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# Tag Info

## Hot answers tagged cc.complexity-theory

26

There are several examples of problems where a parameterized algorithm performs well in practice. Let me mention two such problems. In the $k$-Path problem where we are looking for a simple path of length $k$. Alon, Yuster and Zwick [1] showed that this problem can be solved in $2^{O(k)}\cdot n$ time on $n$-vertex graphs. A weighted version of $k$-Path has ...

13

We can prove $\mathrm{NP\ne QP}$ unconditionally: $\mathrm{QP}$ is closed under quasipolynomial-time reductions, whereas $\mathrm{NP}$ is not (a simple padding argument shows that the closure of $\mathrm{NP}$ under quasipolynomial-time reductions includes all languages computable in nondeterministic quasipolynomial time, and this class strictly contains $\... 8 I think you're right about the first typo. I think the authors are actually fine on the other two questions. Imagine you've applied$f$exactly$\log_2 m$times and that$\varphi$was unsatisfiable. At some first time$k<\log_2 m$, you know$\text{val}(f^{(k)}(\varphi))\leq 1-\epsilon_0$because you double the gap while the value is at least$1-\epsilon_0$... 4 This is quite unlikely to hold, because$\mathrm{EXP_{poly}^{NEXP}}$actually coincides with$\Theta^{\exp}_2$, the exponential analogue of the class$\Theta^P_2$, which is presumably a strict subclass of$\mathrm{EXP^{NP}}$(which is the exponential analogue of$\Delta^P_2$).$\Theta^{\exp}_2$can be variously defined as$\$\Theta^{\exp}_2=\mathrm{EXP^{\|NP}=...

3

I stumbled upon this question now, many years later. In the interim the following paper has appeared: https://dl.acm.org/doi/10.1145/3278158 https://arxiv.org/abs/1704.08705 There the authors do precisely what Kaveh asks for in his question 2: they give a (uniform) TC0 algorithm for balancing, hence obtaining an alternative proof of the main result in Buss '...

1

As far as I have understood, you aim to develop a framework to capture the hardness of combinatorial problems in 3D. However, there are major problems in your question. Your first sentence lacks a couple of technical definitions: For a specific f(), I'm defining a term 'complexity', estimating how difficult is the given function to optimize. First, and the ...

Only top voted, non community-wiki answers of a minimum length are eligible


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##### Conditioned Time-Dilated Convolutions for Sound Event Detection

Sound event detection (SED) is the task of identifying sound events along with their onset and offset times. A recent, convolutional neural networks based SED method, proposed the usage of depthwise separable (DWS) and time-dilated convolutions. DWS and time-dilated convolutions yielded state-of-the-art results for SED, with considerable small amount of parameters. In this work we propose the expansion of the time-dilated convolutions, by conditioning them with jointly learned embeddings of the SED predictions by the SED classifier. We present a novel algorithm for the conditioning of the time-dilated convolutions which functions similarly to language modelling, and enhances the performance of the these convolutions. We employ the freely available TUT-SED Synthetic dataset, and we assess the performance of our method using the average per-frame $\text{F}_{1}$ score and average per-frame error rate, over the 10 experiments. We achieve an increase of 2\% (from 0.63 to 0.65) at the average $\text{F}_{1}$ score (the higher the better) and a decrease of 3\% (from 0.50 to 0.47) at the error rate (the lower the better).

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# Optimal Unbiased Estimator

Feb 2, 2020
Apr 7, 2021 01:49 UTC
Introducing the Minimum Variance Unbiased Estimator and the procedure of deriving it.

The motivation here is to find the optimal estimator of $\theta$. Ideally, we want to find an estimator $\hat\theta_{opt}$ that minimizes the MSE: $$$$\label{eq:min-mse} MSE(\hat\theta_{opt}; \theta) \leq MSE(\hat\theta; \theta) \quad \forall \theta \in \Theta$$$$ for any estimator $\hat\theta$. Unfortunately, this is usually impossible. Recall when we had $Y_i \overset{i.i.d.}{\sim} Unif(0, \theta)$. For proposed estimators $\hat\theta_1 = 1$ and $\hat\theta_2 = \max(Y_1, \cdots, Y_n)$, if we plotted the two MSE curves:

We can see that although $\hat\theta_1$ is a terrible estimator, it beats $\hat\theta_2$ at a narrow range around $1$.

To generalize the above, let $\theta_0$ be any fixed value of $\Theta$. Define estimator $\hat\theta = \theta_0$, then $MSE(\hat\theta; \theta_0) = 0$. For another estimator to beat this, it means $MSE(\hat\theta; \theta) = 0$ would have to be always true for $\theta \in \Theta$ in $\eqref{eq:min-mse}$.

Equation $\eqref{eq:min-mse}$ becomes solvable (and meaningful) when it is coupled with a constraint of unbiasedness.

Definition: An unbiased estimator $\hat\theta$ of $\theta$ is said to be Minimum Variance Unbiased Estimator (MVUE) if $$Var(\hat\theta; \theta) \leq (\hat\theta'; \theta) \quad \forall \theta \in \Theta$$ for any unbiased estimator $\hat\theta$.

## Review of conditional expectation and variance

For two random variables $X$ and $Y$, the expectation of $X$ given $Y$ is given as $$E[X \mid Y] \rightarrow \begin{cases} \text{Compute } E[X \mid Y = y] = \sum_X xp(x \mid y) \text{ or } \int xf(x \mid y) \\ \text{Plug the random variable } Y \text{ into the function above} \end{cases}$$ In the end, this is random and the randomness comes from $Y$. Since $E[X \mid Y]$ is a random variable, we can take its expected value (which is known as the law of total expectation): $$E\Big[E[X \mid Y] \Big] = E[X]$$ The conditional variance is defined similarly as above. The law of total variance states that $$Var(X) = Var(E[X \mid Y]) + E[Var(X \mid Y)]$$

## Rao-Blackwell theorem

Suppose $\hat\theta$ is unbiased for $\theta$, and its variance is finite: $Var(\hat\theta) < \infty$. Let $T$ be a sufficient statistic for $\theta$. Define a new estimator $\hat\theta^*$ where

$$\hat\theta^* = E[\hat\theta \mid T]$$

The Rao-Blackwell theorem states that:

1. $\hat\theta^*$ is still unbiased.
2. $Var(\hat\theta^*; \theta) \leq Var(\hat\theta; \theta)$ for all $\theta \in \Theta$.

In practice, the variance is often much smaller than the original $\hat\theta$. The proof is given as follows.

By the law of total expectation, $$E\left[E[\hat\theta \mid T]\right] = E[\hat\theta] = \theta$$

so $\hat\theta^*$ is unbiased. For the second statement,

$$Var(\hat\theta) = Var\left(E[\hat\theta; T]\right) + E\left[Var(\hat\theta \mid T)\right] \geq Var\left(E[\hat\theta; T]\right) = Var(\hat\theta^*)$$

Note that even if $T$ is not sufficient, the arguments above still hold! So far we’re not using the second condition where $T$ ha to be sufficient. What’s missing here?

The key is that $\hat\theta^*$ has to be a statistic, which shouldn’t depend on the unknown $\theta$. By definition of sufficiency, the conditional distribution $\{(Y_1, \cdots, Y_n) \mid T\}$ doesn’t depend on $\theta$. Since $\hat\theta = f(Y_1, \cdots, Y_n)$, $\{\hat\theta \mid T\}$ doesn’t depend on $\theta$ as well. Therefore, $E[\hat\theta \mid T]$ doesn’t depend on $\theta$.

Remark: if what we’ve described above improves the estimator, why not do it again? $$E[\hat\theta^* \mid T] = ?$$ Recall that if $X$ is a random variable and $f$ is a function, then $$E[f(X) \mid X] = f(X)$$ Since $\hat\theta^*$ is a function of $T$, $E[\hat\theta^* \mid T] = E[\hat\theta \mid T] = \hat\theta^*$. In other words, we can’t further improve $\hat\theta^*$.

## Procedure to obtain good unbiased estimators

1. Find a sufficient statistic $T$ for $\theta$, which can be done efficiently through the factorization theorem.
2. Find a transform/function $f$ such that $\hat\theta = f(T)$ is unbiased for $\theta$, i.e.

$$E[f(T)] = \theta \quad \forall \theta \in \Theta$$

Note that $\hat\theta$ is already a function of $T$, so $E[\hat\theta \mid T] = \hat\theta$. If an unbiased estimator of $\theta$ exists, then such a transform $f$ is step 2 always exists as well. The estimator $\hat\theta^*$ we introduced in the Rao-Blackwell theorem is a function of $T$ and satisfies the conditions above.

### Bernoulli distribution example

Suppose $Y_i \overset{i.i.d.}{\sim} Bern(p)$. We know that $T = Y_1 + \cdots + Y_n$ is sufficient for $p$. We want to find $f$ such that $E[f(T)] = p$.

We know that $E[T] = np$, so it’s straightforward to make $f(T)$ unbiased: $f(t) = \frac{t}{n}$. $$f(T) = \frac{Y_1 + \cdots + Y_n}{n} = \bar{Y}_n$$ and we may conclude that $\bar{Y}_n$ is the good unbiased estimator that we want to find through the procedure.

