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Do cosmological event horizon evaporates like black-hole horizon?

While the inflation phase of the universe, expansion was exponential and so the universe was de Sitter-like. So, for a point-like observer, there was a cosmological event horizon. Do this horizon evaporates due to the emission of particles? If this horizon evaporates does this mean that the expansion tends to be naturally not in a de Sitter phase?

Edit:

What I've tried so far:

The temperature of the de Sitter horizon is: $$$$T=\frac{H}{2\pi}.$$$$ Where $$H=\frac{\dot{a}}{a}$$. The area of the horizon is: $$$$A=\frac{4\pi}{H^2}.$$$$ I don't know if I can do this but let's say that the luminosity inside de horizon due to Hawking radiations is $$L=\sigma A T^4$$, where $$\sigma$$ is the Stefan-Boltzmann constant. Assuming that the energy associated to the Horizon is $$E_\text{dS}=\alpha H^2$$ ($$\alpha$$ is a proportionality constant), we thus have : $$$$\frac{d H^2}{dt}=-\frac{2\sigma}{\alpha (2\pi)^3}H^2.$$$$ Whose solution is: $$$$H(t)=H(0)e^{-\frac{\sigma}{\alpha (2\pi)^3}t}$$$$ Then, we find an $$a$$ of the following form: $$$$a(t)=C\,\text{exp}\left( -\frac{\alpha (2\pi)^3}{\sigma}H(0)e^{-\frac{\sigma}{\alpha (2\pi)^3}t} \right)$$$$ Which tends to $$0$$ at $$t\rightarrow-\infty$$, is exponential when $$t\sim0$$, and tends to give a Minkowski-like metric for $$t\rightarrow +\infty$$

• This discusses event horizons en.wikipedia.org/wiki/Event_horizon#Cosmic_event_horizon . I think you are confusing the Big Bang universe with a black hole event horizon ? Nov 20, 2020 at 19:25
• No I don't, these horizons come from different contexts (even though the Schwatzschild-de Sitter metric exists). I was thinking about the Gibbons-Hawking effect and the particle creation by a de Sitter horizon. Since those particles are created inside the horizon, in opposition to particles created by black holes, I thought that the energy associated to the horizon was increasing with time. So the radius of horizon should increase with time too, thus the expansion tends to not be in a de Sitter phase. But I am not sure about this. Nov 21, 2020 at 9:32
• sorry, it is not my field,so I cannot help you. Nov 21, 2020 at 12:22
• Hi Jeanbaptiste Roux@ I am very much confused about your usage: "inflation phase . . . cosmological event horizon". You seem to be saying that none of several definitions of a "cosmological event horizon" presented in the Wikipedia article match what you mean with respect to your question. It would be very helpful if you would post a reference to a specific definition which you have in mind. The Schwatzschild-de Sitter metric does not apply to cosmological inflation.
– Buzz
Nov 22, 2020 at 20:10
• I know that the de Schwarzschild-de Sitter metric does not apply to cosmological inflation. But I think de Sitter's does since inflation is conjectured to be an exponential expansion. This is why I used "cosmological event horizon" though I have to admit this is not a good naming. I meant by this a causal event horizon arising from this de Sitter phase. The radius of this event horizon should have been very small because of the very fast expansion of the universe in the inflation epoch. Nov 23, 2020 at 7:38

The observable universe, under eternal inflation from dark energy, will asymptotically evolve to a de Sitter spacetime. This spacetime is a vacuum configuration with a cosmological constant $$\Lambda$$. The stationary metric for this spacetime is $$ds^2~=~Adt^2~-~A(r)^{-1}dr^2~-~r^2d\Omega^2,~ A(r)~=~(1~-~\Lambda r^2/3)$$ A radial null geodesic with $$ds^2~=~0$$ and $$d\Omega^2~=~0$$ gives the velocity $${\dot r}~=~dr/dt$$ $$=~A(r)$$, where this pertains to both out and in going geodesics near the cosmological horizon $$r~=~\sqrt{3/\Lambda}$$ as measured from $$r~=~0$$. The total action for the motion of a particle is $$S~=~\int p_r dr$$ $$-~\int Hdt$$. Consider the bare action of massless particles, using methods found in [1], across the horizon from $$r$$ to $$r'$$, $$S~=~\int_r^{r'}p_rdr~=~\int_r^{r'}\int_0^{p_r}dp_rdr.$$ The radial velocity of a particle is $${\dot r}~=~dr/dt$$ $$=~dH/dp_r$$, which enters into the action as, $$S~=\int_r^{r'}\int_0^H{{dH'}\over{\dot r}}dr.$$ The field defines $$H^\prime~=~\hbar\omega'$$. The integration over frequencies is from $$E$$ to $$E~-~\omega$$, for the ADM energy. The action is properly written as $$S~=~-\hbar\int_r^{r'}\int_E^{E-\omega}{{d\omega'}\over{\dot r}}dr,$$ where the negative sign indicates the quanta is tunneling across the horizon to escape the Hubble region with radius The radial velocity $${\dot r}~=~\sqrt{\Lambda/3}r$$ defines the action $$S~=~-\hbar\int_r^{r'}\int_0^\omega{{d\omega dr}\over{\pm 1~-~\sqrt{\Lambda r^2/3} }}~=~\sqrt{3/\Lambda}tanh^{-1}(\sqrt{\Lambda/3}r)$$ The action is then the delay coordinate evaluated as $$r^*~=~\int {{dr}\over{1~-~\Lambda r^2/3}}~=~\sqrt{3/\Lambda}tanh^{-1}(\sqrt{\lambda/3}r).$$ The domain $$[0,~\sqrt{3/\Lambda})$$ defines a real valued action. Since, $$tanh^{-1}(x)~=~{1\over 2}ln((1~+~x)/(1~-~x))$$ for $$r~>~\sqrt{3/\Lambda}$$ the argument of the logarithm is negative. In this case the action is $$S~=~\sqrt{3/\Lambda}ln\Big({{\sqrt{\Lambda/3}r~+~1}\over{\sqrt{\Lambda/3}r~-~1 }}\Big)~+~i\pi\sqrt{3/\Lambda}.$$ The imaginary part represents the action for the quantum field emission as $$r~\rightarrow~\infty$$. The delay coordinate is defined on $$[0,~\infty)$$ which assures an S-matrix is defined on an unbounded causal domain, and this holds in general as well. \vskip.12in The metric elements we need to be concerned about is then $$g_{tt}~=~(1~-~\Lambda r^2/3)$$, $$g_{rr}~=~1/(1~-~\Lambda r^2/3)$$. The extrinsic tensor (Gauss' second fundamental form) is then $$K_{ab}~=~{1\over 2} n^c\partial_c(g_{ab})$$, for $$n^r~=~\sqrt{A(r)}$$. These are rather easy to evaluate this $$K_{tt}~=~n^r\partial_rg_{tt}~=~ {2\over 3}{{\Lambda r}\over 3}(1~-~\Lambda r^2/3)^{1/2}$$ $$K_{rr}~=~n^r\partial_rg_{rr}~=~- {2\over3}{{\Lambda r}\over 3}(1~-~\Lambda r^2/3)^{-3/2}.$$ The curvature $$G^{00}~=~(Tr K)^2~-~Tr(K^2)$$ which is then $$G^{00}~=~(K_{tt}~+~K_{rr})^2~-~K_{tt}^2~-~K_{rr}^2~=~2K_{tt}K_{rr}~=~{8\over 9}\Big({{\Lambda^2 r^2}\over{1~-~\Lambda r^2/3}}\Big).$$ This is the vacuum energy available in the de Sitter vacuum. \vskip.12in We now perform a similar analysis above, but instead consider the radiation production according to the transition $$\Lambda~\rightarrow~\Lambda~+~\delta\Lambda$$. The transition is considered according to the metric back reaction of the de Sitter vacuum. This tunneling defines the imaginary part of the action $$S~=~\int_{\Lambda_0}^{\Lambda}pdr.$$ The velocity term here is the computed from the Hamilton equation $${\dot r}~=~\partial H/\partial p$$ which permits this to be written as $$S~=~\int_{\Lambda_0}^{\Lambda}\int_0^R{{dr}\over{\dot r}}dH.$$ The Hamiltonian used is the $$H~=~G^{00}$$ computed here, with $${\dot r}~=~\pm(1~-~\Lambda r^2/3)$$ with $$S~=~\int_{\Lambda_0}^{\Lambda}\int_{r_i}^{r_f}dr\Big({{4r}\over 3}\Big)^2\int_{\Lambda_0}^{\Lambda}{\Lambda\over{(1~-~\Lambda r^2/3)^2}}\Big(1~+~{{\Lambda r^2/3}\over{1~-~\Lambda r^2/3}}\Big)d\Lambda$$ $$=~16\int_{r_i}^{r_f}dr \Big( {1\over{r^2}}\big(log(\Lambda r^2~-~3)~-~{3\over{\Lambda r^2~-~3}}\big)\Big|_{\Lambda_0}^{\Lambda}$$ $$=~16\big(ln(\Lambda r^2~-~3)/r~-\sqrt{\Lambda/3}~arctanh(\sqrt{\Lambda/3}r)~-~1/r\big) \Big|_{\Lambda_0}^{\Lambda}\Big|_{r_i}^{r_f}.$$ This complicated expression is evaluated for $$\Lambda~=~\Lambda_0~-~\delta\Lambda$$ with $$S~=~8\Big(\sqrt{\Lambda/3}~arctanh({\Lambda/3}r)~+~{r\over{1~-~3/(\Lambda r^2)}}~+~{2\over{\Lambda~-~3/ r^2}}\Big)\Big|_{r_i}^{r_f}\Delta\Lambda$$ Of course this solution exhibits a singularity for the radius extended across $$\sqrt{3/\Lambda}$$. We then consider the integration with respect to the small change in the cosmological constant $$S~=~=~16\int_{r_i}^{r_f}dr \Big( {1\over{r^2}}\Big(log(\Lambda r^2~-~3)~-~{3\over{\Lambda r^2~-~3}}\Big)\Big|_{\Lambda_0}^{\Lambda}$$ $$\simeq~16\Big({1\over{\Lambda r^2~-~3}}~-~{3\over{(\Lambda r^2~-~3)^2}}\Big)\delta\Lambda$$ The calculus of residues and Cauchy integral formula gives this result in a very simple form for $$z^2~=~\Lambda r^2$$ and we then left with the result $$S~=~{\sqrt{3}\over 4}\pi\delta\Lambda$$ \vskip.12in Now that we have that set up we propose a quantum model. The $$G^{00}~=~H$$ which under the Hamiltonian constraint is zero. Yet with quantum mechanics we propose an evolution equation $$i\partial\Psi[g,\phi]/\partial t~=~G^{00}\Psi[g,\phi]$$, where the deviation from zero in the Wheeler DeWitt equation $$NH\Psi[g]~=~0$$ is due to the occurrence of a scalar field in the spacetime. For now on we will substitute $$H$$ for $$G^{00}$$. Now we want to compute a transition rate, which involves both a change in g and the internal scalar field $$\phi$$. So we take a second derivative of $$\Psi~=~\Psi[g,\phi]$$ to get $$-{{\partial^2\Psi}\over{\partial t^2}}~=~ H^2\Psi~+~i{{\partial H}\over{\partial t}}\Psi.$$ We now express this according to $$|\Psi\rangle$$ and $$\Psi'\rangle$$ by $$H^2?~=~H\int d\mu(\Psi')|\Psi'\rangle\langle\Psi'|H|\Psi\rangle,$$ which is just a completeness sum $$1~=~\int d\mu(\Psi')|\Psi'\rangle\langle\Psi'|$$ inserted between the two Hamiltonians. We then have a summation over transition probabilities as $$\int d\mu(\Psi')|\langle\Psi'|H|\Psi\rangle|^2~=~-\langle\Psi|{{\partial^2}\over{\partial t^2}}|\Psi\rangle~-~i\langle\Psi|{{\partial H}\over{\partial t}}|\Psi\rangle.$$ The first term on the right hand side depends only on the transition of the scalar field, which vanishes if the scalar field is initially zero. The last term will then depend upon $$\partial\Lambda/\partial t$$, and may be computed from the $$G^{00}$$ above. These two terms represent the long wavelength Hawking-Gibbon radiation produced by the decay of the cosmological constant and the change in Hamiltonian due to the production of this radiation.


