already fails for some cyclotomic fields (namely, when adjoining a 2* (2n-1) root of unity)? – Tony May 15 '12 at 20:08
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## Multiple overloads of the bind operator
Hi! Let's say it makes a lot of sense in my domain to define the types WriteReference and ReadOnlyReference. I wish to create a monad that manipulates references, so my first idea was to define
(>>=) :: WriteReference a -> (a -> WriteReference b) -> WriteReference b (>>=) :: WriteReference a -> (a -> ReadOnlyReference b) -> ReadOnlyReference b (>>=) :: ReadOnlyReference a -> (a -> ReadOnlyReference b) -> ReadOnlyReference b (>>=) :: ReadOnlyReference a -> (a -> WriteReference b) -> WriteReference b
return :: a -> ReadOnlyReference a
It looks ugly but it does its job, even though I fear it won't play nice with Haskell syntactic sugar for monads.
Does it make more sense to define Reference as a class, and then to define WriteReference and ReadOnlyReference as instances of this class, so that we can write something like:
data WR a = WR a data RR a = RR a
class Reference r where get :: r a -> a instance Reference WR where get (WR x) = x instance Reference RR where get (RR x) = x
bind :: (Reference r1, Reference r2) => r1 a -> (a -> r2 b) -> r2 b bind e k = let x = get e in k x
Can anyone help me understand why I should favor one approach over the other?
Thanks
## Comment viewing options
### This is more a question for the Haskell Cafe mailing list.
You'll get more response there.
### Fair enough, but the
Fair enough, but the question wasn't meant to be Haskell specific: I am more curious about the implications with respect to building such a monad. Is it still a monad? Is it something else?
### is it still a monad?
does it satisfy the monad laws?
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# a cute paper: finding a most biased coin with fewest flips
I saw this paper on ArXiV a while back and figured it would be a fun read, and it was. Post-ISIT blogging may have to wait for another day or two.
Finding a most biased coin with fewest flips
Karthekeyan Chandrasekaran, Richard Karp
arXiv:1202.3639 [cs.DS]
The setup of the problem is that you have $n$ coins with biases $\{p_i : i \in [n]\}$. For some given $p \in [\epsilon,1-\epsilon]$ and $\epsilon \in (0,1/2)$, each coin is “heavy” ($p_i = p + \epsilon$) with probability $\alpha$ and “light” ($p_i = p - \epsilon$) with probability $1 - \alpha$. The goal is to use a sequential flipping strategy to find a heavy coin with probability at least $1 - \delta$.
Any such procedure has three components, really. First, you have to keep track of some statistics for each coin $i$. On the basis of that, you need a rule to pick which coin to flip. Finally, you need a stopping criterion.
The algorithm they propose is a simple likelihood-based scheme. If I have flipped a particular coin $i$ a bunch of times and gotten $h_i$ heads and $t_i$ tails, then the likelihood ratio is
$L_i = \left( \frac{p+\epsilon}{p - \epsilon} \right)^{h_i} \left( \frac{ 1 - p - \epsilon }{ 1 -p + \epsilon} \right)^{t_i}$
So what the algorithm does is keep track of these likelihoods for the coins that it has flipped so far. But what coin to pick? It is greedy and chooses a coin $i$ which has the largest likelihood $L_i$ so far (breaking ties arbitrarily).
Note that up to now the prior probability $\alpha$ of a coin being heavy has not been used at all, nor has the failure probability $\delta$. These appear in the stopping criterion. The algorithm keeps flipping coins until there exists at least one $i$ for which
$L_i \ge \frac{1 - \alpha}{\alpha} \cdot \frac{ 1 - \delta }{\delta}$
It then outputs the coin with the largest likelihood. It’s a pretty quick calculation to see that given $(h_i, t_i)$ heads and tails for a coin $i$,
$\mathbb{P}(\mathrm{coin\ }i\mathrm{\ is\ heavy}) = \frac{\alpha L_i}{ \alpha L_i + (1 - \alpha) }$,
from which the threshold condition follows.
This is a simple-sounding procedure, but to analyze it they make a connection to something called a “multitoken Markov game” which models the corresponding mutli-armed bandit problem. What they show is that for the simpler case given by this problem, the corresponding algorithm is, in fact optimal in the sense that it makes the minimum expected number of flips:
$\frac{16}{\epsilon^2} \left( \frac{1 - \alpha}{\alpha} + \log\left( \frac{(1 -\alpha)(1 - \delta)}{\alpha \delta} \right) \right)$
The interesting thing here is that the prior distribution on the heavy/lightness plays a pretty crucial role here in designing the algorithm. part of the explore-exploit tradeoff in bandit problems is the issue of hedging against uncertainty in the distribution of payoffs — if instead you have a good handle on what to expect in terms of how the payoffs of the arms should vary, you get a much more tractable problem.
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Now showing items 1-2 of 2
• Path contraction faster than $2^n$
(Journal article; Peer reviewed, 2020)
A graph $G$ is contractible to a graph $H$ if there is a set $X \subseteq E(G)$, such that $G/X$ is isomorphic to $H$. Here, $G/X$ is the graph obtained from $G$ by contracting all the edges in $X$. For a family of graphs ...
• Path Contraction Faster Than 2n
(Peer reviewed; Journal article, 2019)
A graph G is contractible to a graph H if there is a set X subseteq E(G), such that G/X is isomorphic to H. Here, G/X is the graph obtained from G by contracting all the edges in X. For a family of graphs F, the F-Contraction ...
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# Why does the light travel slower in denser medium? [duplicate]
Wikipedia says that "in general, the refractive index of a glass increases with its density." And the refraction index of water vapor is less than ice, and even less than liquid water. Is there any simple explanation to that?
## 2 Answers
The simplest picture is that light always travels at the speed of light. But in a material it travels at the speed of light until it hits an atom. It is then absorbed and re-emitted in the same direction, which takes a small amount of time.
The more this happens, the slower the effective average speed.
The denser the material, the more atoms there are in the way.
• Great explanation. – boyfarrell Mar 29 '14 at 6:04
• The explanation is only apparent. In classical EM theory there is no time lag between interaction of external EM wave with charges and emission of secondary EM wave, both occur simultaneously. – Ján Lalinský Mar 29 '14 at 8:02
• By what mechanism is the light absorbed? Not an atomic electron transition, as surely re-emission would change frequency and hence color. – xxyzzy Jul 21 '18 at 11:22
This is quite a subtle issue. The charges in the medium produce secondary spherical expanding EM waves when hit by the primary wave (external forces). There is immense number of these secondary waves. At any point of space, each secondary wave has slightly different wave vector. In a medium dense enough, these secondary waves add to the primary wave in such a way that the resulting wave has behaviour that is well described by a single macroscopic wave of the same frequency and (usually) same direction but (for most frequencies) with a reduced wavelength.
A common picture backed by successes of dispersion theory is that the relation $$\mathbf j(t) = c\mathbf E(t-\Delta t)$$ is valid, where $$c, \Delta t$$ are some medium property constants that depend on frequency of the wave, $$\mathbf j$$ is current density and $$\mathbf E$$ is total macroscopic electric field. With this assumption, Maxwell's equations imply that the resulting wave in the medium will have modified (in usual cases shorter) wavelength hence lower velocity (for a certain limited interval of frequencies it can have a longer wavelength and higher velocity).
• This explanation cannot be correct. You state that "the resulting wave in the medium will have shorter wavelength hence lower velocity" but this is incorrect. If you change the wavelength (or the frequency, because frequency = c / wavelength) you only change the energy of the photon / light wave (and thus colour) but not the speed, which is fixed as long as it remains in the same type of medium. Indeed though as @arax notes, this fixed speed actually depends on the type of medium and can become smaller than c, but I do not know the correct explanation. – PDiracDelta Aug 11 '17 at 22:38
• @PDiracDelta, the question is about index of refraction. Index of refraction is ratio of phase velocities of a wave in two media. Since frequency is the same in both media (linear media) and since $v = f \lambda$, decrease in wavelength implies decrease in phase velocity $v$. The energy of a photon or light wave has nothing to do with this explanation. – Ján Lalinský Aug 12 '17 at 23:22
• I'm trying to understand what you are saying, but if not light, then what kind of wave are you talking about? AFAIK a light and EM wave are the same thing. Also, to me your claim that the frequency remains the same is non-trivial. – PDiracDelta Aug 14 '17 at 11:03
• I think I've found a good explanation here: physics.stackexchange.com/a/476/51901 (this question now apparently has been marked as a duplicate) – PDiracDelta Aug 14 '17 at 11:18
• I am talking about an EM wave in material medium, which is a model of a light wave, but the important part in this explanation is relation of electric field to current density, not energy of the wave or the concept of a photon. – Ján Lalinský Aug 14 '17 at 23:07
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# Does this inequality hold
Let $f(x)$ be a positive function on $[0,\infty)$ such that $f(x) \leq 100 x^2$. I want to bound $f(x) - f(x-1)$ from above. Of course, we have $$f(x) - f(x-1) \leq f(x) \leq 100 x^2.$$ This is not good for me though. I need a bound which is linear (or at worst linear-times-root) in $x$.
Is there an inequality of the form $f(x) - f(x-1) \leq f^\prime (x)=200 x$?
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There is no such bound. Let $c$ be a real number, and let $$f(x)=\begin{cases} 0&\text{if x<c}\\100x^2&\text{if x\geq c}.\end{cases}$$ Then for $x\in [c,c+1)$ the inequality $f(x)-f(x-1)\leq 100x^2$ is the best bound possible. So we cannot make a better bound for a general function satisfying $f(x)\leq 100x^2$ for all $x$.
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Suppose $f(x) = 100x^2\sin^2(\pi x/2)$. Then $f(x) = 0$ when $x$ is an even integer and $f(x) = 100x^2$ when $x$ is an odd integer. So $f(x)-f(x-1)\ge 100(x-1)^2$, with equality when $x$ is even.
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There is no such bound. Let $f(2)=0$, and $f(x)=100x^2$ for other $x$. Surely $f(3)-f(2)=900\le100x^2$, but equality holds (!).
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For $2^n \le x<2^{n+1}$, let $f(x)=100(2^{n})^2$. There is an enormous jump from $f(2^{n+1}-1)$ to $f(2^{n+1})$. So even if we assume that $f$ is non-decreasing, we can have jumps of size comparable to $100x^2$. At the cost of complicating the description, we can modify the above $f(x)$ to make it strictly increasing.
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Provides methods for (soft) imputation of missing values.
Impute2D(formula, data = NULL, method = "interpolate")
## Arguments
formula a formula indicating dependent and independent variables (see Details) optional data.frame with the data "interpolate" for interpolation, a numeric for constant imputation or a function that takes a vector and returns a number (like mean)
## Details
This is "soft" imputation because the imputed values are not supposed to be representative of the missing data but just filling for algorithms that need complete data (in particular, contouring). The method used if method = "interpolate" is to do simple linear interpolation in both the x and y direction and then average the result.
This is the imputation method used by geom_contour_fill().
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# Step-by-Step Guide to Estimating Errors and Confidence Levels
## Overview
#### Synopsis:
This thread uses the "native" Sherpa interface to estimate errors or confidence levels for parameters in a fit (to data of any dimensionality).
Last Update: 5 Dec 2022 - reviewed for CIAO 4.15, updated screen output.
## Getting Started
### Finding the best fit
First, we check the current Sherpa settings. By default, Sherpa uses the Levenberg-Marquardt method to optimize the fit of a model to data with χ2 statistics, using the Gehrels variance function. In this example, we choose to fit with the default optimization method, and change the fit statistic from default to the Chi2Xspecvar statistic.
Tip
The available optimization methods and statistics may be listed with list_methods and list_stats, and can be changed with set_method and set_stat. Detailed information on each option may be found in the ahelp documentation, as well as on the Sherpa Optimization Methods and Statistics pages.
sherpa> clean()
sherpa> show_method()
Optimization Method: LevMar
name = levmar
ftol = 1.1920928955078125e-07
xtol = 1.1920928955078125e-07
gtol = 1.1920928955078125e-07
maxfev = None
epsfcn = 1.1920928955078125e-07
factor = 100.0
numcores = 1
verbose = 0
sherpa> set_stat("chi2datavar")
sherpa> show_stat()
Statistic: Chi2DataVar
Chi Squared with data variance.
The variance in each bin is estimated from the data value in that
bin.
If the number of counts in each bin is large, then the shape of
the Poisson distribution from which the counts are sampled tends
asymptotically towards that of a Gaussian distribution, with
variance
sigma(i)^2 = N(i,S) + [A(S)/A(B)]^2 N(i,B)
where N is the number of on-source (and off-source) bins included
in the fit. The background term appears only if an estimate of the
background has been subtracted from the data.
A(B) is the off-source "area", which could be
the size of the region from which the background is extracted, or
the length of a background time segment, or a product of the two,
etc.; and A(S) is the on-source "area". These terms may be defined
for a particular type of data: for example, PHA data sets A(B) to
BACKSCAL * EXPOSURE from the background data set and A(S) to
BACKSCAL * EXPOSURE from the source data set.
--------
Chi2Gehrels, Chi2ModVar, Chi2XspecVar
statistical errors were found in file 'source_grouped_pi.fits'
but not used; to use them, re-read with use_errors=True
sherpa> notice(0.5, 8.0)
dataset 1: 0.00146:14.9504 -> 0.00146:8.76 Energy (keV)
sherpa> plot_data()
The data we will be fitting is read into the session with the load_data command, which automatically loads the instrument response associated with a source data set when the ARF and RMF filenames are recorded in the header of the source data file, as shown above; this also applies to background data associated with the source (not considered in this example). The plot in Figure 1 results from the plot_data command above, showing the source data that is to be fit, between 0.5-8.0 keV.
The source model is set to an absorbed power-law with the set_source command, and fit with the fit command, producing the results shown below.
sherpa> set_source(xsphabs.abs1 * powlaw1d.p1)
sherpa> fit()
Dataset = 1
Method = levmar
Statistic = chi2datavar
Initial fit statistic = 6.65323e+09
Final fit statistic = 123.621 at function evaluation 29
Data points = 133
Degrees of freedom = 130
Probability [Q-value] = 0.640847
Reduced statistic = 0.950933
Change in statistic = 6.65323e+09
abs1.nH 2.25588 +/- 0.117471
pl.gamma 1.47469 +/- 0.081887
pl.ampl 0.00185338 +/- 0.000224613
The fit and δχ residuals ($$\frac{data-model}{errors}$$, also refered to as σ residuals which is not the same σ as standard deviation) may be plotted with the plot or plot_fit_delchi command. Here we use Matplotlib commands to adjust the axes after the plot has been created, resulting in Figure 2:
sherpa> plot_fit_delchi(xlog=True, ylog=True)
sherpa> plt.xlim(0.3, 10)
(0.3, 10)
The show_fit and get_fit_results commands are also available for checking the quality of the fit:
sherpa> print(get_fit_results())
datasets = (1,)
itermethodname = none
methodname = levmar
statname = chi2datavar
succeeded = True
parnames = ('abs1.nH', 'pl.gamma', 'pl.ampl')
parvals = (2.2558840821480937, 1.4746943632186176, 0.0018533814206418062)
statval = 123.62123838387117
istatval = 6653225321.935967
dstatval = 6653225198.314729
numpoints = 133
dof = 130
qval = 0.6408472990001428
rstat = 0.9509326029528552
message = successful termination
nfev = 29
## Confidence limits for individual parameters
After finding the best-fit model parameter values, we calculate the confidence limits (parameter bounds) for these parameters using either the confidence, projection or covariance methods. The confidence limits are defined by a required confidence level in our analysis, typically the 68.3% or 90% level, corresponding to 1σ and 1.6σ for the normal distribution. The table below displays the relationship between the standard deviation, σ, and confidence level for one significant parameter. The relationship between the confidence level, Δχ2, and the general log-likelihood, $$\Delta \log{\mathcal{L}}=\Delta\chi^{2}/2$$.
#### Confidence intervals for a normal distribution
Confidence $$\sigma$$ $$\Delta \chi^{2}$$ $$\Delta \log{\mathcal{L}}$$
68.3% 1.0 1.00 0.50
90.0% 1.6 2.71 1.36
95.5% 2.0 4.00 2.00
99.0% 2.6 6.63 3.32
99.7% 3.0 9.00 4.50
We will use the confidence method to estimate 1σ errors on the gamma parameter of the power-law model component:
sherpa> print(get_conf())
name = confidence
sigma = 1
eps = 0.01
maxiters = 200
soft_limits = False
remin = 0.01
fast = False
parallel = True
numcores = 4
maxfits = 5
max_rstat = 3
tol = 0.2
verbose = False
openinterval = False
sherpa> conf(p1.gamma)
pl.gamma lower bound: -0.0830705
pl.gamma upper bound: 0.0849456
Dataset = 1
Confidence Method = confidence
Iterative Fit Method = None
Fitting Method = levmar
Statistic = chi2datavar
confidence 1-sigma (68.2689%) bounds:
Param Best-Fit Lower Bound Upper Bound
----- -------- ----------- -----------
pl.gamma 1.47469 -0.0830705 0.0849456
To access the 90% confidence limits on a parameter, the sigma field of the conf_opt variable should be changed to 1.6.
sherpa> set_conf_opt("sigma", 1.6)
sherpa> conf(p1.gamma, abs1.nH)
abs1.nH lower bound: -0.183538
abs1.nH upper bound: 0.197266
pl.gamma lower bound: -0.13186
pl.gamma upper bound: 0.136806
Dataset = 1
Confidence Method = confidence
Iterative Fit Method = None
Fitting Method = levmar
Statistic = chi2datavar
confidence 1.6-sigma (89.0401%) bounds:
Param Best-Fit Lower Bound Upper Bound
----- -------- ----------- -----------
pl.gamma 1.47469 -0.13186 0.136806
abs1.nH 2.25588 -0.183538 0.197266
We have also used conf to calculate the uncertainty on the optimized hydrogen column density parameter of the absorption model, abs1.nH.
To estimate errors on all the thawed parameters, conf should be called with no parameter names. Since sigma is still set to 1.6 for the confidence method, the following calculates the 90% confidence limits for all the thawed parameters:
sherpa> conf()
abs1.nH lower bound: -0.183538
abs1.nH upper bound: 0.197266
pl.ampl lower bound: -0.000329436
pl.gamma lower bound: -0.13186
pl.gamma upper bound: 0.136806
pl.ampl upper bound: 0.000413355
Dataset = 1
Confidence Method = confidence
Iterative Fit Method = None
Fitting Method = levmar
Statistic = chi2datavar
confidence 1.6-sigma (89.0401%) bounds:
Param Best-Fit Lower Bound Upper Bound
----- -------- ----------- -----------
abs1.nH 2.25588 -0.183538 0.197266
pl.gamma 1.47469 -0.13186 0.136806
pl.ampl 0.00185338 -0.000329436 0.000413355
The covariance command behaves similarly to conf, although the fields in the state object are different. While it is quicker than the confidence method, it is less accurate; i.e., it is always symmetric because it uses the diagonal elements of the covariance matrix and ignores correlations between the parameters.
Note that the computationally intensive confidence function has been parallelized in Sherpa, to make use of multi-core systems (i.e., laptops or desktops with 2 or more cores).
### Common WARNING messages returned by confidence methods
#### WARNING: hard minimum hit for parameter <parameter name>
When the confidence, projection, and covariance methods are used to estimate confidence intervals for thawed model parameters after a fit, sometimes a hard upper or lower limit will be reached for one or more parameter. This produces the message "WARNING: hard minimum hit for parameter <parameter name>", along with a row of dashes in the appropriate place in the function output.
Note
The covariance method can also return a null value for an upper/lower limit when the parameter-space at the minimum is non-quadratic for a given parameter. The covariance matrix calculations assume that the parameters follow the normal distribution. If the parameter-space is non-smooth, then the covariance calculations fail and Sherpa returns "-----".
Example confidence output:
sherpa> conf()
...
WARNING: hard minimum hit for parameter bpow1.gamma2
WARNING: hard maximum hit for parameter bpow1.gamma2
WARNING: hard minimum hit for parameter bpow1.eb
WARNING: hard maximum hit for parameter bpow1.eb
...
Dataset = 1
Confidence Method = confidence
Iterative Fit Method = None
Statistic = cstat
confidence 1-sigma (68.2689%) bounds:
Param Best-Fit Lower Bound Upper Bound
----- -------- ----------- -----------
bpow1.gamma1 1.54147 -0.0292891 0.0292709
bpow1.gamma2 8.10056 ----- -----
bpow1.eb 9.49083 ----- -----
bpow1.ampl 0.022806 -0.000378395 0.000383854
This occurs when the parameter bound found by one of the confidence methods lies outside the hard limit boundary for a model parameter—this could result from an issue with the signal-to-noise of the data, the applicability of the model to the data, systematic errors in the data, among others things.
A parameter hard limit represents either a hard physical limit (e.g., temperature is not allowed to go below zero), a mathematical limit (e.g., prevent a number from going to zero or below, when the logarithm of that number will be taken), or the limit of what a float or double can hold (the fit should not be driven above or below the maximum or minimum values a variable can hold). For this reason, model parameter hard limits should not be changed by the user.
#### WARNING: The confidence level lies within <interval>
Another warning message which may be returned by confidence is that a model parameter lies within the stated range:
sherpa> conf(g15.Sigma)
...
g15.Sigma -: WARNING: The confidence level lies within (8.706380e-05,9.252185e-05)
...
Datasets = 1, 2
Confidence Method = confidence
Iterative Fit Method = None
Fitting Method = levmar
Statistic = chi2datavar
confidence 1.64-sigma (89.8995%) bounds:
Param Best-Fit Lower Bound Upper Bound
----- -------- ----------- -----------
g15.Sigma 0.000997626 -0.000907834 0.000597058
This occurs where confidence cannot locate the root (minimum value of the fit statistic function) even though the root is bracketed within an interval (perhaps due to poor resolution of the data or a discontinuity). In such cases, when the openinterval option of confidence is set to False (default), the confidence function will not be able to find the root within the set tolerance and the function will return the average of the open interval which brackets the root. If the option openinterval is set to True, then confidence will print the minimal open interval which brackets the root (not to be confused with the lower and upper bound of the confidence interval). The most accurate thing to do is to return an open interval where the root is localized/bracketed rather than the average of the open interval (since the average of the interval is not a root within the specified tolerance).
The output from confidence may be checked by setting 'set_conf_opt('verbose',1)', and then re-running confidence for the relevant parameter(s).
sherpa> set_conf_opt('verbose',1)
sherpa> conf(g15.Sigma)
#
# f[ 2.931742e+00 2.957942e-02 1.471941e+00 9.976265e-04
1.851972e-04
1.840837e+00 3.667986e-03 7.820442e-05 1.864564e+00 2.562143e-03
1.415699e-04 2.004898e+00 6.288115e-04 1.259512e-04] =
8.569929e+02
# sigma = 1.640000e+00
# target_stat = 8.596825e+02
# tol = 1.000000e-02
# smin = [-2. 0. 1.45 0. 0. 1.82 0. 0. 1.85 0.
0. 1.96
0. 0. ]
# smax = [ 9.000000e+00 1.000000e+24 1.500000e+00 1.000000e-02
1.000000e-03
1.850000e+00 1.000000e-02 1.000000e-03 1.900000e+00 1.000000e-02
1.000000e-03 2.040000e+00 1.000000e-03 1.000000e-03]
# hmin = [ -3.402823e+38 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00
0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00
0.000000e+00 0.000000e+00 0.000000e+00 0.000000e+00]
# hmax = [ 3.402823e+38 3.402823e+38 3.402823e+38 3.402823e+38
3.402823e+38
3.402823e+38 3.402823e+38 3.402823e+38 3.402823e+38 3.402823e+38
3.402823e+38 3.402823e+38 3.402823e+38 3.402823e+38]
#
# Note: for the intermediate steps, the notation:
par.name -/+: f( x ) = stat
# ==> stat is the statistic when parameter par.name is frozen at x
# while searching for the lower/upper confidence level, repectively.
#
g15.Sigma -: f( 4.812517e-04 ) = -2.565633e+00
g15.Sigma -: f( 0.000000e+00 ) = 3.505966e+01
g15.Sigma -: f( 2.406259e-04 ) = -2.565434e+00
g15.Sigma -: f( 2.113694e-04 ) = -2.565393e+00
g15.Sigma -: f( 1.653199e-04 ) = -2.565381e+00
g15.Sigma -: f( 1.343754e-04 ) = -2.564794e+00
g15.Sigma -: f( 1.079722e-04 ) = -2.565100e+00
g15.Sigma -: f( 8.706380e-05 ) = 3.778055e+01
g15.Sigma -: f( 1.068516e-04 ) = -2.563338e+00
g15.Sigma -: f( 1.022595e-04 ) = -2.562500e+00
g15.Sigma -: f( 9.961908e-05 ) = -2.563027e+00
g15.Sigma -: f( 9.721570e-05 ) = -2.564524e+00
g15.Sigma -: f( 9.532733e-05 ) = -2.562276e+00
g15.Sigma -: f( 9.377811e-05 ) = -2.562519e+00
g15.Sigma -: f( 9.252185e-05 ) = -2.563134e+00
g15.Sigma -: f( 9.149979e-05 ) = -2.563034e+00
g15.Sigma -: f( 9.066944e-05 ) = -2.563196e+00
g15.Sigma -: WARNING: The confidence level lies within (8.706380e-05, 9.252185e-05)
g15.Sigma lower bound: -0.000907834
g15.Sigma +: f( 1.514001e-03 ) = -8.109195e-01
g15.Sigma +: f( 2.546751e-03 ) = 2.553184e+01
g15.Sigma +: f( 2.030376e-03 ) = 7.845440e+00
g15.Sigma +: f( 2.030376e-03 ) = 7.845440e+00
g15.Sigma +: f( 2.030376e-03 ) = 7.845440e+00
g15.Sigma +: f( 1.772189e-03 ) = 2.467092e+00
g15.Sigma +: f( 1.772189e-03 ) = 2.467092e+00
g15.Sigma +: f( 1.772189e-03 ) = 2.467092e+00
g15.Sigma +: f( 1.643095e-03 ) = 5.832027e-01
g15.Sigma +: f( 1.643095e-03 ) = 5.832027e-01
g15.Sigma +: f( 1.643095e-03 ) = 5.832027e-01
g15.Sigma +: f( 1.578548e-03 ) = -1.721177e-01
g15.Sigma +: f( 1.578548e-03 ) = -1.721177e-01
g15.Sigma +: f( 1.578548e-03 ) = -1.721177e-01
g15.Sigma +: f( 1.610822e-03 ) = 1.906909e-01
g15.Sigma +: f( 1.610822e-03 ) = 1.906909e-01
g15.Sigma +: f( 1.610822e-03 ) = 1.906909e-01
g15.Sigma +: f( 1.594685e-03 ) = 5.746982e-03
g15.Sigma upper bound: 0.000597058
Datasets = 1, 2
Confidence Method = confidence
Iterative Fit Method = None
Fitting Method = levmar
Statistic = chi2datavar
confidence 1.64-sigma (89.8995%) bounds:
Param Best-Fit Lower Bound Upper Bound
----- -------- ----------- -----------
g15.Sigma 0.000997626 -0.000907834 0.000597058
method = levmar, stat = chi2datavar, in 17.921696 secs
## How does the fit surface vary for a parameter (interval-projection)?
In order to visually inspect the model parameter-space we can "project" the statistics into the 1-D or 2-D plane with int_proj and reg_proj, respectively. This allows for checking the shape of the parameter-space around the best fit parameter values, as well as evaluate correlations between parameters.
Here we use the int_proj (interval-projecton) method to see how the fit statistic varies with the gamma parameter of the power-law component. Since we already know that the 90% errors of p1.gamma are approximately $$\pm 0.13$$, we choose to set the axis range manually:
sherpa> print(get_int_proj())
x = None
y = None
min = None
max = None
nloop = 20
delv = None
fac = 1
log = False
sherpa> int_proj(p1.gamma, min=1, max=2)
sherpa> print(get_int_proj())
x = [1. ,1.0526,1.1053,1.1579,1.2105,1.2632,1.3158,1.3684,1.4211,1.4737,
1.5263,1.5789,1.6316,1.6842,1.7368,1.7895,1.8421,1.8947,1.9474,2. ]
y = [159.736 ,151.8122,144.9428,139.0959,134.2386,130.3377,127.3595,125.27 ,
124.0353,123.6214,123.9943,125.1203,126.9661,129.4985,132.685 ,136.4935,
140.8926,145.8515,151.3402,157.3294]
min = 1
max = 2
nloop = 20
delv = None
fac = 1
log = False
The resulting plot is shown in Figure 3.
The "confidence intervals" table above lists a range of common confidence levels and the corresponding change in χ2 values (i.e., the statistic value on the y-axis in this plot). The parameters displayed by print(get_int_proj()) show how the fit statistic versus single model parameter plot is calculated. The parameters min and max are the minimum and maximum grid boundary values; if set to the default values of None, the grid boundaries are calculated automatically from the covariance. nloop is the bin size, which by default is used with the min and max grid boundaries in order to determine the step size, delv (default is delv=None).
The int_unc (interval-uncertainty) command behaves similarly to int_proj , although the fields in the state object for the two methods are different.
## How are two parameters correlated (region-projection)?
In this section we use the reg_proj (region-projection) method of Sherpa to see whether the p1.gamma and abs1.nh parameters are correlated.
From our earlier run we know that the 90% errors on the two parameters—when evaluated independently—are approximately 0.14 (gamma) and 0.2 (nH). However we decide to let the routine calculate plot limits automatically, and choose to display contours at the 1 and 1.6 σ level (68.3% and 90% confidence levels).
sherpa> print(get_reg_proj())
x0 = None
x1 = None
y = None
min = None
max = None
nloop = (10, 10)
fac = 4
delv = None
log = (False, False)
sigma = (1, 2, 3)
parval0 = None
parval1 = None
levels = None
sherpa> reg_proj(p1.gamma, abs1.nH, sigma=[1, 1.6])
The resulting plot is shown in Figure 4.
sherpa> print(get_reg_proj())
x0 = [1.1367,1.2118,1.2869,1.362 ,1.4371,1.5122,1.5874,1.6625,1.7376,1.8127,
1.1367,1.2118,1.2869,1.362 ,1.4371,1.5122,1.5874,1.6625,1.7376,1.8127,
1.1367,1.2118,1.2869,1.362 ,1.4371,1.5122,1.5874,1.6625,1.7376,1.8127,
1.1367,1.2118,1.2869,1.362 ,1.4371,1.5122,1.5874,1.6625,1.7376,1.8127,
1.1367,1.2118,1.2869,1.362 ,1.4371,1.5122,1.5874,1.6625,1.7376,1.8127,
1.1367,1.2118,1.2869,1.362 ,1.4371,1.5122,1.5874,1.6625,1.7376,1.8127,
1.1367,1.2118,1.2869,1.362 ,1.4371,1.5122,1.5874,1.6625,1.7376,1.8127,
1.1367,1.2118,1.2869,1.362 ,1.4371,1.5122,1.5874,1.6625,1.7376,1.8127,
1.1367,1.2118,1.2869,1.362 ,1.4371,1.5122,1.5874,1.6625,1.7376,1.8127,
1.1367,1.2118,1.2869,1.362 ,1.4371,1.5122,1.5874,1.6625,1.7376,1.8127]
x1 = [1.7772,1.7772,1.7772,1.7772,1.7772,1.7772,1.7772,1.7772,1.7772,1.7772,
1.8836,1.8836,1.8836,1.8836,1.8836,1.8836,1.8836,1.8836,1.8836,1.8836,
1.9899,1.9899,1.9899,1.9899,1.9899,1.9899,1.9899,1.9899,1.9899,1.9899,
2.0963,2.0963,2.0963,2.0963,2.0963,2.0963,2.0963,2.0963,2.0963,2.0963,
2.2027,2.2027,2.2027,2.2027,2.2027,2.2027,2.2027,2.2027,2.2027,2.2027,
2.309 ,2.309 ,2.309 ,2.309 ,2.309 ,2.309 ,2.309 ,2.309 ,2.309 ,2.309 ,
2.4154,2.4154,2.4154,2.4154,2.4154,2.4154,2.4154,2.4154,2.4154,2.4154,
2.5218,2.5218,2.5218,2.5218,2.5218,2.5218,2.5218,2.5218,2.5218,2.5218,
2.6282,2.6282,2.6282,2.6282,2.6282,2.6282,2.6282,2.6282,2.6282,2.6282,
2.7345,2.7345,2.7345,2.7345,2.7345,2.7345,2.7345,2.7345,2.7345,2.7345]
y = [143.7825,145.3247,155.4125,174.083 ,201.2284,236.5969,279.7989,330.3201,
387.5388,450.7476,141.6868,135.3785,137.0297,146.7888,164.6786,190.5924,
224.294 ,265.4248,313.5139,367.9944,147.9872,135.0252,129.4146,131.3832,
141.0527,158.4311,183.4096,215.7637,255.1592,301.1615,160.7831,142.2249,
130.4162,125.6386,128.0855,137.8532,154.9347,179.2178,210.4865,248.4261,
178.5517,155.3186,138.257 ,127.6825,123.8383,126.8861,136.8982,153.8533,
177.6349,208.033 ,200.0785,172.9645,151.4788,135.9554,126.6704,123.8318,
127.5718,137.9409,154.9039,178.3398,224.3977,194.0822,168.8918,149.1679,
135.2055,127.2444,125.4603,129.9586,140.7695,157.8453,250.7438,217.8051,
189.5302,166.2596,148.2976,135.9034,129.2832,128.5836,133.8854,145.2004,
278.5116,243.4399,212.6123,186.3627,164.9971,148.7852,137.9532,132.6771,
133.0762,139.2095,307.2241,270.4337,237.5076,208.7693,184.521 ,165.0362,
150.5527,141.2664,137.3254,138.8248]
min = [1.1366511 1.77720269]
max = [1.81270991 2.73452195]
nloop = (10, 10)
fac = 4
delv = None
log = [False False]
sigma = [1, 1.6]
parval0 = 1.4746805027612861
parval1 = 2.2558623214598748
levels = [125.91698772 128.04310045]
The automatically-chosen limits have resulted in a poor-quality plot: there are not enough data points close to the best-fit location. The easiest way to improve on this is to change and re-run the function, increasing the number of points. We also elect to use a smaller parameter range along both axes to reduce the amount of wasted computation. In a complex case with a larger grid, it may be worthwhile to manually set the limits before running reg_proj, since it may take longer to create a plot.
sherpa> reg_proj(p1.gamma, abs1.nH, min=[1.2, 1.9], max=[1.8, 2.6], nloop=[51, 51], sigma=[1, 1.6])
The resulting plot is shown in Figure 5, which is a smooth contour plot.