### Exponential distribution example

Suppose $Y_i$ are an i.i.d. sample from PDF $$f(y; \theta) = \begin{cases} \frac{1}{\theta}e^{-\frac{y}{\theta}}, & y \geq 0, \\ 0, & \text{otherwise} \end{cases}$$ where $\theta > 0$. \begin{aligned} L(\theta) &= f(Y_1; \theta) \cdots f(Y_n; \theta) \\ &= \frac{1}{\theta^n} \exp \left( \frac{\sum_{i=1}^n Y_i}{\theta} \right) \\ &= g \times 1 \end{aligned} By factorization, $T = \sum_{i=1}^n Y_i$ is a sufficient statistic. Next we want a transform $f$ such that $E[f(T)] = \theta$. $$E[T] = E\left[ \sum_{i=1}^n Y_i \right] = nE[Y_1] = n\theta$$ This structure is exactly like what we saw in the previous example. We propose $f(t) = \frac{t}{n}$, so $f(T) = \bar{Y}_n$ is the good unbiased estimator.

## “Good” is not good enough

Now we know how to find a “good” unbiased estimator. How can we go from here to the best unbiased estimator, namely MVUE?

Definition: A sufficient statistic $T$ for $\theta$ is said to be complete if the transform $f$ found in the procedure is unique.

Lehmann–Scheffé Theorem: If the sufficient statistic $T$ is complete, then $\hat\theta^* = f(T)$ in the procedure is the MVUE.

Proof: Suppose $\hat\theta$ is another unbiased estimator. Let $\tilde{f}(t) = E[\hat\theta \mid T]$. By the Rao-Blackwell theorem, $\tilde{f}(T)$ is unbiased. By the definition of completeness, $\tilde{f} = f$, so $$\tilde{f}(T) = f(T) = \hat\theta^*$$ By the Rao-Blackwell theorem, $$Var(\hat\theta^*) = Var(\tilde{f}(T)) \leq Var(\hat\theta)$$ In this course, we don’t check for completeness. When asked to find the MVUE, just apply the 2-step procedure.

### Example

Suppose $Y_i \overset{i.i.d.}{\sim} Unif(0, \theta)$. We know $T = \max(Y_1, \cdots, Y_n)$ is sufficient for $\theta$, and that $$E[T] = \frac{n}{n+1}\theta$$ We propose $f(t) = \frac{n+1}{n}t$, thus $$f(T) = \frac{n+1}{n}\max(Y_1, \cdots, Y_n)$$ is the MVUE.

Related Posts
 Jan 30 Sufficiency 5 min read Jan 29 Maximum Likelihood Estimator 7 min read Jan 28 The Method of Moments 3 min read Jan 27 Consistency 6 min read Jan 25 Bias and Variance 11 min read


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# In an isothermal process on an ideal gas, the pressure increases by 0.5%. The volume decreases by about

Q: In an isothermal process on an ideal gas, the pressure increases by 0.5%. The volume decreases by about

(a) 0.25 %

(b) 0.50 %

(b) 0.7 %

(d) 1 %

Ans: (b)

Sol: $P’ = P + \frac{0.5 P}{100}$

Since , $\large P V = P’ V’$

$\large P V = P (1 + \frac{0.5}{100}) V’$

$\large \frac{V’}{V} = (1 + \frac{0.5}{100})^{-1}$

$\large \frac{V’}{V} = (1 – \frac{0.5}{100})$

$\large (\frac{V’}{V} -1 )\times 100 = – \frac{0.5}{100}\times 100$

= -0.5 %


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# finding a point with a given direction and distance [help me]

This topic is 2781 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

sorry if this is in the wrong section, i am new here, so anyway here is my problem.
currently I am developing a rendering engine (mostly for making models to insert into a game), inside this engine I have a class (designed for making vector points) when you use this you set a distance and direction relative to a point (the bone joint) it then sets the XYZ points depending on the given rotation and distance. Here is the coding so far
public class Vec { //their dropin us presents... why are they explodin??? int distanceToJoint; int directionToJointY; int directionToJointX; int directionToJointZ; int xpos; int ypos; int zpos; SkelBone skel; public Vec (int ds, int dry, int drx,int drz, SkelBone sk) { distanceToJoint = ds; directionToJointY = dry; directionToJointX = drx; directionToJointZ = drz; skel = sk; } }
basically I need to find a way to set xpos, ypos and zpos

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... Sorry, what?

Please organize your thoughts and ask your question more clearly. What do you know? What don't you know? What are the inputs and outputs to your process?

##### Share on other sites
are drx/dry/drz euler angles? First I think you will definately need to change most of those values to being floats and not ints. My guess is your bone is at (0, 0, 0) initially, you want to rotate it by the given angles and find the end point (storing those in xpos/ypos/zpos). This is not a difficult task if you have some maths things to use (matrices/vectors). Do you have anyhting like that?

I'd look at using a math lib, if not then make a Vector class and a function that takes euler angles and a vector, rotates by those angles and returns a rotated vector (that will be a fair amount of work). Then you can do:

 Vector pos; SkelBone skel; public Vec (int ds, int dry, int drx,int drz, SkelBone sk) { distanceToJoint = ds; directionToJointY = dry; directionToJointX = drx; directionToJointZ = drz; skel = sk; pos = new Vector(0, 0, distanceToJoint); // This would be the default "forward", where the bone points with no rotation pos = RotateFromEulerAngles(drx, dry, drz, pos); }

Vector RotateFromEulerAngles(float xr, float yr. float zr, Vector in);

To implement that you will need to know a bit of maths. Do you know matrix multiplication (there may be other ways)? This shows a general matrix for rotations: http://en.wikipedia.org/wiki/Rotation_matrix#Rotations_in_three_dimensions

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Law of cosines

• Feb 24th 2008, 01:22 AM
kri999
Law of cosines
okay i dont know why, this may be extremely easy but i'm just not getting the right answer :(

Here's the question:
richmond is 200 kilometers due east of Teratown and Hamilton is 150 kilometers directly north of Teratown. Find the shortest distance in kilometers between Hamilton and Richmond.

I use law of cosines right? and so what would the equation look like?
• Feb 24th 2008, 01:27 AM
mr fantastic
Quote:

Originally Posted by kri999
okay i dont know why, this may be extremely easy but i'm just not getting the right answer :(

Here's the question:
richmond is 200 kilometers due east of Teratown and Hamilton is 150 kilometers directly north of Teratown. Find the shortest distance in kilometers between Hamilton and Richmond.

I use law of cosines right? and so what would the equation look like?

Pythagoras' Theorem:

$(200)^2 + (150)^2 = (HR)^2$.
• Feb 24th 2008, 01:34 AM
kri999
but i have to use the law of cosines...how do i do that with this problem? or is just not possible?
• Feb 24th 2008, 01:37 AM
mr fantastic
Quote:

Originally Posted by kri999
but i have to use the law of cosines... Mr F asks: Why?

how do i do that with this problem? or is just not possible?

OK:

$
(HR)^2 = (200)^2 + (150)^2 - 2(200)(150) \cos(90^0) = .....
$

In case you haven't guessed, Pythagoras' Theorem is a special case of the cosine rule.
• Feb 24th 2008, 01:47 AM
kri999
Thank you! I got the same equation but somehow I keep getting the answer wrong, but i just tried a differnet method and it worked out..
yeah i have to use law of cosines because that's what the chapter's on haha

thanks so much!
• Feb 24th 2008, 01:50 AM
angel.white
Quote:

Originally Posted by mr fantastic
OK:

$
(HR)^2 = (200)^2 + (150)^2 - 2(200)(150) \cos(90^0) = .....
$

In case you haven't guessed, Pythagoras' Theorem is a special case of the cosine rule.

I actually didn't know that. I just kinda figured he sat around playing with blocks and rulers for a few weeks until a lightbulb turned on.


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Published:

# Analysis of Bitcoin Volatility

Here I study whether Bitcoin’s volatility has been decreasing. It is widely reported that it has been the case, and indeed if one looks at Eli Dourado’s btcVol.info it seems to be the case. But is it the case?

Here there is a previous analysis by a friend of mine showing that using statistical test (Augmented Dickey-Fuller) to test whether a series is stationary, and the result is that the data is compatible with a stationary process. That is, the underlying distribution from which daily returns are drawn is constant across time.