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# How do you simplify the expression (56a^6)/(8a^4)?

$\frac{{\cancel{56}}^{7} {\cancel{{a}^{6}}}^{{a}^{2}}}{\cancel{8} \cancel{{a}^{4}}} = 7 {a}^{2}$

format: plain


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# Deep Learning: Expression Analysis

A smiling face implies agreeableness and from an angry face, we infer danger! Thus, our minds immediately evaluate the underlying emotion of a person by merely looking at their face. Fascinating! Lets make machines do that.

• To determine the different facial expressions, the variations in each facial features are used. For detection and classification between smile and neutral faces we will use softmax regresion,open source computer vision (OpenCV) and sckit library for training of different set of images.

## Algorithm Description

(a) OpenCV
Uses opencv to read the images in grayscale and resizing them.

(b) Vectorization
While we implement Softmax regression, it need 1 dimension array as an Input, It’s important to convert our 2-D array to 1-D array.

(c) Softmax regression
Softmax regression (or multinomial logistic regression) is a generalization of logistic regression to the case where we want to handle multiple classes. In logistic regression we assumed that the labels were binary: $y(i) ∈ {0,1}$. We used such a classifier to distinguish between two kinds of hand-written digits. Softmax regression allows us to handle $y(i) ∈ {1,…,K}$ where K is the number of classes.

(d) Tensorflow
Tensorflow is a computational framework for building machine learning models. TensorFlow provides a variety of different toolkits that allow you to construct models at your preferred level of abstraction. You can use lower-level APIs to build models by defining a series of mathematical operations. Alternatively, you can use higher-level APIs (like tf.estimator) to specify predefined architectures, such as linear regressors or neural networks.

## Time to analyse

We shall construct a model capable of detecting smiling faces from images.

## 1. Install the dependencies.

• Install Python
• Install pip
• Install sklearn for python: pip3 install scikit-learn
• Install numpy: pip3 install numpy
• Install opencv: pip3 install opencv-python
• Install tensorflow: pip3 install --upgrade tensorflow
• Install matplotlib: pip3 install matplotlib
• Install pandas: pip3 install pandas

• The function load_images_from_folder converts images into grayscale and resize them using opencv.
img = cv2.imread(path, 0)

• Append the images into a new table.

• Also it assign labels to each image depending whether it is under smile or neutral category.

• To build a classification model,split the whole dataset into trainig and test sets.

The reason for splitting the dataset — evaluating the performance of classifier on the same set as it has been trained is unfair or poor practice, because we are not interested in how well the classifier memorizes the training set. Rather, we are interested in how well the classifier generalizes its recognition capability to unseen data.

• Flattening the data-Also we have to convert our image from 2D to 1D vector since input can be taken as only 1 dimension array.

## 3. Model Training

• Building softmax regression model using tensorflow.
• tf.placeholder is used to feed actual training examples.
• tf.Variable for trainable variables such as weights (W) and biases (B) for your model. Your tf.variables will be trained (modified) as the result of this training.
• tf.nn.softmax(tf.matmul(x, W) + b, name="first_operation") funtion produces just the result of applying the softmax function to an input tensor. The traditional matrix multiplication is only used when calculating the model hypothesis as seen in the code to multiply x by W.
• The code y_ * tf.log(y) performs an element-wise multiplication of the original targets => y_ with the log of the predicted targets => y.It is this measure (i.e., the cross-entropy loss) that is minimized by the optimization function.

## How well our model is performing?

• Construct a new gradient descent optimizer.
• tf.Session() initiates a TensorFlow Graph object in which tensors are processed through operations. Also, a session contains variables, global variables, placeholders and ops. These have to be initiated once the session is created. Hence we call sess.run( tf.global_variables_initializer() ).
• Calculate loss and accuracy.

## 4. Prediction

• Load the image in test folder.
• Normalize the image.
• Use softmax regression to predict using weights and bias.
• Print the result according to the label we got as the output.
• To improve its log loss, we need to train our model by adding more images.
• You can try SVM and KNN models also.

For code, see the GitHub repository here.

Ayushi Web Developer, Machine Learning Engineer, Tech Speaker


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VectorMethod Sequence
2016

Problem - 3075
Starting at the origin, a bug crawls 1 unit up, 2 units right, 3 units down and 4 units left. From this new point, the bug repeats this entire sequence of four moves 2015 more times, for a total of 2016 times. The coordinates of the bug’s final location are $(a, b)$. What is the value of $a + b$?


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# How do you theoretical physicists stay motivated?

1. Aug 19, 2010

### camel_jockey

Many of the most fundamental physics problems that theoretical physicists work with are really cool, but they are so abstract that sometimes it just feels that you are sitting with huge tomes/books and looking at details.

How do you stay motivated? How do you bring back the taste for maths and physics?

I try listening to mysterious music such as Shpongle, or playing sci-fi oriented computer games :D

2. Aug 19, 2010

### Norman

I usually don't have to get motivated to work on the physics research (calculations, research code development, modeling) but the other parts of my job (writing papers, reports, administrative stuff) can really drag me down.

I have two coping mechanisms:
1. Sit back and enjoy a cup of coffee. This lets me relax and just think for a bit and helps to get me through the late afternoon quiet that creeps around here which makes me want to sleep.
2. Remember the big picture. I get to work on some really cool projects and they can have profound implications on major space missions. The coolness factor definitely gets lost in the day-to-day grind that can be my job. When I sit back and look at what I get to work on, all interesting results, the implications, the extremely talented people I get to work with- I remember how I fit into the big picture and I can't help but think of how lucky I am. This makes doing the things I like least about my job seem very much worth it.

In the end, it is all about taking a breath and being able to adjust my perspective.

3. Aug 19, 2010

### Katastrofa

Imagine being out of work?

4. Aug 20, 2010

### camel_jockey

This used to work for me. Nowadays it simply doesn't do the trick. :(

5. Aug 21, 2010

### twofish-quant

I sometimes find that the most "boring" parts of my job are the ones that are most relaxing. I'm trying to figure how this #$@$@# paper derived this @#$@#$ formula, and so I work through everything step by step. I get the wrong answer. I do it again. Get the wrong answer. Repeat for a few hours or days, and then I get somewhere. This sort of thing is quite relaxing because it often turns out that there is total chaos around me, and by staring at the problem, I can blot out the chaos.

6. Aug 22, 2010

### Llewlyn

I am used to the same coping mechanisms of Norman.

Ll.

7. Sep 2, 2010

Play games like Halo and remember that physicists probably discovered the slip-space jump and effective materials to build those huge mile-long magnetic accelerator cannons and those gigantor ships. Also the armored body-suit of the Spartans is pretty awesome (they can jump 10 feet and run 55 mph!).

I mean come on. Physics is the foundation that all of engineering and technology develops. Isn't that awesome enough?

8. Sep 2, 2010

### Zhabka

I'm still just a student but I get the same way studying, sometimes I just wonder why I am doing this. Kind of like camel jockey, I love listening to weird or intense music. Metal or psy-trance is good. I do some of my best math when I have Infected Mushroom blasting. It just keeps me awake and driven.

Other than that I go for a lot of walks and talks. I guess I am what they call "kinetic thinkers". Every 10 or 15 minutes I need to get up and jump around, pace in circles, mumble to myself. Sometimes it even gets to the point where I am running back and forth across the room.

Other times I go for a long walk and discuss with a friend either what I am working on, or something completely different. I find thinking about completely different things gives me a fresh perspective and I see the world as new and fascinating again.


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## Topology and Milnor

Hey guys,

So my prof assigned John Milnor's book about topology from a differentiable viewpoint for out topology and geometry class. I was wondering if anyone had a book they could recommend as an introduction into topology because I dont think professor Milnor's book really is an introductory book although I could be wrong.

I do have some basic knowledge of topology that I've picked up from reading books on tensors (tensor geometry by dodson and part of tensors on manifolds by bishop). I was considering Munkres, but does anyone have any other recommendations? On a sidenote I'm primarily a physics major but I dont know if that matters for a topology book on that level.

Thanks.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Try also this as intro for meaning and intuition: http://www.amazon.com/Geometry-Topol...5617658&sr=1-1 and this, rigourus: http://www.amazon.com/Introduction-T...ref=pd_sim_b_2
 Munkres is as good as it gets for UG, though I haven't tried other books of UG Topology.

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Homework Help

## Topology and Milnor

I second Goldbeetle's recommendation of Mendelson's "Introduction to Topology." It is an absolutely crystal clear treatment of the basics of metric spaces and point-set topology. It doesn't cover as much material as Munkres, but for what it does cover I think it's superior. And you can't argue with the price: $6.65 from Amazon.  Recognitions: Homework Help Science Advisor read milnor. that is the best book of its kind in existence. ask questions here for help. when you pay money to be taught by experts, do not resist what they are trying to do for you. take their advice.  Quote by mathwonk read milnor. that is the best book of its kind in existence. ask questions here for help. when you pay money to be taught by experts, do not resist what they are trying to do for you. take their advice. Can't comment on Milnor, but Professors aren't infallible. Case in point, we are using Gamelin and Green for our undergrad Topology. Its a fine book, but even the professor admitted that its geared more towards grad level students. I have supplemented the book with the Medelson text (supplemented, not replaced). Heck, I even supplement with my Ross Analysis text. Not sure if I could handle the class if all I had to rely on was Gamelin and lectures.  Well unfortunately my library doesn't have Mendelson now, so for the time being it is Munkres and I feel MathematicalPhysicist is probably right. I will make sure to check out Mendelson though once I can find it. MathWonk- Believe me if a professor assigns a book I'll get it. I definitely plan on reading Milnor I'm just looking for a book over the summer to help me learn about Topology first. I feel like Milnor's book is more advanced. (he mentions some terms I dont know offhand). Thanks for the advice!  I used Czes Kosniowski's A First Course in Algebraic Topology for my topology class. The first ten chapters of this book are on point-set topology that is suited for anybody who is learning topology for the first time. I thought the book was very well written, and it did a fantastic job interlacing abstract concepts and concrete examples. But since your professor recommends a book that has more emphasis on differential topology, the later chapters in this book might not be helpful.  Quote by LBloom Well unfortunately my library doesn't have Mendelson now, so for the time being it is Munkres and I feel MathematicalPhysicist is probably right. I will make sure to check out Mendelson though once I can find it. MathWonk- Believe me if a professor assigns a book I'll get it. I definitely plan on reading Milnor I'm just looking for a book over the summer to help me learn about Topology first. I feel like Milnor's book is more advanced. (he mentions some terms I dont know offhand). Thanks for the advice! Yeah, I truly believe Mendelson would be be the ideal place to start self-studying over the summer. You can get it new for about$11 at Amazon or your local Barnes and Nobel, or get it used from half.com for about...half that :). The only downside is that there are no solutions to the problems. That upper div Gamelin book I mentioned that we use in an undergrad class is not bad, just to..."polished" (as grad books are) but the plus side is that nearly all answers are in the back. Book cost is about 15 bucks.

Never read Munkrese, but it does seem to be the defacto modem undergrad Topolgy book, so that its available at your library is great!

Dunno what you know about topology or why you'd wanna take it, but Topology is very similar to Analysis. For what ever reason I absolutely hate anlysis, but I am really enjouyinjg topology, as hard as it may, I love topology


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# Math Help - trig-substitution!