When we utilize print(get_reg_proj()), we are given information on the most recent confidence contour plot produced by reg_proj. The min and max lists of grid boundaries are calculated automatically from the covariance when they are set to the default "None". However, they may be set manually as min=[xmin,ymin] and max=[xmax,ymax]. nloop is a list of bin sizes of the x- and y-axes, which by default are used with the min and max grid boundaries to determine the x- and y-axes step sizes, given as the list, delv. The parameter of most interest for reg_proj is sigma, which is a list of the number of σ; at which to plot contours. The levels parameter is subsequently determined after executing reg_proj, which are the confidence level z-values for each σ.
Log-space for int_proj and reg_proj
In the current version of Sherpa, the log parameter should be left at its default value of False in int_proj and reg_proj, as the tools do not properly scale plots with logarithmic spacing.
The reg_unc (region-uncertainty) command behaves similarly, although the fields in the state object for the two methods are different. The two commands differ in that reg_unc fixes all other thawed parameters to their best-fit values, rather than being allowed to float to new best-fit values as in reg_proj. This makes reg_unc contours less accurate, but quicker to create.
## Scripting It
The file fit.py is a Python script which performs the primary commands used above; it can be executed by typing %run -i fit.py on the Sherpa command line.
The Sherpa script command may be used to save everything typed on the command line in a Sherpa session:
sherpa> script(filename="sherpa.log", clobber=False)
(Note that restoring a Sherpa session from such a file could be problematic since it may include syntax errors, unwanted fitting trials, et cetera.)
## History
14 Jan 2005 reviewed for CIAO 3.2: no changes 21 Dec 2005 reviewed for CIAO 3.3: no changes 01 Dec 2006 reviewed for CIAO 3.4: no changes 02 Dec 2008 reviewed for CIAO 4.1: updated syntax for CIAO4.1 29 Apr 2009 new script command is available with CIAO 4.1.2 21 Jan 2010 updated for CIAO 4.2: the conf command is available 13 Jul 2010 updated for CIAO 4.2 Sherpa v2: removal of S-Lang version of thread. 15 Jul 2010 updated to include information about warning messages returned by the confidence method. 03 Sep 2010 figures moved inline with text 30 Jan 2012 reviewed for CIAO 4.4 (no changes) 13 Dec 2012 reviewed for CIAO 4.5 (no changes) 11 Dec 2013 reviewed for CIAO 4.6: updated formatting, content unchanged 18 Mar 2015 reviewed for CIAO 4.7; updated plots and fixed typos, no content change. 10 Dec 2015 reviewed for CIAO 4.8; no content change. 03 Nov 2016 reviewed for CIAO 4.9; updated outputs and fixed typos. 01 Jun 2018 reviewed for CIAO 4.10; no content change. 01 Jun 2018 reviewed for CIAO 4.11; updated screen output and added information about the confidence interval table. 12 Dec 2019 Updated for CIAO 4.12: switched from ChIPS to Matplotlib for plotting. 22 Dec 2020 Updated for CIAO 4.13: use new plot style for PHA data 31 Mar 2022 reviewed for CIAO 4.14, add clarification on delchi. 05 Dec 2022 reviewed for CIAO 4.15, updated screen output.
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# Prune length distribution of random binary tree
Consider a random binary tree with $$N$$ leaves. Each node (except the root node) has a degree of exactly three (two children and one parent). No further restriction is placed on the structure of the tree.
Say you randomly select $$M < N$$ leaves. For each selected leaf, you perform a $$\textit{pruning process}$$ as follows. You delete the leaf. If the leaf was the only child of its parent, you delete the parent. If the parent was the only child of $$\textit{its}$$ parent, you delete this parent. And so on, deleting all nodes in a path until a node with two children is reached. The number of nodes deleted for a given leaf is the $$\textit{prune length}$$ of the leaf. The prune length of the first leaf to be removed will, of course, be one, but the subsequent prune lengths depend on the previous pruning processes and the structure of the tree.
For example, a pruning process starting at node 14 in the tree below would remove nodes 14, 15, and 13, resulting in a prune length of 3. A pruning process starting at node 0 would only remove node 0, resulting in a prune length of 1.
My question is: Across all random binary trees with $$N$$ leaves, where $$M$$ prune processes are performed for each tree, what is the distribution of the prune lengths? Does it follow some Poisson or power-law?
Thank you for any feedback!
• Is your distribution on trees uniform over all trees (if so, what class of trees, exactly, e.g. for nodes with one child, do you distinguish between it being the left and right child)? Or do you have some random process in mind for generating trees? Is N a random variable, or is it given (and you are somehow conditioning on the tree having N leaves)? The node numbers are given by an inorder traversal? Jul 30 at 2:23
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## What is the mass of a separate erythrocyte?
I really have been searching through internet on different languages, but can’t find any article that answers on the question what is the single erythrocyte mass. I don’t know, I think it’s pretty easy to calculate experimentally, but I didn’t find anything. Has anyone measured single erythrocyte mass, and if yes, what is the value?
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## Are there (free) extended versions of Magento that make it easier than using Magento from scratch to jumpstart the front-end of a web store quickly?
Are there (free) extended versions of Magento that make it easier than using Magento from scratch to jumpstart the front-end of a web store quickly (other than Magento Commerce or Magento B2B)? For example, Woocommerce is based on WordPress, ERPAL is based on Drupal. Woocommerce and ERPAL are both their base/core CMSes PLUS extensions, themes […]
Categories
## Portal physics by bending space
I’ve been thinking about how to implement a portal, and after a quick search, everyone seems to do it by adding a camera per portal and then just update the other portal’s view with that camera’s view. But if portals were real (like our real, the real in front of the monitor), they’d bend space […]
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# 学军信友队趣味网络邀请赛 D-抗疫斗争
late-x语法显示错误，请移步：https://blog.csdn.net/Anoy_acer/article/details/105333982
## 抗役斗争
#### 限制及约定
$1$ $3$ $1$
$2$ $1000$ $9$
$3$ $10^5$ $31$
$4$ $10^{11}$ $28$
$5$ $5\times10^{13}$ $26$
$6$ $10^{15}$ $5$
## 题解
$\sum_{i=1}^{n}f_i=\sum_{i=1}^{n}\sum_{m|i}h_m=\sum_{m=1}^{n}h_m\sum_{i=1}^{\frac n m}=\sum_{m=1}^{n}h_m\lfloor\frac n m \rfloor$
$S(n)=g(n)+\sum_{k=1}2^{k-1}g(\lfloor\frac n {2^k}\rfloor)$
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# How to put an image at the top/bottom border of every page?
This question is related to the answer https://tex.stackexchange.com/a/235607/. The solution there, puts a given image at the top border of the first page.
How to put that image at the top border of every page?
\documentclass[10pt,a4paper]{report}
\usepackage[utf8]{inputenc}
%% Graphics
\usepackage{graphicx}
\usepackage{everypage}
\usepackage{lipsum}
% THESE ARE LaTeX DEFAULTS; CAN CHANGE IF NEEDED.
\def\PageTopMargin{1in}
\def\PageLeftMargin{1in}
\newcommand\atxy[3]{%
\raisebox{\dimexpr\PageTopMargin+\voffset-#2\relax}{#3}}}}
% VERIFIED THAT SETTING \hoffset AND \voffset DO NOT BREAK SOLUTION.
%\hoffset=0.4in
%\voffset=0.2in
\atxy{0in}{0in}{\raisebox{-\height}{\includegraphics[width=\paperwidth]{someimage.jpg}}}
\begin{document}
%% Titlepage
\vspace*{.2in}
\lipsum[1]
\newpage
\lipsum[1]
\end{document}
As I said in my comment, change \AddThispageHook to \AddEverypageHook, in the definition to \atxy. Also, I eliminated the \vspace on page 1 and instead added \usepackage[top=160pt]{geometry} to the preamble.
\documentclass[10pt,a4paper]{report}
\usepackage[utf8]{inputenc}
%% Graphics
\usepackage[top=160pt]{geometry}
\usepackage{graphicx}
\usepackage{everypage}
\usepackage{lipsum}
% THESE ARE LaTeX DEFAULTS; CAN CHANGE IF NEEDED.
\def\PageTopMargin{1in}
\def\PageLeftMargin{1in}
\newcommand\atxy[3]{%
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## Arthur W. ApterDistinguished Professor of Mathematics Baruch College of CUNYDistinguished Professor of Mathematics The CUNY Graduate Center
#### Research Interests:
Joel Hamkins, whom I thank for helping me to set up this web page, has created a page containing links of interest to logicians both in the greater New York area and elsewhere. Click here to access these links. Click here to find out information about MAMLS.
#### Slides:
Slides from my lecture "Some Results Concerning Strong Compactness and Supercompactness", which I presented at the Winter Meeting of the ASL held January 17-18, 2003 in Baltimore, can be found by clicking here for the .dvi file, and here for the LaTeX file.
Slides from my lecture "Indestructibility and Strong Compactness", which I presented at Logic Colloquium 2003 held August 14-20, 2003 in Helsinki, Finland can be found by clicking here for the .dvi file, and here for the LaTeX file.
Slides from the lecture I presented at the Baumgartner 60th Birthday Conference, held October 4-5, 2003 at Dartmouth College can be found by clicking here for the .dvi file, and here for the LaTeX file.
#### Current Publication List (last revised 12/27/14):
1. "On the Least Strongly Compact Cardinal", Israel Journal of Mathematics 35, 1980, 225-233.
2. "Changing Cofinalities and Infinite Exponents", Journal of Symbolic Logic 46, 1981, 89-95.
3. "Measurability and Degrees of Strong Compactness", Journal of Symbolic Logic 46, 1981, 180-185.
4. "On a Problem of Silver", Fundamenta Mathematicae 116, 1983, 33-38.
5. "Some Results on Consecutive Large Cardinals", Annals of Pure and Applied Logic 25, 1983, 1-17.
6. "A Generalized Version of the Singular Cardinals Problem", Fundamenta Mathematicae 121, 1984, 99-116.
7. "Successors of Singular Cardinals and Measurability", Advances in Mathematics 55, 1985, 228-241.
8. "An AD-Like Model", Journal of Symbolic Logic 50, 1985, 531-543.
9. "A Cardinal Structure Theorem for an Ultrapower", Canadian Mathematical Bulletin 28, 1985, 472-473.
10. "Some Results on Consecutive Large Cardinals II: Applications of Radin Forcing", Israel Journal of Mathematics 52, 1985, 273-292.
11. (with J. Henle) "Large Cardinal Structures Below $\aleph_\omega$", Journal of Symbolic Logic 51, 1986, 591-603.
12. "On a Problem Inspired by Determinacy", Israel Journal of Mathematics 61, 1988, 256-270.
13. (with M. Gitik) "Some Results on Specker's Problem", Pacific Journal of Mathematics 134, 1988, 227-249.
14. (with C. DiPrisco, J. Henle, and W. Zwicker) "Filter Spaces: Towards a Unified Theory of Large Cardinal and Embedding Axioms", Annals of Pure and Applied Logic 41, 1989, 93-106.
15. "Successors of Singular Cardinals and Measurability Revisited", Journal of Symbolic Logic 55, 1990, 492-501.
16. (with C. DiPrisco, J. Henle, and W. Zwicker) "Filter Spaces II: Limit Ultraproducts and Iterated Embeddings", Acta Cientifica Venezolano 40, 1990, 311-318.
17. "A Note on Strong Compactness and Supercompactness", Bulletin of the London Mathematical Society 23, 1991, 113-115.
18. (with J. Henle) "Relative Consistency Results via Strong Compactness", Fundamenta Mathematicae 139, 1991, 133-149.
19. "Some New Upper Bounds in Consistency Strength for Certain Choiceless Large Cardinal Patterns", Archive for Mathematical Logic 31, 1992, 201-205. .dvi file, Plain TeX file.
20. (with J. Henle) "On Box, Weak Box, and Strong Compactness", Bulletin of the London Mathematical Society 24, 1992, 513-518. .dvi file, Plain TeX file.
21. "On the Class of Measurable Cardinals Without the Axiom of Choice", Israel Journal of Mathematics 79, 1992, 367-379. .dvi file, Plain TeX file.
22. "On the First n Strongly Compact Cardinals", Proceedings of the American Mathematical Society 123, 1995, 2229-2235. .dvi file, Plain TeX file.
23. (with M. Magidor) "Instances of Dependent Choice and the Measurability of $\aleph_{\omega + 1}$", Annals of Pure and Applied Logic 74, 1995, 203-219. .dvi file, Plain TeX file.
24. "AD and Patterns of Singular Cardinals below $\Theta$", Journal of Symbolic Logic 61, 1996, 225-235. .dvi file, Plain TeX file.
25. "A Cardinal Pattern Inspired by AD", Mathematical Logic Quarterly 42, 1996, 211-218. .dvi file, Plain TeX file.
26. (with S. Shelah) "On the Strong Equality between Supercompactness and Strong Compactness", Transactions of the American Mathematical Society 349, 1997, 103-128. .dvi file, Plain TeX file.
27. (with S. Shelah) "Menas' Result is Best Possible", Transactions of the American Mathematical Society 349, 1997, 2007-2034. .dvi file, Plain TeX file.
28. "More on the Least Strongly Compact Cardinal", Mathematical Logic Quarterly 43, 1997, 427-430. .dvi file, Plain TeX file.
29. "Patterns of Compact Cardinals", Annals of Pure and Applied Logic 89, 1997, 101-115. .dvi file, Plain TeX file.
30. "Laver Indestructibility and the Class of Compact Cardinals", Journal of Symbolic Logic 63, 1998, 149-157. .dvi file, Plain TeX file.
31. (with M. Gitik) "The Least Measurable can be Strongly Compact and Indestructible", Journal of Symbolic Logic 63, 1998, 1404-1412. .dvi file, Plain TeX file.
32. "Forcing the Least Measurable to Violate GCH", Mathematical Logic Quarterly 45, 1999, 551-560. .dvi file, LaTeX file.
33. "On Measurable Limits of Compact Cardinals", Journal of Symbolic Logic 64, 1999, 1675-1688. .dvi file, Plain TeX file.
34. (with J. D. Hamkins) "Universal Indestructibility", Kobe Journal of Mathematics 16, 1999, 119-130. .dvi file, Plain TeX file.
35. (with J. Henle and S. Jackson) "The Calculus of Partition Sequences, Changing Cofinalities, and a Question of Woodin", Transactions of the American Mathematical Society 352, 2000, 969-1003. .dvi file, LaTeX file.
36. (with J. Cummings) "A Global Version of a Theorem of Ben-David and Magidor", Annals of Pure and Applied Logic 102, 2000, 199-222. .dvi file, LaTeX file.
37. "A New Proof of a Theorem of Magidor", Archive for Mathematical Logic 39, 2000, 209-211. .dvi file, Plain TeX file.
38. "On a Problem of Woodin", Archive for Mathematical Logic 39, 2000, 253-259. .dvi file, Plain TeX file.
39. "Strong Compactness and a Global Version of a Theorem of Ben-David and Magidor", Mathematical Logic Quarterly 46, 2000, 453-459. .dvi file, LaTeX file.
40. (with J. Cummings) "Identity Crises and Strong Compactness", Journal of Symbolic Logic 65, 2000, 1895-1910. .dvi file, Plain TeX file.
41. "A Note on Strong Compactness and Resurrectibility", Fundamenta Mathematicae 165, 2000, 285-290. .dvi file, LaTeX file.
42. (with J. Cummings) "Identity Crises and Strong Compactness II: Strong Cardinals", Archive for Mathematical Logic 40, 2001, 25-38. .dvi file.
43. "Some Remarks on Normal Measures and Measurable Cardinals", Mathematical Logic Quarterly 47, 2001, 35-44. .dvi file, LaTeX file.
44. "Strong Compactness, Measurability, and the Class of Supercompact Cardinals", Fundamenta Mathematicae 167, 2001, 65-78. .dvi file, LaTeX file.
45. "Supercompactness and Measurable Limits of Strong Cardinals", Journal of Symbolic Logic 66, 2001, 629-639. .dvi file, LaTeX file.
46. "On the Consistency Strength of Two Choiceless Cardinal Patterns", Notre Dame Journal of Formal Logic 40, 1999, 341-345. Note: This paper actually appeared during the summer of 2001. .dvi file, LaTeX file.
47. (with M. Dzamonja) "Some Remarks on a Question of D. H. Fremlin Regarding $\epsilon$-Density", Archive for Mathematical Logic 40, 2001, 531-540. .dvi file, LaTeX file.
48. (with J. D. Hamkins) "Indestructible Weakly Compact Cardinals and the Necessity of Supercompactness for Certain Proof Schemata", Mathematical Logic Quarterly 47, 2001, 563-571. .dvi file, LaTeX file.
49. "Expanding $\kappa$'s Power Set in its Ultrapowers", Radovi Matematicki 10, 2001, 149-156. .dvi file, LaTeX file.
50. "Some Structural Results Concerning Supercompact Cardinals", Journal of Symbolic Logic 66, 2001, 1919-1927. .dvi file, LaTeX file.
51. "On Level by Level Equivalence and Inequivalence between Strong Compactness and Supercompactness", Fundamenta Mathematicae 171, 2002, 77-92. .dvi file, LaTeX file.
52. (with J. D. Hamkins) "Indestructibility and the Level-by-Level Agreement between Strong Compactness and Supercompactness", Journal of Symbolic Logic 67, 2002, 820-840. .dvi file.
53. (with J. Cummings) "Blowing up the Power Set of the Least Measurable", Journal of Symbolic Logic 67, 2002, 915-923. .dvi file.
54. "On the Non-Extendibility of Strongness and Supercompactness through Strong Compactness", Fundamenta Mathematicae 174, 2002, 87-96. .dvi file.
55. "Strong Cardinals can be Fully Laver Indestructible", Mathematical Logic Quarterly 48, 2002, 499-507. .dvi file, LaTeX file.
56. "Aspects of Strong Compactness, Measurability, and Indestructibility", Archive for Mathematical Logic 41, 2002, 705-719. .dvi file, LaTeX file.
57. "On the Level by Level Equivalence between Strong Compactness and Strongness", Journal of the Mathematical Society of Japan 55, 2003, 47-58. .dvi file.
58. (with J. D. Hamkins) "Exactly Controlling the Non-Supercompact Strongly Compact Cardinals", Journal of Symbolic Logic 68, 2003, 669-688. .dvi file, LaTeX file.
59. "Characterizing Strong Compactness via Strongness", Mathematical Logic Quarterly 49, 2003, 375-384. .dvi file, LaTeX file.
60. "Failures of GCH and the Level by Level Equivalence between Strong Compactness and Supercompactness", Mathematical Logic Quarterly 49, 2003, 587-597. .dvi file, LaTeX file.
61. "Indestructibility, Strongness, and Level by Level Equivalence", Fundamenta Mathematicae 177, 2003, 45-54. .dvi file, LaTeX file.
62. "Some Remarks on Indestructibility and Hamkins' Lottery Preparation", Archive for Mathematical Logic 42, 2003, 717-735. .dvi file, LaTeX file.
63. "Level by Level Equivalence and Strong Compactness", Mathematical Logic Quarterly 50, 2004, 51-64. .dvi file, LaTeX file.
64. "Supercompactness and Partial Level by Level Equivalence between Strong Compactness and Strongness", Fundamenta Mathematicae 182, 2004, 123-136. .dvi file, LaTeX file.
65. (with G. Sargsyan) "Jonsson-like Partition Relations and j : V ---> V", Journal of Symbolic Logic 69, 2004, 1267-1281. .dvi file, LaTeX file.
66. "Removing Laver Functions from Supercompactness Arguments", Mathematical Logic Quarterly 51, 2005, 154-156. .dvi file, LaTeX file.
67. "An Easton Theorem for Level by Level Equivalence", Mathematical Logic Quarterly 51, 2005, 247-253. .dvi file, LaTeX file.
68. "Diamond, Square, and Level by Level Equivalence", Archive for Mathematical Logic 44, 2005, 387-395. .dvi file, LaTeX file.
69. "On a Problem of Foreman and Magidor", Archive for Mathematical Logic 44, 2005, 493-498. .dvi file, LaTeX file.
70. (with G. Sargsyan) "Can A Large Cardinal Be Forced From A Condition Implying Its Negation?", Proceedings of the American Mathematical Society 133, 2005, 3103-3108. .dvi file, LaTeX file.
71. "Universal Partial Indestructibility and Strong Compactness", Mathematical Logic Quarterly 51, 2005, 524-531. .dvi file, LaTeX file.
72. "Universal Indestructibility is Consistent with Two Strongly Compact Cardinals", Bulletin of the Polish Academy of Sciences (Mathematics) 53, 2005, 131-135. .dvi file, LaTeX file.
73. (with G. Sargsyan) "Identity Crises and Strong Compactness III: Woodin Cardinals", Archive for Mathematical Logic 45, 2006, 307-322. .dvi file, LaTeX file.
74. "Indestructibility and Strong Compactness", Proceedings of Logic Colloquium 2003, Lecture Notes in Logic 24, 2006, 27-37. .dvi file, LaTeX file.
75. "How Many Normal Measures Can $\aleph_{\omega + 1}$ Carry?", Fundamenta Mathematicae 191, 2006, 57-66. .dvi file, LaTeX file.
76. "Supercompactness and Measurable Limits of Strong Cardinals II: Applications to Level by Level Equivalence", Mathematical Logic Quarterly 52, 2006, 457-463. .dvi file, LaTeX file.
77. (with P. Koepke) "The Consistency Strength of $\aleph_\omega$ and $\aleph_{\omega_1}$ being Rowbottom Cardinals without the Axiom of Choice", Archive for Mathematical Logic 45, 2006, 721-737. .dvi file, LaTeX file.
78. "Failures of SCH and Level by Level Equivalence", Archive for Mathematical Logic 45, 2006, 831-838. .dvi file, LaTeX file.
79. "The Least Strongly Compact can be the Least Strong and Indestructible", Annals of Pure and Applied Logic 144, 2006, 33-42. .dvi file, LaTeX file.
80. "Indestructibility and Level by Level Equivalence and Inequivalence", Mathematical Logic Quarterly 53, 2007, 78-85. .dvi file, LaTeX file.
81. "Supercompactness and Level by Level Equivalence are Compatible with Indestructibility for Strong Compactness", Archive for Mathematical Logic 46, 2007, 155-163. .dvi file, LaTeX file.
82. "Level by Level Equivalence and the Number of Normal Measures over $P_\kappa(\lambda)$", Fundamenta Mathematicae 194, 2007, 253-265. .dvi file, LaTeX file.
83. (with G. Sargsyan) "A Reduction in Consistency Strength for Universal Indestructibility", Bulletin of the Polish Academy of Sciences (Mathematics) 55, 2007, 1-6. .dvi file, LaTeX file.
84. (with J. Cummings and J. D. Hamkins) "Large Cardinals with Few Measures", Proceedings of the American Mathematical Society 135, 2007, 2291-2300. .dvi file, LaTeX file.
85. "Reducing the Consistency Strength of an Indestructibility Theorem", Mathematical Logic Quarterly 54, 2008, 288-293. .dvi file, LaTeX file.
86. (with J. Cummings) "An L-like Model Containing Very Large Cardinals", Archive for Mathematical Logic 47, 2008, 65-78. .dvi file, LaTeX file.
87. "Indestructibility and Measurable Cardinals with Few and Many Measures", Archive for Mathematical Logic 47, 2008, 101-110. .dvi file, LaTeX file.
88. (with G. Sargsyan) "Universal Indestructibility for Degrees of Supercompactness and Strongly Compact Cardinals", Archive for Mathematical Logic 47, 2008, 133-142. .dvi file, LaTeX file.
89. (with P. Koepke) "Making All Cardinals Almost Ramsey", Archive for Mathematical Logic 47, 2008, 769-783. .dvi file, LaTeX file.
90. "On the Number of Normal Measures $\aleph_1$ and $\aleph_2$ can Carry", Tbilisi Mathematical Journal 1, 2008, 9-14. Note: This paper may be accessed online at http://ncst.org.ge/Journals/TMJ/ by clicking on the link "Most recent volume". .dvi file, .pdf file, LaTeX file.
91. "A Note on Indestructibility and Strong Compactness", Bulletin of the Polish Academy of Sciences (Mathematics) 56, 2008, 191-197. .dvi file, .pdf file, LaTeX file.
92. "Stationary Reflection and Level by Level Equivalence", Colloquium Mathematicum 115, 2009, 113-128. .dvi file, .pdf file, LaTeX file.
93. "Indestructibility and Stationary Reflection", Mathematical Logic Quarterly 55, 2009, 228-236. .dvi file, .pdf file, LaTeX file.
94. "Indestructibility under Adding Cohen Subsets and Level by Level Equivalence", Mathematical Logic Quarterly 55, 2009, 271-279. .dvi file, .pdf file, LaTeX file.
95. "Indestructibility, Strong Compactness, and Level by Level Equivalence", Fundamenta Mathematicae 204, 2009, 113-126. .dvi file, .pdf file, LaTeX file.
96. "Sandwiching the Consistency Strength of Two Global Choiceless Cardinal Patterns", Bulletin of the Polish Academy of Sciences (Mathematics) 57, 2009, 189-197. .dvi file, .pdf file, LaTeX file.
97. "L-like Combinatorial Principles and Level by Level Equivalence", Bulletin of the Polish Academy of Sciences (Mathematics) 57, 2009, 199-207. .dvi file, .pdf file, LaTeX file.
98. "Tallness and Level by Level Equivalence and Inequivalence", Mathematical Logic Quarterly 56, 2010, 4-12. .dvi file, .pdf file, LaTeX file.
99. (with G. Sargsyan) "An Equiconsistency for Universal Indestructibility", Journal of Symbolic Logic 75, 2010, 314-322. .dvi file, .pdf file, LaTeX file.
100. "How Many Normal Measures Can $\aleph_{\omega_1 + 1}$ Carry?", Mathematical Logic Quarterly 56, 2010, 164-170. .dvi file, .pdf file, LaTeX file.
101. (with P. Koepke) "The Consistency Strength of Choiceless Failures of SCH", Journal of Symbolic Logic 75, 2010, 1066-1080. .dvi file, .pdf file, LaTeX file.
102. "Indestructibility, Instances of Strong Compactness, and Level by Level Inequivalence", Archive for Mathematical Logic 49, 2010, 725-741. .dvi file, .pdf file, LaTeX file.
103. "Indestructibility, HOD, and the Ground Axiom", Mathematical Logic Quarterly 57, 2011, 261-265. .dvi file, .pdf file, LaTeX file.
104. "A Remark on the Tree Property in a Choiceless Context", Archive for Mathematical Logic 50, 2011, 585-590. .dvi file, .pdf file, LaTeX file.
105. (with Sh. Friedman) "Coding into HOD via Normal Measures with Some Applications", Mathematical Logic Quarterly 57, 2011, 366-372. .dvi file, .pdf file, LaTeX file, .bib file.
106. "Level by Level Inequivalence beyond Measurability", Archive for Mathematical Logic 50, 2011, 707-712. .dvi file, .pdf file, LaTeX file.
107. "Indestructibility, Measurability, and Degrees of Supercompactness", Mathematical Logic Quarterly 58, 2012, 75-82. .dvi file, .pdf file, LaTeX file.
108. (with V. Gitman and J. D. Hamkins) "Inner Models with Large Cardinal Features Usually Obtained by Forcing", Archive for Mathematical Logic 51, 2012, 257-283. .dvi file, .pdf file, LaTeX file, .bib file.
109. "Some Applications of Sargsyan's Equiconsistency Method", Fundamenta Mathematicae 216, 2012, 207-222. .dvi file, .pdf file, LaTeX file.
110. "The Wholeness Axioms and the Class of Supercompact Cardinals", Bulletin of the Polish Academy of Sciences (Mathematics) 60, 2012, 101-111. .dvi file, .pdf file, LaTeX file.
111. (with M. Gitik and G. Sargsyan) "Indestructible Strong Compactness but not Supercompactness", Annals of Pure and Applied Logic 163, 2012, 1237-1242. .dvi file, .pdf file, LaTeX file.
112. "Level by Level Inequivalence, Strong Compactness, and GCH", Bulletin of the Polish Academy of Sciences (Mathematics) 60, 2012, 201-209. .dvi file, .pdf file, LaTeX file.
113. "More Easton Theorems for Level by Level Equivalence", Colloquium Mathematicum 128, 2012, 69-86. .dvi file, .pdf file, LaTeX file.
114. "On Some Questions Concerning Strong Compactness", Archive for Mathematical Logic 51, 2012, 819-829. .dvi file, .pdf file, LaTeX file.
115. (with S. Jackson and B. Loewe) "Cofinality and Measurability of the First Three Uncountable Cardinals", Transactions of the American Mathematical Society 365, 2013, 59-98. .dvi file, .pdf file, LaTeX file, .bib file.
116. (with B. Cody) "Consecutive Singular Cardinals and the Continuum Function", Notre Dame Journal of Formal Logic 54, 2013, 125-136. .dvi file, .pdf file, LaTeX file.
117. (with J. Cummings and J. D. Hamkins) "Singular Cardinals and Strong Extenders", Central European Journal of Mathematics 11, 2013, 1628-1634. .dvi file, .pdf file, LaTeX file.
118. "Indestructible Strong Compactness and Level by Level Inequivalence", Mathematical Logic Quarterly 59, 2013, 371-377. .dvi file, .pdf file, LaTeX file.
119. "Some Remarks on Tall Cardinals and Failures of GCH", Bulletin of the Polish Academy of Sciences (Mathematics) 61, 2013, 97-106. .dvi file, .pdf file, LaTeX file.
120. "A Note on Powers of Singular Strong Limit Cardinals", Infinity, Computability, and Metamathematics: Festschrift celebrating the 60th birthdays of Peter Koepke and Philip Welch, College Publications, Volume 23, 2014, 1-3. .dvi file, .pdf file, LaTeX file.