In what follows, I try to replicate his study, showing some limitations of ADF.

import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
import datetime
sns.set()
%matplotlib inline
from forex_python.bitcoin import BtcConverter
b = BtcConverter()
_=b.get_latest_price('USD')

start_date=datetime.datetime(2011,1,1)
end_date=datetime.datetime(2017,1,1)
r=b.get_previous_price_list('USD', start_date, end_date)

data=pd.DataFrame.from_dict(r,orient='index')
data.columns=['Price']
data.sort_index(inplace=True)
data.index=pd.to_datetime(data.index)
data.reset_index(inplace=True)
data.rename(columns={'index':'Date'},inplace=True)

DatePrice
02011-01-010.300
12011-01-020.300
22011-01-030.295
32011-01-040.299
42011-01-050.299
_=data.plot(x="Date",y="Price")

data['Daily change']=data['Price'].pct_change(1).multiply(100)
_=data.plot(title="Bitcoin daily change (%)",x='Date',y='Daily change')

data['Daily rolling 10']=data['Daily change'].rolling(window=10,center=False).std()
data['Daily rolling 30']=data['Daily change'].rolling(window=30,center=False).std()
data['Daily rolling 60']=data['Daily change'].rolling(window=60,center=False).std()
plt.figure(figsize=(12,6))
_=ax=data.plot(title="Bitcoin daily change (%)",y='Daily rolling 10',x='Date')
_=data.plot(y='Daily rolling 30',x='Date',ax=ax)
_=data.plot(y='Daily rolling 60',x='Date',ax=ax)

<matplotlib.figure.Figure at 0x7f2edc5fde80>

#First, we could do some curvefitting,
data['Date_n']=(data['Date']-data['Date'][0])/ np.timedelta64(1,'D')
fig=plt.figure(figsize=(10,8))
sns.regplot(data=data,x='Date_n',y='Daily rolling 10',scatter_kws={"s": 10,"alpha":0.5},label='Daily rolling 10')
sns.regplot(data=data,x='Date_n',y='Daily rolling 30',scatter_kws={"s": 10,"alpha":0.5},label='Daily rolling 30')
sns.regplot(data=data,x='Date_n',y='Daily rolling 60',scatter_kws={"s": 10,"alpha":0.5},label='Daily rolling 60')
plt.ylabel('Daily change (%)')
plt.legend()
import statsmodels.api as sm

y=data['Daily rolling 10']
X=data['Date_n']
model.summary()

Dep. Variable: R-squared: Daily rolling 10 0.178 OLS 0.178 Least Squares 472.5 Sun, 08 Jan 2017 5.36e-95 12:48:59 -5749.0 2183 1.150e+04 2181 1.151e+04 1 nonrobust
coef std err t P>|t| [95.0% Conf. Int.] 6.9094 0.145 47.579 0.000 6.625 7.194 -0.0025 0.000 -21.738 0.000 -0.003 -0.002
 Omnibus: Durbin-Watson: 812.011 0.072 0 3098.36 1.818 0 7.565 2550

Perhaps it would be better to use a nonlinear function, something like an exponential with a decreasing term like $f(x,a,b,c)=ae^{-xb}+c$ . For optimisation purposes, I noralise the series first.

from scipy.optimize import curve_fit
from sklearn.preprocessing import normalize
def normalize(X,Y):
return (X2-np.mean(X2))/np.std(X2),(y2-np.mean(y2))/np.std(y2)
def func(x,a,b,c):
return a*np.exp(-x*b)+c
def plot(X2,y2,popt,n):
plt.subplot(1,3,n)
plt.plot(X2,func(X2,popt[0],popt[1],popt[2]))
plt.plot(X2,y2)

y=data['Daily rolling 10']
X=data['Date_n']
X1,y1=normalize(X,y)
popt1, pcov =curve_fit(func,X1,y1)
fig=plt.figure(figsize=(12,6))
plot(X1,y1,popt1,1)
plt.title('Exponential fit for Daily rolling 10')
print("Parameters:",popt1)

y=data['Daily rolling 30']
X=data['Date_n']
X2,y2=normalize(X,y)
popt2, pcov =curve_fit(func,X2,y2)
plot(X2,y2,popt2,2)
plt.title('Exponential fit for Daily rolling 30')
print("Parameters:",popt2)

y=data['Daily rolling 60']
X=data['Date_n']
X3,y3=normalize(X,y)
popt3, pcov =curve_fit(func,X3,y3)
plot(X3,y3,popt3,3)
plt.title('Exponential fit for Daily rolling 60')
print("Parameters:",popt3)

plt.figure(figsize=(12,6))
plt.plot(X1,func(X1,popt1[0],popt1[1],popt1[2])-y1,label='Daily rolling 10')
plt.plot(X2,func(X2,popt2[0],popt2[1],popt2[2])-y2,label='Daily rolling 30')
plt.plot(X3,func(X3,popt3[0],popt3[1],popt3[2])-y3,label='Daily rolling 60')
plt.title('Residual plot')
plt.legend()

Parameters: [ 0.26753345  1.07601779 -0.4516436 ]
Parameters: [ 0.36183029  1.01534461 -0.57939952]
Parameters: [ 0.42689696  0.98383072 -0.66565455]

<matplotlib.legend.Legend at 0x7f2edc751748>

I would say that the fit looks right (It seems to be exponentially decreasing, and it seems that variance is decreasing) especially if we remove some of the ‘big’ events like bubbles, etc.

Now we bring on ADF (Which I don’t have experience using), checking the results we get for both a series with and without a trend. Here there is the documentation for this function. First I run some tests to see how the function behaves. The parameter of interest is the second in the returned array, a pseudo p-value. I’ll take it so that if p-value<0.05, we can’t reject the null hypothesis

from statsmodels.tsa.stattools import adfuller
#Here there *is* a trend. Null hypothesis is: constant (The variable follows a random-walk)
#Because there is a trend, the test fails.
plt.plot(np.linspace(0,10, 100)+np.random.randn(100)/10)

(0.030371082537879741,
0.96097283593236305,
5,
94,
{'1%': -3.5019123847798657,
'10%': -2.5834538614757809,
'5%': -2.8928152554828892},
-130.36360151635336)

Here there is a trend. So we tell the test that there is, and ask: does the trend change?

We can also do an ADF test assuming that there is a trend:

plt.plot(np.linspace(0,10, 100)+np.random.randn(100)/10)

(-6.0891054304079377,
1.4245907097885184e-06,
3,
96,
{'1%': -4.0563093932201246,
'10%': -3.1544345187717013,
'5%': -3.4572550874385124},
-161.30436800688932)

In the next example there is a trend that changes, but from a constant underlying function, a sin(x). The test still says the process doesn’t change. Using the ‘ct’ option also says that the process doesn’t change. As the changing trend itself (a cosine) is also stationary around zero

np.random.seed(42)
n=1000
space=np.linspace(0,100,n)
var=np.concatenate([np.sin(space[:n//2]),np.sin(space[n//2:])])+np.random.rand(n)/100
plt.plot(var)

(-14.936838558057197,
2.3543437151057883e-22,
22,
977,
{'1%': -3.9680661492141187,
'10%': -3.1297006047557043,
'5%': -3.4149932715005722},
-8471.9770767955488)

We can now try to see what would the test say if it saw the function we just fitted.

plt.plot(func(X3,popt3[0],popt3[1],popt3[2]))

(-16.794102051514088,
1.2352461745010984e-29,
25,
2107,
{'1%': -3.4334573966160153,
'10%': -2.5675007703726296,
'5%': -2.8629127187998606},
-18524.80171562676)

So even if the variance were decreasing given a very clean equation, the test would still say that the process is stationary, so the test is not that much useful after all.

Ok, let’s go back to our Bitcoins and do the actual test

def nice(data,regtype="c"):

print("P-value for 10 days averaged volatility: ",nice(data['Daily rolling 10']),'***')
print("P-value for 30 days averaged volatility: ",nice(data['Daily rolling 30']),'***')
print("P-value for 60 days averaged volatility: ",nice(data['Daily rolling 60']),'**')

P-value for 10 days averaged volatility:  8.59786631742e-05 ***
P-value for 30 days averaged volatility:  0.000113952000692 ***
P-value for 60 days averaged volatility:  0.034705270592 **

Thus we find that indeed for the 10 days average volatility, the data is consistent with a process with constant variance, and that remains valid for greater windows of averaging.

Let’s now examine what happens if instead, we consider as our hypothesis that the series follows a trend (y=ax+b)

print("P-value for 10 days averaged volatility: ",nice(data['Daily rolling 10'],'ct'),'***')
print("P-value for 30 days averaged volatility: ",nice(data['Daily rolling 30'],'ct'),'***')
print("P-value for 60 days averaged volatility: ",nice(data['Daily rolling 60'],'ct'),'**')

P-value for 10 days averaged volatility:  4.36675365203e-05 ***
P-value for 30 days averaged volatility:  5.24203107756e-05 ***
P-value for 60 days averaged volatility:  0.0309130608266 **

Surprise surprise, the test also says this is true! ** We cannot reject the fact that the series is generated by a process with decreasing variance either! **

For completion, below I show the ADF test for a number of possible lags. The test by default choses one using the Bayesian Information Criterion.

def makeplot(name,pos):
plt.subplot(1,3,pos)
for i in range(n_lags):
plt.plot(range(n_lags),pvalues)
plt.yscale('log', nonposy='clip')

plt.ylabel('P-value')
plt.xlabel('Number of lags for the ADF test')
plt.legend(['Assuming constant','Assuming trend'])
i=1
n_lags=30
pvalues=np.zeros((n_lags,2))
fig=plt.figure(figsize=(12,6))
makeplot('Daily rolling 10',1)
makeplot('Daily rolling 30',2)
makeplot('Daily rolling 60',3)

# Conclusion

The ADF test is not very useful for a time series like the BTC-USD exchange rate. Here I would go with plain OLS,that says that the coefficient of the feature ‘date’ is negative at the 95% confidence level. This trend persist even if we consider only, say, post 2014 or post 2015 data. Using a nonlinear exponential fit is even better.

Thus the data warrants the conclusion that Bitcoin is, after all, becoming less volatile.