1. ## trig-substitution!

i have this $\intk\sqrt{\frac{1-z}{z}}dz$

i set $z=sin^2\theta$
differentiating z w.r.t $\theta$, i have $dz=2sin\theta cos\theta d\theta$
substituting

$2k\int\sqrt{\frac{1-sin^2\theta}{sin^2\theta}}sin\theta cos\theta d\theta$

i recall that $cos^2\theta=1-sin^2\theta$

i tend to be confuse from here onward, pls help!

2. ## Re: trig-substitution!

you'll get cos^2thetha when you'll put 1-sin^2theta=cos^2theta. then you can expand cos^2theta as (cos2theta+1)/2...then integrate cos2theta/2 and 1/2 separately...

3. ## Re: trig-substitution!

$\text{You know that } \int \sqrt{\frac{1-z}{z}}dz = 2\int\sqrt{\frac{1-\sin^{2}(\theta)}{\sin^{2}(\theta)}}\sin(\theta) \cos(\theta)\,\, d\theta$

$\text{ by letting } z = \sin^{2}{(\theta)} \text{ and so } dz = 2\sin{(\theta)}\cos{(\theta)}\,\,d\theta$

Now:

\begin{align*}2\int\sqrt{\frac{1-\sin^{2}(\theta)}{\sin^{2}(\theta)}}\sin(\theta) \cos(\theta)\,\, d\theta =& 2 \int \frac{\sqrt{(\cos^{2}{(\theta)})}}{\sqrt{(\sin^{2} {(\theta)})}} \sin(\theta) \cos(\theta)\,\, d\theta..........[\text{ because } 1 - \sin^{2}(\theta) = \cos^{2}(\theta)] \\ =& 2\int\frac{\cos(\theta)}{\sin(\theta)}\sin(\theta) \cos(\theta)\,\, d\theta \\ =& 2\int\cos^{2}(\theta)\,\, d\theta .....[\text{by using simple algebra}] \\ =& 2\int\frac{\cos(2\theta) + 1}{2}\,\, d\theta ......[\text{because }\cos(2\theta) = 2 \cos^{2}(\theta) - 1] \\ =& 2\left( \int \frac{\cos(2\theta)}{2}d\theta + \int \frac{1}{2}\,\,d\theta\right) \\ =& 2\left(\frac{\sin(2\theta)}{4} + \frac{\theta}{2}\right) + C \\ =& \frac{ \sin(2\theta)}{2} + \theta + C \\ =& \frac{2\sin(\theta)\cos(\theta)}{2} + \theta + C ......[\text{because } \sin(2\theta) = 2\sin(\theta)\cos(\theta)] \\ =& \sin(\theta) \cos(\theta) + \theta + C \end{align*}

By plugging into $\sin(\theta) = \sqrt{z}$, $\cos(\theta) = \sqrt{1 - z}$ and $\theta = \sin^{-1}(\sqrt{z})$ we get:

$\int k\sqrt{\frac{1-z}{z}}dz = \sqrt{z} \sqrt{1 - z} + \sin^{-1}(\sqrt{z}) + C$

And that's it. That's the answer. If you differentiate the new found answer with CAS(like maple) you'll get what you started with which is the given integral in the question. Hope this helps.


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## Introductory Algebra for College Students (7th Edition)

Divide each term in the polynomial by the monomial. Subtract the powers of the variable to simplify each term. (4x$^2$-6x)$\div$x=$\frac{4x^2}{x}$-$\frac{6x}{x}$=4x$^{(2-1)}$-6(1)=4x-6 Check the answer. x(4x-6)=4x(x)-6(x)=4x$^2$-6x


text:
TITLE

# Lower bounds of the size multipartite Ramsey numbers mj(Pn,KjÃ—b)

AUTHOR(S)
PUB. DATE
May 2012
SOURCE
AIP Conference Proceedings;5/22/2012, Vol. 1450 Issue 1, p259
SOURCE TYPE
DOC. TYPE
Article
ABSTRACT
For given two graphs G1 and G2, the size Ramsey multipartite numbers mj(G1,G2) is the smallest integer t such that every factorization of graph KjÃ—t := F1 + F2 satisfies the following condition: either F1 contains G1 as a subgraph or F2 contains G2 as a subgraph. In this paper, we derive some lower bound for the size multipartite Ramsey numbers mj(Pn,KjÃ—b) with integers j,n â‰¥ 3 and b â‰¥ 2.
ACCESSION #
75526981

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An edge-colored graph G is rainbow connected, if any two vertices are connected by a path whose edges have distinct colors. The rainbow connection number of a connected graph G, denoted rc(G), is the smallest number of colors that are needed in order to make G rainbow connected. In this paper we...

• Connected size Ramsey numbers of matchings and stars. Rahadjeng, Budi; Baskoro, Edy Tri; Assiyatun, Hilda // AIP Conference Proceedings;2016, Vol. 1707 Issue 1, p1

Let F, G, and H be finite, simple, and undirected graphs. The notation F â†’ (G,H) means that if the edge set of F is arbitrarily colored by red or blue, then there always exists either a red copy of G or a blue copy of H. The connected size Ramsey number È“c(G,H) is min{|E(F)| : F...

• TREE-PATH SIZE BIPARTITE RAMSEY NUMBERS. Syafrizal Sy // Far East Journal of Mathematical Sciences;May2013, Vol. 76 Issue 1, p139

For graphs G1 and G2, the size bipartite Ramsey number m2(G1, G2 ) is the least natural number t such that any red-blue coloring on the edges of K jxt necessarily forces a red G or a blue H as a subgraph. In this note, we determine the exact values of m2(Pn, Ts ) for all integers n, s â‰¥2,...

• On Some Ramsey Numbers for Quadrilaterals Versus Wheels. Dybizbański, Janusz; Dzido, Tomasz // Graphs & Combinatorics;May2014, Vol. 30 Issue 3, p573

For given graphs G and G, the Ramsey number R( G, G) is the least integer n such that every 2-coloring of the edges of K contains a subgraph isomorphic to G in the first color or a subgraph isomorphic to G in the second color. Surahmat et al. proved that the Ramsey number $${R(C_4, W_n) \leq n +... • Three Results on Cycle-Wheel Ramsey Numbers. Zhang, Yanbo; Broersma, Hajo; Chen, Yaojun // Graphs & Combinatorics;Nov2015, Vol. 31 Issue 6, p2467 Given two graphs$$G_1$$and$$G_2$$, the Ramsey number$$R(G_1, G_2)$$is the smallest integer$$N$$such that, for any graph$$G$$of order$$N$$, either$$G_1$$is a subgraph of$$G$$, or$$G_2$$is a subgraph of the complement of$$G$$. We consider the case that$$G_1 is a cycle and...

• Split Permutation Graphs. Korpelainen, Nicholas; Lozin, Vadim; Mayhill, Colin // Graphs & Combinatorics;May2014, Vol. 30 Issue 3, p633

The class of split permutation graphs is the intersection of two important classes, the split graphs and permutation graphs. It also contains an important subclass, the threshold graphs. The class of threshold graphs enjoys many nice properties. In particular, these graphs have bounded...

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# Past Functional Analysis Seminar

22 January 2013
17:00
to
18:16
David Seifert
Abstract
• Functional Analysis Seminar
15 January 2013
17:00
to
18:12
Chris Heunen
Abstract
• Functional Analysis Seminar
27 November 2012
17:00
to
18:12
David Preiss
Abstract
• Functional Analysis Seminar
20 November 2012
17:00
to
18:10
Pablo Shmerkin
Abstract
• Functional Analysis Seminar
13 November 2012
17:00
to
18:15
Yuri Safarov
Abstract
• Functional Analysis Seminar
30 October 2012
17:00
to
18:23
Vesselin Petkov
Abstract
We study symmetric systems with dissipative boundary conditions. The solutions of the mixed problems for such systems are given by a contraction semigroup $V(t)f = e^{tG_b}f,\: t \geq 0$. The solutions $u = e^{tG_b}f$ with eigenfunctions $f$ of the generator $G_b$ with eigenvalues $\lambda,\: \Re \lambda < 0,$ are called asymptotically disappearing (ADS). If we have (ADS), the wave operators are not complete and the inverse back-scattering problems become complicated. We examine the spectrum of the generator $G_b$ and we show that this spectrum in the open half plane $\Re \lambda < 0$ is formed by isolated eigenvalues with finite multiplicity. The existence of (ADS) is a difficult problem. We establish the existence of (ADS) for the Maxwell system in the exterior of a sphere. We will discuss some other applications related to the existence of (ADS).
• Functional Analysis Seminar
23 October 2012
17:00
to
18:23
Abstract
Let $T_1,\dots,T_n$ be bounded linear operators on a complex Hilbert space $H$. We study the question whether it is possible to find a unit vector $x\in H$ such that $|\langle T_jx, x\rangle|$ is large for all $j$. Thus we are looking for a generalization of the well-known fact for $n = 1$ that the numerical radius $w(T)$ of a single operator T satisfies $w(T)\ge \|T\|/2$.
• Functional Analysis Seminar
29 May 2012
17:00
to
18:12
Yuri Tomilov
Abstract
• Functional Analysis Seminar
15 May 2012
17:00
to
18:10
Fritz Gesztesy
Abstract
We extend the classical trace formula connecting the trace of resolvent dif- ferences of two (not necessarily self-adjoint) operators A and A0 with the logarithmic derivative of the associated perturbation determinant from the standard case, where A and A0 have comparable domains (i.e., one contains the other) to the case where their square root domains are comparable. This is done for a class of positive-type operators A, A0. We then prove an abstract result that permits to compare square root domains and apply this to the concrete case of 2nd order elliptic partial di erential operators in divergence form on bounded Lipschitz domains. This is based on various joint work with S. Hofmann, R. Nichols, and M. Zinchenko.
• Functional Analysis Seminar


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# Getting the actual (not default) value of a node socket that is linked to another socket

I know how to get the default_value from a node socket as described in this answer. But the answer there doesn't answer the question, namely: how to get the actual computed value of a socket.

How can I get the actual value using python?

For example, how would I determine the final Roughness value in this node structure assuming the value is constant for the material?

(I'm thinking this is not possible in general since it wouldn't work for values that change over the model, e.g., trying to obtain the Pointiness value from a Geometry node, which isn't a constant.)

• As a workaround, this answer is helpful. – jdk1.0 Feb 27 '18 at 23:06

## 1 Answer

Basically this cannot be done easily. The reason is:

• The shader node tree is compiled by cycles in SVM or OCL. And here the actual calculation takes place.
• Also every sample in Cycles can potentially have a different Roughness (for example when connected to a TextureImage along the way). So which sample do you want the result from.

If you have a special case you can mimic the needed nodes in python and evaluate for your needs.

• If not as a value, can the result of the shader node tree be baked to an image? E.g. would it be possible to implement an exporter that creates a roughness texture from arbitrary node inputs? – Don McCurdy Mar 14 '18 at 14:23


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## Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole

# Chapter 1 - Review - Exercises - Page 134: 28

#### Answer

$3xy^{2}(3x^{2}+4xy^{2}-y^{3})$

#### Work Step by Step

$12x^{2}y^{4}-3xy^{5}+9x^{3}y^{2}$ Take out common factor $3xy^{2}$: $12x^{2}y^{4}-3xy^{5}+9x^{3}y^{2}=3xy^{2}(4xy^{2}-y^{3}+3x^{2})=...$ Rearrange: $...=3xy^{2}(3x^{2}+4xy^{2}-y^{3})$

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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Fully hydrated spores of the highly radioresistant organism, Clostridium botulinum 33A exposed to 0.8 Mrad60 Co showed a characteristic pattern of survival associated with thermorestoration. When temperature during radiation was controlled at 0-100°C, the lowest viable cell count was at 0°C. Above 0°C radiosurvival gradually increased by 1 to 1 1/2 log cycles, reaching a peak at 75-80°C. Above 80°C, a rapid drop in survival occurred presumably caused by heat itself. The general features of the above survival pattern were observed in water saturated with N2 as well as in water saturated with O2. Radio-survival of spores in oxic water was at all temperatures approximately 1 log cycle lower than that in <tex-math>${\rm N}_{2}\text{-saturated}$</tex-math> water although the general features of the survival curves were approximately the same. An analysis of the shapes of survival curves in N2 and O2 suggested that the gradual radioprotection in the temperature range between 0°C and 80°C was due to annealment of lethal radicals in anoxic water, i.e. probably ·OH. The lethal radicals in oxic water (probably <tex-math>$\cdot {\rm HO}_{2}$</tex-math>) appeared not to be annealed appreciably.