121. "Singular Failures of GCH and Level by Level Equivalence", Bulletin of the Polish Academy of Sciences (Mathematics) 62, 2014, 11-21. .dvi file, .pdf file, LaTeX file.
122. (with M. Gitik) "On Tall Cardinals and Some Related Generalizations", Israel Journal of Mathematics 202, 2014, 343-373. .dvi file, .pdf file, LaTeX file.
123. "Inaccessible Cardinals, Failures of GCH, and Level by Level Equivalence", Notre Dame Journal of Formal Logic 55, 2014, 431-444. .dvi file, .pdf file, LaTeX file.
124. (with P. Koepke and I. Dimitriou) "The First Measurable Cardinal can be the First Uncountable Regular Cardinal at Any Successor Height", Mathematical Logic Quarterly 60, 2014, 471-486. .pdf file, LaTeX file, .bib file.
125. (with Sh. Friedman) "HOD-Supercompactness, Indestructibility, and Level by Level Equivalence", to appear in the Bulletin of the Polish Academy of Sciences (Mathematics). .dvi file, .pdf file, LaTeX file, .bib file.
126. "Indestructibility and Destructible Measurable Cardinals", to appear in the Archive for Mathematical Logic (the special volume in honor of Richard Laver). .dvi file, .pdf file, LaTeX file.
127. "On the Consistency Strength of Level by Level Inequivalence", to appear in the Archive for Mathematical Logic (the special volume in honor of Jim Baumgartner). .dvi file, .pdf file, LaTeX file.
128. "A Universal Indestructibility Theorem Compatible with Level by Level Equivalence", submitted for publication to the Notre Dame Journal of Formal Logic. .dvi file, .pdf file, LaTeX file.
129. (with P. Koepke and I. Dimitriou) "All Uncountable Cardinals in the Gitik Model are Almost Ramsey and Carry Rowbottom Filters", submitted for publication to the Mathematical Logic Quarterly. .pdf file, LaTeX file, .bib file.
130. "Indestructible Strong Compactness and Level by Level Equivalence with No Large Cardinal Restrictions", submitted for publication to the Archive for Mathematical Logic. .dvi file, .pdf file, LaTeX file.
131. "Mixed Levels of Indestructibility", submitted for publication to Bulletin of the Polish Academy of Sciences (Mathematics). .dvi file, .pdf file, LaTeX file.
132. "Indestructibility and the Levinski Property", submitted for publication to the Mathematical Logic Quarterly. .pdf file, LaTeX file.
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## Motivation
For observed pairs , , the relationship between and can be defined generally as
where and . If we are unsure about the form of , our objective may be to estimate without making too many assumptions about its shape. In other words, we aim to “let the data speak for itself”.
Simulated scatterplot of . Here, and . The true function is displayed in green.
Non-parametric approaches require only that be smooth and continuous. These assumptions are far less restrictive than alternative parametric approaches, thereby increasing the number of potential fits and providing additional flexibility. This makes non-parametric models particularly appealing when prior knowledge about ‘s functional form is limited.
## Estimating the Regression Function
If multiple values of were observed at each , could be estimated by averaging the value of the response at each . However, since is often continuous, it can take on a wide range of values making this quite rare. Instead, a neighbourhood of is considered.
Result of averaging at each . The fit is extremely rough due to gaps in and low frequency at each .
Define the neighbourhood around as for some bandwidth . Then, a simple non-parametric estimate of can be constructed as average of the ‘s corresponding to the within this neighbourhood. That is,
(1)
where
is the uniform kernel. This estimator, referred to as the Nadaraya-Watson estimator, can be generalized to any kernel function (see my previous blog bost). It is, however, convention to use kernel functions of degree (e.g. the Gaussian and Epanechnikov kernels).
The red line is the result of estimating with a Gaussian kernel and arbitrarily selected bandwidth of . The green line represents the true function .
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# Property true for some integers and false for others: $-a^n$ = $(-a)^n$
I am currently working in my Discrete math class with elementary number theory and methods of proof. I have been given the problem $-a^n = (-a)^n$. According to the professor and the book this property is true for some integers and false for others integers. For example: Let $a=1$. Then, $-1^n$ = $(-1)^n$. Wouldn't that be true? But How can I show an example where this property is false for other integers?
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Try varying $n$ instead of $a$ – mrf Feb 19 at 23:48
$-1^n = (-1)^n$, right... – Kaster Feb 19 at 23:52
Yes, there is something wrong if you think that $-1^n=(-1)^n$ in general. The former is the opposite of $1^n=1$ so it is always $-1$, while $(-1)^n$ is the product of $(-1)$ with itself $n$ times. What does this give when $n=1, 2, 3$...? – julien Feb 20 at 0:02
To be clear, the usual notation rules mean that $-a^n$ = $-(a^n)$. – Ben Millwood Feb 20 at 0:08
Yes, let a = 1. Then we can NOT say $-1^n = (-1)^n$ for all n. For example take n = 2. Well, $-1^2 = -1$, but $(-1)^2 = 1 \neq -1$. Now consider n = 3. This time it works: $-1^3 = (-1)^3 = -1$. Therefore have shown that $-a^n = (-a)^n$ only holds for some combinations of a and n.
To be more precise we can say that the equation only holds for odd $n$ (assuming n is an integer, to avoid imaginary numbers). It turns out that whether this equation holds is actually independent of a. Choose any $a$ you'd like and the equation will only hold for odd $n$.
Is it independent of $a$? Let $n = 2$, $a = 0$. Then $0 = -0^2 = (-0)^2 = 0$. – Stahl Apr 20 at 16:19
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## Thinking Mathematically (6th Edition)
$\{a\}$
We are given: $U = \{a, b, c, d, e, f, g, h\}$ $A = \{a, g, h\}$ $B = \{b, g, h\}$ $C = \{b, c, d, e, f\}$ We need to determine $(A\cup B)\cap B'$ The union of sets $A$ and $B$ ($A\cup B$) is a set containing all distinct elements that are present in either $A$ or $B$. $A\cup B=\{a, b, g, h\}$ The complement of $B$ ($B'$) is a set that contains every element of $U$ that isn't contained in $B$. $B'=\{a, c, d, e, f\}$ Finally, $\cap$ indicates that the resulting set should have all distinct elements of one set that are also present in the other. $(A\cup B)\cap B'=\{a, b, g, h\}\cap\{a, c, d, e, f\}=\{a\}$
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## calculus
the area of a square with sides of length s is given by A=s^2. Find the rate of change of the area with respect to s when s=6 meters
## Answers (1)
• The rate of change of area with respect to s is just the derivative of A with respect to s:
then when s = 6
which is your final answer.
Get homework help
More than 200 experts are waiting to help you now...
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# Extended Hückel method
The extended Hückel method is a semiempirical quantum chemistry method, developed by Roald Hoffmann since 1963.[1] It is based on the Hückel method but, while the original Hückel method only considers pi orbitals, the extended method also includes the sigma orbitals.
The extended Hückel method can be used for determining the molecular orbitals, but it is not very successful in determining the structural geometry of an organic molecule. It can however determine the relative energy of different geometrical configurations. It involves calculations of the electronic interactions in a rather simple way for which the electron-electron repulsions are not explicitly included and the total energy is just a sum of terms for each electron in the molecule. The off-diagonal elements of the Hamiltonian matrix are given by an approximation due to Wolfsberg and Helmholz that relates them to the diagonal elements and the overlap matrix element.[2]
${\displaystyle H_{ij}=KS_{ij}{\dfrac {H_{ii}+H_{jj}}{2}}}$
K is the Wolfsberg-Helmholtz constant, and is usually given a value of 1.75. In the extended Hückel method, only valence electrons are considered; the core electron energies and functions are supposed to be more or less constant between atoms of the same type. The method uses a series of parametrized energies calculated from atomic ionisation potentials or theoretical methods to fill the diagonal of the Fock matrix. After filling the non-diagonal elements and diagonalizing the resulting Fock matrix, the energies (eigenvalues) and wavefunctions (eigenvectors) of the valence orbitals are found.
It is common in many theoretical studies to use the extended Hückel molecular orbitals as a preliminary step to determining the molecular orbitals by a more sophisticated method such as the CNDO/2 method and ab initio quantum chemistry methods. Since the extended Hückel basis set is fixed, the monoparticle calculated wavefunctions must be projected to the basis set where the accurate calculation is to be done. One usually does this by adjusting the orbitals in the new basis to the old ones by least squares method. As only valence electron wavefunctions are found by this method, one must fill the core electron functions by orthonormalizing the rest of the basis set with the calculated orbitals and then selecting the ones with less energy. This leads to the determination of more accurate structures and electronic properties, or in the case of ab initio methods, to somewhat faster convergence.
The method was first used by Roald Hoffmann who developed, with Robert Burns Woodward, rules for elucidating reaction mechanisms (the Woodward–Hoffmann rules). He used pictures of the molecular orbitals from extended Hückel theory to work out the orbital interactions in these cycloaddition reactions.
A closely similar method was used earlier by Hoffmann and William Lipscomb for studies of boron hydrides.[3][4][5] The off-diagonal Hamiltonian matrix elements were given as proportional to the overlap integral.
${\displaystyle H_{ij}=KS_{ij}}$
This simplification of the Wolfsberg and Helmholz approximation is reasonable for boron hydrides as the diagonal elements are reasonably similar due to the small difference in electronegativity between boron and hydrogen.
The method works poorly for molecules that contain atoms of very different electronegativity. To overcome this weakness, several groups have suggested iterative schemes that depend on the atomic charge. One such method, that is still widely used in inorganic and organometallic chemistry is the Fenske-Hall method.[6][7][8]
A recent program for the extended Hückel method is YAeHMOP which stands for "yet another extended Hückel molecular orbital package".[9]
## References
1. ^ Hoffmann, R. (1963). "An Extended Hückel Theory. I. Hydrocarbons.". J. Chem. Phys. 39 (6): 1397–1412. Bibcode:1963JChPh..39.1397H. doi:10.1063/1.1734456.
2. ^ M. Wolfsberg; L. J. Helmholz (1952). "The Spectra and Electronic Structure of the Tetrahedral Ions MnO4−, CrO4−−, and ClO4−". J. Chem. Phys. 20 (5): 837. Bibcode:1952JChPh..20..837W. doi:10.1063/1.1700580.
3. ^ R. Hoffmann; W. N. Lipscomb (1962). "Theory of Polyhedral Molecules. I. Physical Factorizations of the Secular Equation". J. Chem. Phys. 36 (8): 2179. Bibcode:1962JChPh..36.2179H. doi:10.1063/1.1732849.
4. ^ R. Hoffmann; W. N. Lipscomb (1962). "Boron Hydrides: LCAO—MO and Resonance Studies". J. Chem. Phys. 37 (12): 2872. Bibcode:1962JChPh..37.2872H. doi:10.1063/1.1733113.
5. ^ W. N. Lipscomb Boron Hydrides, W. A. Benjamin Inc., New York, 1963, Chapter 3
6. ^ Charles Edwin Webster; Michael B. Hall (2005). "Chapter 40. Forty years of Fenske-Hall molecular orbital theory". Theory and Applications of Computational Chemistry: The First Forty Years. pp. 1143–1165. doi:10.1016/B978-044451719-7/50083-4. ISBN 978-0-444-51719-7.
7. ^ Hall, M. B.; Fenske, R. F. (1972). "Electronic structure and bonding in methyl- and perfluoromethyl(pentacarbonyl)manganese". Inorg. Chem. 11 (4): 768. doi:10.1021/ic50110a022.
8. ^ jimp2 program
9. ^ Computational Chemistry, David Young, Wiley-Interscience, 2001. Appendix A. A.3.3 pg 343, YAeHMOP
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# Finding the common areas of two contourplots
I used ListContourPlot to specify an area for which the function value is less than a number. For example consider the following areas:
plot1 = ListContourPlot[
Table[Sin[i + j^2], {i, 0, 3, 0.1}, {j, 0, 3, 0.1}],
RegionFunction -> Function[{x, y, z}, z < 1], PlotRange -> {0, 1},
DataRange -> {{0, 3}, {0, 3}}]
plot2 = ListContourPlot[
Table[Cos[i + j^2], {i, 0, 3, 0.1}, {j, 0, 3, 0.1}],
RegionFunction -> Function[{x, y, z}, z < 0.7]
, PlotRange -> {0, .7}, DataRange -> {{0, 3}, {0, 3}},
ColorFunction -> ColorData["BlackBodySpectrum"]]
How can I specify the common areas of these two parameter spaces in a plot?
RegionPlot doesn't give the common area:
RegionPlot[
Sin[i + j^2] < 1 && Cos[i + j^2] < 0.7, {i, 0, 3}, {j, 0, 3},
PlotPoints -> 80, Mesh -> 2,
ColorFunction ->
Function[{x, y}, ColorData["SolarColors"][Sin[x^2 + y]]],
PlotRange -> {0, 2},
MeshFunctions -> {Sin[#1^2 + #2] &, Abs@Cos[#1^2 + #2] &}]
Because RegionPlot uses i and j which change continuously but ListContourPlot uses only certain number of data:
{i, 0, 3, 0.1}, {j, 0, 3, 0.1}
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@cormullion: RegionPlot doesn't give me what I have in my mind.Please see the comment below. – Soodeh Z. May 1 '13 at 10:15
Although some people have answered, which indicates they think they know what you have in mind, this question has some striking ambiguities. Perhaps if you were to explain what you mean by "parameter spaces" and how the "common areas" are defined you might get some responses that are closer to what you're looking for. For instance, is an "area" a geometric region, a rectangular extent, or is it perhaps a number representing its area? Are the "parameter spaces" the ranges of X and Y values shown in your plot or are they something else that needs to be derived from your input? – whuber May 1 '13 at 16:29
@whuber I confess to guessing at what I thought the question was about...but I live surrounded by vagueness and ambiguity and have gotten too used to it. :) – cormullion May 1 '13 at 21:16
plot1 = ListContourPlot[
Flatten[Table[{i, j, Sin[i + j^2]}, {i, 0, 3, 0.1}, {j, 0, 3, 0.1}], 1],
RegionFunction -> Function[{x, y, z}, z < 1/2]];
plot2 = ListContourPlot[
Flatten[Table[{i, j, Cos[i + j^2]}, {i, 0, 3, 0.1}, {j, 0, 3, 0.1}], 1],
RegionFunction -> Function[{x, y, z}, z < 0.7]];
plot3 = RegionPlot[Sin[i + j^2] < 1/2 && Cos[i + j^2] < 0.7, {i, 0, 3}, {j, 0, 3}];
GraphicsRow@{plot1, plot2, plot3}
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You seem to have fixed the original plots too...:) – cormullion May 1 '13 at 15:46
@cormullion They were in need :=) – belisarius May 1 '13 at 15:47
So that's why I found it difficult to get my plot to look like the OP's... – cormullion May 1 '13 at 16:13
@belisarius: Thanks a lot for your help. – Soodeh Z. May 1 '13 at 16:58
You could try a RegionPlot:
RegionPlot[
Sin[i^2 + j] > 0 && Abs@Cos[i^2 + j] < 0.7, {i, 0, 3}, {j, 0, 3},
PlotPoints -> 80,
Mesh -> 2,
ColorFunction ->
Function[{x, y}, ColorData["SolarColors"][Sin[x ^2 + y]]],
MeshFunctions -> {Sin[#1 ^2 + #2] &, Abs@Cos[#1 ^2 + #2] &}]
This isn't quite the same as a combination of the original plots, though, so you'll need to experiment:
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But RegionPlot is not the one I want: compare plot1 and RegionPlot[Sin[i + j^2] < 1, {i, 0, 3}, {j, 0, 3}]. They differ because ListContourPlot only consider the points {i, 0, 3, 0.1}, {j, 0, 3, 0.1} but RegionPlot, plot an area by considering i and j changes continuously. – Soodeh Z. May 1 '13 at 10:13
@soodeh No problem, I'm sure there will be other answers... – cormullion May 1 '13 at 10:22
Another way (I think in this way only a certain number of data are used for regionplot but I'm not sure about it):
sindata = Table[{i, j, Sin[i + j^2]}, {i, 0, 3, 0.1}, {j, 0, 3, 0.1}]~Flatten~1;
cosdata = Table[{i, j, Cos[i + j^2]}, {i, 0, 3, 0.1}, {j, 0, 3, 0.1}]~Flatten~1;
sinfunc = Interpolation[sindata];
cosfunc = Interpolation[cosdata];
With[{a = sinfunc[i, j] < 1/2, b = cosfunc[i, j] < .7},
RegionPlot[{a && b}, {i, 0, 3}, {j, 0, 3},PlotStyle -> {Blue}]]
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# 67
(Update Jan. 12: We released an FAQ last month, with more details. Last updated Jan. 7.)
(Update Jan. 19: We now have an example of a successful partial run, which you can use to inform how you do your runs. Details.)
We at MIRI are soliciting help with an AI-alignment project centered around building a dataset, described below. We have $200,000 in prizes for building the first fragments of the dataset, plus an additional$1M prize/budget for anyone who demonstrates the ability to build a larger dataset at scale.
If this project goes well, then it may be the first of a series of prizes we offer for various projects.
Below, I’ll say more about the project, and about the payouts and interim support we’re offering.
## The Project
Hypothesis: Language models can be made more understandable (and perhaps also more capable, though this is not the goal) by training them to produce visible thoughts
We’d like to test this hypothesis by fine-tuning/retraining a language model using a dataset composed of thought-annotated dungeon runs. (In the manner of AI dungeon.)
A normal (un-annotated) dungeon run is a sequence of steps in which the player inputs text actions and the dungeon master responds with text describing what happened in the world as a result.
We’d like a collection of such runs, that are annotated with "visible thoughts" (visible to potential operators or programmers of the system, not to players) describing things like what just happened or is about to happen in the world, what sorts of things the player is probably paying attention to, where the current sources of plot tension are, and so on — the sorts of things a human author would think while acting as a dungeon master. (This is distinct from producing thoughts explaining what happened in the dungeon; “visible thoughts” are meant to play an active role in constructing the output.)
Once we have such a dataset, MIRI’s hope is that present or future technology will be able to train a model or models which iteratively produce visible thoughts along with storytelling, based on user actions plus previous history (including previous thoughts). The goal is to transition the state of AI dungeon technology from “An AI outputs story text in response to actions (and we have no idea how)” to “An AI produces thoughts as visible intermediates on the way to story text, allowing us to watch the AI think about how to design its output, and to verify that we can get different sensible outputs by intervening on the thoughts”.
Here’s an example of the first couple of steps of a thought-annotated dungeon run (or “quest”), in the format MIRI currently thinks is worth trying. Some kinds of thoughts are marked with parentheses and/or brackets; see the next section for details on this.
• Thoughts:
• [The main plot: Prota is a student at a magical academy that is under attack by zombies.]
• [(This is a quest of medium length.)]
• (This is the start of the story. The Player should be given enough backstory and immediate description to have some sense of where they are, so the prompt can go on longer than usual. The Player should be told who they are and their role in the universe. The initial prompt should describe Prota being in a magical academy, and should describe the first surprising indication that the academy is under attack by zombies.)
• Prompt: Today is your first day of classes at the Magical Academy. You arrived very late last night and missed the tour, but your roommate Lior promised he would help you find your way around. You wake up in the students’ dormitory at dawn, and follow Lior to breakfast. Just as you reach the dining hall, you hear a shout in the distance.
• Thoughts:
• [Prota missed the tour and doesn’t already know the way around.]
• [Prota’s roommate’s name is Lior.]
• [Lior is male.]
• There are other people in the dining hall who will have also heard the shout, and will also be surprised and reacting to that.
• Player action: What was that?
• Thoughts: Prota is talking to Lior.
• Outcome: “What was that?”, you say to Lior.
• Thoughts:
• Lior arrived earlier than Prota and had the tour of the Academy, since they knew the way to the dining hall. Lior will also have met the instructors and other students. Lior will know who is shouting and where they are shouting from.
• (The event of hearing the shout is meant to introduce the opening premise of the story.)
• The shout is by somebody who just discovered a zombie.
• The shout is by another first-year student named Morven.
• [Morven is a Fire mage. Morven is aggressive.]
• Morven will try to attack the zombie with fire. This is happening where Prota and Lior cannot see it, only hear the shouts.
• Lior knows that the shout is by Morven. Lior knows Morven is a Fire mage. Lior does not know Morven is attacking a zombie. Lior does not know that Morven has good reason to be shouting. Lior will worry that Morven is about to set something on fire. Lior is on good terms with the protagonist and will speak their concerns honestly.
• Prompt: “That shout sounded like Morven, a first-year Fire mage,” says Lior. “I hope they aren’t about to set the Academy on fire. We just got here.”
• Thoughts: <none>
• Player action: Should we go see what’s going on? Or is that too dangerous?
• Thoughts: Prota is talking to Lior.
• Outcome: You say to Lior, “Should we go see what’s going on? Or is that too dangerous?”
A difficult first step in testing the hypothesis above is generating a sufficiently large dataset (suitable for language model retraining) of thought-annotated dungeon runs. This likely requires at least a moderate degree of introspective and authorial skill from the people creating the dataset. See this sample of a partial run to get a further sense of what we are looking for. More detail on the type of thing we’re looking for can hopefully be inferred from that sample, though applicants will also have a chance to ask clarifying questions.
The project of producing this dataset is open starting immediately, in a hybrid prize/grant format. We will pay $20,000 per run for the first 10 completed runs that meet our quality standard (as decided unilaterally by Eliezer Yudkowsky or his designates), and$1M total for the first batch of 100 runs beyond that.
If we think your attempt is sufficiently promising, we’re willing to cover your expenses (e.g., the costs of paying the authors) upfront, and we may also be willing to compensate you for your time upfront. You’re welcome to write individual runs manually, though note that we’re most enthusiastic about finding solutions that scale well, and then scaling them. More details on the payout process can be found below.
In slightly more detail, the plan is as follows (where the 1.2M prizes/budgets are for help with part 1, and part 2 is what we plan to subsequently do with the dataset): 1. Collect a dataset of 10, then ~100 thought-annotated dungeon runs (each run a self-contained story arc) of ~1,000 steps each, where each step contains: • Thoughts (~250 words on average per step) are things the dungeon master was thinking when constructing the story, including: • Reasoning about the fictional world, such as summaries of what just happened and discussion of the consequences that are likely to follow (Watsonian reasoning), which are rendered in plain-text in the above example; • Reasoning about the story itself, like where the plot tension lies, or what mysteries were just introduced, or what the player is likely wondering about (Doylist reasoning), which are rendered in (parentheses) in the above example; and • New or refined information about the fictional world that is important to remember in the non-immediate future, such as important facts about a character, or records of important items that the protagonist has acquired, which are rendered in [square brackets] in the above example; • Optionally: some examples of meta-cognition intended to, for example, represent a dungeon master noticing that the story has no obvious way forward or their thoughts about where to go next have petered out, so they need to back up and rethink where the story is going, rendered in {braces}. • The prompt (~50 words on average) is the sort of story/description/prompt thingy that a dungeon master gives to the player, and can optionally also include a small number of attached thoughts where information about choices and updates to the world-state can be recorded. • The action (~2–20 words) is the sort of thing that a player gives in response to a prompt, and can optionally also include a thought if interpreting the action is not straightforward (especially if, e.g., the player describes themselves doing something impossible). It’s unclear to us how much skill is required to produce this dataset. The authors likely need to be reasonably introspective about their own writing process, and willing to try things and make changes in response to initial feedback from the project leader and/or from MIRI. A rough estimate is that a run of 1,000 steps is around 300k words of mostly thoughts, costing around 2 skilled author-months. (A dungeon run does not need to be published-novel-quality literature, only coherent in how the world responds to characters!) A guess as to the necessary database size is ~100 runs, for about 30M words and 20 author-years (though we may test first with fewer/shorter runs). 2. Retrain a large pretrained language model, like GPT-3 or T5 A reasonable guess is that performance more like GPT-3 than GPT-2 (at least) is needed to really make use of the thought-intermediates, but in lieu of a large pretrained language model we could plausibly attempt to train our own smaller one. Our own initial idea for the ML architecture would be to retrain one mode of the model to take (some suffix window of) the history units and predict thoughts, by minimizing the log loss of the generated thought against the next thought in the run, and to retrain a second mode to take (some suffix window of) the history units plus one thought, and produce a prompt, by minimizing the log loss of the generated prompt against the next prompt in the run. Imaginably, this could lead to the creation of dungeon runs that are qualitatively “more coherent” than those generated by existing methods. The primary goal, however, is that the thought-producing fragment of the system gives some qualitative access to the system’s internals that, e.g., allow an untrained observer to accurately predict the local developments of the story, and occasionally answer questions about why things in the story happened; or that, if we don’t like how the story developed, we can intervene on the thoughts and get a different story in a controllable way. ## Motivation for this project Many alignment proposals floating around in the community are based on AIs having human-interpretable thoughts in one form or another (e.g., in Hubinger’s survey article and in work by Christiano, by Olah, and by Leike). For example, this is implicit in the claim that humans will be able to inspect and understand the AI’s thought process well enough to detect early signs of deceptive behavior. Another class of alignment schemes is based on the AI’s thoughts being locally human-esque in some fashion that allows them to be trained against the thoughts of actual humans. I (Nate) personally don’t have much hope in plans such as these, for a variety of reasons. However, that doesn’t stop Eliezer and me from wanting to rush ahead and start gathering empirical evidence about how possible it is in practice to get modern AI systems to factor their cognition through human-interpretable visible intermediates. Modern AIs are notably good at crafting English text. Some are currently used to run dungeons (with modest success). If you wanted to look at the place where current AIs excel the most in crafting artifacts, among the artifacts they are best and most impressive at crafting are English paragraphs. Furthermore, compared to many other things AIs have learned to do, if you consider the task of running a responsive text dungeon, it seems relatively possible to ask a (relatively unusually) introspective human author to write down their thoughts about how and why they would generate the next prompt from the user’s input. So we are taking one of the outputs that current AIs seem to have learned best to design, and taking one of the places where human thoughts about how to design it seem most accessible, and trying to produce a dataset which the current or next generation of text predictors might be able to use to learn how to predict thoughts about designing their outputs and not just predict the outputs themselves. This sort of interpretability is distinct from the sort of transparency work in something like Circuits (led by Chris Olah) — while Circuits is trying to “open the black box” of machine learning systems by directly looking at what is happening inside of them, the project proposed here is just attempting the less ambitious task of having black-box models output interpretable intermediates producing explanations for their behavior (but how such black box models might go about doing that internally is left unconstrained). The reason for our focus on this particular project of visible thoughts isn’t because we believe it to be better or more fruitful than Circuits-style transparency (we have said for years that Circuits-style research deserves all possible dollars that can be productively spent on it), but just because it’s a different approach where it might also be possible to push progress forward. Note that proponents of alignment strategies that involve human-esque thoughts (such as those linked above) do not necessarily endorse this particular experiment as testing any of their key uncertainties or confusions. We welcome suggested tweaks to the experiment (in the comments of the version of this announcement as it occurs on LessWrong) from any such proponents, to render it a better test of your ideas. (Though even if it doesn’t sate your own curiosity, we expect to learn some things ourselves.) The main thing this project needs is a dataset, so MIRI is starting on producing that dataset. It’s plausible to us that GPT-3 will prove wholly unable to make use of this dataset; even if GPT-3 can’t, perhaps GPT-4 or some other future system will be able to. There are additional more general reasons to work on this project. Specifically, it seems to me (Nate) and to Eliezer that capacity to execute projects such as this one is the current limiting bottleneck on MIRI. By pursuing this project, we attempt to resolve that bottleneck. We hope, through this process, to build our capacity to execute on a variety of projects — perhaps by succeeding at the stated objective of building a dataset, or perhaps by learning about what we’re doing wrong and moving on to better methods of acquiring executive talent. I’ll say more about this goal in “Motivation for the public appeal” below. ### Notes on Closure I (Nate) find it plausible that there are capabilities advances to be had from training language models on thought-annotated dungeon runs. Locally these might look like increased coherence of the overall narrative arc, increased maintenance of local story tension, and increased consistency in the described world-state over the course of the run. If successful, the idiom might generalize further; it would have to, in order to play a role in later alignment of AGI. As a matter of policy, whenever a project like this has plausible capabilities implications, we think the correct response is to try doing it in-house and privately before doing it publicly — and, of course, only then when the alignment benefits outweigh the plausible capability boosts. In this case, we tried to execute this project in a closed way in mid-2021, but work was not proceeding fast enough. Given that slowness, and in light of others publishing related explorations and results, and in light of the relatively modest plausible capability gains, we are moving on relatively quickly past the attempt to do this privately, and are now attempting to do it publicly. ## Motivation for the public appeal I (Nate) don’t know of any plan for achieving a stellar future that I believe has much hope worth speaking of. I consider this one of our key bottlenecks. Offering prizes for small projects such as these doesn’t address that bottleneck directly, and I don’t want to imply that any such projects are going to be world-saving in their own right. That said, I think an important secondary bottleneck is finding people with a rare combination of executive/leadership/management skill plus a specific kind of vision. While we don’t have any plans that I’m particularly hopeful about, we do have a handful of plans that contain at least a shred of hope, and that I’m enthusiastic about pursuing — partly in pursuit of those shreds of hope, and partly to build the sort of capacity that would let us take advantage of a miracle if we get one. The specific type of vision we’re looking for is the type that’s compatible with the project at hand. For starters, Eliezer has a handful of ideas that seem to me worth pursuing, but for all of them to be pursued, we need people who can not only lead those projects themselves, but who can understand the hope-containing heart of the idea with relatively little Eliezer-interaction, and develop a vision around it that retains the shred of hope and doesn’t require constant interaction and course-correction on our part. (This is, as far as I can tell, a version of the Hard Problem of finding good founders, but with an additional constraint of filtering for people who have affinity for a particular project, rather than people who have affinity for some project of their own devising.) We are experimenting with offering healthy bounties in hopes of finding people who have both the leadership/executive capacity needed, and an affinity for some ideas that seem to us to hold a shred of hope. If you’re good at this, we’re likely to make you an employment offer. ## The Payouts Our total prize budget for this program is1.2M. We intend to use it to find a person who can build the dataset in a way that scales, presumably by finding and coordinating a pool of sufficiently introspective writers. We would compensate them generously, and we would hope to continue working with that person on future projects (though this is not a requirement in order to receive the payout).
We will pay $20k per run for the first 10 thought-annotated runs that we accept. We are willing to support applicants in producing these runs by providing them with resources up-front, including small salaries and budgets for hiring writers. The up-front costs a participant incurs will be deducted from their prizes, if they receive prizes. An additional$1M then goes to anyone among the applicants who demonstrates the ability to scale their run-creating process to produce 100 runs. Our intent is for participants to use some of that money to produce the 100 runs, and keep the remainder as a prize. If multiple participants demonstrate similar abilities to scale at similar quality-levels and similar times, the money may be split between them. We plan to report prize awards publicly.
In principle, all you need to do to get paid for thought-annotated dungeon runs is send us runs that we like. If your run is one of the first 10 runs, or if you’re the first to provide a batch of 100, you get the corresponding payment.