Tags:


text:
# Performing Second Derivative test on multivariate function

I have two functions $f=xy^2$ and $g=x^2+y^2$. When optimizing $xy^2$ on the circle $x^2+y^2=1$ I get 6 critical points but when I try to perform the second derivative test, it equals 0, meaning that the result is inconclusive. How can I find whether the critical points are in fact a maximum or minimum in this case?

You can go this way. Since $x^2+y^2=1$ then $y^2=1-x^2$ and substituting in $f$ gives the one variable function

$$f(x) = x(1-x^2).$$

Now you can use the derivative test to find max and min.

• I think you need to optimize this equation using lagrange multipliers otherwise not all of the solutions can be found. – andrew749 Mar 18 '15 at 20:52
• Since the tags say "calculus and optimization" so I gave you this alternative technique because you said the test you use was not conclusive. You can use Lagrange multiplier as a different approach. Anyways I gave you a technique that saves your time. – science Mar 18 '15 at 20:57
• It is simply not the case that one must use Lagrange multipliers. I agree however that this solution is incomplete. This approach yields candidate points: $x=\pm 1/\sqrt{3}$, and the second derivative test confirms $+1/\sqrt{3}$ is the proper candidate. We must also consider $x=\pm 1$, since they are extremes, but they are easily eliminated since they require $y=0$. We arrive at two solutions: $(x,y)=(\sqrt{1/3},\pm\sqrt{2/3})$, optimal value $2/(3\sqrt{3})$. No Lagrange multipliers needed. +1 for @science! – Michael Grant Mar 19 '15 at 19:46
• @MichaelGrant: Thank you for providing these details. I appreciate your comment. – science Mar 19 '15 at 20:17

The best way to solve this problem is the method of Lagrange multipliers.Also, by the extreme value theorem,since f is defined on a closed and bounded subset of $R^2$, we know it must contain absolute extrema on the circle.

The method of Lagrange multipliers essentially considers the gradients of both the function f(x,y)=$xy^{2}$ and the level curve constraining function-in this case, the circle g(x,y)= $x^2 + y^2$ =1.(Strictly speaking, a circle isn't a function in the precise definition of the term.But since this is a basic calculus problem, we won't get into the weeds on this.) Since the direction of maximum change occurs along the gradients and f is constrained by g, we would expect these gradients to be parallel to each other. $\nabla {f(x,y}$ = $\lambda\nabla {g(x,y)}$ $\rightarrow$ $(f_x, f_y)$= $\lambda(g_x,g_y)$

So now let's compute each of the partial derivatives. Once we solve the resulting system for x and y, we plug in those values into f and find the extrema.

$f_x$ = $y^2$ $f_y$ = 2xy $g_x$ = 2x $g_y$ = 2y So $y^2$ = $\lambda 2x$ 2xy = $\lambda 2y$

Solving, we get x = $\lambda$ and y = $\pm \sqrt{2}x$.

So plugging into the constraining circle gives:

$x^2 + y^2$ =1 $\rightarrow$ $x^2 + (\pm\sqrt{2}x)^2$ =1 $\rightarrow$ $3x^2$ =1 $\rightarrow$ x = $\lambda$ = $\pm \frac {1}{3}$. So y = $\pm \frac {2}{3}$.

So now we have everything we need to solve the problem by plugging x and y into f(x,y). The level curves of the functions give a good indication of the geometry of the domain and it's local extrema:

The full graph in $R^3$ gives:

You can and should finish it now yourself. I only obtained 4 extrema-what were the other 2 you got? Unless I screwed up,of course, which I'm sure I'll hear about soon if so...............

• You're missing some square roots in there... – Michael Grant Mar 19 '15 at 19:44


text:
# What can time-discount rates tell us about xG and Goals?

Recently, I looked at backtesting and ensembling match prediction models. While working on this, I found a nice result regarding time discounting that I wanted to share.

## What is time-discounting?

Match prediction models tend to produce estimates of teams’ attacking and defensive strength. The more accurate these estimates are, the more accurate the predictions. Factors like teams’ performance levels (or even home-field advantage) are dynamic and change over time. New players come into the team, team tactics are tweaked, and so on. Because of this dynamism, data (matches) from, say, a year ago tends to be less relevant than data from last week.

Time-discounting is a way of weighting input data in match prediction models such that more recent matches are given a higher weight (and therefore more influence on the model’s parameter estimates) than those in the distant past.

## Signal vs staleness

If the most relevant information is data from the last week, the ideal scenario would be to use only the latest match for each team. But, we know that in practice that’s unlikely to work. A single match is quite noisy; anything can happen.

So to combat the noise in a single game, we have to add more data from further back in the past. But that historical data isn’t quite as relevant as the data from last week. How a team played a long time ago doesn’t tell us that much about the team’s performance level now.

So there’s a tension between having more data, and each incremental bit of data being progressively more out of date. Too much old data, and you’re feeding misleading information into the model. Not enough data, and the model can’t separate the signal from the noise.

## More signal = less data

If we use data that has less noise, we don’t have to use as much to extract a reliable signal. So instead of using more data, we can try to use better data.

This is broadly the logic of using expected goals (xG) instead of goals to evaluate teams and players. Goals are really noisy, so over the course of a season, there is still more signal in xG. To reliably evaluate teams on goals with the same confidence, you’d need years of data, at which point the coach has changed, the squad has changed… you’re comparing multiple different teams!

## Exponential weighting

The most common form[1] of time-discount is exponential weighting, where each data point is weighted according to the formula:

where “” is how long ago that match took place (usually in days) and “” (“epsilon”) is a hyperparameter.

Because of the nature of exponential decay, a high discount rate (yellow line) results in a very low weight for matches a long time in the past compared to a low discount rate (green line).

### How do you find the right value of “”?

The simplest solution is to use backtesting. In short, this means training the model for a variety of different values of epsilon and seeing which one results in the best predictions.

## Time-discounting with different models

If we backtest both Goals-based and xG-based models for a variety of “” values, we can compare results.

On the x-axis, I have plotted the weight at day 365 (), instead of just “” because think this is more intuitive. The raw “” values are shown in gray next to each point.

Both plots show the same overall shape: a very low weight at day 365 (i.e. a very high discount weight), leads to very poor performance followed by a sharp rise, a peak, and a slower decline as the discount rate increases (and old data is given more importance).

The peak model performance for the xG-based model comes at with a weight at day 365 of around 25%. On the other hand, the peak performance for the goals-based model comes at around 50% weight-at-day-365.

This goes back to what I mentioned earlier about signal-vs-noise. A goal-based model needs to give matches from the past a higher weight because goals are a noisy data source. Whereas the xG-based model performs best with a higher discount rate, because it’s able to extract more information from each match. Or, to put it another way, it requires less information before the staleness of additional data-points starts to harm the model’s performance.

We already knew xG was a better way to evaluate teams both intuitively and empirically, but I think looking at the comparison of goals and xG through the time-discounting is a nice way to develop our intuition about why xG is better.

## Appendix I: A more entrenched elite?

EDIT (2020-07-19): Originally, I speculated that a change in “” since could be evidence of a more entrenched elite at the top of English football. However, in the original Dixon-Coles paper, they use half-weeks as their time unit (instead of days as I have). Once correcting for this, the difference in “” values is significantly reduced.

Thanks to Diogo for pointing this out to me on Twitter.

Original text (click to expand)

As a final note, the original Dixon-Coles paper from 1997, they estimate an optimal value of “” at around 0.0065, using a goals-based model on Football League data. The above analysis (using Premier League matches between 2016-2021) suggests an optimal value around 0.0023, a less aggressive time-discount.

Assuming I’ve correctly reproduced their method, what might be the cause of this change? Could it be that a more entrenched elite at the top of the Premier League has meant that old (e.g. 6+ months ago) results have more information about the present?

We talk a lot about how the economics of the game have changed and made it harder for small clubs to compete, so I think this is an interesting hypothesis. Especially as this could give us an empirical way to measure those changes. I don’t think we can evaluate from this analysis alone whether it this is driving the change in “”, but this might be an idea worth coming back to…

## Appendix II: Code etc

If you would like to reproduce this analysis, or dig a bit more into the technical details, the code can be found in my wingback (because it’s concerned with back-testing models… I couldn’t help myself, I’m sorry) repository.

This repo is built on top of understat-db, a project for pulling data from Understat.com into a database. It uses a Python library called mezzala for modelling team-strengths and match predictions.

## Appendix III: Data scarves

I still have some Liverpool 2019/20 “data scarves” left for sale (shipping to the UK). If football+data is your thing, take a look:

1. The most common form that I know of. If you know of any others that you think might work better, let me know! I’d love to try them out. ↩


text:
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

2 edited body

$\def\QQ{\mathbb{Q}}$This question seems unmotivated to me, but I might as well answer it: Yes, for any finite extension $K \supset \QQ$, there is some $\alpha \in K$ such that $\QQ(\alpha^n) = K$ for all nonzero integers $n$.

$\def\p{\mathfrak{p}}$Let $p$ be a prime which splits completely in $K$, say $p = \p_1 \p_2 \ldots \p_d$ with $d=[K:\QQ]$. We know that $p$ exists by CebatarovCebotarov. By the Chinese remainder theorem, there is some $\alpha$ such that the $d$ valuations $v_{\p_i}(\alpha)$ are all distinct. So, for any $n$, the $d$ valuations $v_{\p_i}(\alpha^n)$ are all distinct.