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# How do find the second derivative for dy/dx = 5x^2 - 6/(y-2)?

Refer to explanation

#### Explanation:

We know that $y ' = \frac{\mathrm{dy}}{\mathrm{dx}} = 5 {x}^{2} - \frac{6}{y - 2}$ hence

$\frac{{d}^{2} y}{{d}^{2} x} = \frac{d}{\mathrm{dx}} \left(5 {x}^{2} - 6 \cdot {\left(y - 2\right)}^{- 1}\right) = 25 x + 6 {\left(y - 2\right)}^{- 2} \cdot \left(y\right) '$

But replacing the value of the first derivative we have that

$\frac{{d}^{2} y}{{d}^{2} x} = 25 x + 6 {\left(y - 2\right)}^{- 2} \cdot \left(5 {x}^{2} - \frac{6}{y - 2}\right) = 25 x + \frac{30 {x}^{2}}{{\left(y - 2\right)}^{2}} - \frac{36}{{\left(y - 2\right)}^{3}}$


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# NCERT Solutions For Class 10 Maths Chapter 14

## NCERT Solutions Class 10 Maths Statistics Types of statistics

### Ncert Solutions For Class 10 Maths Chapter 14 PDF Free Download

NCERT solutions class 10 maths chapter 14 statistics is one of the most important topics in Class 10 mathematics syllabus. Students must practice this chapter well to prepare maths more effectively. The class 10 maths chapter 14 statistics NCERT solutions is provided here to help students understand the concepts of the chapter in details. Students must practice the NCERT Solutions for class 10 maths chapter 14 thoroughly to score good marks in their class 10 board examination. Check the class 10 maths NCERT Solutions for chapter 14 pdf given below.

The topics covered in the chapter is given below

 Section Number Topic 14.1 Introduction 14.2 Mean of Grouped Data 14.3 Mode of Grouped Data 14.4 Median of Grouped Data 14.5 Graphical Representation of Cumulative Frequency Distribution 14.6 Summary

### NCERT Solutions Class 10 Maths Chapter 14 Exercises

Exercise – 1

Question 1:

A group of pupil has conducted a survey as a as the part of their college program, They collected examined data’s of plants from 20 houses and nearby areas. Find the mean of the collected plants from the given houses.

 No. of Plants Taken from 20 houses 0-2 2-4 4-6 6-8 8-10 10-12 12-14 Given number of houses 2 1 5 1 2 6 3

Name the method you use for finding the mean and why?

Solution:

 Class Interval fi xi fi xi 0-2 2 1 1 2-4 1 3 6 4-6 5 5 5 6-8 1 7 35 8-10 2 9 54 10-12 6 11 22 12-14 3 13 40 Sum fi=20 Sum fixi=163

Mean can be calculated as follows:

$\bar{x}= \frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{163}{20}=8.15$

In this issues the $f_{i} and x_{i}$ are small so the direct method is given to solve the problem.

Question2:

The distributed wages is given for 50 workers working in the factory.

 Daily wages given to workers (in Rs) 100-120 120-140 140-160 160-180 180-200 Number of wage workers in factory 14 12 6 8 10

Find the mean of workers of daily wages given by the workers.

Solution:

In this case, value of xi is quite large and hence we should select the assumed mean method.

 Class given Interval fi xi di = xi – a fi di 100-120 14 110 -40 -480 120-140 12 130 -20 -280 140-160 6 150 0 0 160-180 8 170 20 120 180-200 10 190 40 400 Sum fi=50 Sum fi di = -240

Now, mean of deviations of daily wagesis calculated as follows:

$\bar{d}=\frac{\sum f_{i}d_{i}}{\sum f_{i}}=\frac{-240}{50}$

x = d + a = 150+(-4.8)= 145.20

Question 3:

This distribution shows that the money daily pocket given to students in the given area. The mean pocket money is Rs. 18. Find the f which is the missing frequency.

 Daily money allowance (in Rs) 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Number of children 7 6 9 13 f 5 4

Solution:

 Class interval fi xi fixi 11-13 7 12 84 13-15 6 14 84 15-17 13 16 144 17-19 9 18 234 19-21 f 20 20f 21-23 4 22 110 23-25 5 24 96 Sum fi=44+f Sum fixi= 752+20f

We have;

$\bar{x}= \frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}$ $18= \frac{752+20f}{44+ f_{i}}$

18(44+18f) = 752+20f

792+18f = 752+20f

2f = 40

Missing frequency f= 20

Question4:

The number of 30 women were checked by one doctor and the distributed the heartbeat of them in following distribution table. Find the mean of heart beats of the given per minute for these thirty women, by choosing a suitable following method.

 Heartbeat per minute in given number 65-68 68-71 71-74 74-77 77-80 80-83 83-86 Number of 30 women 2 3 4 7 8 2 4

Solution:

 Class Interval fi xi di = xi – a fi di 65-68 2 66.5 -9 -18 68-71 3 69.5 -6 -24 71-74 4 72.5 -3 -9 74-77 7 75.5 0 0 77-80 8 78.5 3 21 80-83 2 81.5 6 24 83-86 4 84.5 9 18 Sum fi= 30 Sum fi di = 12

Now, mean can be calculated as follows:

$\bar{d}=\frac{\sum f_{i}d_{i}}{\sum f_{i}}=\frac{12}{30}=0.4$ $\bar{x}=\bar{d}+a=0.4+75.5=75.9$

Question 5:

The oranges were packed and were sold by the fruit seller in a  market. The numbers of oranges are arranged in different boxes by the fruit seller. Through the number of oranges the number of oranges are distributed in following manner.

 No.of oranges 50-52 53-55 56-58 89-61 62-64 No. of boxes 15 110 135 115 25

In the distributed pattern the number of oranges. Find the mean. Choose the method to find the mean:

Solution:

 Class interval fi xi di=x-a fi di 50-52 110 54 -6 90 53-55 15 51 -3 -330 56-58 135 60 0 0 59-61 25 57 3 345 62-64 115 63 6 150 Sum fi= 400 Sum fi di=75

Mean can be calculated as follows:

$\bar{d}=\frac{\Sigma f_{i}d_{i}}{\Sigma f_{i}}=\frac{75}{400}=0.1875$ $\bar{x}=\bar{d}+a=0.1875+57=57.1875$

In this case, there are wide variations in $f_{i}$ and hence assumed mean method is used.

Question 6:

The following distribution shows the food expenditure of the given households from the locality:

 Daily food expenditure (in Rs) 50-100 100-150 150-200 200-250 250-300 Number of given households 4 5 12 2 2

Find the mean daily expenditure on food by a suitable method.

Solution:

 Class Interval fi xi di= xi – a ui = di/h fiui 50-100 4 125 -100 -2 -8 100-150 5 175 -50 -1 -5 150-200 12 225 0 0 0 200-250 2 275 50 1 2 250-300 2 325 100 2 4 Sum fi = 25 Sum fiui = -7

Mean can be calculated as follows:

$\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h$ $=225+\frac{-7}{25}\times 50 =225$

Question 7:

To find concentration in sulphur dioxide (in parts per million, i.e. ppm),The parts of data are collected by different areas from cities are given in distributed table:

 Concentration in SO2 (in ppm) Frequency of F 0.00-0.04 4 0.04-0.08 9 0.08-0.12 9 0.12-0.16 2 0.16-0.20 4 0.20-0.24 2

Give the mean of the sulphur dioxide collected from the given cities.

Solution:

 Class Interval fi xi fixi 0.00-0.04 9 0.02 0.08 0.04-0.08 4 0.06 0.54 0.08-0.12 9 0.10 0.90 0.12-0.16 2 0.14 0.28 0.16-0.20 2 0.18 0.72 0.20-0.24 4 0.22 0.44 Sum fi=30 Sum fixi=2.96

Mean can be calculated as follows:

$\bar{x}=\frac{\Sigma f_{i}x_{i}}{\Sigma f_{i}}=\frac{2.96}{30}=0.099 ppm$

Question 8:

The absentee record of 40 students is handled by the teacher for one complete term. Find the mean of absent days of the given students:

 Number of following days 0-6 6-10 10-14 14-20 20-28 28-38 38-40 Number of following students 10 11 4 7 3 4 1

Solution:

 Class Interval fi xi fixi 0-6 11 3 33 6-10 10 8 80 10-14 4 12 84 14-20 7 17 68 20-28 3 24 96 28-38 4 33 99 38-40 1 39 39 Sum fi = 40 Sum fixi= 499

Mean can be calculated as follows:

$\bar{x}=\frac{\sum f_{i}x_{i}}{\sum f_{i}}$= $\frac{499}{40}$=12.4

Question 9:

The data of the literacy digit from the 35 cities is given in the following table. Find the mean of the 35 literacy cities.

 Literacy rate from 35cities (in %) 40-45 50-55 60-65 70-75 80-85 Cities in values 10 3 8 11 3

Solution:

 Class Interval fi xi di=xi-a ui=di/h fiui 40-45 10 50 -20 -2 -6 50-55 3 60 -10 -1 -10 60-65 8 70 0 0 0 70-75 11 80 10 1 8 80-85 3 90 20 2 6 Sum fi=35 Sum fiui= -2

Mean can be calculated as follows:

$\bar{x}=a+\frac{\Sigma f_{i}u_{i}}{\Sigma f_{i}}\times h$ $= 70+\frac{-2}{35}\times 10=69.42$

EXERCISE – 2

Question 1:

The age of patients admitted in a hospitals are given in the following table for a complete year.

 Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65 Number of patients 6 11 21 23 14 5

Calculate the mode and mean of the data above. Compare and interpret the two measures of central tendency.

Solution:

Modal class = 35-45, length = 35, height = 10, $a_{1} = 23, a_{0} = 21\, and\, a_{3} = 14$

Therefore, Mode = $l +\frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}}\times h$

= $35+\left ( \frac{23-21}{2\times 23-21-14} \right )\times 10$

= $35+\frac{2}{11}\times 10=36.8$

Calculating Mean

 CLASS INTERVAL ai zi aizi 5-15 6 10 60 15-25 11 20 220 25-35 21 30 630 35-45 21 30 920 45-55 14 50 700 55-65 5 60 300 Sum ai = 80 Sum aizi= 2830

$\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a_{i}}=\frac{2830}{80}=35.37$

$\ therefore$The mode of data displays that the maximum number of patients is in the age group of 26.8, whereasthe average age of all the patients is 35.37.

Question 2:

The data given below provides the information for the observed lifetime (in hours) of 225 electrical components:

 Lifetime ( in hours) 0 – 20 20 – 40 40 – 60 60 – 80 80 – 100 100  – 120 Frequency 10 35 52 61 38 29

Find the modal lifetimes of the components.

Solution:

Modal class =  60-80, l = 60,$a_{1} = 61, a_{0} = 52, a_{2} = 38$ and h = 20

$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

$= 60 + \left ( \frac{61 – 52}{2 \times 61 – 52 – 38 } \right ) \times 20$

$= 60 + \frac{9}{32} \times 20 = 65.62$

Question 3:

The total monthly household expenses of 200 families of a town are given in the following table. Calculate the modal monthly expenses of the families.  Also, calculate the mean monthly expenses.