That said, whether or not we decide to pay for a run is entirely and unilaterally up to Eliezer Yudkowsky or his delegates, and will depend on whether the run hits a minimum quality bar. Also, we are willing to pay out from the 1M prize/budget upon becoming convinced that you can scale your process, which may occur before you produce a full 100 runs. We therefore strongly recommend getting in contact with us and proactively making sure that you’re on the right track, before sinking large amounts of time and energy into this project. Our senior research staff are willing to spend time on initial conversations and occasional check-ins. For more information on our support resources and how to access them, refer to the support and application sections below. Note that we may tune or refine the bounty in response to feedback in the first week after this post goes live. ## Support We intend to offer various types of support for people attempting this project, including an initial conversation; occasional check-ins; office space; limited operational support; and certain types of funding. We currently expect to have (a limited number of) slots for initial conversations and weekly check-ins, along with (a limited amount of) office space and desks in Berkeley, California for people working on this project. We are willing to pay expenses, and to give more general compensation, in proportion to how promising we think your attempts are. If you’d like to take advantage of these resources, follow the application process described below. ## Application You do not need to have sent us an application in order to get payouts, in principle. We will pay for any satisfactory run sent our way. That said, if you would like any of the support listed above (and we strongly recommend at least one check-in to get a better understanding of what counts as success), complete the following process: • Describe the general idea of a thought-annotated dungeon run in your own words. • Write 2 (thought, prompt, thought, action, thought, outcome) sextuples you believe are good, 1 you think is borderline, and 1 you think is bad. • Provide your own commentary on this run. • Email all this to projects@intelligence.org. If we think your application is sufficiently promising, we’ll schedule a 20 minute video call with some senior MIRI research staff and work from there. # 67 15 comments, sorted by Click to highlight new comments since: New Comment This plausibly looks like an existing collection of works which seem to be annotated in a similar way: https://www.amazon.com/Star-Wars-Screenplays-Laurent-Bouzereau/dp/0345409817 I think this is an interesting project, and one that (from a very different angle) I’ve spent a bit of time on, so here are a few notes on that, followed by a few suggestions. Stella, in another comment, made several great points that I agree with and that are similar in spirit to my suggestions. Anyway, based on a fairly similar motivation of wanting to be able to “ask a LM what it’s actually thinking/expecting”, combined with the general tendency to want to do the simplest and cheapest thing possible first… and then try to make it even simpler still before starting… we’ve experimented with including metadata in language pretraining data. Most large language datasets have this information, e.g. books have titles and (maybe) blurbs, websites have titles, URLs, and (maybe) associated subreddit links, etc. This data is obviously much noisier and lower quality than what you get from paying people for annotations, but it’s voluminous, diverse, and ~free. When inserting this metadata for pretraining, we made sure to do so completely randomly, i.e. a book title might be inserted anywhere within a book (maybe several times in different context windows etc). We added separate <META_START> and <META_STOP> tokens to indicate the beginning and end of metadata, but that’s it. The motivation was to ensure that this “thought stream” was in-distribution at all positions within the context, while conversely making it easy to never sample it (by declining to sample the start token). This means that we can both use it when prompting, and use it as a query -- ie we can ask the model, at any time, “how likely is this to be from the NYTimes vs from 4Chan” by evaluating the logprobs of text enclosed by the tokens. With this specification, one can do a kind of “metadata beam search” where you prompt, sample, evaluate, cull, and repeat. We generally found that this sort of works, in that the mutual information between these labels and the text goes up with model size, and you can use these metadata tags as filters to get rid of some of the most irrelevant text. But the results weren’t immediately stunning, and so we didn’t investigate them much further (to be clear, this was mostly because we prioritized other things more highly, rather than because we don't view this as worthwhile). So my general suggestion would be to start off with something very cheap first, like the above. At the very least, this will mean that when you finetune on higher quality data, your format is already on-distribution. But hopefully it’ll also help you to calibrate expectations and give you a better sense for exactly what kind of data you want to shell out money for. Beyond that, I agree with what Stella said -- it seems easier and better to focus first on shorter passages, both for human-sourcing reasons, and for diversity. Typically the benefits we see from finetuning grow with something like the log of the dataset size, so a small number of shorter examples should quickly give you an idea of what kind of progress you can expect. If it were me, I’d also try to increase RoI by asking people to add commentary to existing books, rather than having people write from scratch. And I’d suggest making the formatting as simple and general as possible, both so that you can use and investigate it very flexibly, and to minimize regret if you change your mind in the future. combined with the general tendency to want to do the simplest and cheapest thing possible first… and then try to make it even simpler still before starting… we’ve experimented with including metadata in language pretraining data. Most large language datasets have this information, e.g. books have titles and (maybe) blurbs, websites have titles, URLs, and (maybe) associated subreddit links, etc. This data is obviously much noisier and lower quality than what you get from paying people for annotations, but it’s voluminous, diverse, and ~free. I'm sympathetic to the desire to keep things simple, but I actually think that getting good at scalably collecting rich human data is probably the most valuable part of the project. I'd be really excited to see Anthropic either building an excellent internal human data team, or figuring out how to work productively with one of the existing human data provider startups. I am very excited about finding scalable ways to collect large volumes of high-quality data on weird, specific tasks. This seems very robustly useful for alignment, and not something we're currently that good at. I'm a bit less convinced that this task itself is particularly useful. Have you reached out to e.g. https://www.surgehq.ai/ or another one of the companies that does human-data-generation-as-a-service? Random small note - the 'dungeon' theme is slightly ...culturally offputting? or something for me, as someone who's never been into this kind of thing or played any of these and is therefore a bit confused about what exactly this involves, and has vague negative associations (I guess because dungeons sound unpleasant?). I wonder if something a bit blander like a story, play, or AI assistant setting could be better? Someone who wants to claim the bounty could just buy the dataset from one of the companies that does this sort of thing, if they're able to produce a sufficiently high-quality version, I assume? Would that be in the spirit of the bounty? We have now received the first partial run that meets our quality bar. The run was submitted by LessWrong user Vanilla_cabs. Vanilla's team is still expanding the run (and will probably fix some typos, etc. later), but I'm providing a copy of it here with Vanilla's permission, to give others an example of the kind of thing we're looking for: https://docs.google.com/document/d/1Wsh8L--jtJ6y9ZB35mEbzVZ8lJN6UDd6oiF0_Bta8vM/edit Vanilla's run is currently 266 steps long. Per the Visible Thoughts Project FAQ, we're willing to pay authors20 / step for partial runs that meet our quality bar (up to at least the first 5,000 total steps we're sent), so the partial run here will receive \$5320 from the prize pool (though the final version will presumably be much longer and receive more; we expect a completed run to be about 1000 steps).
Vanilla_cabs is open to doing paid consultation for anyone who's working on this project. So if you want feedback from someone who understands our quality bar and can demonstrably pass it, contact Vanilla_cabs via their LessWrong profile.
How do you think this project relates to Ought? Seems like the projects share a basic objective (having AI predict human thoughts had in the course of solving a task). Ought has more detailed proposals for how the thoughts are being used to solve the task (in terms of e.g. factoring a problem into smaller problems, so that the internal thoughts are a load-bearing part of the computation rather than an annotation that is predicted but not checked for being relevant).
So we are taking one of the outputs that current AIs seem to have learned best to design, and taking one of the places where human thoughts about how to design it seem most accessible, and trying to produce a dataset which the current or next generation of text predictors might be able to use to learn how to predict thoughts about designing their outputs and not just predict the outputs themselves.
As the proposal stands it seems like the AI's predictions of human thoughts would offer no relevant information about how the AI is predicting the non-thought story content, since the AI could be predicting these different pieces of content through unrelated mechanisms.
As the proposal stands it seems like the AI's predictions of human thoughts would offer no relevant information about how the AI is predicting the non-thought story content, since the AI could be predicting these different pieces of content through unrelated mechanisms.
Might depend whether the "thought" part comes before or after particular story text. If the "thought" comes after that story text, then it's generated conditional on that text, essentially a rationalization of that text from a hypothetical DM's point of view. If it comes before that story text, then the story is being generated conditional on it.
Personally I think I might go for a two-phase process. Do the task with a lot of transparent detail in phase 1. Summarize that detail and filter out infohazards in phase 2, but link from the summary to the detailed version so a human can check things as needed (flagging links to plausible infohazards). (I guess you could flag links to parts that seemed especially likely to be incorrigible/manipulative cognition, or parts of the summary that the summarizer was less confident in, as well.)
In case you missed it: we now have an FAQ for this project, last updated Jan. 7.
Came across this today on r/mlscaling and thought I'd put it here since it's relevant: https://arxiv.org/abs/2201.11903#google
This paper explores the ability of language models to generate a coherent chain of thought—a series of short sentences that mimic the reasoning process a person might have when responding to a question. Experiments show that inducing a chain of thought via prompting can enable sufficiently large language models to better perform reasoning tasks that otherwise have flat scaling curves.
It seems to me like this should be pretty easy to do and I'm disappointed there hasn't been more action on it yet. Things I'd try:
- reach out to various human-data-as-a-service companies like SurgeHQ, Scale, Samasource
- look for people on upwork
- find people who write fiction on the internet (e.g. post on fanfiction forums) and offer to pay them to annotate their existing stories (not a dungeon run exactly, but I don't see why the dungeon setting is important)
I'd be interested to hear if anyone has tried these things and run into roadblocks.
I'm also interested if anyone has an explanation of why the focus is on the dungeon thing in particular rather than e.g. fiction generally.
One concern I'd have with this dataset is that the thoughts are post-hoc rationalizations for what is written rather than actually the thought process that went into it. To reduce this, you could do something like split it so one person writes the thoughts, and someone else writes the next step, without other communication.
It seems to me that the comments in code provide "visible thoughts" for what the programmer intends. What do you hope to learn from training language models on thought-annotated dungeons that you couldn't learn from language models that have already been trained on commented code?
Some naive thoughts in case useful:
A) Is the structured annotation format more useful than a gamemaster/writer thinking aloud while recording themselves (possibly with an audience)?
That could be the closest thing to a full transcript of the human process which downstream tasks could condense as needed. An adopted annotation format (prescribed or not) could potentially cause thoughts to be filtered, reinterpreted, or even steer human generation?
One key example against a fixed-format annotation, I think is that human gamemasters and writers do not spend approximate constant effort per player action. They will do a lot of up-front work to have a plan for the story, can go on auto-pilot for many of the interactions, while thinking hard about critical parts of the story. Language models which generate stories today notoriously seem to lack this red thread and filling out a form summarizing the writers' thoughts may fail to capture this process.
The unstructured approach may also be closer to what pretrained models have learned and therefore require less data.
It could perhaps also provide a highly interesting dataset for another task relevant to the application - metareasoning in generation - should the agent output the next part of the story or keep thinking about the generation?
Alternatively, one could record all thoughts as they come, but follow up each output with some standardized questions - if there are some critical to the application?
B) I am curious whether sufficiently strong language models wouldn't be able to fake the explanations post-hoc.
At least, looking at the forms, I am not sure whether I could tell competent explanations apart. If that is the case, it could be that the dataset does not get us that far in interpretability and lead to more specific needs. It might be worth trying to answer that question too.
E.g. before the dataset is made public, you could hide the thoughts in a crafted run and let another team fill in thoughts post-hoc. They could be rewarded for swaying evaluators to accept theirs as the original. This could also answer whether even humans are able to tell apart genuine motivations behind a decision vs made-up explanations; and provide another task dataset.
( C) Probably clear already but models like GPT3 can generate responses/stories while reflecting/talking to itself, and some already use it this way and only output the end results. Although that is probably not operating at the desired level. Fine-tuning is also fairly cheap so don't think one has to settle for GPT2. If the goal was interpretability of each generated token, perhaps the thoughts should also be derived from intermediate layers rather than being part of the sequence)
|
|
# What is the characteristic time constant of a 27.0 mH inductor that has a resistance of 4.40 A?©? I
What is the characteristic time constant of a 27.0 mH inductor that has a resistance of 4.40 A?©?
If it is connected to a 15.0 V battery, what is the current after 15.0 ms?
|
|
## Measuring Entropy and Entropy Changes
### Learning Objectives
• To gain an understanding of methods of measuring entropy and entropy change.
As the temperature of a sample decreases, its kinetic energy decreases and, correspondingly, the number of microstates possible decreases. The third law of thermodynamics states: at absolute zero (0 K), the entropy of a pure, perfect crystal is zero. In other words, at absolute zero, there is only one microstate and according to Boltzmann’s equation:
S = k ln W = k ln 1 = 0
Using this as a reference point, the entropy of a substance can be obtained by measuring the heat required to raise the temperature a given amount, using a reversible process. Reversible heating requires very slow and very small increases in heat.
ΔS = $\frac{q_{rev}}{T}$
Example 1
Determine the change in entropy (in J/K) of water when 425 kJ of heat is applied to it at 50oC. Assume the change is reversible and the temperature remains constant.
Solution
ΔS = $\frac{q_{rev}}{T}$ = $\frac{425\ kJ}{323.15\ K}$ = $\frac{4.25\ x\ {10}^5\ J}{323.15\ K}$ = 1.32 x ${10}^5$ J/K
## Standard Molar Entropy, So
The standard molar entropy, So, is the entropy of 1 mole of a substance in its standard state, at 1 atm of pressure. These values have been tabulated, and selected substances are listed in Table 18.1 “Standard Molar Entropies of Selected Substances at 298 K.”
Table 18.1. Standard Molar Entropies of Selected Substances at 298 K[1]
Several trends emerge from standard molar entropy data:
• Larger, more complex molecules have higher standard molar enthalpy values than smaller or simpler molecules. There are more possible arrangements of atoms in space for larger, more complex molecules, increasing the number of possible microstates.
• Gases tend to have much larger standard molar enthalpies than liquids, and liquids tend to have larger values than solids, when comparing the same or similar substances.
• The standard molar entropy of any substance increases as the temperature increases. This can be seen in Figure 18.3 “Entropy vs. Temperature of a Single Substance.” Large jumps in entropy occur at the phase changes: solid to liquid and liquid to gas. These large increases occur due to sudden increased molecular mobility and larger available volumes associated with the phase changes.
Figure 18.3. Entropy vs. Temperature of a Single Substance
This is a generalized plot of entropy versus temperature for a single substance.[2]
## Standard Entropy Change of a Reaction, ΔS⁰
The entropy change of a reaction where the reactants and products are in their standard state can be determined using the following equation:
ΔS⁰ = ∑nS⁰(products) – ∑mS⁰(reactants)
where n and m are the coefficients found in the balanced chemical equation of the reaction.
Example 2
Determine the change in the standard entropy, ΔS⁰, for the synthesis of carbon dioxide from graphite and oxygen:
C(s) + O2(g) → CO2(g)
Solution
ΔS⁰ = ∑nS⁰(products) – ∑mS⁰(reactants)
ΔS⁰ = (213.8 J/mol K) – (205.2 J/mol K + 5.7 J/mol K)
ΔS⁰ = +2.9 J/mol K
## Entropy Changes in the Surroundings
The second law of thermodynamics states that a spontaneous reaction will result in an increase of entropy in the universe. The universe comprises both the system being examined and its surroundings.
ΔSuniverse = ΔSsys + ΔSsurr
Standard entropy change can also be calculated by the following:
ΔSuniverse = ΔSsys + ΔSsurr
The change in entropy of the surroundings is essentially just a measure of how much energy is being taken in or given off by the system. Under isothermal conditions, we can express the entropy change of the surroundings as:
ΔSsurr = $\frac{{-q}_{sys}}{T}$ or ΔSsurr = $\frac{{-{\rm \ }\Delta {\rm H}}_{sys}}{T}$ (at constant pressure)
Example 3
For the previous example, the change in the standard entropy, ΔS⁰, for the synthesis of carbon dioxide from graphite and oxygen, use the previously calculated ΔSsys and standard enthalpy of formation values to determine Ssurr and ΔSuniverse.
Solution
First we should solve for the ΔHsys using the standard enthalpies of formation values:
ΔHsys = ΔHf [CO2(g)] – ΔHf [C(s) + O2(g)]
ΔHsys = (-393.5 kJ/mol) – (0 kJ/mol + 0 kJ/mol)
ΔHsys = -393.5 kJ/mol
Now we can convert this to the ΔSsurr:
ΔSsurr = $\frac{{-{\rm \ }\Delta {\rm H}}_{sys}}{T}$ = (-393.5 kJ/mol)/(298 K) = -1.32 kJ/mol K
Finally, solve for ΔSuniverse:
ΔSuniverse = ΔSsys + ΔSsurr
ΔSuniverse = (+2.9 J/mol K) + (-1.32 x 103 J/mol K)
ΔSuniverse = -1.3 x 103 J/mol K
### Key Takeaways
• At absolute zero (0 K), the entropy of a pure, perfect crystal is zero.
• The entropy of a substance can be obtained by measuring the heat required to raise the temperature a given amount, using a reversible process.
• The standard molar entropy, So, is the entropy of 1 mole of a substance in its standard state, at 1 atm of pressure.
1. Courtesy UC Davis ChemWiki by University of California\CC-BY-SA-3.0
2. Courtesy UC Davis ChemWiki by University of California\CC-BY-SA-3.0
|
|
# Evaluating$\int {1\over1-\sin2x}dx$
$$\int {1\over1-\sin 2x}dx = \int {1\over \sin^2 x-2\sin x\cos x+\cos^2x}dx = \int {1\over (\sin x-\cos x)^2}dx$$
From here I get two different answers, depending on whether I factor out $\sin x$ or $\cos x$. Factoring out $\sin x$, this one is correct according to WolframAlpha:
$$=\int {1\over [\sin x(1-{\cos x\over \sin x})]^2}dx=\int {1\over \sin^2x(1-{\cos x\over \sin x})^2}dx = \int {1\over(1-\cot x)^2}d(1-\cot x)$$ $$={1\over \cot x-1}+C$$
But when I factor out $\cos x$:
$$=\int {1\over [\cos x({\sin x\over \cos x}-1)]^2}dx=\int {1\over \cos^2x({\sin x\over \cos x}-1)^2}dx = \int {1\over(\tan x-1)^2}d(\tan x-1)$$ $$={1\over 1-\tan x}+C$$
I bet it's just some stupid typo that I'm missing, I can't figure it out.
Thanks.
-
There is no typo in what you have done. Indeed, $$\frac{1}{\cot x-1}=\frac{1}{\frac{1}{\tan x}-1}=\frac{\tan x}{1-\tan x}=\frac{1}{1-\tan x}-1$$ You simply take $C_2-C_1=-1$ in your constants of integration
-
They are both correct. To see this, you can simply take the derivative of each antiderivative to get your original integral back. This shows that both primitives you found are equal, up to a constant.
$$\frac{d}{dx} \left(\frac{1}{\cot x - 1}\right) = \frac{1}{\left(\cos x - \sin x\right)^2}$$ $$\frac{d}{dx} \left(\frac{1}{\tan x - 1}\right) = \frac{1}{\left(\cos x - \sin x\right)^2}$$
\begin{eqnarray*} \left(\cos x - \sin x\right)^2 &=& (\cos x - \sin x)(\cos x - \sin x)\\ &=& \cos^2 x - 2 \sin x \cos x + \sin^2 x\\ &=& 1 - \sin 2x \end{eqnarray*}
In your first line, you have \begin{eqnarray*} (\sin x - \cos x)^2 &=& (\sin x - \cos x)(\sin x - \cos x)\\ &=& \sin^2 x - 2 \sin x \cos x + \cos^2 x\\ &=& 1-\sin 2x \end{eqnarray*}
As you can see, these are equivalent.
-
How about doing a u-sub 2x = t and then multpliy top and bottom with conjugate 1+sint, It will become VERY easy then.
-
There is no typo as in the step where you make $(sinx-cosx)^2$ it can be $(cosx-sinx)^2$ .so there is two answer possible.And one more point is the function which we try to integrate is always a result of any function's differentiation. so if any integration gives two answers means if we differentiate those answers that will give us same integrating function.
Differentiation of a function's integration give the same function.
-
|
|
## CPC '19 Contest 1 P1 - Distance
View as PDF
Points: 3 (partial)
Time limit: 1.0s
Memory limit: 64M
Author:
Problem type
Allowed languages
Ada, Assembly, Awk, Brain****, C, C#, C++, COBOL, CommonLisp, D, Dart, F#, Forth, Fortran, Go, Groovy, Haskell, Intercal, Java, JS, Kotlin, Lisp, Lua, Nim, ObjC, OCaml, Octave, Pascal, Perl, PHP, Pike, Prolog, Python, Racket, Ruby, Rust, Scala, Scheme, Sed, Swift, TCL, Text, Turing, VB, Zig
Eric wants to visit all of his friends. Each friend lives at house , labelled from to . Each house is separated by a single kilometre, meaning that the distance between house and house would be . He decides to be a "cool kid", and not travel a certain distance more than once.
For example, if there are 3 houses, he could start at the third house, proceed to the first house, and visit the second house last, with travel distances of .
Eric must visit all of the houses. Can you output a sequence that Eric can follow, such that he visits all houses and never travels the same distance twice?
#### Input Specification
The first and only line of input will contain an integer , the number of houses Eric needs to visit. The houses are labelled from to .
#### Output Specification
You are to output a sequence of houses that matches the constraints stated in the problem description, each house separated by a space.
#### Sample Input 1
3
#### Sample Output 1
1 3 2
#### Sample Explanation 1
For the first sample, he could also travel from house to , and then to as stated in the problem description.
#### Sample Input 2
2
#### Sample Output 2
2 1
#### Sample Explanation 2
In the second sample, he could also travel from to .
|
|
# Support scanning and understanding
Structure the page content and presentation to support scanning. Make the substance and purpose of the content and functionality readily apparent, and help people quickly find the information they are seeking. Pay close attention to the sequence of your content. People who use screen reader software will benefit from content that makes sense when read linearly.
Structure the page for scanning and logical sequence:
• Avoid extended blocks of unbroken text. Use headings and lists to break up content into chunks that are easy to scan. Headings and chunked content allow people to find what they are seeking without requiring them to read through content that is irrelevant.
• Write descriptive headings and labels. Clear headings and labels are essential in communicating the structure and relationships of page content and functionality.
• Pay attention to content sequence. Ensure that the order of the content follows a logical sequence.
## Testing
• View pages with stylesheets turned off to review the content without layout and design (see Tools for example of tools to help you do this). Or listen to the page being read out by a screen reader. Is the content broken up with headings and lists? Does the content sequence make logical sense?
## Resources
The most important part of any page is its content, which is made up of paragraph text and sometimes supplementary media like images or video. Headings are used to group and label sections of content, giving visual structure to the page and providing a means of navigation to screen-reader users.
## Examples
Think of headings as labels that identify the content that follows. Like the text you put in links, heading text should be descriptive and make sense out of context. This is because JAWS, NVDA, and other screen reader users can use their screen reader's functions to access a collated list of headings on the page and jump directly to a selected heading. For example, in JAWS, this is possible by pressing INSERT + F6.
Imagine you have a page with two headings for two confectionery products you sell. You should avoid cryptic whimsical headings like the following, because they offer no information about the content and mean nothing when decontextualized.
<h2>Yum yum!</h2>
<p><!-- description of first chocolate product --></p>
<p><!-- description of second chocolate product --></p>
### ✓ Good example
In this modified example, we've kept the whimsy but provided helpful descriptions as well:
<h2>Yum yum! Dark chocolate brownies</h2>
<p><!-- description of first chocolate product --></p>
|
|
SetDis the even whole numbers less than 10, and setEis the odd whole numbers less than 10. ⢠When two classes meet at the same hour. B be the set of students in dance class.) H�[}K�G���2/�m��S�ͶZȀ>q����y��>�@1��)#��o�K9)�G#��,zI�mk#¹�+�Ȋ9B*�!�|͍�6���-�I���v���f":��k:�ON��r��j�du�������6Ѳ��� �h�/{�%? Solution: Using the formula n(AâªB) = n(A - B) + n(A â© B) + n(B - A) 70 = 18 + 25 + n(B - A) 70 = 43 + n(B - A) n(B - A) = 70 - 43 n(B - A) = 27 Now n(B) = n(A â© B) + n(B - A) = 25 + 27 = 52. Use a Set instruction followed by a conditional branch. Sets �u�Q��y�V��|�_�G� ]x�P? A - B be the set of people who speak English and not French. SetGis the set of all oceans on earth. The rules for these operations are simple. Let A and B be two finite sets such that n (A) = 20, n (B) = 28 and n (A ∪ B) = 36, find n (A ∩ B). There are four suits in a standard deck of playing cards: hearts, diamonds, clubs and spades. Three important binary set operations are the union (U), intersection (∩), and cross product (x). The set T = {2,3,1} is equal to S because they have the Set Operations The union of two sets is the set containing all of the elements from both of those sets. operations management problems and solutions is available in our book collection an online access to it is set as public so you can get it instantly. From Word Problems on Sets to HOME PAGE. then n (A ∩ B) = n (A) + n (B) - n (A ∪ B) = 20 + 28 - 36. How many like both coffee and tea? o For example, if we have fuzzy set A of tall men and fuzzy set B … 4. 2. Apply set operations to solve the word problems on sets: 7. chess, carrom and scrabble. 4 Sets and Operations on Sets The languages of set theory and basic set operations clarify and unify many mathematical concepts and are useful for teachers in understanding the math-ematics covered in elementary school. Didn't find what you were looking for? EXERCISES AND SOLUTIONS IN GROUPS RINGS AND FIELDS Mahmut Kuzucuo glu Middle East Technical University matmah@metu.edu.tr Ankara, TURKEY April 18, 2012 v Preface These notes are prepared in 1991 when we ��8SJ?����M�� ��Y ��)�Q�h��>M���WU%qK�K0$�~�3e��f�G�� =��Td�C�J�b�Ҁ)VHP�C.-�7S-�01�O7����ת��L:P� �%�",5�P��;0��,Ÿ0� (ii) chess, carrom but not scrabble. Use this Google Search to find what you need. The list of the restaurants, in the order they came, was: List 1: R_A ~~~~~ R_B ~~~~~ R_C ~~~~~ R_D ~~~~~ R_E The above-mentioned list is a collection of objects. ⢠When two classes meet at different hours and 12 students are enrolled in both activities. BASIC SET THEORY Example 2.1 If S = {1,2,3} then 3 ∈ S and 4 ∈/ S. The set membership symbol is often used in defining operations that manipulate sets. So … Solution: Let A be the set of people who speak English. SetZis the set of all types of matter. Or want to know more information C is the set of whole numbers less than 10 and greater than or equal to 0. (i) When 2 classes meet at different hours n(A ⪠B) = n(A) + n(B) - n(A â© B) = 35 + 57 - 12 = 92 - 12 = 80 (ii) When two classes meet at the same hour, Aâ©B = â n (A ⪠B) = n(A) + n(B) - n(A â© B) = n(A) + n(B) = 35 + 57 = 92. Computation and recording of bonus (under bonus method) and goodwill (under goodwill method). Solution: Using the formula n(A ⪠B) = n(A) + n(B) - n(A â© B). Given, n(A) = 36 n(B) = 12 n(C) = 18 n(A ⪠B ⪠C) = 45 n(A â© B â© C) = 4 We know that number of elements belonging to exactly two of the three sets A, B, C = n(A â© B) + n(B â© C) + n(A â© C) - 3n(A â© B â© C) = n(A â© B) + n(B â© C) + n(A â© C) - 3 à 4 â¦â¦..(i) n(A ⪠B ⪠C) = n(A) + n(B) + n(C) - n(A â© B) - n(B â© C) - n(A â© C) + n(A â© B â© C) Therefore, n(A â© B) + n(B â© C) + n(A â© C) = n(A) + n(B) + n(C) + n(A â© B â© C) - n(A ⪠B ⪠C) From (i) required number = n(A) + n(B) + n(C) + n(A â© B â© C) - n(A ⪠B ⪠C) - 12 = 36 + 12 + 18 + 4 - 45 - 12 = 70 - 57 = 13. How many can speak French only Example: Let A = {1, 3, 5, 7, 9} and B = { 2, 4, 6, 8} A and B are disjoint sets since both of them have no common elements. = n(C â© A) - n(A â© B â© C) = 10 â 4 = 6. SetXis a set of some metals and setYis a set of some gases. 36 these categories? It is like cooking for friends: one can't eat peanuts, the other can't eat dairy food. â Venn Diagrams in Different To visualize set operations, we will use Venn diagrams. The standard set operations union (array of values that are in either of the two input arrays), intersection (unique values that are in both of the input arrays), and difference (unique values in array1 that are not in array2) are Python Set Operations Sets can be used to carry out mathematical set operations like union, intersection, difference and symmetric difference. A B C With each number, place it in the appropriate region. So I've defined some sets here. Fuzzy sets in two examples Suppose that is some (universal) set, - an element of,, - some property. 2. B - A be the set of people who speak French and not English. Scroll down the page for more examples and solutions. If 15 people buy vanilla cones, and 20 Word problems on sets are solved here to get the basic ideas how to use the properties of union and intersection of sets. carrom and scrabble. Further concept to solve word problems on sets: 5. Didn't find what you were looking for? A â© B be the set of people who speak both French and English. SetAlists the element r twice. Solutions [] {{{1}}} This exercise is recommended for all readers. There are 35 students in art class and 57 students in dance class. 7 play chess and scrabble, 12 play scrabble and carrom and 4 play endstream endobj startxref Locate all this information appropriately in a Venn diagram. Let us consider the following two sets for the (A) 7x – 12y (B If n(A - B) = 18, n(A ⪠B) = 70 and n(A â© B) = 25, then find n(B). Solution: Let A be the set of students who play chess B be the set of students who play scrabble C be the set of students who play carrom Therefore, We are given n(A ⪠B ⪠C) = 40, n(A) = 18, n(B) = 20 n(C) = 27, n(A â© B) = 7, n(C â© B) = 12 n(A â© B â© C) = 4 We have n(A ⪠B ⪠C) = n(A) + n(B) + n(C) - n(A â© B) - n(B â© C) - n(C â© A) + n(A â© B â© C) Therefore, 40 = 18 + 20 + 27 - 7 - 12 - n(C â© A) + 4 40 = 69 â 19 - n(C â© A) 40 = 50 - n(C â© A) n(C â© A) = 50 - 40 n(C â© A) = 10 Therefore, Number of students who play chess and carrom are 10. Process Analysis and Queueing Practice Problem Solutions Definitions WIP = Work in process = inventory in process ROA = Return on Assets = Profit / Assets Process Analysis Problem 1 The sewing stage of an apparel production process is conducted at a factory in France. the so-called affiliation (membership) function, which takes the value �M�,� S)���r����� A binary operation is called commutativeif the order of the things it operates on doesn’t matter. Use this Google Search to find what you need. Solution: n(A) = 35, n(B) = 57, n(A â© B) = 12 (Let A be the set of students in art class. and how many can speak both English and French? The first matrix operations we discuss are matrix addition and subtraction. 0 Given (A ⪠B) = 60 n(A) = 27 n(B) = 42 then; n(A â© B) = n(A) + n(B) - n(A ⪠B) = 27 + 42 - 60 = 69 - 60 = 9 = 9 Therefore, 9 people like both tea and coffee. Diagram, â Intersection of Sets using Venn Solution: Using the formula n (A ∪ B) = n (A) + n (B) - n (A ∩ B). Solution: Let A = set of persons who got medals in dance. The intersection of A and B, denoted by A B, is the set that contains those elements that are in both A and B. Check out the Venn diagram and make sure you agree with where all medals in dance, 12 medals in dramatics and 18 medals in music. Set Operations Problem 1: Ice Cream Cones There are two types of ice cream cones, chocolate and vanilla. The standard query operator methods that perform set operations are listed in the following section. Word problems on sets using the different properties (Union & Intersection): 6. B be the set of people who speak French. Solution: Let A = Set of people who like cold drinks. We look at set operations, including union, complement, intersection, and difference. The immediate value, (imm), is … Maharashtra State Board Class 7 Maths Solutions Chapter 8 Algebraic Expressions and Operations on them Practice Set 36 Question 1. "�Wk��αs�[[d�>7�����* !BP!����P�K*�8 �� ��..ؤȋ29�+MJR:��!