If $L$ is a proper subfield of $K$, then there is some pair $\p_i$ and $\p_j$ such that $\p_i \cap L = \p_j \cap L$. (Because $p$ can only break into at most $[L:\QQ]$ factors in $L$.) So, if $\alpha^n$ were in $L$, then $v_{\p_i}(\alpha^n)$ would equal $v_{\p_j}(\alpha^n)$. This contradiction shows that $\alpha^n$ is not contained in any subfield of $K$.

Remark: One can make the same argument work using the archimedean place, if it happens that $K$ is totally real. This is what is happening in the case of $\QQ(\sqrt{2})$.

1

$\def\QQ{\mathbb{Q}}$This question seems unmotivated to me, but I might as well answer it: Yes, for any finite extension $K \supset \QQ$, there is some $\alpha \in K$ such that $\QQ(\alpha^n) = K$ for all nonzero integers $n$.

$\def\p{\mathfrak{p}}$Let $p$ be a prime which splits completely in $K$, say $p = \p_1 \p_2 \ldots \p_d$ with $d=[K:\QQ]$. We know that $p$ exists by Cebatarov. By the Chinese remainder theorem, there is some $\alpha$ such that the $d$ valuations $v_{\p_i}(\alpha)$ are all distinct. So, for any $n$, the $d$ valuations $v_{\p_i}(\alpha^n)$ are all distinct.

If $L$ is a proper subfield of $K$, then there is some pair $\p_i$ and $\p_j$ such that $\p_i \cap L = \p_j \cap L$. (Because $p$ can only break into at most $[L:\QQ]$ factors in $L$.) So, if $\alpha^n$ were in $L$, then $v_{\p_i}(\alpha^n)$ would equal $v_{\p_j}(\alpha^n)$. This contradiction shows that $\alpha^n$ is not contained in any subfield of $K$.

Remark: One can make the same argument work using the archimedean place, if it happens that $K$ is totally real. This is what is happening in the case of $\QQ(\sqrt{2})$.


text:
The figure below gives $F'(x)$ for some function $F$.

Use this graph and the facts that the area labeled A is 18.5, that labeled B is 20, that labeled C is 5, and that$F(2) = 3$ to sketch the graph of $F(x)$. Label the values of at least four points.

Then, using your graph, give four $(x,y)$ points on the curve:


text:
# What's the 3d equation of the Alcubierre warp?

Tags:
1. Oct 19, 2015

### Teslanumber1

What I mean by this is what's the 3d equation of the alcubierrie warp effect? You know when space is contracted infront of the ship, and inflated behind it in terms of z like this eqaurion for a rain drop z=(sin(2x+y)+sin(5y+2x))/5? Here's the photo if you don't understand what I'm talking about https://upload.wikimedia.org/wikipedia/commons/c/c4/Alcubierre.png

Last edited: Oct 19, 2015
2. Oct 19, 2015

### Staff: Mentor

There isn't a "3d equation" for the Alcubierre warp spacetime. Spacetime is 4-dimensional, not 3-dimensional.

A brief summary of the math is given on the same Wikipedia page that shows the picture you linked to. Have you read that page? Do you have questions about what is said there?

3. Oct 19, 2015

### Teslanumber1

I'm not acfually asking for the eqautions that represent warp drive just the picture I linked which is only 3d dimensional, and which only is supposed to help visualize what happens with a alcubierre drive.

4. Oct 19, 2015

### Staff: Mentor

That picture is not a 3-dimensional picture of anything. It's a way of trying to very heuristically represent what the warp drive does to spacetime, not space.

5. Oct 19, 2015

### Teslanumber1

Yes but the actual picture shown is 3 dimensional is it not? On the x,y, and z plane? I'm simply asking what 3d function would represent that particular picture that just so happens tries to helps us visualize warp. Maybe it would have been better, and more fruitful to ask for examples of some 3d wave functions.

6. Oct 19, 2015

### Staff: Mentor

Not really. It's an attempt to show the geometry of a 2-dimensional "slice" out of a 4-dimensional spacetime, and since the slice is curved, it is shown embedded in a 3-dimensional space. But the 3-dimensional space is not physically real.

7. Oct 19, 2015

### Teslanumber1

I'm really not asking anything about the mathematics, or physics, or intricacies of warp drive as quite frankly that's completely beyond me at my current education level, I am asking what's the 3 dimensional equation used to plot the photo that I linked.

8. Oct 19, 2015

### Staff: Mentor

Well, this is a forum for discussing physics, so we usually assume that's what people come here to discuss.

I don't know. Since that equation isn't physically meaningful, I don't know if it is going to appear in any sources that discuss the physics of the Alcubierre drive.

9. Oct 20, 2015

### PWiz

It's just the intersection of a plane with an elliptical disc at a non-zero angle with some special effects added around the intersection points. There is no one specific set of functions; for example, $\frac {x^2}{4^2} + \frac{y^2}{2^2} =1$ being defined at $z=0$ only and the plane $z=x$ would produce the same thing (except for the special effects).


text:
# Lesson 3

Half an Equilateral Triangle

• Let’s investigate the properties of altitudes of equilateral triangles.

### 3.1: Notice and Wonder: Triangle Slices

Sketch an equilateral triangle and an altitude from any vertex in the equilateral triangle.

What do you notice? What do you wonder?

### 3.2: Decomposing Equilateral Triangles

1. Here is an equilateral triangle with side length 2 units and an altitude drawn. Find the values of $$x$$ and $$y$$.

2. Measure several more of these “half equilateral triangles” by drawing equilateral triangles and altitudes. Compute the ratios of the side lengths of these new triangles.
3. Make a conjecture about side lengths in “half equilateral triangles.”

### 3.3: Generalize Half Equilateral Triangles

Calculate the lengths of the 5 unlabeled sides.

Here is a collection of triangles which all have angles measuring 30, 60 and 90 degrees.

What is the total area enclosed by the 5 triangles?

### Summary

Drawing the altitude of an equilateral triangle decomposes the equilateral triangle into 2 congruent triangles. They are right triangles with acute angles of 30 and 60 degrees. These congruent angles make all triangles with angles 30, 60, and 90 degrees similar by the Angle-Angle Triangle Similarity Theorem.

If we consider a right triangle with angle measures of 30, 60, and 90 degrees, and with the shortest side 1 unit long, then the hypotenuse must be 2 units long since the triangle can be thought of as half of an equilateral triangle. Call the length of the altitude $$a$$. By the Pythagorean Theorem, we can say $$a^2+1^2=2^2$$ so $$a=\sqrt3$$.

Now, consider another right triangle with angle measures of 30, 60, and 90 degrees, and with the shortest side $$y$$ units long. By the Angle-Angle Triangle Similarity Theorem, it must be similar to the right triangle with angles 30, 60, and 90 degrees and with sides 1, $$\sqrt3$$, and 2 units long. The scale factor is $$y$$, so a triangle with angles 30, 60, and 90 degrees has side lengths $$y, y\sqrt3,$$ and $$2y$$ units long.

In triangle $$ABC, 2y=5$$ so $$y=\frac 52$$. That means $$DB$$ is $$\frac 52$$ units and $$DC$$ is $$\frac 52 \sqrt 3$$ units.

In triangle $$EGH, y\sqrt 3=4$$ so $$y=\frac{4}{\sqrt 3}$$. That means $$FG$$ is $$\frac{4}{\sqrt 3}$$ units and $$EG$$ is $$2 \frac{4}{\sqrt 3}$$ or $$\frac{8}{\sqrt 3}$$ units.


text:
# Summation symbol and fraction in LaTeX

The expressions of fraction containing a summation in the display and inline modes are slightly different like the below.

$\frac{ \sum_{k=1}^{n} k^2 }{ 5 }$
$\frac{ \sum_{k=1}^{n} k^2 }{ 5 }$

In the inline equation, the subscript of summation is so close to the fraction bar. The display styles of sum and frac make an equation more beautiful. From the below example, making both frac and sum displaystyle may be a good way if you want a denominator to look larger.

$\frac{ \displaystyle\sum_{k=1}^{n} k^2 }{ 5 }$
$\frac{ \displaystyle\sum_{k=1}^{n} k^2 }{ 5 }$ \$15pt] \displaystyle\frac{ \displaystyle\sum_{k=1}^{n} k^2 }{ 5 }  ## Fraction in summation \[ \sum_{k=1}^{n} \frac{ k^2 }{ 5 }$
$\sum_{k=1}^{n} \frac{ k^2 }{ 5 }$ \\[15pt]
$\displaystyle\sum_{k=1}^{n} \frac{ k^2 }{ 5 }$ \\[15pt]
$\displaystyle\sum_{k=1}^{n} \displaystyle\frac{ k^2 }{ 5 }$


text:
# Evil Hangman (Project)

Learning Objectives: By the end of this learning module, you should understand how to...

1. use the map and set interfaces in an actual, real-life project.
2. write a basic AI for the game of hangman.

## Introduction

Everyone has played hangman at one point or another during their childhood. The idea is that a player (or perhaps in this case, a computer) chooses a random word from a dictionary and another player (in this case, a user) has to progressively figure out what that random word was through guessing characters before they run out of incorrect guesses.