 Expenses Number of families 1000 – 1500 24 1500 – 2000 40 2000 – 2500 33 2500 – 3000 28 3000 – 3500 30 3500 – 4000 22 4000 – 4500 16 4500 – 5000 7

Solution:

Modal class = 1500-2000, l = 1500,$a_{1} = 40, a_{0} = 24, a_{2} = 33$ and h = 500

$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

$= 1500 + \left ( \frac{40 – 24}{2 \times 40 – 24 – 33} \right ) \times 500$

$= 1500 + \left ( \frac{16}{23} \right ) \times 500 = 1847.82$

Calculation of mean:

 Class Interval ai zi di =zi – b ui= di/h aiui 1000 – 1500 24 1250 -1500 -3 -72 1500 – 2000 40 1750 -1000 -2 -80 2000 – 2500 33 2250 -500 -1 -33 2500 – 3000 28 2750 0 0 0 3000 – 3500 30 3250 500 1 30 3500 – 4000 22 3750 1000 2 44 4000 – 4500 16 4250 1500 3 48 4500 – 5000 7 4750 2000 4 28 ai = 200 aiui = -35

$\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a{i}}\times h$ $= 2750 – \frac{35}{200} \times 500$

= 2750 – 87.5 = 2662.50

Question 4:

The state – wise teacher- student ratio in a college in India are given in the following table. Calculate the mean and mode of the given data. Interpret the two measures.

 Number of students per teacher Number of states/UT 15 – 20 3 20 – 25 8 25 – 30 9 30 – 35 10 35 – 40 3 40 – 45 0 45 – 50 0 50 – 55 2

Solution:

Modal class = 30-35, l = 30, $a_{1} = 10, a_{0} = 9, a_{2} = 3$ and h = 5

$Mode = l +\left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

$= 30 – \left ( \frac{20 – 9}{2 \times 10 – 9 – 3 } \right ) \times 5$

$= 30 – \frac{1}{8} \times 5 = 30.625$

Calculating mean:

 Number of students per teacher ai zi di= zi –b ui = ui/h aiui 15 – 20 3 17.5 -15 -3 -9 20 – 25 8 22.5 -10 -2 -16 25 – 30 9 27.5 -5 -1 -9 30 – 35 10 32.5 0 0 0 35 – 40 3 37.5 5 1 3 40 – 45 0 42.5 10 2 0 45 – 50 0 47.5 15 3 0 50 – 55 2 52.5 20 4 8 Sum ai = 35 sum aiui= -23

$\bar{p}=\frac{\sum a_{i}x_{i}}{\sum a_{i}}\times h$

$= 30.5 – \frac{23}{35}\times 5$

$= 30.5 – \frac{23}{7} = 29.22$

Therefore, from the mode we know that the maximum number of states has 30 – 35 students per teacher, and from the mean we know that the average ratio of students per teacher is 29.22

Question 5:

Provided table shows the total runs scored by some of the top batsman of the world in one-day cricket matches.

 Runs Scored Number of Batsman 3000 – 4000 4 4000 – 5000 18 5000 – 6000 9 6000 – 7000 7 7000 – 8000 6 8000 – 9000 3 9000 – 10000 1 10000 – 11000 1

Calculate the mode of the given information.

Solution:

Modal class = 4000-5000, l = 4000, $a_{1} = 18, a_{0} = 4, a_{2} = 9$ and h = 1000

$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$ $= 4000 + \left ( \frac{18 – 4}{2 \times 18 – 4 -9} \right )\times 1000$ $= 4000 + \left ( \frac{14}{23} \right )\times 1000 = 4608.70$

Question 6:

A complete observation of the number of cars on the roads for 100 periods of 3 minutes are summarized in the following table. Calculate the mode of the data.

 Number of cars 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 70 – 80 Frequency 7 14 13 12 20 11 15 8

Solution:

Modal class = 40 – 50, l = 40, $a_{1} = 20, a_{0} = 12, a_{2} = 11$ and h = 10

$Mode = l + \left ( \frac{a_{1}-a_{0}}{2a_{1}-a_{0}-a_{2}} \right )\times h$

$= 40 + \left ( \frac{20 – 12}{2 \times 20 – 12 – 11} \right )\times 10$

$= 40 + \left ( \frac{8}{17} \right )\times 10 = 44.70$

Exercise 3

Question 1:

The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.

 Monthly consumption(in units) No. of customers 65-85 4 85-105 5 105-125 13 125-145 20 145-165 14 165-185 8 185-205 4

Solution:

 Class Interval Frequency Cumulative frequency 65-85 4 4 85-105 5 9 105-125 13 22 125-145 20 42 145-165 14 56 165-185 8 64 185-205 4 68 N=68

Where, n = 68 and hence $\frac{n}{2}=34$

Hence, the median class is 125-145 with cumulative frequency = 42

Where, l = 125, n = 68, cf = 22, f = 20, h = 20

Median is calculated as follows:

Median=$l+\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$

=$125+\left ( \frac{34-22}{20} \right )X20$

=125+12=137

Mode = Modal class = 125-145, $f_{1} = 20, f_{0} = 13, f_{2} = 14$ & h = 20

Mode=$l+\left ( \frac{f_{1}-f_{0}}{2Xf_{1}-f_{0}-f_{2}} \right )Xh$

=$125+\left ( \frac{20-13}{2X20-13-14} \right )X20$

=$125+\frac{7}{13}X20$

=125+10.77

=135.77

Calculate the Mean:

 Class Interval fi xi di=xi-a ui=di/h fiui 65-85 4 75 -60 -3 -12 85-105 5 95 -40 -2 -10 105-125 13 115 -20 -1 -13 125-145 20 135 0 0 0 145-165 14 155 20 1 14 165-185 8 175 40 2 16 185-205 4 195 60 3 12 Sum fi= 68 Sum fiui = 7

$\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}X h$ $=135+\frac{7}{68}X20$

=137.05

Mean, median and mode are more/less equal in this distribution.

Question 2:

If the median of a distribution given below is 28.5 then, find the value of an x &y.

 Class Interval Frequency 0-10 5 10-20 x 20-30 20 30-40 15 40-50 y 50-60 5 Total 60

Solution:

n = 60

Where,$\frac{n}{2}$ = 30

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class = 20, cf = 5 + x , f = 20 & h = 10

Median=l+$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$

Or,28.5=20+$\left ( \frac{30-5-x}{20} \right )X10$

Or,$\frac{25-x}{2}$

Or,   25-x=17

Or,    x= 25 – 17   =8

Now, from cumulative frequency, we can identify the value of x + y as follows:

60=5+20+15+5+x+y

Or, 45+x+y=60

Or, x + y=60-45=15

Hence, y = 15 – x= 15 – 8 = 7

Hence, x = 8  &  y = 7

Question 3:

The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the  persons whose age is 18 years onwards but less than the 60 years.

 Age(in years) Number of policy holder Below 20 2 Below 25 6 Below 30 24 Below 35 45 Below 40 78 Below 45 89 Below 50 92 Below 55 98 Below 60 100

Solution:

 Class interval Frequency Cumulative frequency 15-20 2 2 20-25 4 6 25-30 18 24 30-35 21 45 35-40 33 78 40-45 11 89 45-50 3 92 50-55 6 98 55-60 2 100

Where, n = 100 and $\frac{n}{2}$ = 50

Median class = 35-45

Then, l = 35, cf = 45, f = 33 & h = 5

Median=l+$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$ $=35+\left ( \frac{50-45}{33} \right )X5$ $=35+\left ( \frac{25}{33} \right )$

=35.75

Question 4:

The lengths of 40 leaves in a plant are measured correctly to the nearest millimetre, and the data obtained is represented as in the following table:

 Length(in mm) Number of leaves 118-126 3 127-135 5 136-144 9 145-153 12 154-162 5 163-171 4 172-180 2

Find the median length of leaves.

Solution:

 Class Interval Frequency Cumulative frequency 117.5-126.5 3 3 126.5-135.5 5 8 135.5-144.5 9 17 144.5-153.5 12 29 153.5-162.5 5 34 162.5-171.5 4 38 171.5-180.5 2 40

Where, n = 40 and $\frac{n}{2}$ = 20

Median class = 144.5-153.5

then, l = 144.5, cf = 17, f = 12 & h = 9

Median=l+$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$ $=144.5+\left ( \frac{20-17}{12} \right )X9$ $=144.5+\left ( \frac{9}{4} \right )$

=146.75

Question 5:

The following table gives the distribution of a life time of 400 neon lamps.

 Lifetime(in hours) Number of lamps 1500-2000 14 2000-2500 56 2500-3000 60 3000-3500 86 3500-4000 74 4000-4500 62 4500-5000 48

Find the median lifetime of a lamp.

Solution:

 Class Interval Frequency Cumulative 1500-2000 14 14 2000-2500 56 70 2500-3000 60 130 3000-3500 86 216 3500-4000 74 290 4000-4500 62 352 4500-5000 48 400

Where, n = 400 &$\frac{n}{2}$ = 200

Median class = 3000 – 3500

Therefore, l = 3000, cf = 130, f = 86 & h = 500

Median=l+$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$ $=3000+\left ( \frac{200-130}{86} \right )X500$

=3000+406.97

=3406.97

Question 6:

In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows:

 Number of letters 1-4 4-7 7-10 10-13 13-16 16-19 Number of surnames 6 30 40 16 4 4

Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution: To Calculate median:

 Class Interval Frequency Cumulative Frequency 1-4 6 6 4-7 30 36 7-10 40 76 10-13 16 92 13-16 4 96 16-19 4 100

Where, n = 100 &$\frac{n}{2}$ = 50

Median class = 7-10

Therefore, l = 7, cf = 36, f = 40 & h = 3

Median=l+$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$ $=7+\left ( \frac{50-36}{40} \right )X3$ $=7+\left ( \frac{14}{40} \right )X3$

=8.05

Calculate the Mode:

Modal class = 7-10,

Where, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3

Mode=$l+\left ( \frac{f_{1}-f_{0}}{2Xf_{1}-f_{0}-f_{2}} \right )Xh$

=$7+\left ( \frac{40-30}{2X40-30-16} \right )X3$

=$7+\frac{10}{34}X3$

=7.88

Calculate the Mean:

 Class Interval fi xi fixi 1-4 6 2.5 15 4-7 30 5.5 165 7-10 40 8.5 340 10-13 16 11.5 184 13-16 4 14.5 51 16-19 4 17.5 70 Sum fi = 100 Sum fixi = 825

$\bar{x}=a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}X h$ $=\frac{825}{100}$

=8.25

Q7. The distributions of below gives a weight of 30 students of a class. Find the median weight of a students.

 Weight(in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75 Number of students 2 3 8 6 6 3 2

Solution:

 Class Interval Frequency Cumulative frequency 40-45 2 2 45-50 3 5 50-55 8 13 55-60 6 19 60-65 6 25 65-70 3 28 70-75 2 30

Where, n = 30 and $\frac{n}{2}$= 15

median class = 55-60

therefore, l = 55, cf = 13, f = 6 & h = 5

Median=l+$\left ( \frac{\frac{n}{2}-cf}{f} \right )Xh$ $=55+\left ( \frac{15-13}{6} \right )X5$ $=55+\left ( \frac{5}{3} \right )$

=56.67

Question 7:

The distribution shows the pocket money of students in an area. The mean pocket money is Rs 18. Find the missing frequency f.

 Daily pocket allowance (in Rs) 11-13 13-15 15-17 17-19 19-21 21-23 23-25 Number of children 07 06 09 013 F 05 04

Solution:

 Class interval fi xi fixi 11-13 07 012 84 13-15 06 014 84 15-17 09 016 144 17-19 013 018 234 19-21 f 020 20f 21-23 05 022 110 23-25 04 024 96 Sum fi = 44+f Sum fixi = 752+20f

We have,

Mean pocket money, $x_{i}$ = 18(given)

Hence substituting for fixi,

18(44+f) = 752 + 20f

Or, 792 + 18f = 752 + 20f

Or, 2f = 40

$\rightarrow f=20$

The modern world is highly data-oriented and it becomes necessary to know to represent data in a meaningful way. The branch of mathematics which deals with the representation of data in a meaningful way is known as statistics. Let us look at a few examples of how statistics is useful in our daily life

• The data that you see in the weather forecast section of the news is due to the goodwill of statistics.
• The selection of a player in any sport is mostly made based on the player’s statistical performance report.