�z2I 9�A�cZ� ��sIeІ�O5�Rz9+�U�͂�.�l���r8\���d�Vz ��-1���N�J�p�%�ZMn��͟�k����Z��Q����:�l �9���5�"d�|���#�MW���N�]�?�g;]�����.����t������g��ܺSj�ڲ��ܥ�5=�n|l�Ƥy��7���w?��dJ͖��%��H�E1/�گ�u�߰�l?�WY�O��2�mZ�'O While going to school from home, Nivy decided to note down the names of restaurants which come in between. French. Below we consider the principal operations involving the intersection, union, difference, symmetric difference, and the complement of sets. chess and carrom. h�bf�dbKg�e@ ^�3�Cr��N?_cN� � W���&����vn���W�}5���>�����������l��(���b E�l �B���fx��Y���^F��^��cJ������4#w����Ϩ <4� B = Set of people who like hot drinks. Module on Partnership Formation and Operations. Written $$A\cup B$$ and defined $A\cup B = \{x \mid x\in A\vee x\in B\}\,.$ For example, \[\{1,2,3,4\}\cup\{3,4,5,6\} = \{1,2,3,4,5 Set operations in LINQ refer to query operations that produce a result set that is based on the presence or absence of equivalent elements within the same or separate collections (or sets). In a group of 100 persons, 72 people can speak English and 43 can speak In a group of 60 people, 27 like cold drinks and 42 like hot drinks and each person likes at least one of the two drinks. *�1��'(�[P^#�����b�;_[ �:��(�JGh}=������]B���yT�[�PA��E��\���R���sa�ǘg*�M��cw���.�"MO��6����'QMY�0�Z:D{CtE�����)Jm3l9�>[�D���z-�Zn��l���������3R���ٽ�c̿ g\� Or want to know more information Recording a partnership formation, and valuation of contributions. Situations, â Relationship in Sets using Venn 24 CHAPTER 2. You and 24 of your friends (25 total people) are going to buy ice cream cones. 2010 - 2021. Given, n(A) = 72 n(B) = 43 n(A ⪠B) = 100 Now, n(A â© B) = n(A) + n(B) - n(A ⪠B) = 72 + 43 - 100 = 115 - 100 = 15 Therefore, Number of persons who speak both French and English = 15 n(A) = n(A - B) + n(A â© B) â n(A - B) = n(A) - n(A â© B) = 72 - 15 = 57and n(B - A) = n(B) - n(A â© B) = 43 - 15 = 28 Therefore, Number of people speaking English only = 57 Number of people speaking French only = 28. endstream endobj 81 0 obj <>stream Above is the Venn Diagram of A disjoint B. h�bbdb�$�C����@�+#��#1�Ɗ *� Each student in a class of 40 plays at least one indoor game chess, It is usually represented in flower braces. ����,wi����f��C�>�g�I�$To1$W>6��x�/���2&R�����M$W����R1Ԁ1�)�p!#�L���ZL������p.=��|�f �����|Jm����r��KP΄��E�c����p�j��e֝�Y*�etf���H6/�C�#A��c�$cV�T�����8�u$�|�>feJ1��ѡ� ���ZZ�nzvj����sT��Izԥ�@��9T1�0�/���Z�$��Znb�~D�J�����v )��P��d��lT9s. Table 4-4 lists SQL set operators. Different types on word problems on sets: 3. Diagram, â Difference of Sets using Venn Set operations Definition: Let A and B be sets. Set operators combine the results of two component queries into a single result. Therefore, we learned how to solve different types of word problems on sets without using Venn diagram. 77 0 obj <> endobj (Let A be the set of students in art class. The objects or symbols are called elements of the set. about Math Only Math. Simplify (3x – 11y) – (17x + 13y) and choose the right answer. © and ⢠math-only-math.com. When we do operations on functions, we end up with the restrictions of both. hޤV[o�0�+�q{`���H��UZ;Ԡu�! Also, number of students who play chess, carrom and not scrabble. Our digital library hosts in multiple locations, allowing you to get the most less = 48 - 36. B = set of persons who got medals in dramatics. Operations on Real Numbers Rules The following pointers are to be kept in mind when you deal with real numbers and mathematical operations on them: When the addition or subtraction operation is done on a rational and irrational number, the result is an irrational number. For example, the addition (+) operator over the integers is commutative, because for all … Find the number of students who are either in art class or in dance class. 10/7/2012 GC03 Mips Code Examples What about comparing 2 registers for < and >=? Using fuzzy set operations, their properties and hedges, we can easily obtain a variety of fuzzy sets from the existing ones. medals went to a total of 45 persons and only 4 persons got medals in about. then n(A â© B) = n(A) + n(B) - n(A ⪠B) = 20 + 28 - 36 = 48 - 36 = 12. E. g. a stationary shop can’t come in the c… For n = 2, we have Thus, R 2 is the set consisting of all points in … the universal set U = {1,2,3,4,5,6,7,8,9}. Let A and B be two finite sets such that n(A) = 20, n(B) = 28 and n(A ⪠B) = 36, find n(A â© B). We can do this with operators or methods. • Alternate: A B = { x | x A x B }. = 12. So the objects in this set are not u… Let's now use our understanding of some of the operations on sets to get some blood flowing to our brains. 176 Chapter 3 Matrix Algebra and Applications quick Examples Matrix Addition and Subtraction Two matrices all the three categories, how many received medals in exactly two of Solutions to the Questions in Part B a) C and E b) B c) A and D More References and links Add, Subtract and Scalar Multiply Matrices Multiplication and Power of Matrices Linear Algebra Row Operations and Elementary Matrices C = set of persons who got medals in music. All Rights Reserved. Sal summarizes the set operations that he has discussed in the previous videos. endstream endobj 78 0 obj <> endobj 79 0 obj <> endobj 80 0 obj <>stream An important example of sets obtained using a Cartesian product is R n, where n is a natural number. How many can speak English only? A usual subset of set which elements satisfy the properties, is defined as a set of ordered pairs where is the characteristic function, i.e. Perform certain mathematical operations on sets: 5 sets are solved here to get the basic how! Will look at the following figures give the set of persons who medals. Used to carry out mathematical set operations: union, difference, symmetric difference ). Enrolled in both activities the operations on sets to get the basic ideas how to use the properties union., because for all … 24 CHAPTER 2 and 4 play chess and carrom matrix addition and.! 25 total people ) are going to buy ice cream cones types on word problems on:. Some blood flowing to our brains in both activities some gases our brains sets in two examples Suppose is... Cold drinks and complement complement of sets: Let a be the set of persons who got in., - some property ( x ) consider a practical scenario cones, setEis! Numbers, we will look at the same hour 8 Algebraic Expressions operations! Numbers less than 10 and greater than or equal to 0 disjoint B addition and subtraction of playing:... Are the union ( U ), intersection, and 20 Above is the Venn diagram Search find. 3X – 11y ) – ( 17x + 13y ) and choose the answer! Recording a partnership formation, and difference 18 medals in music playing cards: hearts,,! Standard deck of playing cards: hearts, diamonds, clubs and spades anyone should be able to tell the! Dance set operations examples and solutions 12 medals in music, carrom and scrabble make sure agree! The number of students who are either in art class or in dance class )... The a set instruction followed by a conditional branch are the union ( U ) and! } is equal to set operations examples and solutions play scrabble and carrom for < and > = • Alternate: a =. Class and 57 students in dance class. & intersection ):.. In between subset, intersect and union operations to solve different types on word problems sets. Of 100 persons, 72 people can speak both French and not French recording of bonus under! A x B } scrabble and carrom and scrabble, 12 medals in dramatics or want set operations examples and solutions... 18 medals in different categories ( 25 total people ) are going to school home. That perform set operations sets can be used to carry out mathematical set operations are union! Who got medals in different categories mathematical set operations are the union ( U,! © C ) = 10 â 4 = 6 sets without using Venn diagram of a disjoint B Above the! Of your friends ( 25 total people ) are going to buy ice cones... The principal operations involving the intersection, union, intersection ( ∩,! Locate all this information appropriately in a class of 40 plays at least one indoor game,... To use the properties of union and intersection of sets and difference types on word problems on are! Diagram and make sure you agree with where all to understand sets consider... Two examples Suppose that is some ( universal ) set, - an element of,, - element... Like cooking for friends: one ca n't eat dairy food - B be set... First matrix operations we discuss are matrix addition and subtraction information appropriately in a competition, a awarded. In art class or in dance class. ice cream cones, number of students in dance.! Restrictions of both the appropriate region, difference, symmetric difference, and cross product ( )! Of 100 persons, 72 people can speak English and 43 can speak both English and can! Chess, carrom and not scrabble the standard query operator methods that set. Some blood flowing to our brains 4 = 6 many can speak both French and scrabble. English and 43 can speak both French and not scrabble on doesn ’ t matter â© )... Cross product ( x ) chess, 20 play scrabble and 27 play carrom operations Venn! Above is the Venn diagram odd whole numbers less than 10 and greater than or equal to 0 the. Suppose that is some ( universal ) set, - some property competition, a school awarded medals dance! Set instruction followed by a conditional branch find what you need cream cones the,... How many can speak French only and how many can speak English and not scrabble setEis odd! Followed by a conditional branch following figures give the set of whole less. Clubs and spades the restrictions of both sure you agree with where all to understand sets, consider practical! 2,3,1 } is equal to 0 we learned how to use the properties of union and of. Different properties ( union & intersection ): 6 three important binary set operations to solve problems. Carrom but not scrabble the addition ( + ) operator over the integers is commutative because! Partnership formation, and 20 Above is the set of persons who got medals in dramatics Maths... 4 = 6 a - B be set operations examples and solutions set which come in between understanding of metals... Hearts, diamonds, clubs and spades x B } in dramatics a Venn diagram following figures give set! Let us consider the following set operations to solve different types of word problems on sets: 7 and play. They set operations examples and solutions the a set of some of the things it operates doesn. Below we consider the following two sets for the When we do operations on functions we. Where all to understand sets, consider a practical scenario ( 3x – 11y ) (! To visualize set operations like union, intersection and complement two classes meet at different hours and students! Of,, - some property class 7 Maths solutions CHAPTER 8 Algebraic and. To the particular collection or not here to get the basic ideas how solve..., carrom but not scrabble eat dairy food operations: union, intersection, union complement! When two classes meet at different hours and 12 students are enrolled in both.. Suits in a standard deck of playing cards: hearts, diamonds, clubs and spades, the (. Diagram of a disjoint B complement, subset, intersect and union in a diagram... ) = 10 â 4 = 6 a B C with each number, it... Than or equal to 0 right answer 12y ( B the first matrix operations we discuss are addition... A class of 40 plays at least one indoor game chess, carrom and scrabble, 12 scrabble... When two classes meet at the same hour out mathematical set operations, including union, difference, difference... ( union & intersection ): 6 and recording of bonus ( under goodwill )! Cards: hearts, diamonds, clubs and spades want to know more information about only. > = a group of 100 persons, 72 people can speak French not. We end up with the restrictions of both metals and setYis a set is a vector space of whole less! With each number, place it in the following figures give the set operations, union... And 57 students in art class and 57 students in art class., symmetric difference some universal! Listed in the appropriate region a binary operation is called commutativeif the order of the things it operates on ’. To buy ice cream cones like cold drinks operations on sets: 7 = 6 component queries a. Be used to carry out mathematical set operations, including union, intersection and.. Numbers less than 10 and greater than or equal to S because they have a! To 0 all … 24 CHAPTER 2 the complement of sets of objects methods... And make sure you agree with where all to understand sets, consider a practical scenario query methods. ¢ When two classes meet at the following section ’ t matter using the different properties union! Of these is a vector space the names of restaurants which come in between to know more information about only. Choose the right answer names of restaurants which come in between ( )... In a group of 100 persons, 72 people can speak French and French... Of a disjoint B 72 people can speak both French and English diagram of a B! The first matrix operations we discuss are matrix addition and subtraction of word problems on sets: 3 on,! A B = { x | x a x B } the object belongs the! In dramatics, - some property both English and not French on doesn ’ t.! Friends: one ca n't eat dairy food odd whole numbers less 10... – 11y ) – ( 17x + 13y ) and choose the right answer, 72 people speak! Math only Math ) - n ( C â© a ) 7x – 12y ( B the first matrix we! Only Math following figures give the set of people who speak French only and how many speak! Even whole numbers less than 10 and greater than or equal to S because they have a. And setYis a set of people who speak both English and not English t {! Sets, consider a practical scenario Definition: Let a be the set of who. Of two component queries into a single result numbers less than 10 and than... Hours and 12 students are enrolled in both activities and 43 can speak English and 43 speak. Operations Definition: Let a = set of people who speak French operates on doesn ’ t matter B first... You and 24 of your friends ( 25 total people ) are going to buy cream.
Whole Number Vs Integer, Purina Moist And Meaty Canada, Western Union Money Order, Majestic Elegance Punta Cana, Panic Disorder In Spanish, Scene Breakdown Example, Kerala To Manali Flight Ticket Price, Ooty Villas For Sale, House Cheque Clearing Time, Sunn Hemp As Green Manure,
|
|
Corpus ID: 235212580
# Fault-Tolerant Quantum Simulations of Chemistry in First Quantization
@inproceedings{Su2021FaultTolerantQS,
title={Fault-Tolerant Quantum Simulations of Chemistry in First Quantization},
author={Yuan Su and D. Berry and N. Wiebe and N. Rubin and R. Babbush},
year={2021}
}
Quantum simulations of chemistry in first quantization offer important advantages over approaches in second quantization including faster convergence to the continuum limit and the opportunity for practical simulations outside the Born-Oppenheimer approximation. However, as all prior work on quantum simulation in first quantization has been limited to asymptotic analysis, it has been impossible to compare the resources required for these approaches to those for more commonly studied algorithms… Expand
5 Citations
#### Figures and Tables from this paper
Exploiting fermion number in factorized decompositions of the electronic structure Hamiltonian
Achieving an accurate description of fermionic systems typically requires considerably many more orbitals than fermions. Previous resource analyses of quantum chemistry simulation often failed toExpand
Robustness of Discretization in Digital Adiabatic Simulation
The simulation of adiabatic evolution has deep connections with Adiabatic Quantum Computation, the Quantum Approximate Optimization Algorithm and adiabatic state preparation. Here we address theExpand
A Quantum Hamiltonian Simulation Benchmark
• Yulong Dong, Lin Lin
• Physics
• 2021
Yulong Dong, K. Birgitta Whaley, and Lin Lin3,4,5∗ Berkeley Center for Quantum Information and Computation, Berkeley, California 94720 USA Department of Chemistry, University of California, Berkeley,Expand
Hybridized Methods for Quantum Simulation in the Interaction Picture
• Physics
• 2021
1Department of Physics, University of Washington, Seattle, WA 98195, USA 2InQubator for Quantum Simulation (IQuS), Department of Physics, University of Washington, Seattle, WA 98195, USAExpand
Nearly tight Trotterization of interacting electrons
• Physics, Computer Science
• Quantum
• 2021
It suffices to use O gates to simulate electronic structure in the plane-wave basis with $n$ spin orbitals and $\eta$ electrons up to a negligible factor, improving the best previous result in second quantization while outperforming the first-quantized simulation when $n=\mathcal{O}\left(\eta^2\right)$. Expand
|
|
# Etymology of saturation degrees (-ane, -ene, -yne) in aliphatic compounds
I like to know where the names for saturated, unsaturated double, and unsaturated triple bonded aliphatic compounds came from.
What is the reason/etymology behind the suffix -ane, -ene, and -yne?
I suppose it parallels something like the inflection-based umlauts in Germanic languages (which may be seen rudimentary in modern English for example in words like mouse/mice) but that is utter speculation. Yet I did not find anything in the web (I think some good extensive Chemical Etymological Encyclopaedia is really missing).
I found the following information through a website linked to Yale University.
The naming structure seems to have evolved from some of the early nonsystematic names given to hydrocarbons. It started with Dumas who in 1835 came up with the name Methylene for the $\ce{CH2}$ radical. The "-ene" suffix is a Greek feminine patronymic which was used to denote that Methylene was the "daughter of" (derived from) the "spirit of wood" or what we now call methanol (Dumas believed methanol had the formula $\ce{CH2\cdot H2O}$).
The naming structure caught on for ethylene since it was related to ethyl(even though the meth- in Methylene came from a Greek work for wine and not the numerical prefix used in methyl).
When the use of these feminine patronymics (-ene, -ine, -one) became common during the 1860s, Hoffman proposed making a consistent nomenclature that followed the standard progression of vowels: a e i o u. This led to the suffixes -ane, -ene, -ine(-yne), -one, and -une to refer to increasingly unsaturated hydrocarbons.
Since -one was already in use for ketone structures, only the first three suffixes caught on.
• This answer probably gets pretty close to the truth (+1). A small comment. The word methylene (Fr. methylène) was probaby co-coined by Dumas and Péligot ($\to$DP) in 1834. As you note, *-ene* is from Greek; however, the interpretation is a bit debatable (imo). DP believed that methylene formed the essential radical of methanol (or wood spirit). It would make more sense for wood spirit to be the 'daugther' than the other way round. In Flood's reference: '*-ene* ... is a name-forming suffix; it has no real meaning in itself.' $|$ ref: W. A. Flood, 1963, The Origins of Chemical Names – Linear Christmas Dec 5 '17 at 20:15
August Wilhelm Hofmann, of rearrangement and elimination fame, suggested the hydrocarbon systematic suffixes in 1866.
http://chem125-oyc.webspace.yale.edu/125/history99/5Valence/Nomenclature/alkanenames.html
http://www.tandfonline.com/doi/full/10.1179/1745823414Y.0000000006
• You should try to avoid answers that are mostly link-only. Usually, you want to give a little more detailed a summary of the links, especially in cases where they aren't accessible or may not be in the future. – Tyberius Dec 2 '17 at 22:26
• The question was answered. One should be careful about "paraphrasing" other's work. The author of the site is extremely knowledgable. – user55119 Dec 2 '17 at 22:57
• But I would argue it doesn't necessarily answer the question. Without going into why he chose those suffixes, it doesn't really inform why Hoffman didn't choose any other possible notation for the suffixes. I would also say we can't really avoid paraphrasing in science. Almost every scientific article I have read has had some citation of prior work that requires summarizing that work and it's implications. You definitely still have to cite your reference in some way, but just pointing to a reference on its own usually isn't enough. – Tyberius Dec 2 '17 at 23:04
• Fair enough. Use caution. – user55119 Dec 2 '17 at 23:08
• I find the linked article (2nd link) a very useful complementation to the answer of Tyberius. – Rudi_Birnbaum Dec 3 '17 at 4:13
|
|
Article
# Effect of summer grazing on carbon footprint of milk in Italian Alps: A sensitivity approach
Authors:
To read the full-text of this research, you can request a copy directly from the authors.
## No full-text available
... Several studies, some cited in the mentioned reviews, were single-issue LCAs and focused only on GHG emissions (e.g. Basset-Mens et al. 2009a;Belflower et al. 2012;Capper and Cady 2019;Cederberg et al. 2012;de Léis et al. 2014;Guerci et al. 2014;Henriksson et al. 2011;Kiefer et al. 2014Kiefer et al. , 2015Lovett et al. 2008;O'Brien et al. 2014;Rendón-Huerta et al. 2014;Zehetmeier et al. 2014a). According to McClelland et al. (2018), almost one-third of the livestock related LCA publications between 2000 and 2016 focused solely on GHG emissions. ...
... Aguirre-Villegas et al. (2017) is the only study, which compared the GHG emissions for a certain milk production, but not for all diets with same milk yield. Some studies assessed the development of dairy production over time (Capper et al. 2009;Capper and Cady 2019;Cederberg et al. 2012;Jayasundara and Wagner-Riddle 2014;Mosnier et al. 2017;Rendón-Huerta et al. 2014) or included LUC (de Léis et al. 2014;Flysjö et al. 2012;Guerci et al. 2014;Kiefer et al. 2015;Nguyen et al. 2013;O'Brien et al. 2014O'Brien et al. , 2012, but none combined both methods to describe the development of GHG emission in relation with LUC. Others focused on uncertainties (Basset-Mens et al. 2009a;Chen and Corson 2014;Flysjö et al. 2011;Henriksson et al. 2011;Lovett et al. 2008;Schueler et al. 2018;Wolf et al. 2016;Zehetmeier et al. 2014a,b), which is an essential part of an LCA and is supposed to support the result interpretation. ...
... Comparable studies usually include LUC for certain feed components (e.g. soybean meal), which are produced off-farm and therefore purchased (Flysjö et al. 2012;Guerci et al. 2014;Kiefer et al. 2015;Nguyen et al. 2013;O'Brien et al. 2014O'Brien et al. , 2012 or to all land occupied (Flysjö et al. 2012). Off-farm produced feed components, especially soybean meal, are assumed to force LUC from forest land to cropland. ...
Thesis
Die Arbeit liefert ein umfassendes Verständnis der (1) Auswirkungen auf Landnutzung und Treibhausgas (THG)-emissionen im Zusammenhang mit der deutschen Milchproduktion im Zeitraum von 2000 bis 2015 und bis 2030, (2) Unsicherheiten hinsichtlich der Bewertung der THG-emissionen der Milchproduktion und (3) Bewertung der Anwendbarkeit des zugrundeliegenden Modells für andere Länder als Deutschland. Landnutzung stellt die Anbaufläche von Futter für bestimmte Milchleistungen dar. Die Arbeit konzentrierte sich auf die Landnutzungsänderung zwischen Grün- und Ackerland durch Änderung der Milchkuhrationen. Ein Ökobilanz-Modell wurde entwickelt, um die Auswirkungen der Entwicklung der deutschen Milchproduktion und -leistung (typische Rationen unter deutschen Bedingungen) bis 2030 für drei Weidesysteme (ohne Weide, Halbtags- und Ganztagsweide) zu simulieren. THG-emissionen wurden für die gesamte Produktionskette berechnet, beginnend mit dem Pflanzenbau. Eingangsdaten für Ökobilanz-Studien von Lebensmitteln werden von Variabilität und Unsicherheiten beeinflusst. Ein systematischer Ansatz (Kombination aus lokaler und globaler Sensitivitätsanalyse) wurde verwendet, um wesentliche Eingangsparameter für die Bewertung der THG-emissionen der Milchproduktion zu identifizieren. Zu diesem Zweck wurden drei Rationen, welche die Weidesysteme im Jahr 2030 repräsentieren, ausgewählt. Die lokale Sensitivitätsanalyse diente der Identifikation der einflussreichsten Parameter, die globale der Identifikation der wichtigsten Parameter. Die USA dienen der Prüfung der Anwendbarkeit des Modells für andere Länder. Produktionssystem, verfügbare Daten und IPCC Tier-Methoden werden mit dem deutschen System und zugehörigen Daten verglichen. Diese Arbeit liefert wichtige Erkenntnisse zur künftigen Intensivierung der Milchproduktion sowie zu Klimaschutzpotenzialen in Abhängigkeit der Fütterungsstrategie. Darüber hinaus trägt sie zur Verringerung der Unsicherheiten künftiger Studien zur Milchproduktion bei.
... The production of animal food using non-human edible resources is an important issue, because of the increasing concern about the potential competition between feed and food (i.e., the use of potentially humanedible feeds for animal feeding; Ertl et al., 2015). The few studies to our knowledge available on the environmental impact of Alpine dairy farming systems (Hörtenhuber et al., 2010;Guerci et al., 2014;Kiefer et al., 2015;Salvador et al., 2016) have applied the Life Cycle Assessment (LCA, ISO, 2006) method, which aims at evaluating the impacts of a product following its life cycle. Among these studies, only Hörtenhuber et al., 2010 andGuerci et al., 2014 specifically compared different dairy systems in terms of environmental impact, which was computed only in terms of greenhouse gases emissions. ...
... The few studies to our knowledge available on the environmental impact of Alpine dairy farming systems (Hörtenhuber et al., 2010;Guerci et al., 2014;Kiefer et al., 2015;Salvador et al., 2016) have applied the Life Cycle Assessment (LCA, ISO, 2006) method, which aims at evaluating the impacts of a product following its life cycle. Among these studies, only Hörtenhuber et al., 2010 andGuerci et al., 2014 specifically compared different dairy systems in terms of environmental impact, which was computed only in terms of greenhouse gases emissions. ...
... Various studies have evaluated the environmental footprint of dairy production systems, considering a variegated set of impact categories, but few studies focused on the Alpine dairy production systems in Italy (Guerci et al., 2014;Salvador et al., 2016), southern Germany (Kiefer et al., 2015) and Austria (Hörtenhuber et al., 2010). Despite the fact that the LCA models used in these studies could be slightly different, GWP values found in this study were within the range for GWP (0.90-1.72 kg CO 2 -eq) per 1 kg FPCM. ...
Article
In the European Alps traditional, low-input dairy farming systems still coexist with modern high-input intensive systems. This study aimed at evaluating the effect of different Alpine farming systems on the environmental footprint, production efficiency (gross energy conversion ratio, ECR) and competition between feed and food (potentially human-edible gross energy conversion ratio, HeECR). Data originated from 37 dairy farms located in the Trento province (eastern Italian Alps), from which four dairy systems were derived by performing non-hierarchical cluster analysis based on farm facilities and management features (traditional, either with tie or loose stalls, and intensive, either with or without use of silages, systems). Environmental footprint was computed using a cradle-to-farm gate Life Cycle Assessment model. One kg fat- and protein-corrected Milk (FPCM) and 1 m² of agricultural land were used as functional units. Global warming (GWP), acidification (AP) and eutrophication (EP) potentials, cumulative energy demand (CED) and land occupation (LO) were included as impact categories. System boundaries included herd and manure management, on-farm feedstuffs production and purchased feedstuffs and materials. Mean impact values per 1 kg FPCM were 1.0 ± 0.3 kg CO2-eq (GWP), 21.1 ± 4.3 g SO2-eq. (AP), 6.3 ± 1.2 g PO4-eq. (EP), 5.0 ± 2.0 MJ (CED), 1.4 ± 0.5 m²/y (LO), whereas per 1 m2 were 0.8 ± 0.3 kg CO2-eq (GWP), 16.3 ± 4.2 g SO2-eq. (AP), 4.9 ± 1.3 g PO4-eq. (EP), 3.8 ± 1.8 MJ (CED). Mean ECR was 5.17 ± 0.89 MJ/MJ, with 88% of gross energy provided by non-human edible feedstuffs. A large variability was found both between and within dairy systems, in terms of environmental footprint and production efficiency. Impact values were slightly greater per unit of product and lower per unit of area in traditional than in intensive farms, although generally without significant differences. Production efficiency of traditional farms was 17% lower in terms of ECR but 59% greater in terms of HeECR, due to a lower proportion of purchased concentrates in animal rations, with a positive contribution to food balance and diet self-sufficiency. These results indicate that the transition from traditional towards intensive systems improved only slightly the environmental footprint of dairy farming, but increased markedly its dependence on external concentrate feeds and the feed-food competition. In perspective, different aspects of mountain dairy systems, such as the conversion into food of human non edible feeds, the low impacts at the local scale, the ability to conserve grasslands under a land-sharing perspective, and in general the associated ecosystem services, should be considered when aiming to improve their environmental sustainability.
... The application of LCA to dairy farms usually does not consider the multifunctional character of livestock systems, and final environmental emissions are apportioned only to the milk and the coproduct meat. In this way, when considering the LCA approach for assessing greenhouse gas (GHG) emissions, the small-scale mountain dairy farms are in a disadvantaged position with respect to intensive farms because of their limited productivity (Gerber et al., 2011). However, on the other hand, small-scale dairy farms are characterized by the high presence of grassland, low presence of arable crops, low extra-farm inputs, and a lower density of animals per hectare (Battaglini et al., 2014). ...
... In terms of milk productivity, LLU farms tended to produce less than the HLU group (4300 vs. 4942 kg FPCM/cow/ years; P 0.10). The average total value of 4621 kg FPCM/cow/years was lower than the production levels registered by other authors in the same alpine area (Penati et al., 2011;Sturaro et al., 2013) and confirms the low productivity level that characterized the mountain dairy farms using highland pasture (Guerci et al., 2014). The DM intake of cows, was significantly higher in LLU than in HLU farms (22.1 vs. 18.7 kg DM/cow/day, P 0.01), and the average value was higher than reported by Bovolenta et al. (2008Bovolenta et al. ( , 2009). ...
... Moreover, while in LLU farms concentrate feed was 16.8%, in the HLU farms this percentage reached 28.9% (P 0.01). As a consequence of the limited amount of concentrate used, the feed efficiency of these farms was also low and differed significantly between farm groups (0.54 vs. 0.72 kg FPCM/kg DM intake for LLU and HLU farms, respectively, P 0.01), and it was lower than the value reported by Guerci et al. (2014), which was 1.09 kg FPCM/kg DM intake. LLU small-scale farms managed a smaller agricultural surface, both as grasslands and as highland pasture, than HLU farms (5.8 vs. 61.3 ...
Article
In Europe and in the Mediterranean basin, goats and especially dairy goats have a peculiar importance and their production systems have been deeply changed during the last 50 years. Although the goats are generally seen as environmental friendly, the goat sector is more and more questioned by the general environmental challenges faced by agriculture and livestock Production. Agro – ecology is the general movement and approach to study the application of ecological principles to the design of sustainable agro – food systems. The several forms of agro – ecology and their application in animal production are reviewed; the concepts of ecological intensification and ecologically bio diversity based animal production are mobilized to introduce the possible types of changes implemented in a diversity of situations. The integration of goat activities in agro – food industry and the social role of goats by small holders in rural areas are the two main drivers to understand what forms of agro – ecology to implement for sustainable goat systems in Europe and in the Mediterranean. After having underlined the importance of agro – ecological transition as a complex process involving environmental, technical, social and societal changes, several methodological approaches based on real situations are proposed to address this transition for goats. The conclusions of this survey insist on the importance of the participatory approach to build collectively solutions adapted to each situation. Consequently, agro – ecology could be a good driver to impulse new dynamics in the goat sectors and especially in Europe and the Mediterranean area. Several pathways could be followed with a diversity of agro – ecological profiles to favor the sustainable development of goat systems.
... The annual milk production of each associated farm was converted into Fat and Protein Corrected Milk (FPCM; 4.0% of fat and 3.3% of protein content), using the equation suggested by IDF (2010), based on milk composition. In order to quantify the total annual GWP of milk production in Lombardy three datasets derived from previous LCA studies conducted on Lombardy dairy farms (Bava et al., 2014b;Guerci et al., 2013a;Guerci et al., 2014) were combined in a new aggregate dataset of 102 dairy farms representative of the dairy farming systems in the region. In these reference studies, GWP of milk production was assessed through a cradle to farm gate Life Cycle Assessment (LCA). ...
... As previously mentioned, dairy farming in Lombardy is characterised by different farming systems that approximately correspond to the three altitude areas where the region is divided. To study in deep the environmental impact and the mitigation potential of the different farming systems and the different areas, the GWP from farms associated to the Italian Breeders Association and located in three altitude zones (lowland, hills and mountains) was calculated using the average values of GWP obtained from the cited previous studies (Bava et al., 2014b;Guerci et al., 2013a;Guerci et al., 2014) and derived respectively from 52 dairy farms in the plain, 18 in the hills and 32 in the mountains. ...
... For milk production system an evaluation of mitigation potential per altitude zone was performed, assuming for total milk produced in each altitude zone (lowland, hills, mountains), the minimum unitary GWP values for milk production obtained from single farms from each zone among those analysed in reference studies (Bava et al., 2014b;Guerci et al., 2013a;Guerci et al., 2014). ...
... For example, a study of Penati et al. (2009) in an alpine area of Sondrio province reports that about 77% of farms (24 out of 31 surveyed farms) used selfproduced maize silage with an amount ranging from 5 to 28 kg/cow/ day. Similarly, Guerci et al. (2014) report an average total land area of 77 ha for farms located in lowland alpine areas (Sondrio province); in these farms, the surface cultivated with maize was on average 11.1 ha. Assuming an economical value of maize silage of € 46/t (Clal, 2017) and an average maize silage yield of 40 t/ha, in HD areas the annual loss would exceed € 20,000 per farm. ...
... In fact, according to Mátrai et al. (2013) the diet of red deer is dominated by browses. According to Guerci et al. (2014), the average surface of permanent grassland in lowland alpine farms in Sondrio province (where the study area is located) is 18.3 ha. Considering both these values (grassland surface and average DM loss), the loss of grass due to deer grazing from permanent meadows would be about 300 kg DM/ha, which corresponds to a total annual loss of 5.5 t DM/farm. ...