However, what if the computer wasn’t forced to choose their random word from the dictionary at the beginning? Instead the computer coould maintain a list of every word in the English language and then continuously trim it down to try to dodge the player’s guesses as much as possible. While this is technically cheating in hangman, we can implement such a computer strategy very cleverly to prevent guesser from finding out that the computer is cheating. It turns out that by making the computer a tad bit smarter (and perhaps more devious), it becomes incredibly difficult to correctly guess the word from the dictionary!

Let’s delve into the specifics a bit more carefully. A fundamental component of hangman is that the computer accurately represents a word that was chosen beforehand using the letters that have been guessed. However, if the computer doesn’t do this, then it gives the player who chooses the hidden word an enormous advantage. As an example, support that you’re trying to guess the word which currently looks like do-ble. There are only two words in the English dictionary that follow this pattern: doable and double. However, it doesn’t matter which one you choose here because the computer will just ‘pick’ the other option, since they are cheating and haven’t actually committed to any word. If this is your last allowed guess, you will certainly lose the game.

In this programming exercise, you will program an “Evil Hangman” computer whose goal is to try to make it as difficult as possible for the player to guess the word correctly through using this described cheating technique.

## Example

Suppose that we are playing hangman against the computer and there are only nine words in the entire English dictionary: ally, beta, cool, deal, else, flew, good, hope, and ibex. Suppose the we guess the letter e. The computer will then sort these words of the english dictionary into different classes dependeing on how many letters of e each of these words contain, and where they are located in the word:

• ----: ally, cool, and good
• -e--: beta and deal
• --e-: flew and ibex
• e--e: else
• ---e: hope

The computer wants to keep the class of words as big as possible, and the biggest class of words here is --- with three words in the class. Therefore, the computer will tell us that there was no e in our word, and our set of possible words has not been reduced down to ally, cool, and good. In order for the game to remain consistent, none of the other six words in the English dictionary can be the “correct” word in this game.

Now that you understand the idea behand the evil hangman, it’s your turn to implement it! Navigate to the repl linked here, which is also shown below, and follow the following steps.

### Problem 1

Read through the InnocentHangman, Hangman, and Main Java implementations to try and understand the structure of the program. How does everything work with each other? Assuming the boolean EVIL is set to false in Main, try running the program and playing a game of hangman using the “innocent” AI.

### Problem 2

Navigate to the EvilHangman source code and first read through what’s already given. First, implement the getArchetypes() and countCinS() helper functions.

• getArchetypes() takes in a character c as an input and attempts to construct the equivalence classes as shown in the Example section above. The idea is that it creates a Map that takes the “archetype” of a class of words as keys, and the Set of words with that archetype as a value. For example, going with our earlier example, ---- might be a key in the map and the set { alley, cool, good } would be the value in the map. -e-- would be another key in the map and the set { beta, deal } would be the associated value. The archetype keys should include not only the character c, but also all of the previous “correctly” guessed characters as well that are in those class of words.
• countCinS() takes in a character c and counts the number of unique times that this character appears in a String s.

### Problem 3

Implement the getAllWords(), getGuesses(), getKey(), getTurnsRemaining(), and isGameOver() methods. If your functions get too long or complex, you’re probably doing something wrong.

### Problem 4

Finally, implment the makeGuess() function. This is the crux of this programming project. This function takes in a character c that represents the character that the player guessed, and the computer will update the fields of the EvilHangman.java class appropriately. Here are some things that you should think about when writing this function.

1. Make sure to choose the archetype/key with the most number of words with that archetype. You can get all of the archetypes and associated sets of words through using the getArchetypes() function that you wrote in Problem 2 above.
2. If two archetypes have the same number of words that are associated with them, then pick the archetype that reveals the least number of new characters.
3. Make sure to increment the number of turns if the character guessed doesn’t show up in the archetype chosen. To check whether the character guessed is in the archetype chosen, check if your original key and updated key are the same.
4. Add the guessed character to the set of guesses. Simultaneously, make sure that the argument character is a valid guess, meaning that it hasn’t been guessed before and that it’s also a letter from a to z.
5. This function should return the updated archetype/key, or it should return null if the guess wasn’t a valid guess per step 4 above.

### Problem 5

After you’ve completed Problems 2 through 4, change EVIL to true in Main.java and try to run your code. You’ll probably have at least a couple of bugs to work through first, but once you iron those out, try playing with your “evil” hangman AI. Is it more difficult to win than when you were playing against the “innocent” AI?

Credits: This project description was largely cited from the online document linked here.