In this chapter, students are introduced to histograms, graphical representations of data in the form of bar charts, cumulative frequency tables and frequency polygons. Students will be introduced to the three measures of central tendency. The three measures of central tendency are mean, median and mode.

Byju’s is India’s largest k-12 learning app with top-notch teachers from across the nation with excellent teaching skills. Find notes, question papers for various other subjects like Mathematics, Biology, Chemistry and many competitive exams as well. Enjoy learning with great experience. Stay tuned to know more about NCERT Solutions and preparation tips. To know more about NCERT solutions you can register to BYJU’s or download our app for simple and interesting content.


text:
Hello,

I am new to ROS and I am trying to complete the tf2 tutorials. While running this particular tutorial (http://wiki.ros.org/tf2/Tutorials/Wri... a tf2 broadcaster (Python)) I created the turtle_tf2_broadcaster.py file and the start_demo.launch file exactly as described in the tutorial and in the directory as described in the tutorial. After this, I invoked catkin_make and I source devel_isolated/setup.bash in the terminal (I also tried sourcing devel/setup.bash). However, the following error comes up and it closes:

    RLException: Unable to launch [turtle1_tf2_broadcaster-4].
If it is a script, you may be missing a '#!' declaration at the top.
The traceback for the exception was written to the log file

The full output of the command I have attached at the end.

I am not able to figure out a way to get through this. Please suggest me what should I do. Please also find below my code for turtle_tf2_broadcaster.py and start_demo.launch. On lines 5 and 22 in turtle_tf2_broadcaster.py, I tried changing tf (as mentioned on the tutorial http://wiki.ros.org/tf2/Tutorials/Wri... a tf2 broadcaster (Python)) to tf2 and it still didn't work.

   #!/usr/bin/env python
import rospy

# Because of transformations
import tf2_conversions

import tf2_ros
import geometry_msgs.msg
import turtlesim.msg

def handle_turtle_pose(msg, turtlename):
t = geometry_msgs.msg.TransformStamped()

t.child_frame_id = turtlename
t.transform.translation.x = msg.x
t.transform.translation.y = msg.y
t.transform.translation.z = 0.0
q = tf2_conversions.transformations.quaternion_from_euler(0, 0, msg.theta)
t.transform.rotation.x = q[0]
t.transform.rotation.y = q[1]
t.transform.rotation.z = q[2]
t.transform.rotation.w = q[3]

br.sendTransform(t)

if __name__ == '__main__':
turtlename = rospy.get_param('~turtle')
rospy.Subscriber('/%s/pose' % turtlename,
turtlesim.msg.Pose,
handle_turtle_pose,
turtlename)
rospy.spin()

Code for start_demo.launch

  <launch>
<node pkg="turtlesim" type="turtlesim_node" name="sim"/>
<node pkg="turtlesim" type="turtle_teleop_key" name="teleop" output="screen"/>

<param name="turtle" type="string" value="turtle1" />
</node>
<param name="turtle" type="string" value="turtle2" />
</node>

</launch>

Full out put when lunching the start_demo.launch.

suyash@ROS:~/tutorial_ws$roslaunch learning_tf2 start_demo.launch ... logging to /home/suyash/.ros/log/b7a6d97a-f1bb-11e9-a228-08002776bdd9/roslaunch-ROS-2614.log Checking log directory for disk usage. This may take awhile. Press Ctrl-C to interrupt Done checking log file disk usage. Usage is <1GB. started roslaunch server http://ROS:36971/ SUMMARY ======== PARAMETERS * /rosdistro: melodic * /rosversion: 1.14.3 * /turtle1_tf2_broadcaster/turtle: turtle1 * /turtle2_tf2_broadcaster/turtle: turtle2 NODES / sim (turtlesim/turtlesim_node) teleop (turtlesim/turtle_teleop_key) turtle1_tf2_broadcaster (learning_tf2/turtle_tf2_broadcaster.py) turtle2_tf2_broadcaster (learning_tf2/turtle_tf2_broadcaster.py) auto-starting new master process[master]: started with pid [2625] ROS_MASTER_URI=http://localhost:11311 setting /run_id to b7a6d97a-f1bb-11e9-a228-08002776bdd9 process[rosout-1]: started with pid [2636] started core service [/rosout] process[sim-2]: started with pid ... edit retag close merge delete ## Comments Always include links to tutorials, source code, pages etc you're reading, following or copy-pasting. Without seeing what you're actually doing, we cannot help you. Seeing the error message you received, please also show us what exactly you copied (ie: the contents of your copy of turtle_tf2_broadcaster.py). ( 2019-10-18 04:50:25 -0600 )edit Thank you for your response, sorry I missed those details. I have edited my question and included the details.Also one more update is I tried doing the tf tutorial as well (http://wiki.ros.org/tf/Tutorials/Writing a tf broadcaster (Python), and I am still getting the exact same error, so there is something fundamental that I am missing. ( 2019-10-18 10:21:39 -0600 )edit Please show us the exact contents of your turtle_tf2_broadcaster.py. And also the output of ls -al turtle_tf2_broadcaster.py (run it in the directory that contains the file). ( 2019-10-18 10:28:09 -0600 )edit I ran the command and the output was: suyash@ROS:~/tutorial_ws/src/learning_tf2/nodes$ ls -al turtle_tf2_broadcaster.py
-rwxr-xr-x 1 suyash suyash 1218 Oct 17 18:47 turtle_tf2_broadcaster.py
suyash@ROS:~/tutorial_ws/src/learning_tf2/nodes\$

I have also copied and pasted the exact contents of turtle_tf2_broadcaster.py. Also I have included full output of roslaunch learning_tf2 start_demo.launch.

( 2019-10-18 10:44:37 -0600 )edit

Hi there. Is there any update for me on this question?

( 2019-10-21 10:28:10 -0600 )edit

Sort by » oldest newest most voted

Hi, I have figured out the solution. There was around 3 space characters between the start and #!/usr/bin/env python. When I removed the spaces for this line and all subsequent lines, it worked.

more

There was around 3 space characters between the start and #!/usr/bin/env python

This was indeed the reason why I asked you to copy the contents of your script verbatim.

Good to hear you figured it out.

( 2019-10-24 02:43:58 -0600 )edit

Hi Suyash,

Without more data, and assuming that you are following the ROS wiki tutorials, here: http://wiki.ros.org/tf2/Tutorials/Wri...

The most probable cause for your error is that you missed the 1st line in your script, this one:

#!/usr/bin/env python

Basically, this line makes sure that your script is executed as a Python script. Try to add it at the top of your script and the error should be gone.

more

Hi Alberto.Thank you for your response and pardon me for the delayed response. I am using the same tutorial as mentioned by you. I actually found that the line of code is included in my script. Please take a look at my edited question ( I have added my codes). Also one more update is I tried doing the tf tutorial as well (http://wiki.ros.org/tf/Tutorials/Writing a tf broadcaster (Python)), and I am still getting the exact same error, so there is something fundamental that I am missing.

( 2019-10-18 10:24:17 -0600 )edit


text:
# Homework Help: Light angles measured in a moving reference frame

1. Oct 29, 2014

I've managed to get through all of this question without trouble until part d).

The full question is given here:

I've calculated the "true" angles of Star A and Star B as 71.57 degrees and 45 degrees respectively in Frame S, and the "light" angles should be the same, since the stars are stationary in frame S.

For part c), the "true" angles of the stars are found using Lorentz transformations to find x

$x$′$_A = \frac{x_A}{\gamma} = 0.527$

And similarly for B, yielding angles of 80.04 degrees and 62.21 degrees respectively.

However, I can't figure out how to calculate the angles to the stars when the light measured was emitted (part d). The answers are 23.2 degrees for Star A, and 13.4 degrees for Star B.

I should be able to do this calculation, but there's something I'm just not getting, I've been sitting on it for a while and need to move on. Any help is appreciated.

2. Oct 29, 2014

### Staff: Mentor

Are you familiar with relativistic aberration? In any case, this thread belongs in one of the homework forums, so I'm moving it there.


text:
XXIX Workshop on Geometric Methods in Physics 27.06-03.07.2010
 Home First Announcement Second Announcement Travel Participants Photos Proceedings Email

# Andrey Bytsenko

## Deformation of complex structures in topological field theory

We study a Lie algebra of formal vector fields $W_n$ with it application to the perturbative deformed holomorphic symplectic structure in the A-model, and a Calabi-Yau manifold with boundaries in the B-model. We show that equivalent classes of deformations are describing by a Hochschild cohomology theory of DG-algebra $\cA = (A, Q)$, $Q =\overline{\partial}+\partial_{\rm deform}$, which is defind to be the cohomology of $(-1)^n Q +d_{\rm Hoch}$. Here $\overline{\partial}$ is the initial non-deformed BRST operator while $\partial_{\rm deform}$ is the deformed part whose algebra is a Lie algebra of linear vector fields ${\rm gl}_n$. We discuss the identification of the harmonic structure $(HT^\bullet(X); H\Omega_\bullet(X))$ of affine space $X$ and the group ${\rm Ext}_{X?}^n(\cO_{\triangle}, \cO_{\triangle})$ (the HKR isomorphism), and bulk-boundary deformation pairing.


text:
## News Forum

### Happy MAR10 Day! (Weekly Brain Potion)

Happy MAR10 Day! (Weekly Brain Potion)

This week's brain potion is a version of a classic puzzle adapted to honor Mario Day (MAR10, i.e. March 10th).

The seven Koopalings (youngest to oldest) are named:

1. Larry
2. Morton
3. Wendy
4. Iggy
5. Roy
6. Lemmy
7. Ludwig

They were tired of working for Bowser without getting paid, so they decided to steal $100$ of Mario's coins and escape in their airship.

They agreed that they would divide the coins as follows:

• Ludwig, as the oldest, would suggest a number of coins to give to everyone.
• The others (not including Ludwig) would then vote on whether or not to accept the proposed distribution of the coins.
• If $\geq 50\%$ of them vote to accept, then they give out the coins according to the disposal.
• If $< 50\%$ of them vote to accept, they throw Ludwig from the airship and then repeat the procedure with the next oldest suggesting a distribution.

Assume that each of the Koopalings is rational, good at logical thinking, and, in order of importance,

1. Would not like to be thrown overboard.
2. Would like as many coins as possible.
3. Enjoys throwing someone else overboard.

How many coins can Ludwig get? Explain your reasoning and include how many coins each of the Koopalings will receive.


text:
# The One Mistake Rule

post by Zvi · 2020-04-10T14:50:00.413Z · score: 45 (16 votes) · LW · GW · 15 comments

## Contents

  If a model gives a definitely wrong answer anywhere, it is useless everywhere.
None

Epistemic Status: The Bed of Procrustes

If a model gives a definitely wrong answer anywhere, it is useless everywhere.

This principle is doubtless ancient, and has doubtless gone by many names with many different formulations.

All models are wrong. That does not make them useless. What makes them useless is when they are giving answers that you know are definitely wrong. You need to fix that, if only by having the model more often spit out “I don’t know.”

As an example of saying “I don’t know” that I’m taking from the comments, if you want to use Newtonian Physics, you need to be aware that it will give wrong answers in relativistic situations, and therefore slightly wrong answers in other places, and introduce the relevant error bars.

Of course, a wrong prediction of what is probably going to happen is not definitely wrong, in this sense. An obviously wrong probability is definitely wrong no matter the outcome.

The origin of this particular version of this principle was when me and a partner were, as part of an ongoing campaign of wagering, attempting to model the outcomes of sporting events.

He is the expert on sports. I am the expert on creating models and banging on databases and spreadsheets. My specialty was assuming the most liquid sports betting market odds were mostly accurate, and extrapolating what that implied elsewhere.