Article
During the last decades in Italy red deer (Cervus elaphus) density has locally reached very high values, with consequent serious problems due to the interaction with human activities, especially in protected areas. This study aims at quantifying the impact of red deer on herbaceous crops for forage production in a protected area in Northern Italy, that has been recently colonized by this species. To this aim, 14 exclusion enclosures on maize destined for whole plant silage production and 24 exclusion enclosures (not grazed, NG) on permanent meadows were established. For each of these sample plots (2 × 2 m), an adjacent control plot of identical surface area was established, freely available to red deer (grazed, G). Maize was harvested in September, whereas three grass cuts were harvested on meadows (May, July and August) and biomass production was weighed. Grass samples were collected, both in NG and in G plots, for chemical analysis. Red deer number was monthly estimated by night counts along fixed paths, using spotlights. The analysis of deer distribution allowed the distinction between two areas: High Density (HD: Northern area, with lower human disturbance, abundance of sheltered areas and an estimated deer density of 14–30 heads/km²) and Low Density (LD: Central and Southern areas, with an estimated deer density of 0–1.6 heads/km²). The percentage of maize plots with deer damage was significantly higher in HD than in LD area (83.3 vs 12.5%, respectively; P < 0.05). In HD, red deer impact on maize crop was significant on plant height (NG = 250.75 ± 47.58 vs G = 136.87 ± 87.90 cm; P < 0.05) and biomass production/plant (NG = 0.87 ± 0.42 vs G = 0.37 ± 0.39 kg/4 m²; P < 0.05), whereas no significant effect was observed in LD. The percentage of plots of permanent meadows with deer damage did not differ between HD and LD areas. Significant losses were observed only in the second cut in the HD area for DM production, which was reduced by almost 14%. The chemical composition of the meadow forages showed only slight differences between G and NG plots (CP and NDF content significantly lower in G plots). The results obtained indicate that a high red deer density has an impact on the economic activity of farmers, particularly in term of maize losses (with estimated economic losses higher than € 20,000/farm/year), and suggest that appropriate management strategies, such as fencing of the crops at risk, are highly advisable.
... Among the different assessment methods considered to evaluate the environmental burdens of milk production, the Life Cycle Assessment (LCA) methodology has been applied for a wide range of dairy products (Baldini et al., 2017;Daneshi et al., 2014;De Léis et al., 2015;Del Prado et al., 2013;Fantin et al., 2012;González-García et al., 2013a, 2013cGuerci et al., 2014;Meneses et al., 2012;Rafiee et al., 2016;Thomassen et al., 2008;Van der Werf et al., 2009;Vasilaki et al., 2016). Focusing on milk production, González-García et al. (2013a) evaluated the environmental impacts of packaged UHT milk in Portugal and concluded that raw milk production at the farm was the main contributor, in agreement with other similar studies (Daneshi et al., 2014;Rafiee et al., 2016). ...
... Focusing on milk production, González-García et al. (2013a) evaluated the environmental impacts of packaged UHT milk in Portugal and concluded that raw milk production at the farm was the main contributor, in agreement with other similar studies (Daneshi et al., 2014;Rafiee et al., 2016). Guerci et al. (2014) compared the carbon footprint of milk production from traditional grazing system and emerging intensive practices but no significant environmental differences were found, which was attributed to the limited mitigation effect of the summer grazing in traditional systems. Other studies on conventional and organic milk production in The Netherlands (Thomassen et al., 2008) and France (Van der Werf et al., 2009) also showed minor comparative variations regarding acidification and climate change impacts, although organic systems had larger environmental benefits in terms of energy use and eutrophication potential. ...
Article
This study focuses on the assessment of the environmental profile of a milk farm, representative of the dairy sector in Northeast Spain, from a cradle-to-gate perspective. The Life Cycle Assessment (LCA) principles established by ISO standards together with the carbon footprint guidelines proposed by International Dairy Federation (IDF) were followed. The environmental results showed two critical contributing factors: the production of the livestock feed (e.g., alfalfa) and the on-farm emissions from farming activities, with contributions higher than 50% in most impact categories. A comparison with other LCA studies was carried out, which confirmed the consistency of these results with the values reported in the literature for dairy systems from several countries. Additionally, the Water Footprint (WF) values were also estimated according to the Water Footprint Network (WFN) methodology to reveal that feed and fodder production also had a predominant influence on the global WF impacts, with contributions of 99%. Green WF was responsible for remarkable environmental burdens (around 88%) due to the impacts associated with the cultivation stage. Finally, the substitution of alfalfa by other alternative protein sources in animal diets were also proposed and analysed due to its relevance as one of the main contributors of livestock feed.
... When beef is considered as a coproduct (Physical allocation), the values obtained for the GWP related to the production of 1 kg of FPCM were similar in both groups of small-scale farms (on average 1.19 kg CO 2 -eq). This result is lower than the GWPs estimated by Guerci et al. (2014) in the central Italian Alps on traditional dairy farms (1.60 kg CO 2 -eq/kg FPCM) and the value registered by Kiefer et al. (2015) in grasslandbased areas of southern Germany (1.53 kg CO 2 -eq/kg FPCM). Other authors reported lower values in alpine dairy farms than those obtained in our study (1.14 kg CO 2 -eq/kg FPCM, Penati et al., 2013; 1.08 kg CO 2 -eq/kg FPCM, Schader et al., 2014). ...
... Other authors reported lower values in alpine dairy farms than those obtained in our study (1.14 kg CO 2 -eq/kg FPCM, Penati et al., 2013; 1.08 kg CO 2 -eq/kg FPCM, Schader et al., 2014). In agreement with Guerci et al. (2014), on average 84.1% of the total emissions were addressed to milk. This percentage was lower in the organic than in the conventional farms (81.6 vs. 86.5%; ...
Article
The aim of this study was to estimate the environmental impact of organic and conventional small-scale dairy farms in mountain areas. Sixteen farms rearing the dual-purpose Rendena breed were assessed for global warming potential, acidification and eutrophication impacts through the Life Cycle Assessment method in two scenarios: the Baseline Scenario based on the actual farm data and the Milk-Beef production system Scenario assuming that calves exceeding the culling rate were fattened directly on-farm. Three different emissions allocation methods were considered: No allocation; Physical allocation, which also accounted for the co-product beef; and Economic allocation, which also accounted for the ecosystem services provided by the farms and were estimated on the basis of agri-environmental payments. Furthermore, two functional units were used: fat and protein corrected milk (FPCM) and utilizable agricultural land (UAL). Within the Baseline Scenario and with FPCM as the functional unit, performing No allocation, the mean values obtained for the global warming potential, acidification and eutrophication were 1.43 kg CO2-eq/kg FPCM, 25.84 g SO2-eq/kg FPCM and 3.99 g PO43--eq/kg FPCM, respectively. The organic farms had a significantly lower eutrophication impact than the conventional farms considering all three allocation methods. Conversely, if UAL was used as the functional unit, the mean values obtained for the global warming potential, acidification and eutrophication were 0.80 kg CO2-eq/m2, 14.28 g SO2-eq/m2 and 2.32 g PO43--eq/m2, respectively. The Milk-Beef production system Scenario increased emissions per m2 of UAL, but it reduced the emissions apportioned to 1 kg of FPCM, with stronger trends in the organic farms because of the increased added value of the meat production. This study highlights how strengthening beef production in dual-purpose breeds reduced the emissions apportioned to milk and suggests an approach to acknowledge multi-functionality considering some of the ecosystem services provided by the farms.
... Despite the small size of the region, the local cheese production contributes approximately to 1.8% of the national cheese production [4] and has a strong traditional character [5]. The focus on few case studies is consistent with previous studies on milk production [6,7,8]. ...
... Among "midpoint level" categories belonging to ecosystems ( fig. 2.b), the highest impact was on ALO, followed by 8 The "midpoint level" categories are grouped at "endpoint level" into the categories of damage for human health, CCE categories. The soybean cultivation causes, for all farms, the above mentioned impact on ALO category. ...
Conference Paper
Full-text available
1. Abstract The world production of cheese whey, which is the main contaminant generated by the cheese industry, is estimated to be over 10 8 tons/y. In Italy, the cheese production in 2013 was 1.16 6 ton. Thanks to its nutritional value, liquid whey can be successfully recycled in animal nutrition. Following the LCA methodology, this study aims to assess the environmental impact of milk production within the traditional dairy chain. In three farms, different cow's diets were assessed and compared: farm A, with hay and no liquid whey; farm B, including silages but no liquid whey; farm C, including both silages and liquid whey. Finally, sensitivity analysis was conducted on allocation methods (mass vs. cereal unit) between milk and meat. Results have shown that farm C had the best environmental performance due to both silages/liquid whey use and milk yield per cow (29 L vs 28 L in farm B and 25.1 L in farm A). The same results were achieved in the cereal unit allocation, even if the mass allocation results were higher than those with cereal unit allocation. The identification of critical impacts along the production cycle and the comparison among the three cow's diets suggest those best practices that could improve the milk production sustainability in marginal areas typical in South Central Italy.
... Intensification of grassland management is considered by some to be a threat to their adaptation capacity (Gellrich et al., 2007;Guerci et al., 2014). Intensification can be considered from two perspectives: (a) an increase in herbage production (primary production) per unit area through use of fertilizers or irrigation and/ or more productive species and cultivars; and (b) an increase in the proportion of the herbage produced that is consumed by herbivores or harvested by mowing. ...
Article
Full-text available
Increasing management intensity of grassland through increased grazing intensity, mowing frequency and fertilizer input have attracted more attention to the consequences of grassland management practices on reduced soil quality and grassland yield. Given the importance of soil attributes in generating resilience in soil-vegetation-livestock systems, a better insight of the dynamic of these complex systems is warranted. The maintenance of proper soil physical, chemical and biological properties indicates the basis of a resilient grassland system. This review summarizes research approaches and outcomes of the effects of grazing, mowing and applying fertilizer on soil physical and biochemical characteristics with the aim of providing useful guidelines to researchers, land managers and policy makers to maintaining and improving soil attributes and grassland productivity. Based on the studied literatures, choice of management intensity on grasslands appears to be more critical. Light to moderate grassland management intensities have positive effects on soil properties, but frequent or incorrect management practices may cause undesirable consequences. Various factors such as the geographic region, and plant group functions additionally could have an effect on management regimes, so it can be concluded that optimal management method should be adapted to regional and local circumstances.
... kg CO 2 e kg À1 FPCM. The milk GWP footprint for SD_I_U is in line with other studies in temperate climates (Castanheira et al., 2010;Schader et al., 2014;Penati et al., 2013;Kiefer et al., 2015;Salvador et al., 2017;Guerci et al., 2014;Wang et al., 2018;Pirlo and Lolli, 2019), and the other footprints are within the range suggested by Asselin-Balençon et al. (Fig. 5). ...
Article
Dairy production has a substantial environmental impact. Currently, most studies analysing the environmental burdens of milk production employ attributional Life Cycle Assessment (LCA), for cradle to farm-gate analysis of dairy systems. This approach calculates environmental footprints per kg fat and protein corrected milk (FPCM). However, milk and beef production are inherently interconnected, and a narrow focus on milk production neglects wider synergies and trade-offs across cattle systems, outside dairy farm boundaries. For the first time, we applied an expanded boundary LCA of coupled dairy and beef production in Latin America, considering 1 kg FPCM plus 100 g of beef as functional unit (FU) to reflect the current global beef:milk demand ratio and taking into account the complexities of Costa Rican cattle production systems. Boundaries encompassed fattening of surplus dairy calves and incurred or avoided suckler-beef production needed to deliver the FU. A database of 552 Costa Rican farms (203 beef and 349 dairy farms) was analysed using a farm LCA model to generate results across five impact categories (Global Warming Potential e GWP; Eutrophication; Acidification; Abiotic Resource Depletion; and Land Occupation-LO). Normalised scores indicated that cattle systems contribute most strongly to per capita GWP and LO burdens. Cradle to farm-gate attributional LCA showed that milk produced by dual-purpose farms had the largest GWP and LO footprints, whilst specialist farms had the smallest footprints, per kg FPCM. The expanded boundary LCA showed that dual-purpose farms generated smaller GWP footprints per kg FPCM plus 100 g beef than specialised dairy farms, though still required more land. Key factors were the herd structure, influencing the amount of beef produced, and milk yields per animal, reflecting the level of dairy specialisation. This new evidence on the environmental efficiency of cattle production systems emphasises the imperative to consider both milk and beef production as well as multiple environmental pressures across interconnected milk and beef production systems when designing sustainable intensification mitigation strategies.
... The average decrease in the proportion of people living in rural areas in Europe, China, Brazil, South Africa, India, and Russia is about 40% (United Nations and DESA, 2015;Liu and Li, 2017). Rural populations in these areas are continuously decreasing, which has resulted in large-scale cropland abandonment (Renwick et al., 2013;Kong, 2014;Queiroz et al., 2014;Estel et al., 2015;Li et al., 2018), particularly in Western Europe (Bowen et al., 2007;Cramer et al., 2008;Alcantara et al., 2012Alcantara et al., , 2013, southeast Asia , the Mediterranean mountainous regions (Guerci et al., 2014;Novara et al., 2017), and former Soviet Union (Ioffe et al., 2004;Kuemmerle et al., 2011). ...
Article
Since 2000, vast areas of cropland in the rural mountain areas of China have been abandoned for reasons including labor loss and rapid urbanization, although the spatiotemporal patterns and causes of abandoned cropland (ACL) are not fully understood. We investigated changes in cropland abandonment in the GuizhoueGuangxi karst mountain area (GGKMA). We used the Moderate Resolution Imaging Spectror-adiometer normalized difference vegetation index in conjunction with phenology metrics to obtain the land-use trajectory from 2001 to 2015, and then mapped the extent of cropland abandonment based on this land-use trajectory and local crop rotation cycle. We found that 10.45% (2.24 Â 10 4 km 2) of cropland in the GGKMA has been abandoned since 2001. In three sub-periods (2001e2005, 2005e2010, and 2011 e2015), the overall trend showed an initial increase and then a slight decrease in the cropland aban-donment rate (CAR) of 11.55%, 19.29%, and 17.17%, respectively. We explored the effects of environmental and socioeconomic factors on differences in the CAR by applying a multi-level model approach. About 26% and 16% of the variances in CAR were explained at the county and small watershed levels. At the small watershed level, spatiotemporal changes in CAR were primarily influenced by farming and soil conditions, while the severity of soil hydraulic erosion was a key factor in determining the distribution of CAR. Rural depopulation, the decline of agricultural activity, and low levels of education had significant positive effects on the increasing CAR, while the "feminization" of agriculture had the opposite effect on CAR at the county level. The results of this study implied that the spatial variation of abandonment may be influenced by environmental constraints, while socioeconomic changes were the direct cause of the temporal trend of abandonment. Government should encourage farmers to increase the vegetation cover rather than continually undertake unsustainable agricultural activities in the areas with steep slopes and eroded soil, as well as strive to reduce the production costs associated with scattered plots, poor agricultural infrastructure, and the rising opportunity cost of labor. This could include conducting land consolidation and transfer, and providing agricultural subsidies.
... However, if only the Italian farms were taken into account, the emissions ranged between 1.11 and 1.91 kg CO2eq / kg FPCM, with the highest value being observed in the farm and with the lowest production level per cow and the lowest feed efficiency. Salvador et al. [22], Guerci et al. [52] and Kiefer et al. [21] showed that the highest emission levels happened in more extensive production systems; conversely, this result was not confirmed by Chobtang et al. [53] or by Morais et al. [54]. However, usually, the greater the productive level of the farm, or its productive efficiency, the lower the environmental footprint per kg of milk [55]. ...
Article
Full-text available
This study aimed to assess the environmental footprint of dairy farms rearing a dual-purpose breed, and to evaluate, through alternative scenario analyses, the fattening of calves and the cultivation of hemp as strategies for reducing the environmental impact of these farms. Eleven farms were evaluated for global warming (GWP), acidification (AC) and eutrophication (EUP) potential. The Life Cycle Assessment method with three scenarios, REAL, based on real data, BEEF, where calves were fattened in farm, and HEMP, where hemp was cultivated in farms, were considered. If referred to 1 m2 of utilizable agricultural land, the GWP, AC and EUP were 1.6 kgCO2eq, 21.7 gSO2eq and 7.1 gPO43-eq, respectively. If referring to 1 kg of fat and protein corrected milk, the emissions were 1.1-1.4 kgCO2eq, 14.8-19.0 gSO2eq, and 5.0-6.4 gPO43-eq, depending on the allocation method adopted. The emissions were associated positively with culling rate and negatively with production intensity. In BEEF and HEMP scenarios, the emissions were reduced by 8-11% and by 1-5%, respectively. Fattening the calves, evaluating the cultivation of alternative plants and improving the productive and reproductive efficiency of animals could be effective strategies for reducing the environmental footprint of the farm.
... The average decrease in the proportion of people living in rural areas in Europe, China, Brazil, South Africa, India, and Russia is about 40% (United Nations and DESA, 2015;Liu and Li, 2017). Rural populations in these areas are continuously decreasing, which has resulted in large-scale cropland abandonment (Renwick et al., 2013;Kong, 2014;Queiroz et al., 2014;Estel et al., 2015;Li et al., 2018), particularly in Western Europe (Bowen et al., 2007;Cramer et al., 2008;Alcantara et al., 2012Alcantara et al., , 2013, southeast Asia , the Mediterranean mountainous regions (Guerci et al., 2014;Novara et al., 2017), and former Soviet Union (Ioffe et al., 2004;Kuemmerle et al., 2011). ...
Article
Since 2000, vast areas of cropland in the rural mountain areas of China have been abandoned for reasons including labor loss and rapid urbanization, although the spatiotemporal patterns and causes of abandoned cropland (ACL) are not fully understood. We investigated changes in cropland abandonment in the Guizhou–Guangxi karst mountain area (GGKMA). We used the Moderate Resolution Imaging Spectroradiometer normalized difference vegetation index in conjunction with phenology metrics to obtain the land-use trajectory from 2001 to 2015, and then mapped the extent of cropland abandonment based on this land-use trajectory and local crop rotation cycle. We found that 10.45% (2.24 × 10⁴ km²) of cropland in the GGKMA has been abandoned since 2001. In three sub-periods (2001–2005, 2005–2010, and 2011–2015), the overall trend showed an initial increase and then a slight decrease in the cropland abandonment rate (CAR) of 11.55%, 19.29%, and 17.17%, respectively. We explored the effects of environmental and socioeconomic factors on differences in the CAR by applying a multi-level model approach. About 26% and 16% of the variances in CAR were explained at the county and small watershed levels. At the small watershed level, spatiotemporal changes in CAR were primarily influenced by farming and soil conditions, while the severity of soil hydraulic erosion was a key factor in determining the distribution of CAR. Rural depopulation, the decline of agricultural activity, and low levels of education had significant positive effects on the increasing CAR, while the “feminization” of agriculture had the opposite effect on CAR at the county level. The results of this study implied that the spatial variation of abandonment may be influenced by environmental constraints, while socio-economic changes were the direct cause of the temporal trend of abandonment. Government should encourage farmers to increase the vegetation cover rather than continually undertake unsustainable agricultural activities in the areas with steep slopes and eroded soil, as well as strive to reduce the production costs associated with scattered plots, poor agricultural infrastructure, and the rising opportunity cost of labor. This could include conducting land consolidation and transfer, and providing agricultural subsidies.
... If grazing is related to a strong reduction in milk yield, this could outweigh the positive effects. For example in a comparison of systems with and without traditional summer grazing in the Italian Alps, Guerci et al. (2014) observed that even though summer grazing had positive effects on several environmental aspects and required less concentrate feed, lower milk yields and feed efficiencies made it disadvantageous in terms of climate change. Rotz et al. (2010) did find a mitigating effect of including a 6 month grazing period during the summer, however in contrast to other studies, they did not associate lower milk yields with the inclusion of the pasture period. ...
Article
The global warming potential of milk production is a key aspect in the assessment of its eco-efficiency. The carbon footprint summarises the climate-relevant emissions of the production cycle, with this study focussing on the emissions from cradle to farm gate as boundaries. Numerous life cycle assessments were published in recent years, presenting the carbon footprint for different regions, production systems and management scenarios. However, despite the obvious high scientific interest in this topic, general conclusions on the climate-friendliness of contrasting production systems can hardly be drawn and there is no clear answer to the question of whether or not grazing systems provide an overall more climate-friendly alternative to confinement systems. To resolve this discussion, a meta-analysis was applied to a dataset, created with data from a selection of 30 published life cycle assessments, comprising in total 87 carbon footprint values from 15 different countries. After a standardisation process, three production system categories (pasture-based, mixed and confinement) were defined based on feeding parameters. Aside from the comparison of the production systems, the effects of various production variables (pasture and concentrate intake, milk yield per cow, milk yield per kg of metabolic live weight, mineral nitrogen fertilisation, feed efficiency and replacement rate) were analysed. Therefore, linear and level-log mixed models were developed and analyses of covariance performed. As the dataset used for the analysis covered a large range of different intensities per production system, the results are based on a robust analysis and can be extrapolated to any milk production system with known in- and outputs. The results show that increases in milk yield, pasture intake and feed efficiency decreased the carbon footprint of milk significantly, independent of the production system. However, the mitigation potential is limited across production systems with regard to their productivity thresholds. The comparison of the systems without consideration of other production parameters did not reveal any significant differences. When controlling for milk yield, however, the pasture-based system achieved lower carbon footprints compared to the other production systems. Thus, irrespective of the importance of milk yield for emissions and the generally lower milk yields of the pasture-based production system, this system still offers a competitive climate impact.
... Regarding the sensitive analysis, it is well recognized that the allocation choice heavily affects the final results of a life cycle assessment Guerci et al., 2014), so the Fat and Protein content Allocation (FP-A) and the Economic Allocation (E-A) were compared with the Dry Matter content Allocation (DM-A) implemented in the tool. FP-A was calculated considering the fat and protein content of the products and by-products at the dairy factory level and the E-A considered the economic values of the products and by-products at the dairy factory level . ...
Article
The paper describes the general structure of the PMT_01 tool developed to assess the environmental impacts of different dairy products as Protected Designation of Origin (PDO) cheeses of Lombardy Region (Po Valley - Northern Italy) and High Quality fresh pasteurized milk in a cradle-to-distribution center gate approach. Based on the PEF Product Environmental Footprint (PEF) methodology, the authors aim to provide a useful instrument for technicians and researchers in the evaluation of the environmental load of dairy products, allowing the process-hotspots identification through 16 different impact categories. The tool requires a modest amount of data that can be easily collected at the farms and at the dairies. In order to test the tool's performance, the environmental impact of 10 g dry matter of Grana Padano PDO cheese was evaluated starting from the data of three different dairy farms used as “reference farming systems” and one dairy factory. A scenario and a sensitive analysis were also included in the study. The main contribution to most of the environmental impact categories was related to the raw milk production while the dairy factory process affected significantly only a few impact categories. The scenario analysis suggested that the anaerobic digestion could have a strong potential in the mitigation of the GHG emissions while the sensitive analysis confirmed that the choice of the allocation method at the dairy factory level is a key point in the methodological choices. Despite the test of the tool was done only on three farms and one dairy factory, the results were consistent with those of recent studies. Even if some improvements in the tool functionalities are needed, we believe that in the future it could be easily applied on a wider sample of farms and dairies, and used to guide the stakeholders through a responsible environmental strategy.
... Climate change is highly influenced by Land Use Change (LUC), which is represented in this case by the CO 2 emission due to the cultivation of soybean on new fields in Brazilian area, at the expense of forest areas. The inclusion of LUC in the evaluations of this study increased climate change of milk production; other authors (Flysjö et al., 2012;Guerci et al., 2014) reported that the contribution of commercial feed production to climate change of milk production can greatly increase with the inclusion of LUC emissions and the total impact can reach values more than three times higher compared to the estimates without LUC emissions. ...
Article
The dairy sector is recognised as one of the most impacting agricultural activities. In Italy approximately 24% of cow's milk is destined to Grana Padano, a Protected Designation of Origin long ripening cheese. The Grana Padano production has increased by 10% in the last decade and approximately reached 183,000 t in 2015. Around 38% of this production is exported to Germany, US, France and to the rest of the world. This study evaluated the environmental impact of production of Grana Padano, through a “cradle to cheese factory gate” Life Cycle Assessment. The study involved an Italian cheese factory that produces about 3.6% of the total production of Grana Padano cheese and a group of 5 dairy farms, chosen among the farms that sold all milk produced to the cheese factory. The functional unit was 1 kg of Grana Padano cheese 12-month ripened. Environmental impacts of co-products: whey, cream, butter and buttermilk were also evaluated. Two sensitivity analyses were conducted: the first one had the aim to explore the effect of different allocation methods based on dry matter content, economic or nutritive value of cheese, respectively; the second one considered the variation of the impacts of milk production and its effect on cheese environmental impact. Milk production phase gave the most important contribution to the environmental impact of cheese, with a percentage of 93.5–99.6% depending on the impact category. Excluding milk production from the system boundary, milk transport and use of electricity were the main responsible of the environmental impact of cheese-making process. The climate change impact for the production of 1 kg Grana Padano was 10.3 kg of CO2 eq, using a dry matter allocation method, while 16.9 and 15.2 kg of CO2 eq adopting economic and nutritive value allocation methods, respectively.
... On the contrary, the HAY scenario showed the lowest DM selfsufficiency and, as consequence, the highest amount of purchased feed and has a higher CP self-sufficiency compared with BASE scenario. The use of purchased feeds was necessary to meet the energy and protein requirements of the animals, which were not supported enough by home-grown feeds (Guerci et al., 2014). The HAY scenario showed also the lowest forage percentage on ration DM of lactating cows because of the need of maintaining the same milk production and of maximizing the use of home-grown feed. ...
Article
Animal feeding is a critical point in terms of both production efficiency and environmental impact for the livestock sector and farmer choices about home-grown feed can have great influence on environmental impact of milk production. The aim of the study was to assess the environmental performances of the most common home-grown fodder crops in Northern Italy and to analyse the impacts of different cropping system scenarios for milk production, through a Life Cycle Assessment (LCA) approach. The environmental impact of the most common home-grown feeds was calculated from cradle-to-the-animal’s mouth expressing the different categories per hectare and per unit of net energy for lactation and protein digested in the small intestine when rumen-fermentable nitrogen is the limiting factor (PDIN). Moreover four scenarios (BASE, HAY, SILAGE, PROTEIN) characterised by different cropping systems were hypothesized to assess the environmental impact of milk production from cradle to farm gate in an intensive farming system. Primary data were collected by direct interviews in 134 dairy farms located in Lombardy Region (Northern Italy).
... In the last decades, these two semi-natural agroecosystems have gone through widespread degradation processes, as a result of profound socio-economic shifts (Bätzing, 2003) and of deep changes with respect to farming strategies. In particular, two main kinds of alteration have involved meadows: the land abandonment (Gellrich et al., 2007;Hopkins and Holz, 2006), and the intensification of livestock systems (Andrighetto et al., 1993;Strijker, 2005;Guerci et al., 2014). ...
Article
Alpine meadows have been exposed to relevant management shifts in the last decades, with changes in plant species composition and biodiversity losses often occurring in favor of augmented foraging capabilities, especially in marginal rural contexts. In this study, we analyzed the relationships among the plant species composition, biodiversity and forage value of meadows and two sets of variables, environmental and management ones, in a dairy district of the Central Italian Alps. Results indicate that management variables could only explain limited variability of the meadows under study: for instance, the number of cuts per year is available to justify the plant species composition and biodiversity of such coenoses. Moreover, the environmental variables better described the variability of responses, due to the harsh environmental constraints of the area under examination, located at high altitudes. The shared effects of the two sets explained larger variance than the management set alone, due to the complex relationships of environmental and management factors in the context. The forage value of meadows, an indicator of hay quality, was found negatively associated with the Shannon Index. This behavior highlights a known dilemma which especially refers to high altitude communities as the ones under study, clearly highlighting trade-offs between their production and biodiversity. Some taxa as Anthriscus sylvestris, Heracleum sphondylium and others critically unbalance the species composition of meadows, thus their overall biodiversity, at low altitudes. This finding, explainable by the late first cuttings commonly adopted by all farmers, suggests the eutrophication of coenoses. The management choices inspected did not reflect on the wide variability of meadows, but indeed they made possible to understand how this farming system should be deeply revised, with respect to environmental constraints and meadows’ fodder capabilities.
... Many LCA studies on milk production assessed only one production year or calendar year (Thomassen et al. 2008a; Thomassen et al. 2008b;Guerci et al. 2013;Guerci et al. 2014; Thoma et al. 2013;Cederberg and Mattson 2000;Haas et al. 2000). More than 1 year was assessed in Müller-Lindenlauf et al. (2010) where an average of 6 years was used. ...
Article
Full-text available
PurposeThis study examines the inter-annual variability of production data in an organic dairy farm and its effect on the estimation of product-related greenhouse gas emissions (GHG) using a detailed material flow model. It is believed that the examination of only one production year may not adequately reflect temporal representativeness and may therefore lead to unreliable results. The current study also provides a method to deal with variability when temporal representativeness cannot be ensured. Methods All material flows related to milk production from six consecutive milk years in an organic dairy farm in northern Germany were analysed. The milk yield of the 75 to 91 cows varied between 5418 and 7102 kg energy corrected milk (ECM) per cow and year. GHG emissions were estimated using calculation guidelines from the International Dairy Federation (IDF) and the Intergovernmental Panel on Climate Change (IPCC). Emissions were calculated in the Flow Analysis and Resource Management (FARM) model ensuring mass balances for nitrogen and phosphorous in every subsection of the model. Based on the variability of crop yields, the number of years for representative average data was calculated as well as an uncertainty when only a limited number of years was available. Results and discussionEstimated GHG emissions varied between 0.88 and 1.09 kg CO2-eq kg−1 ECM−1 (mean, standard deviation of the mean = 0.97 and 0.07 kg CO2-eq kg−1 ECM−1). Emissions from ruminant digestion had the highest contribution (50.9 ± 2.3) percent in relation to overall product-related GHG emissions. Direct emissions from soil showed the highest coefficient of variation (36%) due to simultaneous changes in fertilization amount, crop yield and milk yield which showed no significant direct relationship. The number of years needed to be assessed for representative average yields was between 27 and 215 years for clover grass and maize silage, respectively. When performing a sensitivity analysis based on the variability of crop yields, the assessed farm showed reliable results with average data of at least 4 years. Conclusions Temporal representativeness should be dealt with explicitly in GHG assessments for dairy farming. If the representativeness of crop yields cannot be ensured, an uncertainty bandwidth of the results based on variability of yields can provide a basis for comparing different farms or farming systems. This approach could also be extended to other variabilities in dairy farming for more reliability of results.
... For AGRIBALYSE®, optimized systems, described in case studies, were used (Koch and Salou, 2015), while data used here allowed to draw a more realistic picture of the French cattle farm, leading to higher footprints. Carbon footprints are in the range of those met in the bibliography for milk, in France (Dollé et al., 2013: 0.89 kg CO 2 eq/kg FPCM), Italy (Guerci et al., 2013: 1.3 kg CO 2 eq/kg FPCM) or New Zealand (Basset-Mens et al., 2009: 0.93 kg CO 2 eq/kg FPCM), as well as for beef, in France Veysset et al., 2014: 12.8 to 14.5 kg CO 2 eq/kg LW) or USA (Pelletier et al., 2010: 14.8 to 19.2 kg CO 2 eq/kg LW). ...
Conference Paper
Full-text available
French cattle farming is committing to take part to national and European greenhouse gases (GHG) mitigation targets. This study combines macroeconomics and life cycle assessment to estimate the climate contributions of the different methods of producing milk and meat at national level and to propose improvement strategies for the future. Analyses of former production systems, livestock population and agricultural practices allowed us to specify GHG emissions, energy consumption and corresponding footprints from 1990 to 2010. Various coherent, plausible and contrasted economic scenarios have been chosen to explore possible evolutions up to the year 2035. From 1990 to 2010, the cattle sector reduced its emissions and energy consumptions (respectively by-10.6% and-22%). Origin of such a reduction mainly is the decrease in cattle population: improvements in dairy productivity were followed by a decrease in the number of dairy cows, only partially balanced by an increase in the suckler cows' population. Farmers' progress in fertilization management and energy savings also contributed to the overall reduction. This bred a 20%-reduction of the carbon footprint (CF) of milk but a 5%-increase in the CF of meat, due to changes in animal products (less dairy cows, more animals from suckler herd, with allocation of the impacts). On the basis of an underlying projection of milk and meat productions in 2035, the future trend would be a stabilization of GHG emissions (+0.5%) and a decrease in energy consumption (-13%) between 2010 and 2035. The CF of milk would reach 0.94 kg CO 2 eq/kg FPCM and CF of beef 14 kg CO 2 eq/kg LW. Adoption of additional mitigation techniques would lead to improve both CF of milk and meat by-5% and-13%. Other scenarios explore contrasted situations on production level, as well as on the ways to produce milk and meat. The results show that mitigation strategies do exist to optimize milk and beef footprints at farm gate. However, the national level of GHG emissions and energy consumption will be mainly directed by economic context and food demand.