text:
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 21 - 30 / 4512345 21.Energy response of ISS-CREAM calorimeter with attenuation effectJ. P. Lundquist, H.G. Zhang, 2020, other component partsAbstract: The NASA mission, Cosmic Ray Energetic And Mass experiment for the International Space Station (ISS-CREAM) is to measure individual cosmic-ray particle energy spectra from protons to iron nuclei, with an energy range from ~1 TeV to the so-called "knee", near 1015eV. Energies of cosmic-ray particles are measured from electromagnetic showers induced by particles in the calorimeter. As a pioneer mission, the balloon-borne CREAM instrument has successfully flown seven times over the Antarctica for a cumulative exposure of 191 days. The CREAM calorimeter has shown sufficient capability to measure energies of cosmic-ray particles by capturing the electromagnetic shower profile within the interested energy range. The ISS-CREAM calorimeter is expected to have a similar performance and, before it was launched, an engineering-unit calorimeter was shipped to CERN for a full beam test. The full performance test includes position, energy, and angle scans of electron and pion beams together with a high voltage scan for calibration and characterization. In addition to the regular analysis for performance test, we also applied an additional step to generate the universal energy responses by correcting the attenuation effect in the calorimeter readout. The general energy responses could be obtained after shifting the incident beam positions to a reference position near the center of the calorimeter, which provided improved energy resolutions. The result of this analysis will be used to determine the incident energies of the cosmic-ray particles in the flight data.Found in: ključnih besedahSummary of found: ...cosmic rays, high-energy, particle physics, detectors...Keywords: cosmic rays, high-energy, particle physics, detectorsPublished: 29.04.2020; Views: 2149; Downloads: 126 Fulltext (2,54 MB) 22.The Cosmic Ray Energy Spectrum between 2 PeV and 2 EeV Observed with the TALE Detector in Monocular ModeJ. P. Lundquist, R.U. Abbasi, 2018, original scientific articleAbstract: We report on a measurement of the cosmic ray energy spectrum by the Telescope Array Low-Energy Extension (TALE) air fluorescence detector (FD). The TALE air FD is also sensitive to the Cherenkov light produced by shower particles. Low-energy cosmic rays, in the PeV energy range, are detectable by TALE as Cherenkov events. Using these events, we measure the energy spectrum from a low energy of ~2 PeV to an energy greater than 100 PeV. Above 100 PeV, TALE can detect cosmic rays using air fluorescence. This allows for the extension of the measurement to energies greater than a few EeV. In this paper, we describe the detector, explain the technique, and present results from a measurement of the spectrum using ~1000 hr of observation. The observed spectrum shows a clear steepening near 10^17.1 eV, along with an ankle-like structure at 10^16.2 eV. These features present important constraints on the origin of galactic cosmic rays and on propagation models. The feature at 10^17.1 eV may also mark the end of the galactic cosmic ray flux and the start of the transition to extragalactic sources.Found in: ključnih besedahSummary of found: ...astroparticle physics, cosmic rays, UHECR, energy spectrum...Keywords: astroparticle physics, cosmic rays, UHECR, energy spectrumPublished: 30.04.2020; Views: 1921; Downloads: 0 Fulltext (4,40 MB) 23.Study of muons from ultrahigh energy cosmic ray air showers measured with the Telescope Array experimentJ. P. Lundquist, R.U. Abbasi, 2018, original scientific articleAbstract: One of the uncertainties in the interpretation of ultrahigh energy cosmic ray data comes from the hadronic interaction models used for air shower Monte Carlo (MC) simulations. The number of muons observed at the ground from ultrahigh energy cosmic ray–induced air showers is expected to depend upon the hadronic interaction model. One may therefore test the hadronic interaction models by comparing the measured number of muons with the MC prediction. In this paper, we present the results of studies of muon densities in ultrahigh energy extensive air showers obtained by analyzing the signal of surface detector stations which should have high muon purity. The muon purity of a station will depend on both the inclination of the shower and the relative position of the station. In seven years’ data from the Telescope Array experiment, we find that the number of particles observed for signals with an expected muon purity of ∼65% at a lateral distance of 2000 m from the shower core is 1.72± 0.10(stat)±0.37(syst) times larger than the MC prediction value using the QGSJET II-03 model for proton-induced showers. A similar effect is also seen in comparisons with other hadronic models such as QGSJET II-04, which shows a 1.67±0.10±0.36 excess. We also studied the dependence of these excesses on lateral distances and found a slower decrease of the lateral distribution of muons in the data as compared to the MC, causing larger discrepancy at larger lateral distances.Found in: ključnih besedahKeywords: UHECR, cosmic rays, muons, particle physicsPublished: 30.04.2020; Views: 1781; Downloads: 0 Fulltext (2,16 MB) 24.Search for correlations between the arrival directions of IceCube neutrino events and ultrahigh-energy cosmic rays detected by the Pierre Auger Observatory and the Telescope ArrayJ. P. Lundquist, M. G. Aartsen, 2016, original scientific articleAbstract: This paper presents the results of different searches for correlations between very high-energy neutrino candidates detected by IceCube and the highest-energy cosmic rays measured by the Pierre Auger Observatory and the Telescope Array. We first consider samples of cascade neutrino events and of high-energy neutrino-induced muon tracks, which provided evidence for a neutrino flux of astrophysical origin, and study their cross-correlation with the ultrahigh-energy cosmic ray (UHECR) samples as a function of angular separation. We also study their possible directional correlations using a likelihood method stacking the neutrino arrival directions and adopting different assumptions on the size of the UHECR magnetic deflections. Finally, we perform another likelihood analysis stacking the UHECR directions and using a sample of through-going muon tracks optimized for neutrino point-source searches with sub-degree angular resolution. No indications of correlations at discovery level are obtained for any of the searches performed. The smallest of the p-values comes from the search for correlation between UHECRs with IceCube high-energy cascades, a result that should continue to be monitored.Found in: ključnih besedahSummary of found: ...UHECR, cosmic rays, neutrinos, multi-messenger physics, anisotropy...Keywords: UHECR, cosmic rays, neutrinos, multi-messenger physics, anisotropyPublished: 30.04.2020; Views: 1853; Downloads: 0 Fulltext (2,49 MB) 25.PHYSICS II - ExercisesJurij Urbančič, other educational materialAbstract: The following script is the first part of selected exercises and examples from the topics of Hydrostatics and Hydrodynamics. They were collected during the academic year 2019/20 when I was holding Physics II tutorials for the study program Physics and Astrophysics I. level within the School of Science at the University of Nova Gorica. The source material I used is given in the Bibliography at the end. Hopefully, these exercise fill help in providing a more in-depth understanding of physical phenomena discussed during the lectures.Found in: ključnih besedahKeywords: Hydrostatics, Hydrodynamics, Physics 2Published: 07.05.2020; Views: 2191; Downloads: 0 Fulltext (404,17 KB) 26.Correlations of the arrival directions of ultra-high energy cosmic ray with extragalactic objects as observed by the telescope array experimentJ. P. Lundquist, T. Abu-Zayyad, 2013, original scientific articleAbstract: We search for correlations between the positions of extragalactic objects and the arrival directions of ultra-high energy cosmic rays (UHECRs) with primary energy E ≥ 40 EeV as observed by the surface detector array of the Telescope Array (TA) experiment during the first 40 months of operation. We examine several public astronomical object catalogs, including the Veron-Cetty and Veron catalog of active galactic nuclei. We count the number of TA events correlated with objects in each catalog as a function of three parameters: the maximum angular separation between a TA event and an object, the minimum energy of the events, and the maximum redshift of the objects. We determine the combination of these parameters that maximizes the correlations, and we calculate the probability of having the same levels of correlations from an isotropic distribution of UHECR arrival directions. No statistically significant correlations are found when penalties for scanning over the above parameters and for searching in several catalogs are taken into account.Found in: ključnih besedahSummary of found: ...acceleration of particles, astroparticle physics, cosmic rays...Keywords: acceleration of particles, astroparticle physics, cosmic raysPublished: 19.05.2020; Views: 2083; Downloads: 112 Fulltext (779,07 KB) 27.Search for anisotropy of ultrahigh energy cosmic rays with the Telescope Array experimentJ. P. Lundquist, T. Abu-Zayyad, 2012, original scientific articleAbstract: We study the anisotropy of Ultra-High Energy Cosmic Ray (UHECR) events collected by the Telescope Array (TA) detector in the first 40 months of operation. Following earlier studies, we examine event sets with energy thresholds of 10 EeV, 40 EeV, and 57 EeV. We find that the distributions of the events in right ascension and declination are compatible with an isotropic distribution in all three sets. We then compare with previously reported clustering of the UHECR events at small angular scales. No significant clustering is found in the TA data. We then check the events with E > 57 EeV for correlations with nearby active galactic nuclei. No significant correlation is found. Finally, we examine all three sets for correlations with the large-scale structure (LSS) of the universe. We find that the two higher-energy sets are compatible with both an isotropic distribution and the hypothesis that UHECR sources follow the matter distribution of the universe (the LSS hypothesis), while the event set with E > 10 EeV is compatible with isotropy and is not compatible with the LSS hypothesis at 95% CL unless large deflection angles are also assumed. We show that accounting for UHECR deflections in a realistic model of the Galactic magnetic field can make this set compatible with the LSS hypothesis.Found in: ključnih besedahSummary of found: ...acceleration of particles, astroparticle physics, cosmic rays, magnetic fields, methods: statistical, relativistic...Keywords: acceleration of particles, astroparticle physics, cosmic rays, magnetic fields, methods: statistical, relativistic processesPublished: 19.05.2020; Views: 2078; Downloads: 0 Fulltext (1,04 MB) 28.Evidence for a Supergalactic Structure of Magnetic Deflection Multiplets of Ultra-High Energy Cosmic RaysJ. P. Lundquist, R. Abbasi, 2020, original scientific articleAbstract: Evidence for a large-scale supergalactic cosmic ray multiplet (arrival directions correlated with energy) structure is reported for ultra-high energy cosmic ray (UHECR) energies above 10^19 eV using seven years of data from the Telescope Array (TA) surface detector and updated to 10 years. Previous energy-position correlation studies have made assumptions regarding magnetic field shapes and strength, and UHECR composition. Here the assumption tested is that, since the supergalactic plane is a fit to the average matter density of the local Large Scale Structure (LSS), UHECR sources and intervening extragalactic magnetic fields are correlated with this plane. This supergalactic deflection hypothesis is tested by the entire field-of-view (FOV) behavior of the strength of intermediate-scale energy-angle correlations. These multiplets are measured in spherical cap section bins (wedges) of the FOV to account for coherent and random magnetic fields. The structure found is consistent with supergalactic deflection, the previously published energy spectrum anisotropy results of TA (the hotspot and coldspot), and toy-model simulations of a supergalactic magnetic sheet. The seven year data post-trial significance of this supergalactic structure of multiplets appearing by chance, on an isotropic sky, is found by Monte Carlo simulation to be 4.19σ. The ten years of data post-trial significance is 4.09σ. Furthermore, the starburst galaxy M82 is shown to be a possible source of the TA Hotspot, and an estimate of the supergalactic magnetic field using UHECR measurements is presented.Found in: ključnih besedahSummary of found: ...astroparticle physics, cosmic rays, UHECR, supergalactic plane, multiplets, magnetic...Keywords: astroparticle physics, cosmic rays, UHECR, supergalactic plane, multiplets, magnetic deflection, large-scale structure of universePublished: 19.05.2020; Views: 1922; Downloads: 0 Fulltext (4,40 MB) 29.REFLECTIONS OF A VANISHED TIMEGiustina Selvelli, original scientific articleAbstract: In this paper, I focus on the role played by material objects in the evocation of a specific imaginary of the recent past (mainly the 1970s and 1980s) in the literary works of Bulgarian writer Georgi Gospodinov (in the novels “The Physics of Sorrow” and “Natural Novel”) and of Turkish novelist Orhan Pamuk (in “The Museum of Innocence”). I analyze the presence of feelings of melancholy and loss accompanied by a fixation on objects in the fictional works and describe their possible overcoming through an externalization in the form of museum exhibitions in the city of Sofia and Istanbul such as the “Inventory depository of Socialism” and the “Museum of Innocence”. By viewing objects as actors capable of creating meaningful social networks, I consider their use in the narration of personal and collective histories and their transformation as powerful symbols of a bygone era.Found in: ključnih besedahSummary of found: ...writer Georgi Gospodinov (in the novels “The Physics of Sorrow” and “Natural Novel”) and of...Keywords: melancholy, objects, socialism, Georgi Gospodinov, Istanbul, Orhan Pamuk, Physics of Sorrow, The Museum of InnocencePublished: 19.06.2020; Views: 2001; Downloads: 0 Fulltext (341,59 KB) 30.On the GeV Emission of the Type I BdHN GRB 130427ALaura Beccera, She Sheng Xue, Yu Wang, Narek Sahakyan, Mile Karlica, Yen-Chen Chen, Simonetta Filippi, Christian Cherubini, Carlo Luciano Bianco, Jorge Armando Rueda, Rahim Moradi, Remo Ruffini, 2019, original scientific articleAbstract: We propose that the inner engine of a type I binary-driven hypernova (BdHN) is composed of Kerr black hole (BH) in a non-stationary state, embedded in a uniform magnetic field B_0 aligned with the BH rotation axis and surrounded by an ionized plasma of extremely low density of 10^−14 g cm−3. Using GRB 130427A as a prototype, we show that this inner engine acts in a sequence of elementary impulses. Electrons accelerate to ultrarelativistic energy near the BH horizon, propagating along the polar axis, θ = 0, where they can reach energies of ~10^18 eV, partially contributing to ultrahigh-energy cosmic rays. When propagating with $\theta \ne 0$ through the magnetic field B_0, they produce GeV and TeV radiation through synchroton emission. The mass of BH, M = 2.31M ⊙, its spin, α = 0.47, and the value of magnetic field B_0 = 3.48 × 10^10 G, are determined self consistently to fulfill the energetic and the transparency requirement. The repetition time of each elementary impulse of energy ${ \mathcal E }\sim {10}^{37}$ erg is ~10^−14 s at the beginning of the process, then slowly increases with time evolution. In principle, this "inner engine" can operate in a gamma-ray burst (GRB) for thousands of years. By scaling the BH mass and the magnetic field, the same inner engine can describe active galactic nuclei.Found in: ključnih besedahSummary of found: ...black hole physics, binaries, gamma-ray burst, neutron stars, supernovae, Astrophysics...Keywords: black hole physics, binaries, gamma-ray burst, neutron stars, supernovae, Astrophysics - High Energy Astrophysical PhenomenaPublished: 20.07.2020; Views: 1984; Downloads: 0 Fulltext (1,09 MB)
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# Translations:Angles/39/en

Large half angles are of the form ${\displaystyle 360 \times 2^n - \theta}$, with ${\displaystyle \theta \in [0,90]}$ and ${\displaystyle n \in \{0, ... ,14\}}$. They are all increasing until ${\displaystyle n \geq 8}$, at which point they may either be increasing or decreasing.