First we would talk and he would explain how things worked. Then I would look at the data lots of different ways and create a spreadsheet that modeled things. Then, he would vary the inputs to that spreadsheet until he got it to give him a wrong answer, or at least one that seemed wrong to him.

Then he’d point out the wrong answer and explain why it was definitely wrong. I could either argue that the answer was right and change his mind, or I could accept that it was wrong and go back and fix the model. Then the cycle repeated until he couldn’t find a wrong answer.

Until this cycle stopped, we did not use the new model for anything at all, anywhere, no matter what. If a new wrong answer was found, we stopped using the model in question until we resolved the problem.

Two big reasons:

If we did use the model, even if it was only wrong in this one place, then the one place it was wrong would be the one place we would disagree with the market. Fools and their money would be soon parted.

Also, if the model was obviously wrong here, there’s no reason to trust anything else the model says, either. Fix your model.

This included tail risk style events that were extremely unlikely. If you can’t predict the probability of such events in a reasonable way, even if those outliers won’t somehow bankrupt you directly, you’re going to get the overall distributions wrong.

This also includes the change in predictions between different states of the world. If your model predictably doesn’t agree with itself over time, or changes its answer based on things that can’t plausibly matter much, then it’s wrong. Period. Fix it.

You should be deeply embarrassed if your model outputs an obviously wrong or obviously time-inconsistent answer even in a hypothetical situation. You should be even more embarrassed if it gives such an answer to the actual situation.

The cycle isn’t bad. It’s good. It’s an excellent way to improve your model: Build one, show it to someone, they point out a mistake, you figure out how it happened and fix it, repeat. And in the meantime, you can still use the model’s answers to help supplement your intuitions, as a sanity check or very rough approximation, or as a jumping off point. But until the cycle is over, don’t pretend you have anything more than that.

comment by Donald Hobson (donald-hobson) · 2020-04-10T16:10:51.739Z · score: 35 (19 votes) · LW(p) · GW(p)
You should be deeply embarrassed if your model outputs an obviously wrong or obviously time-inconsistent answer even in a hypothetical situation.

Suppose you have a particle accelerator that goes up to half the speed of light. You notice an effect whereby faster particles become harder to accelerate.

You curve fit this effect and get that . and both fit the data, well the first one fits the data slightly better. However, when you test your formula on the case of a particle travelling at twice the speed of light, you get back nonsensical imaginary numbers. Clearly the real formula must be the second one. (The real formula is actually the first one)

A good model will often give a nonsensical answer when asked a nonsensical question, and nonsensical questions don't always look nonsensical.

comment by Dagon · 2020-04-10T17:29:03.568Z · score: 8 (5 votes) · LW(p) · GW(p)

This is worth being explicit about. I took the advice as applying to hypothetical situations consistent with the intended usage of the model. A sports prediction model probably doesn't need to work for low-gravity situations, most physics models don't need to include FTL particles.

It would be nice to say more formally that you can fix it by improving the model OR by specifying which subset of the imagination space the model is intended for.

edit based on replies (thank you for making me think further): A LOT hinges on the meta-model and the theories behind your belief that the model is useful in the first place. It makes sense to be VERY skeptical of correlations found in data with no explanation of what it means. For these, I agree that any failure is grounds to reject. For models that start with a hypothesis, it makes a lot of sense to use the theory to identify reasonable exclusions, where you don't throw out the model for weird results, because you don't expect it to apply.

comment by Donald Hobson (donald-hobson) · 2020-04-10T19:38:03.917Z · score: 6 (4 votes) · LW(p) · GW(p)

I was imagining doing this before the speed of light limit was known. In which case you can find yourself saying that the subset where the model produces sensible results is the subset of imagination space the model is intended for.

comment by FeepingCreature · 2020-04-11T10:57:55.370Z · score: 1 (1 votes) · LW(p) · GW(p)

But to be fair, if you then fixed the model to output errors once you exceeded the speed of light, as the post recommends, you would have come up with a model that actually communicated a deep truth. There's no reason a model has to be continuous, after all.

comment by orthonormal · 2020-04-10T18:58:34.597Z · score: 4 (2 votes) · LW(p) · GW(p)

Similarly, a linear regression to some data will often behave well for interpolation but less well for distant extrapolation.

comment by orthonormal · 2020-04-10T19:05:10.811Z · score: 22 (12 votes) · LW(p) · GW(p)

I think the need for this is relatively higher in an anti-inductive environment like trying to beat the market than in an inductive environment like materials science or disease modeling. In the former, you don't dare deploy until you're confident it's missing nothing; in the latter, what you have mid-iteration is still very useful, especially if you're making big time-sensitive decisions.

comment by Unnamed · 2020-04-10T21:20:12.055Z · score: 12 (7 votes) · LW(p) · GW(p)

This seems important.

Another feature of competitive markets is that "not betting" is always available as a safe default option. Maybe that means waiting to bet until some unknown future date when your models are good enough, maybe it means never betting in that market. In many other contexts (like responding to covid-19) there is no safe default option.

comment by Dagon · 2020-04-10T22:26:55.069Z · score: 4 (2 votes) · LW(p) · GW(p)

"Not betting" is an illusion. In all cases, choosing not to take any given bet is itself a bet: you're betting that you'll find something better to do with that chunk of resources.

comment by orthonormal · 2020-04-10T23:55:51.792Z · score: 7 (4 votes) · LW(p) · GW(p)

Not betting, in the sense of keeping the money in index funds or somewhere else on the risk/reward Pareto frontier of easy strategies, at least limits your expected downside compared to entering shark-infested waters in an imperfect cage.

comment by Unnamed · 2020-04-14T09:08:38.033Z · score: 4 (2 votes) · LW(p) · GW(p)

One obviously mistaken model that I got a lot of use out of during a stretch of Feb-Mar is the one where the cumulative number of coronavirus infections in a region doubles every n days (for some globally fixed, unknown value of n).

This model has ridiculous implications if you extend it forward for a few months, as well as various other flaws. I was aware of those ridiculous implications and some of those other flaws, and used it anyways for several days before trying to find less flawed models.

I'm glad that I did, since it helped me have a better grasp of the situation and be more prepared for what was coming. And I don't think it would've made much difference at the time if I'd learned more about SEIR models and so on.

It's unclear how examples like this are supposed to fit with the One Mistake Rule or the exceptions in the last paragraph.

comment by Chris_Leong · 2020-04-10T21:34:11.637Z · score: 4 (2 votes) · LW(p) · GW(p)

I think you're making this argument a bit strongly. Now, I've written a number of posts arguing that most people are too dismissive of flaws in models that only occur in hypothetical or unrealistic situations, but I don't think perfection is realistic. It seems that a model with no flaws would have to approach infinite complexity in most cases. The only reason why this rule might work is eventually your model will become complex enough that you can't find the mistake. Additionally, you will be limited by the data you have. It's no good knowing that prediction X is wrong because you ignore factor F if you don't have data related to factor F.

comment by Decius · 2020-04-12T11:51:33.319Z · score: 3 (2 votes) · LW(p) · GW(p)

What are good methods of determining when the model is giving a definitely wrong answer, and when your not-a-model is grossly overconfident and wrong?

comment by TAG · 2020-04-12T13:36:26.047Z · score: 1 (1 votes) · LW(p) · GW(p)

All models go wrong somewhere , and they are still useful if you use them where they don't go wrong.

comment by Yandong Zhang (yandong-zhang) · 2020-04-14T00:37:00.645Z · score: 1 (1 votes) · LW(p) · GW(p)

Also, people can design a specific application environment to reduce the bad effects resulting from the error of the model.

comment by jmh · 2020-04-11T14:45:46.807Z · score: 1 (1 votes) · LW(p) · GW(p)

Thinking about Donald's and Dagon's comments I do think there is something that can be more fully acknowledged (though was also reasonably assumed I think).

I walked away from the OP, but those two comments further clarified for me my own thinking, the one strike rule is a good way to refine our maps when we know the territory we are mapping out.

However, that is not always the case so when making call on a strike we should make sure that we're actually in the territory or home plate and have not wondered somewhere else. I think that was supposed to be address as part of the discussion about was it a mistake or not.

I think this makes for a useful heuristic when modeling but clearly other heuristics should also be applied to ensure you do stay in the territory intended. So is there a good one for helping to keep one on the right sports field? Is that needed, if no why and if so when?


text:
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SERVING THE QUANTITATIVE FINANCE COMMUNITY

• 1
• 2

rwang716
Topic Author
Posts: 12
Joined: July 18th, 2019, 6:18 am

### Explicit Formula for computing IV

Working on the fast computation of BS implied volatility. I have read some papers regarding the explicit formula ofr BS IV. Like the Li(2005) paper: A new formula for computing implied volatility. However, I met a very serious problem in their formulas because the terms in the square root bracket can goes to negative is the option price is small enough, the simple formula by Corrado and Miller(1996) also suffers from the same problem. How can I solve this problem?
BTW, the explicit formula only serves as the initial guess for the subsequent numerical methods.
I don't know how to insert image for the formula, sorry.

Cuchulainn
Posts: 62126
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: Explicit Formula for computing IV

There was a long-running discussion on this a while back

https://forum.wilmott.com/viewtopic.php?f=34&t=97812&p=732891&hilit=all+for+iv#p732891

The most robust was least squares AFAIR. The posts have unfortunately been become almost unreadable.

Since this is a nonlinear problem I would be surprised if there is an explicit formula. It would be the first in history.

terms in the square root bracket can goes to negative is the option price is small enough,
Can happen. It means the (implicit) assumptions/constraints on which the algorithm is based no longer hold.

rwang716
Topic Author
Posts: 12
Joined: July 18th, 2019, 6:18 am

### Re: Explicit Formula for computing IV

There are some approximation IV formulas that are 99.9% accurate. However, the problem is that those formula only works for near-ATM options, for options that are, e.g., 25% ITM/OTM, the formula becomes nonsense. So I just wonder if there are some approximation formulas that does not have this kind of issue.

FaridMoussaoui
Posts: 507
Joined: June 20th, 2008, 10:05 am
Location: Genève, Genf, Ginevra, Geneva

### Re: Explicit Formula for computing IV

Isnt' Li formula (15) designed for the deep I(O)TM calls?

rwang716
Topic Author
Posts: 12
Joined: July 18th, 2019, 6:18 am

### Re: Explicit Formula for computing IV

Do you mean Li(2005) formula?
I don't think so. I have tried this formula and it produces imaginary numbers.
Try the input: S = 2800, K = 3050, T = 1/12, r = 0.02, divi = r, call price = 15.36.
The result is a negative value in the square root bracket.

Cuchulainn
Posts: 62126
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: Explicit Formula for computing IV

I tried c = 15.3559 and get iv = 0.10979904 using both Differential Evolution (DE) and Brent's method in Boost C++.
A benign problem, numerically.

//
The approach taken in LI 2005 (Taylor series, >>) does not appeal to me at all.
I've lost count of the number of time that 'explicit formula' is mentioned in the text..
Last edited by Cuchulainn on July 29th, 2019, 7:18 pm, edited 1 time in total.

FaridMoussaoui
Posts: 507
Joined: June 20th, 2008, 10:05 am
Location: Genève, Genf, Ginevra, Geneva

### Re: Explicit Formula for computing IV

May be because he treated only two cases (as shown in the appendix derivation) for the "asymptotic" behaviour of the IV.

You can have a look to Quantlib function "blackFormulaImpliedStdDevApproximation" for approximations of IV.

rwang716
Topic Author
Posts: 12
Joined: July 18th, 2019, 6:18 am

### Re: Explicit Formula for computing IV

As far as the paper I have read, this is an inevitable problem for all formulas that stem from Taylor expansion. All of these formulas works for a small range of option prices and moneyness, which makes them much less useful in practice.
I guess I might just have to read a lot more papers to find the solution.
Last edited by rwang716 on July 30th, 2019, 12:08 am, edited 1 time in total.

rwang716
Topic Author
Posts: 12
Joined: July 18th, 2019, 6:18 am

### Re: Explicit Formula for computing IV

I tried c = 15.3559 and get iv = 0.10979904 using both Differential Evolution (DE) and Brent's method in Boost C++.
A benign problem, numerically.