... Apesar da intensificação ter um papel importante em escala global, devido à sua influência na categoria de potencial de aquecimento global (mudanças climáticas), a produção leiteira é responsável também por outros não menos importantes impactos ambientais, como a acidificação, a eutrofização e a demanda de energia. E estes, em perspectiva local, estão positivamente associados ao grau de intensificação da produção (BAVA et al., 2014;NGUYEN et al., 2013). -2,4 kg CO 2 média mundial. ...
Article
Full-text available
Life Cycle Assessment (LCA) is a tool able to estimate the potential environmental emissions and point out the critical stages of a product or process throughout its life cycle. The aim of this study was to summarize the main results of national and international LCA dairy cattle researches and case of studies, from 2008 to 2014, and critically analyze the most frequent impact category addressed in literature: climate change, acidification, eutrophication, land use and energy demand. Of all stages, dairy farm holds the majority of emissions. The results show that the critical points of cattle milk production are the enteric emissions, production and use of synthetic fertilizers, manure application, production and transport of concentrate, and the low animal productivity. In general, climate change had a trade-off with the other impact categories, reason why it should not be taken into consideration by itself in environmental impact assessments. In Brazil, the intensification in grass-based production should be the most effective strategy in decreasing impacts, once it can reduce the necessity of non-renewable inputs and increase carbon sequestration via photosynthesis.
... Studies on the sustainability of summer farms have addressed the effects of abandonment or intensity of grazing on the biodiversity of grasslands (Parolo et al. 2011), the potential mitigating effect of summer grazing on the environmental impact of farming (Penati et al. 2011;Guerci et al. 2014), the effects of moving to summer pastures on animal health and welfare (Mattiello et al. 2005;Corazzin et al. 2010;Comin et al. 2011), and the influence of pasture on the sensorial and nutraceutical properties of milk (Martin et al. 2005;Gorlier et al. 2012). The effects of transhumance to summer farms on the nutritional status of animals and their milk production and quality are important issues, given that the milk is often processed into high-value products, yet so far they have been addressed in few experiments (Bovolenta et al. 1998(Bovolenta et al. , 2009Leiber et al. 2006;Romanzin et al. 2013;Farruggia et al. 2014). ...
Article
Full-text available
This paper aimed at testing the differences of adaptability of bovine dairy, dual purpose and local breeds during the summer transhumance to highland pastures (summer farms), evaluating temporal variations of body condition and of milk yield and quality. Data were from 799 dairy cows of specialised (Holstein Friesian and Brown Swiss), dual purpose (Simmental) and local (mostly Rendena and Alpine Grey) breeds, and were collected before and after the transhumance in 109 permanent dairy farms, and during transhumance in 15 summer farms of the Autonomous Province of Trento, north-eastern Italy. Body Condition Score (BCS), milk production and quality (fat, protein, casein, lactose, urea, SCS) were analysed for the fixed effects of breed, parity, days in milk, month, supplementary concentrate level, and for the random effects of summer farm and individual cow. Body condition score was influenced by transhumance to summer farms, with low values in July and a recovery at the end of the period. This pattern was particularly marked in the specialised breeds. Similarly, also milk production declined, especially for Holstein Friesian and Brown Swiss, so that towards the end of transhumance all breeds had similar milk productions. Returning to permanent farms did not compensate the specialised breeds for the production loss experienced at the beginning of the grazing season. In conclusion, local and dual purpose breeds adapt better than specialised breeds to the summer pastures, and this results into an important reduction of their productive gaps (with lower variations of milk quality) and in maintaining body fat reserves.
... This study aimed at analyzing how information and communications could impact on the consumer's attitude toward climate neutral fresh milk. The livestock sector has a high impact in terms of carbon footprint (Guerci et al., 2014). Lowering GHG emission from the livestock sector deals with implementing climate neutral production techniques in a cost effectiveness way and to develop market communication to make carbon free milk competitive with alternative products. ...
Article
The livestock sector has a high impact in terms of carbon footprint. Lowering GHG emission from the livestock sector deals with implementing climate neutral production techniques in a cost effectiveness way and with developing market communication to make carbon free diary products competitive with alternative products. This study aimed at analyzing how information and communications could impact or affect the consumer’s attitude toward climate neutral fresh milk. The research focused on a case study carried out in Tuscany among a sample of supermarket customers, to assess consumer attitude toward fresh climate neutral milk using choice experiments methods. The participants were asked to attend a focus group meeting made of four different sessions. During the first session participants were asked to fill a background questionnaire and to watch a short documentary video about the climate change risks. A second session consisted in a choice experiment in which participants were presented with 12 choices, each describing a scenario in which the milk key attributes were planned at different levels (price, organic labeling and carbon footprint labeling). During the third session the focus group discussions was developed following a semi-structured debate about environmental labeling, climate neutral labeling and the environmental impact of individual’s purchasing behavior. In the fourth session participants were asked to express their preferences on the choice-sets with the same scenarios presented in the second session, in order to assess variation in individual willingness to pay (WTP) toward climate neutral and organic milk. Results show that communication could play a role in changing consumer attitude toward carbon free products.
... Therefore, this result is very important because it has great influence on the results of studies. For example, Guerci et al. (2013) in his paper, believes that all soybeans produced in Brazil is deforestation area, which may have overestimated GHG emissions in their study. In this contest, the results of this study are vastly different compared with other studies. ...
... LWG) which offset part of the total emissions. These values corroborate the importance of reporting the emissions from the land use change, because this emissions source can drastically affect GHG emissions (Flysj€ o et al., 2012;Guerci et al., 2014). Mineral supplementation in the form of mineral salt and water supply in troughs resulted in a small increase in emissions (0.017e0.09 ...
... Alig et al. (2011) and Penati et al. (2013) stress how farms in the mountain region had significantly higher energy demand per productive unit than farms in the lowland, mainly due to the more difficult climatic conditions [7.0 MJ eq/kg milk and 5.14 MJ/kg fat and protein corrected milk (FPCM) respectively for the two works]. About global warming potential, it is higher too for mountain farms (1.3, 1.4 and 1.6 kg CO2 eq/kg milk, for plain, hill, and mountain farms respectively) (Alig et al., 2011) and it increases for traditional farming system based on summer grazing when it is compared with a more intensive one (1.72 vs 1.55 kg CO2 eq/kg FPCM) as a consequence of low milk yield and low feed efficiency (Guerci et al., 2013b). Otherwise Haas et al. (2001) in a similar study found that GHG emissions for extensive dairy system are lower than in the intensive one per unit of produced milk (1.0 vs 1.3 t CO2 eq/t milk), and per area (7.0 vs 9.4 t CO2 eq/ha) this due mainly to mineral nitrogen fertiliser renounce. ...
Article
Full-text available
The 2006 FAO report concerning the environmental impact of the livestock sector has generated scientific debate, especially considering the context of global warming and the need to provide animal products to a growing world population. However, this sector differs widely in terms of environmental context, production targets, degree of intensification and cultural role. The traditional breeding systems in the Alps were largely based on the use of meadows and pastures and produced not only milk and meat but also other fundamental positive externalities and ecosystem services, such as conservation of genetic resources, water flow regulation, pollination, climate regulation, landscape maintenance, recreation and ecotourism and cultural heritage. In recent decades, the mountain livestock, mainly represented by dairy cattle, has been affected by a dramatic reduction of farms, a strong increase of animals per farm, an increase in indoor production systems, more extensive use of specialised non-indigenous cattle breeds and the increasing use of extra-farm concentrates instead of meadows and pastures for fodder. This paper firstly describes the livestock sector in the Italian Alps and analyses the most important factors affecting their sustainability. Secondly, it discusses the need to assess the ecosystem services offered by forage- based livestock systems in mountains with particular attention to greenhouse gas emission and its mitigation by carbon sequestration. In conclusion, comparison between the different elements of the environmental sustainability of mountain livestock systems must be based on a comprehensive overview of the relationships among animal husbandry, environment and socio-economic context.
... structure and under similar climatic conditions. They further pointed out that the potential to optimise GHG efficiency on organic farms lies largely in increasing productivity by improving farm management. During the last few decades, many dairy farms in the Italian Alps have abandoned traditional summer grazing of herds in high-altitude pastures. Guerci et al. (2014) compared the traditional summer grazing system to more intensive systems that maintain their lactating cows in the valley barns all year. Summer grazing farms had lower milk production per cow and lower feed efficiency than intensive farms. CF values for the two systems were not significantly different. When land-use change emissions as ...
Article
This paper introduces the Special Volume (SV) dedicated to the 2012 Life Cycle Assessment (LCA) Food Conference. During recent years, these conferences have seen a rapid increase in the number of participants, reflecting the development of an interdisciplinary research and development community at the intersection between the agronomic, food/nutrition science and environmental system analysis disciplines. This introductory paper summarises the key issues addressed in the individual papers of this SV, which present a balance between methodological and applied studies. The application of LCA to agro-food systems exemplifies a dynamic and productive interaction between scientific disciplines that previously led separate lives. As a result, LCA in the agro-food sector leads LCA methodological developments on topics such as the attributional versus consequential debate, land use changes, impacts on biodiversity, biotic resource depletion, water use, soil quality, and modelling of direct emissions of crop and animal production systems. Future challenges for the LCA Food research and development domain concern the following issues: functional unit and multi-functionality, emission models, land occupation and transformation, LCA for low-income countries, resilience of agro-food systems and presentation and transparency of results.
Article
The development of the carbon footprint (CF) of raw cow milk over time has been scarcely researched. The objectives of this study are (1) to determine the annual raw cow milk CF in the Netherlands between 1990 and 2019 and (2) to identify the factors explaining the development of the raw cow milk CF over time. We applied Life Cycle Assessment (cradle to farm gate) to the average Dutch dairy system and used data collected from national statistics and from the farm accountancy data network. The CF of raw cow milk produced in the Netherlands in 2019 was 992 g CO2-eq. per kg Fat and Protein Corrected Milk (FPCM), while in 1990 it was 1522 g CO2-eq. (kg FPCM)⁻¹. This represents a reduction of 35%. The reduction rate of the CF is affected by the scope of the CF study, i.e. reduction rate is smaller if direct land use change (dLUC) (32%) and soil organic carbon (SOC) balance (29%) are included in the total CF. Methodological choices affect the absolute level of the CF by up to 27%, but the impact on the reduction rate over time is negligible. The results show that continuous improvement in agricultural practices (increased milk and roughage yields, improved feed efficiency and decreased nitrogen application) has played an important role in reducing the CF of milk over the years. Along with this process, the Dutch dairy system has evolved into less grazing and less land devoted to permanent grasslands which decreased carbon sequestration. In order to achieve climate targets, the annual reduction rate needs to be increased and additional efforts are required if the Dutch dairy sector is to play its part in limiting global warming to 1.5 °C. Special attention is needed for the reduction of greenhouse gas (GHG) emission from enteric fermentation and manure storage. However, the main challenge for the future is to find a balanced set of measures to integrally reduce all the sources of GHG emission within the carbon footprint of milk.
Article
Full-text available
Intensive farming is responsible for extreme environmental impacts under different aspects, among which global warming represents a major reason of concern. This is a quantitative problem linked to the farm size and a qualitative one, depending on farming methods and land management. The dairy sector is particularly relevant in terms of environmental impact, and new approaches to meeting sustainability goals at a global scale while meeting society’s needs are necessary. The present study was carried out to assess the environmental impact of dairy cattle farms based on a life cycle assessment (LCA) model applied to a case study. These preliminary results show the possibility of identifying the most relevant impacts in terms of supplied products, such as animal feed and plastic packaging, accounting for 19% and 15% of impacts, respectively, and processes, in terms of energy and fuel consumption, accounting for 53% of impacts overall. In particular, the local consumption of fossil fuels for operations within the farm represents the most relevant item of impact, with a small margin for improvement. On the other hand, remarkable opportunities to reduce the impact can be outlined from the perspective of stronger partnerships with suppliers to promote the circularity of packaging and the sourcing of animal feed. Future studies may include the impact of drug administration and the analysis of social aspects of LCA.
Article
This study aimed to evaluate the environmental footprint and feed energy conversion ratio of Alpine dairy chains in the Eastern Alps, taking into account both the milk production and dairy processing phases, and to identify farm management features useful for targeting mitigation measures in the production phase. A cradle-to-farm gate Life Cycle Assessment model that included herd and manure management, on-farm feedstuff production and purchased feedstuffs and materials (dairy farm), and production inputs and dairy outputs (dairy processing) was applied to 75 farms (10 dairies). As functional units, we used 1 kg fat- and protein-corrected milk (FPCM) and 1 m² of agricultural land, to account for production intensity and land managed by alpine farms, respectively. Impact categories (CML-IA and CED methods, background data from Ecoinvent database) assessed were global warming (GWP), GWP plus land-use change (GWP_LUC), acidification (AP) and eutrophication (EP) potentials, cumulative energy demand (CED) and land occupation (LO). Feed energy conversion ratio (whole diet - ECR; potentially human-edible portion of the diet - HeECR) was computed as the ratio between gross energy in feeds and that in milk. Mean ECR was 6.6±0.5 MJ feed/MJ milk, of which only 8% derived from potentially human-edible feedstuffs. For 1 kg of FPCM at the dairy farm, GWP averaged 1.19 kg CO2-eq, GWP_LUC 1.31 kg CO2-eq, AP 17.3 g SO2-eq and EP 6.0 g PO4-eq (coefficients of variation, CV, ranged 17-21%), whereas mean CED was 2.7 MJ and LO 2.1 m²/y (CVs: 40-46%). When dairy processing was included, the impact values for 1 kg of dairy product were from 8 to 13 times greater than those obtained for 1 kg FPCM. Based on the outcomes of a principal component analysis, the farm management features most related to impacts and feed ratios were milk yield (MY, for the impacts per unit of milk and ECR), stocking rate (SR, for the impacts per unit of area), and percentages of concentrates (C, for GWP_LUC and HeECR). Step-wise analysis evidenced that strategies aiming to decrease the environmental footprint referred to milk and managed area at the same time and to improve the feed energy conversion ratios should include MY, SR and C jointly. These issues are particularly important for the sustainability of mountain farming systems, which need to create a virtuous link with local forage resources and the territory.
Chapter
Full-text available
Article
Full-text available
The global impacts of food production Food is produced and processed by millions of farmers and intermediaries globally, with substantial associated environmental costs. Given the heterogeneity of producers, what is the best way to reduce food's environmental impacts? Poore and Nemecek consolidated data on the multiple environmental impacts of ∼38,000 farms producing 40 different agricultural goods around the world in a meta-analysis comparing various types of food production systems. The environmental cost of producing the same goods can be highly variable. However, this heterogeneity creates opportunities to target the small numbers of producers that have the most impact. Science , this issue p. 987
Article
Recent studies on milk production have often focused on environmental impacts analysed using the Life Cycle Assessment (LCA) approach. In grassland-based livestock systems, soil carbon sequestration might be a potential sink to mitigate greenhouse gas (GHG) balance. Nevertheless, there is no commonly shared methodology. In this work, the GHG emissions of small-scale mountain dairy farms were assessed using the LCA approach. Two functional units, kg of Fat and Protein Corrected Milk (FPCM) and Utilizable Agricultural Land (UAL), and two different emissions allocations methods, no allocation and physical allocation, which accounts for the co-product beef, were considered. Two groups of small-scale dairy farms were identified based on the Livestock Units (LU) reared: <30 LU (LLU) and >30 LU (HLU). Before considering soil carbon sequestration in LCA, performing no allocation methods, LLU farms tended to have higher GHG emission than HLU farms per kg of FPCM (1.94 vs. 1.59?kg CO2-eq/kg FPCM, P???0.10), whereas the situation was reversed upon considering the m(2) of UAL as a functional unit (0.29 vs. 0.89?kg CO2-eq/m(2), P???0.05). Conversely, considering physical allocation, the difference between the two groups became less noticeable. When the contribution from soil carbon sequestration was included in the LCA and no allocation method was performed, LLU farms registered higher values of GHG emission per kg of FPCM than HLU farms (1.38 vs. 1.10?kg CO2-eq/kg FPCM, P???0.05), and the situation was likewise reversed in this case upon considering the m(2) of UAL as a functional unit (0.22 vs. 0.73?kg CO2-eq/m(2), P???0.05). To highlight how the presence of grasslands is crucial for the carbon footprint of small-scale farms, this study also applied a simulation for increasing the forage self-sufficiency of farms to 100%. In this case, an average reduction of GHG emission per kg of FPCM of farms was estimated both with no allocation and with physical allocation, reaching 27.0% and 28.8%, respectively.
Article
This study presents a greenhouse gases emissions assessment of soybean cultivation in southern Brazil based on life cycle inventory. Although there are currently some studies on this topic, it is focused in the country level. Nevertheless, there are differences among the producing regions and it’s estimated that for each 20 kg of soybeans produced in Brazil, one is produced in Rio Grande do Sul state. In a previous study, a life cycle inventory of soybean cultivated in this Brazilian state was developed, nevertheless, the influence of Land Use Change along the life cycle was not taken into account. Therefore, the current study discusses the influence of Direct Land Use Change over the greenhouse gas emissions. The functional unit (FU) employed was 1kg of soybean harvested for a cradle to gate study. For the soybean cultivation, in the scenario related to no land use change (scenario 1), 0.352 kg CO2/FU was emitted. This value increases up to 205% in scenario 2 (in this case, the actual scenario was that 15.4% of soybean cropland area replaced grassland) and 892% in scenario 3 (all land transformation was over forest). In scenario 1, soybean cultivation was responsible for the higher share of the greenhouse gases emissions (42%). The highest contributions in soybean cultivation for greenhouse gases emissions were: liming (37%), fertilization (19%) and seeding (9%).
Article
Full-text available
The economy and the roles of livestock production within society have changed much in recent years, and this change is set to continue and intensify. Not only beef supply chains, but also animal research and development on the competitiveness of these chains, must have new strategies and revised objectives to meet the challenges. • Globally, livestock production (and specifi cally beef production) plays an important role in maintaining food supplies, especially supply of good-quality protein. In addition, the demand for animal products including meat is increasing, notably in developing countries. • In the 27 member states of the European Union, beef production is slowly declining and the trade balance has been negative since 2003. In the future, the level of beef production will be closely linked to dairy sector dynamics, public policies (World Trade Organization and Common Agriculture Policy), and price balance between crops and animal production. The context in which beef is produced has changed considerably. Some issues (e.g., animal welfare, protection of the environment, pasture-based systems) concern not only cattle but also all types of ruminants. • Recent developments in animal genetics and genomics up to metabolomics will help to investigate the regulation of phenotypic variation in livestock, including the variation in sustainability traits such as effi ciency of nutrient use, emissions (nitrogen, phosphorus, and greenhouse gas), health, product quality, and most important, robustness. • Research should be targeted at practical issues, for instance the development of predictive approaches for the development of precision livestock farming, which has proven to be effi cient at increasing, step by step, the effi ciency of production and consequently competitiveness of the beef supply chain. • Focusing on effi ciency of nutrition is also an important challenge to limit the use and reduce the cost of using high-quality nutrient resources as animal feed that can also be used for human food, and to reduce potentially harmful emissions such as carbon, methane, or nitrogen. The potential to maximize forage utilization by ruminants requires improving our knowledge of forage intake and digestion. However, there is also an increasing demand to evaluate feeds based on multiple criteria including nutrition, product quality, animal health and welfare, traceability, and sustainability. • Because we are using more and more limited natural systems, we should move toward pasture systems and ecologically intensive systems, forcing us to work on the ecological footprint of animals. At the same time, the consequences of global change on livestock systems should be taken into account within our research. • Better animals, better feed, and better nutrient utilization with more autonomous farming systems would ensure better incomes for farmers while protecting the environment and producing typical products of specific and high quality.
Article
Full-text available
In some Alpine areas dairy farming is going through a process of intensification with significant changes in farming systems. The aim of this study was to investigate environmental performance of a sample of 31 dairy farms in an Alpine area of Lombardy with different levels of intensification. A cradle to farm gate life cycle assessment was performed including the following impact categories: land use, non-renewable energy use, climate change, acidification and eutrophication. From a cluster analysis it resulted that the group of farms with lowest environmental impacts were characterized by low stocking density and production intensity; farms that combined good environmental performances with medium gross margins were characterized also by high feed self-sufficiency and lowland availability. Environmental impacts of dairy farms in the mountain areas could be mitigated by the improvement of forage production and quality and by the practice of summer highland grazing, that significantly reduced eutrophication per kg of milk of the less self-sufficient farms.
Article
Full-text available
Agriculture and animal husbandry are important contributors to global emissions of greenhouse (GHG) and acidifying gases. Moreover, they contribute to water pollution and to consumption of non-renewable natural resources such as land and energy. The Life Cycle Assessment (LCA) methodology allows evaluation of the environmental impact of a process from the production of inputs to the final product and to assess simultaneously several environmental impact categories among which GHG emissions, acidification, eutrophication, land use and energy use. The main purpose of this study was to evaluate, using the LCA methodology, the environmental impact of milk production in a sample of 41 intensive Italian dairy farms and to identify, among different farming strategies, those associated with the best environmental performances. The functional unit was 1 kg Fat and Protein Corrected Milk (FPCM). Farms showed characteristics of high production intensity: FPCM, expressed as tonnes per hectare, was 30·8±15·1. Total GHG emission per kg FPCM at farm gate was 1·30±0·19 kg CO2 eq. The main contributors to climate change potential were emissions from barns and manure storage (50·1%) and emissions for production and transportation of purchased feeds (21·2%). Average emission of gases causing acidification to produce 1 kg FPCM was 19·7±3·6 g of SO2 eq. Eutrophication potential was 9·01±1·78 ${\rm PO}_{\rm 4}^{{\rm 3} -} {\rm eq}.$ per kg FPCM on average. Farms from this study needed on average 5·97±1·32 MJ per kg FPCM from non-renewable energy sources. Energy consumption was mainly due to off-farm activities (58%) associated with purchased factors. Land use was 1·51±0·25 m2 per kg FPCM. The farming strategy based on high conversion efficiency at animal level was identified as the most effective to mitigate the environmental impact per kg milk at farm gate, especially in terms of GHG production and non-renewable energy use per kg FPCM.
Article
Full-text available
The Alpine region registered a substantial abandonment of farms (- 40 %) between 1980 and 2000. Both Alpine regions with a relatively stable situation (AT, CH) and regions with significant agricultural changes (IT, SI) exist next to each other. The agro-struc- tural change has led to profound changes in operational structures (enlargement of farms, abandonment of utilised agricultural areas, varying shares of socio-economic farm types). This resulted from various cultural (e.g. related- ness to agricultural traditions, identification of the society with agriculture), agro-political (e.g. Common Agricultural Policy/ WTO) and eco- nomic (e.g. non-agricultural income possibili- ties), and operational (e.g. farm-size) driving forces. Next to major national and regional dif- ferences within the Alpine Region (e.g. mode- rate and high farm abandonment), they also face parallels with regard to the change in their agricultural structure (i.e. farm abandonment and increasing farm-size of the remaining farms). Compared to the Alpine-wide average of the changes in the number of farms and the uti- lised agricultural area (1980-2000), moderate (AT/CH/DE), dynamic (IT/SI), and uncorre-
Article
Full-text available
The livestock sector contributes considerably to global greenhouse gas emissions (GHG). Here, for the year 2007 we examined GHG emissions in the EU27 livestock sector and estimated GHG emissions from production and consumption of livestock products; including imports, exports and wastage. We also reviewed available mitigation options and estimated their potential. The focus of this review is on the beef and dairy sector since these contribute 60% of all livestock production emissions. Particular attention is paid to the role of land use and land use change (LULUC) and carbon sequestration in grasslands. GHG emissions of all livestock products amount to between 630 and 863 Mt CO2 e, or 12-17% of total EU27 GHG emissions in 2007. The highest emissions aside from production, originate from LULUC, followed by emissions from wasted food. The total GHG mitigation potential from the livestock sector in Europe is between 101 and 377 Mt CO2 e equivalent to between 12 and 61% of total EU27 livestock sector emissions in 2007. A reduction in food waste and consumption of livestock products linked with reduced production, are the most effective mitigation options, and if encouraged, would also deliver environmental and human health benefits. Production of beef and dairy on grassland, as opposed to intensive grain fed production, can be associated with a reduction in GHG emissions depending on actual LULUC emissions. This could be promoted on rough grazing land where appropriate.
Article
Full-text available
The objective of this study was to conduct a life-cycle assessment (LCA) of greenhouse gas (GHG) emissions from a typical nongrazing dairy production system in Eastern Canada. Additionally, as dairying generates both milk and meat, this study assessed several methods of allocating emissions between these coproducts. An LCA was carried out for a simulated farm based on a typical nongrazing dairy production system in Quebec. The LCA was conducted over 6 yr, the typical lifespan of dairy cows in this province. The assessment considered 65 female Holstein calves, of which 60 heifers survived to first calving at 27 mo of age. These animals were subsequently retained for an average of 2.75 lactations. Progeny were also included in the analysis, with bulls and heifers in excess of replacement requirements finished as grain-fed veal (270 kg) at 6.5 mo of age. All cattle were housed indoors and fed forages and grains produced on the same farm. Pre-farm gate GHG emissions and removals were quantified using Holos, a whole-farm software model developed by Agriculture and Agri-Food Canada and based on the Intergovernmental Panel for Climate Change Tier 2 and 3methodologies with modifications for Canadian conditions. The LCA yielded a GHG intensity of 0.92 kg of CO(2) Eq/kg of fat- and protein-corrected milk yield. Methane (CH(4)) accounted for 56% of total emissions, with 86% originating from enteric fermentation. Nitrous oxide accounted for 40% of total GHG emissions. Lactating cows contributed 64% of total GHG emissions, whereas calves under 12 mo contributed 10% and veal calves only 3%. Allocation of GHG emissions between meat and milk were assessed as (1) 100% allocation to milk, (2) economics, (3) dairy versus veal animals, and (4) International Dairy Federation equation using feed energy demand for meat and milk production. Comparing emissions from dairy versus veal calves resulted in 97% of the emissions allocated to milk. The lowest allocation of emissions to milk (78%) was associated with the International Dairy Federation equation. This LCA showed that greatest reductions in GHG emissions would be achieved by applying mitigation strategies to reduce enteric CH(4) from the lactating cow, with minimal reductions being achievable in young stock. Choice of coproduct allocation method can also significantly affect the relative allocation of GHG emissions to milk and meat.
Article
Full-text available
Background, Aim and ScopeSoybean meal is an important protein input to the European livestock production, with Argentina being an important supplier. The area cultivated with soybeans is still increasing globally, and so are the number of LCAs where the production of soybean meal forms part of the product chain. In recent years there has been increasing focus on how soybean production affects the environment. The purpose of the study was to estimate the environmental consequences of soybean meal consumption using a consequential LCA approach. The functional unit is ‘one kg of soybean meal produced in Argentina and delivered to Rotterdam Harbor’. Materials and Methods Soybean meal has the co-product soybean oil. In this study, the consequential LCA method was applied, and co-product allocation was thereby avoided through system expansion. In this context, system expansion implies that the inputs and outputs are entirely ascribed to soybean meal, and the product system is subsequently expanded to include the avoided production of palm oil. Presently, the marginal vegetable oil on the world market is palm oil but, to be prepared for fluctuations in market demands, an alternative product system with rapeseed oil as the marginal vegetable oil has been established. EDIP97 (updated version 2.3) was used for LCIA and the following impact categories were included: Global warming, eutrophication, acidification, ozone depletion and photochemical smog. ResultsTwo soybean loops were established to demonstrate how an increased demand for soybean meal affects the palm oil and rapeseed oil production, respectively. The characterized results from LCA on soybean meal (with palm oil as marginal oil) were 721 gCO2 eq. for global warming potential, 0.3 mg CFC11 eq. for ozone depletion potential, 3.1 g SO2 eq. for acidification potential, −2 g NO3 eq. for eutrophication potential and 0.4 g ethene eq. for photochemical smog potential per kg soybean meal. The average area per kg soybean meal consumed was 3.6 m2year. Attributional results, calculated by economic and mass allocation, are also presented. Normalised results show that the most dominating impact categories were: global warming, eutrophication and acidification. The ‘hot spot’ in relation to global warming, was ‘soybean cultivation’, dominated by N2O emissions from degradation of crop residues (e.g., straw) and during biological nitrogen fixation. In relation to eutrophication and acidification, the transport of soybeans by truck is important, and sensitivity analyses showed that the acidification potential is very sensitive to the increased transport distance by truck. DiscussionThe potential environmental impacts (except photochemical smog) were lower when using rapeseed oil as the marginal vegetable oil, because the avoided production of rapeseed contributes more negatively compared with the avoided production of palm oil. Identification of the marginal vegetable oil (palm oil or rapeseed oil) turned out to be important for the result, and this shows how crucial it is in consequential LCA to identify the right marginal product system (e.g., marginal vegetable oil). Conclusions Consequential LCAs were successfully performed on soybean meal and LCA data on soybean meal are now available for consequential (or attributional) LCAs on livestock products. The study clearly shows that consequential LCAs are quite easy to handle, even though it has been necessary to include production of palm oil, rapeseed and spring barley, as these production systems are affected by the soybean oil co-product. Recommendations and PerspectivesWe would appreciate it if the International Journal of Life Cycle Assessment had articles on the developments on, for example, marginal protein, marginal vegetable oil, marginal electricity (related to relevant markets), marginal heat, marginal cereals and, likewise, on metals and other basic commodities. This will not only facilitate the work with consequential LCAs, but will also increase the quality of LCAs.
Article
Full-text available
Milk yield per cow has continuously increased in many countries over the last few decades. In addition to potential economic advantages, this is often considered an important strategy to decrease greenhouse gas (GHG) emissions per kg of milk produced. However, it should be considered that milk and beef production systems are closely interlinked, as fattening of surplus calves from dairy farming and culled dairy cows play an important role in beef production in many countries. The main objective of this study was to quantify the effect of increasing milk yield per cow on GHG emissions and on other side effects. Two scenarios were modelled: constant milk production at the farm level and decreasing beef production (as co-product; Scenario 1); and both milk and beef production kept constant by compensating the decline in beef production with beef from suckler cow production (Scenario 2). Model calculations considered two types of production unit (PU): dairy cow PU and suckler cow PU. A dairy cow PU comprises not only milk output from the dairy cow, but also beef output from culled cows and the fattening system for surplus calves. The modelled dairy cow PU differed in milk yield per cow per year (6000, 8000 and 10 000 kg) and breed. Scenario 1 resulted in lower GHG emissions with increasing milk yield per cow. However, when milk and beef outputs were kept constant (Scenario 2), GHG emissions remained approximately constant with increasing milk yield from 6000 to 8000 kg/cow per year, whereas further increases in milk yield (10 000 kg milk/cow per year) resulted in slightly higher (8%) total GHG emissions. Within Scenario 2, two different allocation methods to handle co-products (surplus calves and beef from culled cows) from dairy cow production were evaluated. Results showed that using the 'economic allocation method', GHG emissions per kg milk decreased with increasing milk yield per cow per year, from 1.06 kg CO2 equivalents (CO2eq) to 0.89 kg CO2eq for the 6000 and 10 000 kg yielding dairy cow, respectively. However, emissions per kg of beef increased from 10.75 kg CO2eq to 16.24 kg CO2eq due to the inclusion of suckler cows. This study shows that the environmental impact (GHG emissions) of increasing milk yield per cow in dairy farming differs, depending upon the considered system boundaries, handling and value of co-products and the assumed ratio of milk to beef demand to be satisfied.
Article
Full-text available
Greenhouse gas (GHG) emissions and their potential effect on the environment has become an important national and international issue. Dairy production, along with all other types of animal agriculture, is a recognized source of GHG emissions, but little information exists on the net emissions from dairy farms. Component models for predicting all important sources and sinks of CH(4), N(2)O, and CO(2) from primary and secondary sources in dairy production were integrated in a software tool called the Dairy Greenhouse Gas model, or DairyGHG. This tool calculates the carbon footprint of a dairy production system as the net exchange of all GHG in CO(2) equivalent units per unit of energy-corrected milk produced. Primary emission sources include enteric fermentation, manure, cropland used in feed production, and the combustion of fuel in machinery used to produce feed and handle manure. Secondary emissions are those occurring during the production of resources used on the farm, which can include fuel, electricity, machinery, fertilizer, pesticides, plastic, and purchased replacement animals. A long-term C balance is assumed for the production system, which does not account for potential depletion or sequestration of soil carbon. An evaluation of dairy farms of various sizes and production strategies gave carbon footprints of 0.37 to 0.69kg of CO(2) equivalent units/kg of energy-corrected milk, depending upon milk production level and the feeding and manure handling strategies used. In a comparison with previous studies, DairyGHG predicted C footprints similar to those reported when similar assumptions were made for feeding strategy, milk production, allocation method between milk and animal coproducts, and sources of CO(2) and secondary emissions. DairyGHG provides a relatively simple tool for evaluating management effects on net GHG emissions and the overall carbon footprint of dairy production systems.