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# zbMATH — the first resource for mathematics

A study on quasi-duo rings. (English) Zbl 0938.16002
A ring $$R$$ (associative with identity) is called right (left) duo if every right (left) ideal of $$R$$ is two-sided. A ring $$R$$ is called right (left) quasi-duo if every right (left) maximal ideal of $$R$$ is two-sided. A ring $$R$$ is called weakly right (left) duo if for each $$a\in R$$ there exists a positive integer $$n$$, depending on $$a$$, such that $$a^nR$$ ($$Ra^n$$) is a two-sided ideal of $$R$$. As another generalization of commutative rings, there are 2-primal rings. A ring $$R$$ is called 2-primal if the prime radical of $$R$$ is equal to the set of all nilpotent elements in $$R$$. In this paper the authors present some examples to give negative answers to the following questions (they also give positive answers in very special cases): 1) Are right quasi-duo rings 2-primal? 2) Are formal power series rings over weakly right duo rings also weakly right duo? 3) Are 2-primal rings right quasi-duo?

##### MSC:
 16D25 Ideals in associative algebras 16N40 Nil and nilpotent radicals, sets, ideals, associative rings 16U80 Generalizations of commutativity (associative rings and algebras)


text:
Journal article Open Access

# Outlines of a probabilistic evaluation of possible SARS-CoV-2 origins

Demaneuf, Gilles; De Maistre, Rodolphe

### Citation Style Language JSON Export

{
"publisher": "Zenodo",
"DOI": "10.5281/zenodo.4067919",
"language": "eng",
"title": "Outlines of a probabilistic evaluation of possible SARS-CoV-2 origins",
"issued": {
"date-parts": [
[
2020,
9,
26
]
]
},
"abstract": "
We explore the relative probabilities of a lab-related accident against a non-lab-related zoonotic event being at the root of the current COVID-19 pandemic. We show that, based on present knowledge, the relative probability of a lab-related accident against a non-lab related zoonotic event is not negligible across a wide range of defensible input probabilities.
\n\n
For instance, under a reference set of input probabilities, the relative probabilities are at least 55% for a lab-related event against 45% at most for a non-lab-related zoonotic event. Even under a particularly conservative set of assumptions the relative probability of the lab-related accident is still 6% (to 94% for the non-lab related zoonotic event).
\n\n
Through a review of the Chinese specialised literature, we further show that our underlying estimate for the probability of lab-acquired infection is consistent with risk assessments from Chinese authorities and specialists. As part of this study, we list 112 individual BSL-3 labs in China, across 62 lab complexes.
\n\n
We then review a list of common probabilistic misunderstandings that are often associated with discussions about COVID-19 origins and conclude by discussing how such a probabilistic treatment can also offer a way to properly guide an investigation into the causes of the pandemic while being able to embrace different estimates of the underlying probabilities.
\n\n

\nNote: This paper is produced by the authors strictly in their personal capacity, and should not be constructed as representing in any way the opinion of any of the institutions the authors are employed or associated to.
",
"author": [
{
"family": "Demaneuf, Gilles"
},
{
"family": "De Maistre, Rodolphe"
}
],
"note": "This paper comes with a spreadsheet listing 112 individual BSL-3 labs in China across 62 lab-complexes.",
"version": "1.0.1",
"type": "article-journal",
"id": "4067919"
}
5,788
2,944
views


text:
# The value of x

• April 1st 2012, 03:10 AM
Mhmh96
The value of x
• April 1st 2012, 03:17 AM
Prove It
Re: The value of x
• April 1st 2012, 03:24 AM
princeps
Re: The value of x
Quote:

Originally Posted by Mhmh96

$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{1}{\sqrt n +\sqrt{n+1}}$
• April 1st 2012, 03:37 AM
Mhmh96
Re: The value of x
• April 1st 2012, 07:01 AM
Wilmer
Re: The value of x
FOR k = 0 to 2006 STEP 2
x = x + 1 / [SQRT(k) + SQRT(k+2)]
NEXT k

x = 22.4053565...

SURPRISE(?): x^2 = 502 (integer) (Wondering)
• April 1st 2012, 08:22 PM
browni3141
Re: The value of x
It appears that $\sum\frac{1}{\sqrt{2n}+\sqrt{2(n+1)}}$from n = 0 to k is $\sqrt{(k+1)/2}$
It is almost certain that $x = \sqrt{502}$
• April 2nd 2012, 08:16 AM
Wilmer
Re: The value of x
Well, for all the good this'll do:

1 / [SQRT(n) + SQRT(n + 2)] = SQRT{[n + 1 - SQRT(n^2 + 2n)] / 2}
• April 9th 2012, 06:43 PM
browni3141
Re: The value of x
I'm surprised no one was able to figure this out until now (including me). It's actually fairly simple.
$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{1}{\sqrt n +\sqrt{n+1}}$
$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \frac{\sqrt n -\sqrt{n+1}}{(\sqrt n -\sqrt{n+1})(\sqrt n +\sqrt{n+1})}$
$x=\frac{1}{\sqrt 2} \cdot \displaystyle \sum_{n=0}^{1003} \sqrt{n+1} -\sqrt n$
Now it is a telescoping series.
$x=\frac{1}{\sqrt 2}(\sqrt{1003+1} -\sqrt 0)$
$x= \sqrt{502}$


text:
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If tanA = 12/5 and pi lt A lt ...

# If tanA = 12/5 and pi lt A lt (3pi)/(2), find the values of the following:
 i) sin2A
 ii) cos2A
 iii) tan2A

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Updated On: 27-06-2022

Text Solution

Solution : i) 120/169, ii) -119/169, iii) -120/119

Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

Transcript

Hindi station we have this time this is given as 12 by 5 so we have to find the value of the sign of 2A and this is cause of 2A and this is 10 of today show in the first part we have this is sin 2A so as we know that the formula of this sign of to a so this is the formula to 10 of a / we have this is 1 + 10 square of a sore from where we can put the value of this destiny we have this is well by 5 so we can put this and again simplify this so this will be come to into we have this is 12 by 5 and this is divided by 1 + 12 by 5 souls care of this so from where we can simplify this again so therefore this value will be this is we can write 24 this is 5 and by simple

this we will get this S1 plus 144 and this is 25 so again by simplifying we will get this as this is we had 24 by 5 and this will be divided by this is 25 + 144 and this is divided by 25 Again by simplifying we will get this is 24 by 5 and thus we have 16925 so from where we can simplify that this is we had 24 by 5 and thus we have 169 and this is 25 so therefore this will be come we have this is this will be 5A and this is 24 and 25 so this will become 120 and this is divided by 169 so this is the value of this sign off to a now for the cause of two way so as we know that this course of two way we have so we can simplify this again this is I

have cost of two way as we know that formula of this course of to a so this is we have 1 - of this 10 square of a divide by this is oneplus this is 10 square of a sore from where we can simplify after putting the value of this 10 of a this is we had 12 by 5 so we can simplify the so this will become 1 - of this is 1285 scared of this and this will be divided by 1 + of this is 12 by 5 square of this again by simplifying we will get this as this value we have 1 - of this is 144 divided by 25 and this is we have oneplus 14425 so from where we can get this as this will be 25 - of this is 144 this will become 25 and there we have this is 25 + 144 this is divided

25 so this will be cancel out there and we have this is 25 - of this is 144 / this is 25 + 144 therefore this will be we can write this as - of 11 9 and this is we have 169 so this is the value of cos of to a now for the 10 of two way as we have this is the third part this is we have third part of this is turn off to a bi know that formula of this turn off to a n term of 10 so we can write this as this will be to 10 of a / we have this is one minus of 10 square of a sore from where we can write this is as we have just enough this is equal to 12 by 5 by putting this value we have the source to into 12 by 5 and this will be divided by 1 minus of this is we have

1285 scared of this so from here we can simplify that so this value will be this is we had 24 by 5 and this will be divided by this is we have 25 - of this is 144 / this is 25 so again by simplifying we will get this value as this value will be 24 this is divided by 5 and this is we have had this is -119 and we have this is 25 year so this value will be ki will become this is this will be five year and this is 24 and 25 so this will become 120 and this is we have -119 so we can do this so this is the answer of 10 off to a thank you