//
The approach taken in LI 2005 (Taylor series, >>) does not appeal to me at all.
I've lost count of the number of time that 'explicit formula' is mentioned in the text..

The iteration methods for computing IV is definitely applicable in all cases, and accurate. However, since in my cases we are looking for the extreme speed-to reduce the time consumption to less than or equal to 1 BS formula time. This makes most numerical solvers methods inappropriate for us.

Alan
Posts: 10167
Joined: December 19th, 2001, 4:01 am
Location: California
Contact:

### Re: Explicit Formula for computing IV

Please justify the need for extreme speed. What is the application?

rwang716
Topic Author
Posts: 12
Joined: July 18th, 2019, 6:18 am

### Re: Explicit Formula for computing IV

I have pretty much solved the problem myself. Anyway, thanks guys for the help.

BTW, by extreme speed, I mean I hope the time for computing the BS implied vol is less than 0.02us, hopefully 0.01us. My current speed is 0.022us on my laptop but I think with further code optimization I can reduce it to 0.015us.

Cuchulainn
Posts: 62126
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: Explicit Formula for computing IV

I tried c = 15.3559 and get iv = 0.10979904 using both Differential Evolution (DE) and Brent's method in Boost C++.
A benign problem, numerically.

//
The approach taken in LI 2005 (Taylor series, >>) does not appeal to me at all.
I've lost count of the number of time that 'explicit formula' is mentioned in the text..

The iteration methods for computing IV is definitely applicable in all cases, and accurate. However, since in my cases we are looking for the extreme speed-to reduce the time consumption to less than or equal to 1 BS formula time. This makes most numerical solvers methods inappropriate for us.
Prove it! Sounds a bit waffly...

haslipg
Posts: 5
Joined: July 11th, 2013, 9:03 am

### Re: Explicit Formula for computing IV

I used found this method very reliable https://www.ssrn.com/abstract=2908494

Cuchulainn
Posts: 62126
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: Explicit Formula for computing IV

I used found this method very reliable https://www.ssrn.com/abstract=2908494
Looks good. I am just wondering about multiple calls to exp() and testing for y < 0 etc. AFAIR if-else tests in loops are an issue?

Cuchulainn
Posts: 62126
Joined: July 16th, 2004, 7:38 am
Location: Amsterdam
Contact:

### Re: Explicit Formula for computing IV

Another choice is a bisection/Newton combi (reliability/speed) that I find useful if the nonlinear function is unimodal:

1. We can pre-compute the number of iterations (and hence number of function evaluations!) needed for a given tolerance (e.g. 1/10?) with Bisection. The value will be the seed for Newton. Espen Haug in his book how to compute the initial interval/bracket for the solution.

https://en.wikipedia.org/wiki/Bisection_method

2. Since the seed is in the neighbourhood of the true solution, then 2-3 iterations of Newton is enough. all things being equal.

3. More upmarket, we might need to compute the 'initial bracket'.
https://en.wikipedia.org/wiki/Kantorovich_theorem


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# cotangent bundle is a bundle

Verifying the first criterion is simply a matter of writing it out:

 $(x^{1},\ldots,x^{2n})\mapsto(\pi(x^{1},\ldots,x^{2n}),\phi_{\alpha}(x^{1},% \ldots,x^{n}))=((x^{1},\ldots,x^{n}),(x^{n+1},\ldots,x^{2n}))$

This is obviously a homeomorphism.

As for the second criterion,

 $\displaystyle\sum_{j=1}^{n}{g_{\alpha\beta}}^{i}_{j}(x^{1},\ldots,x^{2n}){\phi% _{\beta}}^{j}(x^{1},\ldots,x^{2n})$ $\displaystyle=$ $\displaystyle\sum_{j=n}^{2n}{\partial\big{(}\sigma_{\alpha\beta}(x^{1},\ldots x% ^{n})\big{)}^{i}\over\partial x^{j}}x^{j+n}$ $\displaystyle=$ $\displaystyle{{\sigma^{\prime}}_{\alpha\beta}}^{i+n}(x^{1},\ldots,x^{2n})$ $\displaystyle=$ $\displaystyle{\phi_{\alpha}}^{i}(x^{1},\ldots,x^{2n})$

The third criterion follows from the chain rule:

 $\displaystyle{g_{\alpha\beta}}^{i}_{j}{g_{\beta\gamma}}^{j}_{k}={\partial\big{% (}\sigma_{\alpha\beta}(x^{1},\ldots x^{n})\big{)}^{i}\over\partial x^{j}}{% \partial\big{(}\sigma_{\beta\gamma}(x^{1},\ldots x^{n})\big{)}^{j}\over% \partial x^{k}}$
Title cotangent bundle is a bundle CotangentBundleIsABundle 2013-03-22 14:54:31 2013-03-22 14:54:31 rspuzio (6075) rspuzio (6075) 6 rspuzio (6075) Proof msc 58A32


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# Gauss formula to add number of sequence for arbitrary range

Gauss formula to add numbers from $1-100$ is:

$$\frac{n(n+1)}{2}$$

How can this be made applicable for arbitrary range, lets say $3-30$? Is there an easy way of doing that rather then linearly adding the numbers up?

• Hint: add numbers 1-30, 1-2, subtract. – dxiv Sep 7 '16 at 4:56

The sum of the natural numbers from $a$ to $b$ inclusive is $$\frac{(a+b)(b-a+1)}{2}.$$
$\begin{eqnarray} 2S_n & = & S_n + S_n \\ & = & a & + (a + d) & + \ldots & + (a + (n-1)d) & + (a + (n-1)d) & + \ldots & + (a + d) & + a \\ & = & a & + (a + d) & + \ldots & + (a + (n-1)d) \\ & + & (a + (n-1)d) & + (a + (n-2)d) & + \ldots & + a \\ & = & (2a + (n-1)d) & + (2a + (n-1)d) & + \ldots & + (2a + (n-1)d) \\ & = & n(2a + (n-1)d) \\ \therefore S_n & = & \frac{n}{2}(2a + (n-1)d)\end{eqnarray}$


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# Elementary substructure and elementary equivalence

So the subject I'm stumbling around a bit is elementary equivalence and elementary substructures. So maybe you could help me sort things out.

So the definitions as I know them.

Structures $M, N$ over signature $S$ are called elementary equivalent if for every formula $\phi$ over $S$, $M \vDash \phi \iff N \vDash \phi$. $M, N$ being isomorphic implies they are elementary equivalent (but not vice versa).

A substructure $N$ of structure $M$ is called an elementary substructure of $M$, if for every formula $\phi$, and for every $b_1, \cdots, b_n \in N: N \vDash \phi (b_1, \cdots, b_n) \iff M \vDash \phi(b_1, \cdots, b_n)$

All is cool up to now. My problem is understanding the examples given to show that a substructure might be elementary equivalent with its structure, but is not an elementary substructure. To my understanding, in order to prove a substructure isn't elementary, you need to find a formula $\phi(v_1 \cdots v_n)$ so that for every $b_1, \cdots, b_n \in N$, so that $M$ won't satisfy $\phi (b_1, \cdots ,b_n)$, and $N$ will, or vice versa.

So an example: $M=(\mathbb Z; +,0)$, and its substructure $N=(2\mathbb Z;+,0)$, and we need to show that $N$ is not an elementary substructure of $M$. So before that, we can see that they are elementary equivalent, as they are isomorphic.

So, for the rebuttal: Lets take $\phi: \forall x \exists y: y+y=x$. But this formula is 'independent' of choosing variables (there are not free variables), so by being vacuously truth it seems that for every $b_1, \cdots, b_n \in N$, we have $N \vDash \phi(b_1, \cdots, b_n)$ and $M \not\vDash \phi(b_1, \cdots, b_n)$ so $N$ is not elementary equivalent.

My question is: Is using this type of formula correct for refutal? Or should I only use formulas with free variables, such as $\phi \exists y: y+y=x$, and then $M \vDash \phi(6)$, but $N$ doesn't.

The second, smaller question: Doesn't $\phi: \forall x \exists y: y+y=x$ show they are not actually elementary equivalent?

I deeply apologize for the wall of text. Any help will be greatly appreciated! Thank you.

The sentence $\forall x\exists y(y+y=x)$ is true in neither $\mathbb{Z}$ nor $2\mathbb{Z}$. So the sentence does not refute elementary equivalence.
But there is no $y$ in $2\mathbb{Z}$ such that $y+y=2$, while there is in $\mathbb{Z}$, so $2\mathbb{Z}$ is not an elementary substructure of $\mathbb{Z}$. There is no real difference between using formulas with free variables, for which we substitute elements of $N$, and sentences in an extended language with symbols for elements of $N$.


text:
# Proof that the number of diagonals of a polygon is $\frac{n(n-3)}{2}$

For $n \geq 3$ proof that the number of diagonals of a polygon is $\frac{n(n-3)}{2}$ using induction.

I don't know how to start this problem, can you give me a hint?

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How many diagonals meet at each vertex? Hm, but that way doesn't use induction. –  Daniel Fischer Feb 3 '14 at 21:03

Let $d_n$ be the number of diagonals of a polygon.

$d_3=0$ (polygon is triangle) Assume that the number of diagonals of a $n$-polygon is $\frac{n(n−3)}{2}.$

Сonsider $n+1-$poligon $A_1 A_2 \ldots A_{n+1}.$ Under the assumption the number of diagonals of $n-$poligon $A_1 A_2 \ldots A_{n}$ is $\frac{n(n−3)}{2}.$ Besides there are $n-1$ diagonals $A_{n+1} A_2,\,\,$ $A_{n+1} A_3,\ldots , A_{n+1} A_{n-1},$ and $A_1A_n$ for $n+1-$poligon $A_1 A_2 \ldots A_{n+1}.$ Therefore the number of diagonals of a $n+1$-polygon $A_1 A_2 \ldots A_{n+1}$ equal to $d_{n+1}=\frac{n(n−3)}{2}+n-1=\frac{n^2−3n+2n-2}{2}=\frac{(n+1)((n+1)-3)}{2}.$ Sorry for my English.

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I don't know what you mean by induction. But it's simple: For each vertex (n), there is one diagonal linking for all vertex, except the same vertex and the two neighbours (n-3). Thus you have n(n-3), that you got to divide by two or else each diagonal will be counted going and coming back.

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"Induction" stands for a basic logical way of proving something. Basically it is used, when something need to be proven $\forall n \in \mathhbb{N}. So, e.g. here, where polygon can have arbitrary number of vertices, it is good to use induction. Not enough room for details, but basically: proof by induction is done in two steps, the first base, and then the induction step. For example look at my answer, for more, search the internet;) – quapka Apr 9 '14 at 6:08 Suppose you have an$n$-gon with vertices$V_1, \ldots, V_n$which by induction hypothesis has$n (n - 3)/2$diagonals. Now let us add another vertex$V_{n+1}$. How many new diagonals have appeared? Add that to the existing diagonals and verify that you get$(n + 1) (n + 1 - 3)/2$. (Do not forget that the side$V_n V_1$becomes a diagonal when you add$V_{n+1}$beween$V_n$and$V_1$.) - Hint Well start with the first step: prove, that$\frac{n(n-3)}{2}$is the number of diagonals for the smallest possible polygon. In this case it is$n$-polygon, where$n=3$. In the second step assume, that$\frac{k(k-3)}{2}$is the number of diagonals for$k$-polygon and you must show that$\frac{(k+1)(k+1-3)}{2}$is the number of diagonals for$(k+1)\$-polygon. It might be helpful to draw a simple picture with arbitrary number of vertices and add one vertex and find out, what it does with the number of diagonals.

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