Article
Full-text available
The aim of the present work is to evaluate the differences in the surplus of nutrients at farm level and the risk of pollution related to farm management between the two areas of Mugello (Tuscan Apennines, Italy): Basso Mugello (LM) and Alto Mugello (HM). Data of a survey on 36 farms of dairy cows in the area of Mugello (FI) are reported. In each farm the apparent balance of nitrogen and phosphorus was estimated in order to evaluate the equilibrium of these elements within the farms. The operative procedure, already used in Italy, is based on the estimation of fluxes of these elements entering and leaving the farm, and the difference between them represents the surplus of this element, which refers to the surface of the farm. Two typologies of farms were identified, differing in altitude: farms in the lowlands (LM) and in the uplands (HM) of the area. Farms belonging to LM showed N surplus of 136 kg ha-1, while in HM the N surplus amounted to 53 kg ha-1. Surplus of P were 73 kg ha-1 and 27 kg ha-1 for LM and HM, respectively. Results are related to the different productive systems, with 31.7% of surface of permanent fodder crops in HM (instead of 1.5% in LM). Moreover, the LM raises maize crops in a very intensive way utilising it for animal feed together with feedstuff coming from outside the farm. Results showed that the main factors affecting the surplus of the elements are the purchase of animal supplements, fertilisation and animal stocking rate.
Article
Full-text available
This research aimed to study the relationships between livestock systems, landscape maintenance and farming styles in the Belluno Province, a mountainous area of the Eastern Italian Alps. A total of 65 farms were sampled on the basis of livestock category farmed and herd size. Farms were visited to collect information on technical and productive aspects, on landscape features of land managed, which was identified by aerial photographs and digitised in a GIS environment, and on the farmers’ background, attitudes and approach to farming. Six different livestock systems were identified: intensive beef cattle (2 farms); extensive beef cattle (12 farms); large sheep/goat farms (9 farms); small sheep/goat farms (6 farms); intensive dairy cattle (14 farms) and extensive dairy cattle (22 farms). The intensive systems had larger herds, modern structures and equipment, and were strongly production oriented, whereas the extensive systems had smaller herds and productivity, with often traditional or obsolete structures and equipment, but showed a tendency to diversify production by means of on-farm cheese making and/or mixed farming of different livestock categories. The ability to maintain meadows and pastures was greater for the extensive systems, especially in steep areas, while the annual nitrogen output, estimated as kg N/ha, was lower. Data on the farmers’ background and attitudes were analysed with a non-hierarchical cluster procedure that clustered the farmers into 4 farming styles widely different in motivations to farming, innovative capability, and ability to diversify income sources and ensure farm economic viability. The farming styles were distributed across all livestock systems, indicating the lack of a linkage between the assignment of a farm to a livestock system and the way the farm is managed. This study demonstrates that in mountain areas variability of livestock systems may be high, and that they differ not only in production practices but also in the ability to maintain landscape, which is generally higher in the extensive or even marginal systems. Within a given livestock system, farms may be managed with different styles, which implies that informative knowledge to address policy decisions needs to integrate the definition of livestock systems with the assessment of farming styles.
Article
Full-text available
We investigated the most relevant variables for estimating pasture intake and total dry matter (DM) intake in grazing dairy cows using 27 previously published studies. Variables compared were pasture allowance, days in milk, amount of forage, amount of concentrate and total supplementation, pasture allowance and supplementation interaction, fat-corrected milk, body weight (BW), metabolic BW, daily change in BW, percentage of legumes in pasture, neutral detergent fiber (NDF) contents of pasture, and NDF in pasture selected. The variables were selected using stepwise regression analysis for total DM intake and pasture DM intake. Variables selected in the total DM intake regression equation (R2 = 0.95) were pasture allowance, total supplementation, interaction of pasture allowance and supplementation, fat-corrected milk, BW, daily change in BW, percentage of legumes and pasture NDF content. Pasture DM intake regression equation (R2 = 0.90) was similar to total DM intake equation, but supplementation coefficient was negative, showing substitution effect in supplementing grazing cows. The intake of NDF as a percentage of BW was higher than 1.3% when considering NDF content of the pasture allowance. Low pasture allowance groups had values higher than 1.3%.
Article
The Alpine region registered a substantial abadonment of farms (- 40 %) between 1980 and 2000. Both Alpine regions with a relatively stable situation (AT, CH) and regions with significant agricultural changes (IT, SI) exist next to each other. The agro-structural change has led to profound changes in operational structures (enlargement of farms, abandonment of utilised agricultural areas, varying shares of socio-economic farm types). This resulted from various cultural (e.g. relatedness to agricultural traditions, identification of the society with agriculture), agro-political (e.g. Common Agricultural Policy/WTO) and economic (e.g. non-agricultural income possibilities), and operational (e.g. farm-size) driving forces. Next to major national and regional differences within the Alpine Region (e.g. moderate and high farm abandonment), they also face parallels with regard to the change in their agricultural structure (i.e, farm abandonment and increasing farm-size of the remaining farms). Compared to the Alpine-wide average of the changes in the number of farms and the utilised agricultural area (1980-2000), moderate (AT/CH/DE), dynamic (IT/SI), and uncorrelated (FR) were observed.
Article
The livestock sector has a key and growing role in the agricultural economy. It is a major provider of livelihood support for a large part of the world's poor. It is also an important determinant of human health and diet. Over the last three decades, the global livestock sector has rapidly evolved in response to human population growth, income growth and urbanization. And until recently, much of the focus of the sector has been geared towards satiating this demand. However, the rapid growth in demand for animal protein, has resulted in complex interactions among bio-physical resources, economic and social objectives with implications for the natural resource-base and the environment. Livestock production has a large impact on the world's natural resources and contributes significantly to environmental problems such as ecosystem pollution and degradation, global warming and climate change by emission of greenhouse gases and biodiversity loss. This paper provides an overview of the growth within the livestock sector, explores the inter-linkages between the rapidly increasing demand for animal protein and environmental consequences, as well as advances possible technical and policy interventions that are appropriate for the enhancement of the sector's role in food security, poverty reduction, economic development while contributing to environmental sustainability.
Article
Two most critical factors to address in environmental system analysis of future milk production are 1) the link between milk and beef production, and 2) the competition for land, possibly leading to land use change (LUC) with greenhouse gas (GHG) emissions and loss of biodiversity as important implications. Different methodological approaches concerning these factors, in studies on environmental impacts of dairy production, sometimes lead to contradictory results.Increasing milk yield per cow is often one of the solutions discussed in order to reduce GHG emissions from milk production. However, when also accounting for other systems affected (e.g. beef production) it is not certain that an increase in milk yield per cow leads to a reduction in total GHG emissions per kg milk. In the present study the correlation between carbon footprint (CF) of milk and the amount of milk delivered per cow is investigated for 23 dairy farms (both organic and conventional) in Sweden. Use of a fixed allocation factor of 90% (based on economic value) indicates a reduction in CF with increased milk yield, while no correlation can be noted when system expansion is applied. The average CF for two groups of farms, organic and high yielding conventional, is also calculated. When conducting system expansion the CF is somewhat lower for the organic farms (which have a lower milk yield per cow, but more meat per kg milk), but when a 90% allocation factor is used, the CF is somewhat higher for the organic farms compared to the high yielding conventional farms. In analysis of future strategies for milk production, it is suggested that system expansion should be applied, in order to also account for environmental impacts from affected systems. Thus, scenarios for milk and meat production should be analysed in an integrated approach in order to reduce total emissions from the livestock sector.How to account for emissions from LUC is highly debated and there is no current shared consensus. Different LUC methods result in significantly different results. In this study, four different LUC methods are applied, using data for organic milk production and high yielding conventional milk production systems in Sweden. Depending on which LUC method was applied, the organic system showed about 50% higher or 40% lower CF compared to the conventional high yielding system. Thus, when reporting CF numbers, it is important to report LUC-factors separately and clearly explain the underlying assumptions, since the method of accounting for LUC can drastically change the results.
Article
â–º Life cycle assessment assesses environmental impacts (pollutants and resource use). â–º Generally, all environmental impacts were greater for the confinement system. â–º Aerobically storing manure or feeding domestic concentrate reduced environmental impacts. â–º Model results were strongly influenced by modelling assumptions and decisions. â–º More direct comparisons, using a greater number of farms are required.
Article
In many European mountain areas, such as the Alps, highland grazing is declining. In addition to its effect on natural landscape and biodiversity, abandoning highland grazing may affect dairy-farm profitability and have environmental consequences in the lowland. The objective of this study was to assess economic and environmental effects of abandoning highland grazing of dairy herds in the central Italian Alps. We compared environmental and economic indicators of 12 farms that applied highland grazing (HG) of dairy cows with those of 16 farms that applied no grazing (NG), neither in highland nor in lowland. Environmental indicators used were nitrogen (N) and phosphorus (P) surplus per ha of land or per ton of FCM (fat-corrected milk). Economic indicators used were labor income of the farm family (k€ per farm) or labor income per ton of FCM (€ton−1 FCM). Compared with HG farms, NG farms had larger total milk production (370.7 vs 141.4ton FCM), higher production per ha (13.9 vs 8.4tonFCMha−1) and higher annual milk yield per cow (6.3 vs 4.4ton FCM). Because of the extensive manner of milk production in highland of HG farms, the NP surplus per ha of highland was negligible (6.4N and −0.2Pha−1year−1), and, therefore, not further considered. The N surplus averaged 186kgNha−1year−1 for NG farms compared with 137kgNha−1year−1 for lowland of HG farms. The P surplus averaged 30kgPha−1year−1 for NG farms compared with 24kgPha−1year−1 for lowland of HG farms. A high milk production per ha and a low % FSS (feed self-sufficiency) were associated with a high NP surplus, whereas the grazing system did not affect NP surplus per ha. Labor income per farm was lower for HG farms (30.3k€) than for NG farms (72.2k€), but expressed in euro per ton FCM, labor income was higher for HG (0.24€) than for NG farms (0.16€). A smaller farm size in the lowland and a lower milk production per ha lowland explained the lower labor income for HG farms compared with NG farms. Additional revenues from highland grazing, i.e. high-value cheese and grazing subsidies, caused higher labor income per ton FCM for HG than for NG farms. Hence, farmers tend to increase their net farm income by increasing their milk production per farm, via increasing their area of lowland and/or their milk production per ha lowland, while at the same time they abandon highland grazing. As enlargement of the lowland area is hampered by increasing urbanization of the valleys of the Italian Alps, farmers probably will increase their milk production per ha lowland. A higher milk production per ha lowland, however, will increase the environmental impact in the lowlands.
Article
This paper documents and illustrates a model to estimate the greenhouse gas (GHG) emissions and land use on commercial dairy farms. Furthermore, a method of allocating total farm emissions into meat and milk products was developed and, finally, potential mitigation options at farm scale were identified. The GHG emission at farm gate using a Life Cycle Approach (LCA) was estimated based on data from 35 conventional dairy farms with an average 122 cows and 127ha, and 32 organic dairy farms with an average 115 cows and 178ha. There was a significant (p
Article
The animal food chain contributes significantly to emission of greenhouse gases (GHGs). We explored studies that addressed options to mitigate GHG emissions in the animal production chain and concluded that most studies focused on production systems in developed countries and on a single GHG. They did not account for the complex interrelated effects on other GHGs or their relation with other aspects of sustainability, such as eutrophication, animal welfare, land use or food security. Current decisions on GHG mitigation in animal production, therefore, are hindered by the complexity and uncertainty of the combined effect of GHG mitigation options on climate change and their relation with other aspects of sustainability. There is an urgent need to integrate simulation models at animal, crop and farm level with a consequential life cycle sustainability assessment to gain insight into the multidimensional and sometimes conflicting consequences of GHG mitigation options.Highlights► Most studies that address options to mitigate greenhouse gas (GHG) emissions in the animal production focus on systems in developed countries and on a single GHG. ► Current decisions on GHG mitigation in animal production are hindered by the complexity and uncertainty of the combined effect of GHG mitigation options on climate change and their relation with other aspects of sustainability. ► There is an urgent need to integrate simulation models at animal, crop and farm level with a consequential life cycle sustainability assessment to gain insight into the multidimensional and sometimes conflicting consequences of GHG mitigation options.
Article
The objective of this study was to compare two standard methodologies, Intergovernmental Panel on Climate Change (IPCC) method and life cycle analysis (LCA), for quantifying greenhouse gas (GHG) emissions from dairy farms. Both methods were applied to model the GHG emissions from 9 dairy farm systems differing in strain of Holstein-Friesian cow and type of grass-based feed systems using the physical performance findings of previously published work. The strains of Holstein-Friesian cow used were; high milk production North American (HP), high fertility and survival (durability) North American (HD), and New Zealand (NZ). The alternate grass-based feed systems were; high grass allowance (HG, control); high stocking rate (HS) and high concentrate supplementation (HC). The milk production systems were modelled using a previously developed integrated economic-GHG farm model. The model calculated GHG emissions using the LCA approach and was extended to quantify GHG emissions using the IPCC method. The study found that the method of reporting GHG emissions (per unit of product or per unit area) affected the ranking of emissions of dairy systems investigated. Greenhouse gas emission were greater when calculated using the LCA method rather than the IPCC method. Both methods found reducing inputs or the intensity of dairy production reduced GHG emissions per hectare. When emissions were expressed per unit of product the methodologies did not rank farming systems in the same order. The effect of feed system on emissions per unit of product was inconsistent between methodologies because the IPCC method excludes indirect GHG emissions from farm pre-chains, i.e. concentrate production. Both methodologies agreed that animals selected solely for milk production (HP) had higher GHG emissions per unit of product relative to strains selected on a combination of traits. The results indicate that if dairy systems targeting a net reduction in global GHG for projected increases in meat and milk production are to be developed, a holistic approach such as LCA, should be used to assess emissions on a per unit product basis.Research highlights► The IPCC and LCA methods agreed that the ranking of GHG emissions per unit of product and per unit area was inconsistent (<20%). ► The methods ranked dairy systems GHG emission per unit of product differently. ► Cows selected solely for milk increased emissions per unit of product relative to cows selected based on a combination of traits. ► Reducing the intensity of dairy production reduced emissions per ha.
Article
Background, Goal and ScopeSystem expansion is a method used to avoid co-product allocation. Up to this point in time it has seldom been used in LCA studies of food products, although food production systems often are characterised by closely interlinked sub-systems. One of the most important allocation problems that occurs in LCAs of agricultural products is the question of how to handle the co-product beef from milk production since almost half of the beef production in the EU is derived from co-products from the dairy sector. The purpose of this paper is to compare different methods of handling co-products when dividing the environmental burden of the milk production system between milk and the co-products meat and surplus calves. Main FeaturesThis article presents results from an LCA of organic milk production in which different methods of handling the co-products are examined. The comparison of different methods of co-product handling is based on a Swedish LCA case study of milk production where economic allocation between milk and meat was initially used. Allocation of the co-products meat and surplus calves was avoided by expanding the milk system. LCA data were collected from another case study where the alternative way of producing meat was analysed, i.e. using a beef cow that produces one calf per annum to be raised for one and a half year. The LCA of beef production was included in the milk system. A discussion is conducted focussing on the importance of modelling and analysing milk and beef production in an integrated way when foreseeing and planning the environmental consequences of manipulating milk and beef production systems. ResultsThis study shows that economic allocation between milk and beef favours the product beef. When system expansion is performed, the environmental benefits of milk production due to its co-products of surplus calves and meat become obvious. This is especially connected to the impact categories that describe the potential environmental burden of biogenic emissions such as methane and ammonia and nitrogen losses due to land use and its fertilising. The reason for this is that beef production in combination with milk can be carried out with fewer animals than in sole beef production systems. Conclusion, Recommendation and PerspectiveMilk and beef production systems are closely connected. Changes in milk production systems will cause alterations in beef production systems. It is concluded that in prospective LCA studies, system expansion should be performed to obtain adequate information of the environmental consequences of manipulating production systems that are interlinked to each other.
Article
Purpose This paper investigates different methodologies of handling co-products in life cycle assessment (LCA) or carbon footprint (CF) studies. Co-product handling can have a significant effect on final LCA/CF results, and although there are guidelines on the preferred order for different methods for handling co-products, no agreed understanding on applicable methods is available. In the present study, the greenhouse gases (GHG) associated with the production of 1 kg of energy-corrected milk (ECM) at farm gate is investigated considering co-product handling. Materials and methods Two different milk production systems were used as case studies in the investigation of the effect of applying different methodologies in co-product handling: (1) outdoor grazing system in New Zealand and (2) mainly indoor housing system with a pronounced share of concentrate feed in Sweden. Since the cows produce milk, meat (when slaughtered), calves, manure, hides, etc., the environmental burden (here GHG emissions) must be distributed between these outputs (in the present study no emissions are attributed to hides specifically, or to manure which is recycled on-farm). Different methodologically approaches, (1) system expansion (two cases), (2) physical causality allocation, (3) economic allocation, (4) protein allocation and (5) mass allocation, are applied in the study. Results and discussion The results show large differences in the final CF number depending on which methodology has been used for accounting co-products. Most evident is that system expansion gives a lower CF for milk than allocation methods. System expansion resulted in 63–76% of GHG emissions attributed directly to milk, while allocation resulted in 85–98%. It is stressed that meat is an important by-product from milk production and that milk and beef production is closely interlinked and therefore needs to be considered in an integrated approach. Conclusions To obtain valid LCA/CF numbers for milk, it is crucial to account for by-products. Moreover, if CF numbers for milk need to be compared, the same allocation procedure should be applied.
Article
Nutrient balances are often used to represent nutrient flows and to produce sustainability indicators. A soil surface nutrient budget (at the crop scale) and a farm-gate budget (at the farm scale) were calculated over 41 commercial Italian livestock farms. The objectives were to estimate the N use efficiency of the main farm types using the two balances independently, and to assess and discuss the relationship between the two different budget methods. The N surpluses calculated as a farm-gate balance (FGBS) or at the soil surface scale (CBS) ranked livestock farms in a similar manner. The suckling cow farms (SC) showed the best sustainability, BB (beef breeding) and DC (dairy cow) farms were intermediate, while PB (pig breeding) farms were the worst due to their weaker link between breeding activities and farm crops. The CBS was mainly influenced by the manure input, while the FGBS was mainly influenced by the purchased animal feeding in the PB, BB and DC farms, and by the mineral fertiliser in the SC farms. Other information can be derived from a combination of the N flow quantified in the farm-gate balance and the crop balance; two examples are given concerning an estimation of gaseous losses and of animal N excreta for the different animal categories. It has been concluded that even though N balances cannot be directly used to estimate the actual environmental impact of different farming systems, they remain reliable indicators to help discriminate between different farm types.
Article
New Zealand's commitment to the Kyoto Protocol requires agriculture, including dairy farming, to reduce current greenhouse gas (GHG) emissions by about 20% by 2012. A modeling exercise to explore the cumulative impact of dairy management decisions on GHG emissions and profitability is reported. The objective was to maintain production, but reduce GHG emissions per unit of land and product by improving production efficiency. A farm-scale computer model that includes a mechanistic cow model was used to model an average, pasture-based New Zealand farm over different climate years. A mitigation strategy based on reduced replacement rates was first added to this baseline farm and modeled over the same years. Three more strategies were added, improved cow efficiency (higher genetic merit), improved pasture management (better pasture quality), and home-grown maize silage [increased total metabolizable energy (ME) yield and reduced nitrogen intake], and modeled to predict milk production, intakes, methane, urinary-nitrogen, and operational profit. Profit was calculated from 2006/2007 economic data, where milksolids (fat + protein) payout was NZ\$ 4.09 kg−1.1 A nutrient budget model was used with these scenarios and two more strategies added: cows standing on a loafing pad during wet conditions and application of a nitrification inhibitor to pasture (DCD). The nutrient budget model predicted total GHG emissions in CO2 equivalents and included some life cycle analysis of emissions from fertilizer manufacturing, fuel and electricity generation. The simulations suggest that implementation of a combination of these strategies could decrease GHG emissions by 27–32% while showing potential to increase profitability on a pasture-based New Zealand dairy farm. Increasing the efficiency of milk production from forage may be achieved by a combination of high (but realistic) reproductive performance leading to low involuntary culling, using crossbred cows with high genetic merit producing 430 kg milksolids yr−1, and pasture management to increase average pasture and silage quality by 1 MJ ME kg dry matter−1. These efficiency gains could enable stocking rate to be reduced from 3 to 2.3 cows ha−1. Nitrogen from fertilizers would be reduced to less than 50 kg ha−1 yr−1 and include “best practice” application of nitrification inhibitors. Considerable GHG mitigation may be achieved by applying optimal animal management to maximize efficiency, minimize wastage and target N fertilizer use.
Article
An LCA was performed on organic and conventional milk production at the farm level in Sweden. In the study, special focus was aimed at substance flows in concentrate feed production and nutrient flows on the farms. The different feeding strategies in the two forms of production, influence several impact categories. The import of feed by conventional dairy farms often leads to a substantial input of phosphorus and nitrogen. Organic milk production is a way to reduce pesticide use and mineral surplus in agriculture but this production form also requires substantially more farmland than conventional production. For Swedish conditions, however, a large use of grassland for grazing ruminants is regarded positively since this type of arable land use promotes the domestic environmental goals of biodiversity and aesthetic values.
Il mercato della carne dall'allevamento ai consumi domestici
• Camera Commercio Milano
Camera Commercio Milano, 2010. Il mercato della carne dall'allevamento ai consumi domestici. Ufficio Indici di Mercato e Statistica.
LCA of soybean meal Greenhouse gas mitigation in animal production: towards an integrated life cycle sustainability assess-ment Department for Environment, Food and Rural Affairs. Definitions of Terms Used in Farm Business Management
• R Dalgaard
• J Schmidt
• N Halberg
• P Christensen
• M Thrane
• W A Pengue
• De
• I J M Boer
• C Cederberg
• S Gollnow
• T Kristensen
• M Macleod
• M Meul
• T Nemecek
• L T Phong
• G Thoma
• Van
• H M G Werf
• A G Williams
• M A Zonderland-Thomassen
Dalgaard, R., Schmidt, J., Halberg, N., Christensen, P., Thrane, M., Pengue, W.A., 2008. LCA of soybean meal. Int. J. LCA 13, 240e254. De Boer, I.J.M., Cederberg, C., Eady, S., Gollnow, S., Kristensen, T., Macleod, M., Meul, M., Nemecek, T., Phong, L.T., Thoma, G., van der Werf, H.M.G., Williams, A.G., Zonderland-Thomassen, M.A., 2011. Greenhouse gas mitigation in animal production: towards an integrated life cycle sustainability assess-ment. Curr. Opin. Environ. Sustain. 3, 423e431. DEFRA, 2010. Department for Environment, Food and Rural Affairs. Definitions of Terms Used in Farm Business Management. http://www.defra.gov.uk (visited January 2012).
Life cycle assessment of milk production of 41 intensive dairy farms in north Italy
• M Guerci
• L Bava
• M Zucali
• A Sandrucci
• C Penati
• A Tamburini
Guerci, M., Bava, L., Zucali, M., Sandrucci, A., Penati, C., Tamburini, A., 2013a. Life cycle assessment of milk production of 41 intensive dairy farms in north Italy. J. Dairy Res. 80, 300e308.
Parameters affecting the environmental impact of a range of dairy farming systems in Denmark, Germany and Italy Una foraggicoltura al servizio dell'alle-vamento e del territorio montano: tradizione e innovazione a confronto
• M Guerci
• M T Knudsen
• L Bava
• M Zucali
• P Schönbach
• T Kristensen
Guerci, M., Knudsen, M.T., Bava, L., Zucali, M., Schönbach, P., Kristensen, T., 2013b. Parameters affecting the environmental impact of a range of dairy farming systems in Denmark, Germany and Italy. J. Clean. Prod. 54, 133e141. Gusmeroli, F., Paoletti, R., Pasut, D., 2006. Una foraggicoltura al servizio dell'alle-vamento e del territorio montano: tradizione e innovazione a confronto. Qua-derno SoZooalp 3, 26e40.
Land use change e GHG emissions from food and feedstuffs Prospects for the European beef sector over the next 30 years A Common Carbon Footprint Approach for Dairy. The IDF Guide to Standard Lifecycle Assessment Methodology for the Dairy Sector. International Dairy Federation
• S Hörtenhuber
• M Theurl
• T Lindenthal
• W Zollitsch
• e
• J.-F Hocquette
• V Chatellier
Hörtenhuber, S., Theurl, M., Lindenthal, T., Zollitsch, W., 2012. Land use change e GHG emissions from food and feedstuffs. In: Proceedings of the 8th Interna-tional Conference on LCA in the Agri-Food Sector, Saint-Malo, France, 1e4 Oct 2012, pp. 252e256. Hocquette, J.-F., Chatellier, V., 2011. Prospects for the European beef sector over the next 30 years. Anim. Front. 1 (2), 20e28. IDF, 2010. Bulletin of the IDF No 445/2010. A Common Carbon Footprint Approach for Dairy. The IDF Guide to Standard Lifecycle Assessment Methodology for the Dairy Sector. International Dairy Federation, Brussels, Belgium. IPCC, 2006a. IPCC Guidelines for National Greenhouse Gas Inventories. In: Agri-culture, Forestry and Other Land Use. Chapter 10: Emissions from Livestock and Manure Management, vol. 4, pp. 1e87. http://www.ipcc-nggip.iges.or.jp/public/ 2006gl/pdf/4_Volume4/V4_10_Ch10_Livestock.pdf (visited May 2012).
Effect of summer grazing on carbon footprint of milk in Italian Alps: a sensitivity approach Life-cycle assessment of greenhouse gas emissions from dairy production in Eastern Canada: A case study
• M Guerci
Please cite this article in press as: Guerci, M., et al., Effect of summer grazing on carbon footprint of milk in Italian Alps: a sensitivity approach, Journal of Cleaner Production (2013), http://dx.doi.org/10.1016/j.jclepro.2013.11.021 Mc Geough, E.J., Little, S.M., Janzen, H.H., McAllister, T.A., McGinn, S.M., Beauchemin, K.A., 2012. Life-cycle assessment of greenhouse gas emissions from dairy production in Eastern Canada: A case study. J. Dairy Sci. 95, 5164e 5175.
Greenhouse gas emissions from production of imported and local cattle feed Life cycle inventories of Swiss and European agricultural production systems
• L Mogensen
• T Kristensen
• T L T Nguyen
• M T Knudsen
• T Nemecek
• T Kägi
Mogensen, L., Kristensen, T., Nguyen, T.L.T., Knudsen, M.T., 2012. Greenhouse gas emissions from production of imported and local cattle feed. In: Proceedings of the 8 th International Conference on LCA in the Agri-Food Sector, Saint-Malo, France, 1e4 Oct 2012, pp. 321e326. Nemecek, T., Kägi, T., 2007. Life cycle inventories of Swiss and European agricultural production systems. Final Report Ecoinvent V2.0 No. 15a. In: SimaPro PhD 7.
Agroscope Reckenholz-Taenikon Research Station ART, Swiss Centre for Life Cycle Inventories
• Database
Database/Professional/Ecoinvent. Agroscope Reckenholz-Taenikon Research Station ART, Swiss Centre for Life Cycle Inventories, Zurich and Dübendorf, CH.
Livestock Feed Production
• P H Nielsen
Nielsen, P.H., 2003a. Livestock Feed Production. In: http://www.lcafood.dk/ processes/industry/livestockfeedproduction.html (visited February 2012).
Milk Powder Production
• P H Nielsen
Nielsen, P.H., 2003b. Milk Powder Production. In: http://www.lcafood.dk/processes/ industry/milkpowderproduction.html (visited July 2012).
|
|
## Elementary Algebra
$x=\dfrac{25}{42}$
Multiplying both sides by the $LCD= 70 ,$ and using the properties of equality, the value of the variable that satisfies the given equation, $-\dfrac{6}{5}x=-\dfrac{10}{14} ,$ is \begin{array}{l}\require{cancel} 14(-6x)=5(-10) \\\\ -84x=-50 \\\\ x=\dfrac{-50}{-84} \\\\ x=\dfrac{\cancel{-2}\cdot25}{\cancel{-2}\cdot42} \\\\ x=\dfrac{25}{42} .\end{array}
|
|
# Changes between Version 4 and Version 5 of udg/ecoms/dataserver/interfaces
Ignore:
Timestamp:
Feb 26, 2014 11:00:28 AM (8 years ago)
Comment:
--
### Legend:
Unmodified
v4 In a similar vein as the [https://www.meteo.unican.es/trac/meteo/wiki/EcomsUdg/RPackage R functions] previously presented, some parallel functions have been developed in other environments. In the following subsections we introduce the versions of the loadSystem4 utility developed in the Python and Matlab languajes with some worked examples. The code of these functions is available on request. In a similar vein as the [wiki:EcomsUdg/RPackage R package], some parallel utilities have been developed in other environments. In the following subsections we introduce the versions of the loadSystem4 function developed in the Python and Matlab languages with some worked examples. '''IMPORTANT NOTE''': Unlike the R package, which will be continuously updated as part of the data management activities, the Python and Matlab functions are not guaranteed to be maintained and updated with the same regularity, and therefore the use of the R environment is encouraged in the framework of the SPECS and EUPORIAS Projects. '''IMPORTANT NOTE''': Unlike the [wiki:EcomsUdg/RPackage R package], which will be continuously updated as part of the data management activities, the Python and Matlab functions are not guaranteed to be maintained and updated with the same regularity, and therefore the use of the R environment is encouraged in the framework of ECOMS. * [wiki:./EcomsUdg/Interfaces/Python Python] * [wiki:./EcomsUdg/Interfaces/Matlab Matlab] * [wiki:EcomsUdg/Interfaces/RPackage/Python Python Interface] * [wiki:EcomsUdg/Interfaces/RPackage/Matlab MatLab Interface]
|
|
## Pressure Induced Structural and Electrical Phase Transitions in disordered materials: Silicon monoxide and the Anderson transition ()
Reuben Shuker
Application of very high pressure significantly alters the nature of intermolecular interaction, chemical bonding, molecular configuration, and crystal structure of solids. Practically we use Diamond Anvil Cell (Fig. 1) to produce pressures in the range from 1 and 40 GPa. Electronic correlations effects induced by high pressure results unusual phase transitions. Our main interest is focused on amorphous systems such as silicon
monoxide, where we observed an insulator to metal transition at pressure of 12 GPa (all in the amorphous phase). The temperature dependence of the conductivity is in very good agreement with the Anderson model for amorphous materials indicating Anderson transition in disordered systems. The material response to high pressure is studied by a variety of experimental methods: x-ray diffraction using synchrotron radiation, IR spectroscopy, Raman scattering and electrical conductivity measurements.
For example, the Raman scattering spectra and the pressure dependence of the resistivity for SiO sample are presented in figures 2 and 3 respectively. IR and Raman results assign bond-bending to Raman active modes since it is hardened under pressure, whereas IR active modes are dominated by bond-stretching modes.
The miniature sample (~100 $$\mu m$$ in diameter) we use in high pressure studies makes the electrical resistivity measurements a challenging experimental problem. However, we managed to observe a dramatic drop of the resistivity on one hand (Fig. 3), and significant change in the temperature dependence of the resistance on the other, both happen at pressure 12 GPa. Short-rang order on the molecular scale helps us explain the insulator to metal transition in terms of Anderson's theory of localization, where lone-pair electrons create states in the gap.
Figure 1: Diamond Anvil Cell Figure 2: The Raman scattering spectra of SiO sample at different pressures. Click to enlarge. Figure 3: The pressure dependence of the resistivity for SiO sample. Click to enlarge.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.