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# Seriation Classification: sc-1¶

The goal of experiment sc-1 is to validate that the Laplacian eigenvalue spectral distance can be useful in k-Nearest Neighbor classifiers for seriation output. In this experiment, I take a supervised learning approach, starting with two regional metapopulation models, simulating unbiased cultural transmission with 50 replicates across each model, sampling and time averaging the resulting cultural trait distributions in archaeologically realistic ways, and then seriating the results using our IDSS algorithm. Each seriation resulting from this procedure is thus "labeled" as to the regional metapopulation model from which it originated, so we can assess the accuracy of predicting that label based upon the graph spectral similarity.

In [2]:
import numpy as np
import networkx as nx
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns
%matplotlib inline
import cPickle as pickle
from copy import deepcopy
from sklearn.metrics import classification_report, accuracy_score, confusion_matrix

Couldn't import dot_parser, loading of dot files will not be possible.

In [3]:
train_graphs = pickle.load(open("train-freq-graphs.pkl",'r'))

sklearn-mmadsen is a python package of useful machine learning tools that I'm accumulating for research and commercial work. You can find it at http://github.com/mmadsen/sklearn-mmadsen.

In [4]:
import sklearn_mmadsen.graphs as skm

## Initial Classification Attempt¶

Let's just see if the graph spectral distance does anything useful at all, or whether I'm barking up the wrong tree. I imagine that we want a few neighbors (to rule out relying on a single neighbor which might be anomalous), but not too many. So let's start with k=5.

The approach here is to essentially do a "leave one out" strategy on the dataset. The KNN model isn't really "trained" in the usual sense of the term, so we don't need to separate a test and train set, we just need to make sure that the target graph we're trying to predict is not one of the "training" graphs that we calculate spectral distances to, otherwise the self-matching of the graph will always predict zero distance. So we first define a simple function which splits a graph out of the training set and returns the rest. I'd use scikit-learn functions for this, but our "data" is really a list of NetworkX objects, not a numeric matrix.

In [5]:
gclf = skm.GraphEigenvalueNearestNeighbors(n_neighbors=5)

In [6]:
def leave_one_out_cv(ix, train_graphs, train_labels):
"""
Simple LOO data sets for kNN classification, given an index, returns a train set, labels, with the left out
graph and label as test_graph, test_label
"""
test_graph = train_graphs[ix]
test_label = train_labels[ix]
train_loo_graphs = deepcopy(train_graphs)
train_loo_labels = deepcopy(train_labels)
del train_loo_graphs[ix]
del train_loo_labels[ix]
return (train_loo_graphs, train_loo_labels, test_graph, test_label)

In [7]:
test_pred = []
for ix in range(0, len(train_graphs)):
train_loo_graphs, train_loo_labels, test_graph, test_label = leave_one_out_cv(ix, train_graphs, train_labels)
gclf.fit(train_loo_graphs, train_loo_labels)
test_pred.append(gclf.predict([test_graph])[0])

In [8]:
cm = confusion_matrix(train_labels, test_pred)
cmdf = pd.DataFrame(cm)
cmdf.columns = map(lambda x: 'predicted {}'.format(x), cmdf.columns)
cmdf.index = map(lambda x: 'actual {}'.format(x), cmdf.index)

print cmdf
print(classification_report(train_labels, test_pred))
print("Accuracy on test: %0.3f" % accuracy_score(train_labels, test_pred))

          predicted 0  predicted 1
actual 0           41            9
actual 1           14           35
precision    recall  f1-score   support

0       0.75      0.82      0.78        50
1       0.80      0.71      0.75        49

avg / total       0.77      0.77      0.77        99

Accuracy on test: 0.768

In [9]:
sns.heatmap(cm.T, square=True, annot=True, fmt='d', cbar=False)

Out[9]:
<matplotlib.axes._subplots.AxesSubplot at 0x10642a350>

For a first try, this is pretty darned good, I think. Almost 77% of the time, we can correctly predict whether a seriation solution from one of two models belongs to the correct model. It would be nice to get that accuracy to near perfect if possible, howeve, because the goal here is to examine the fit between an empirical solution and a number of models, and the empirical solution will never have arisen from one of our pure theoretical models.

## Leave-One-Out Cross Validation for Selecting Optimal K¶

Before working on more complex approaches, let's simply make sure we're choosing the optimal number of neighbors for the k-Nearest Neighbors classifier.

In [10]:
knn = [1, 3, 5, 7, 9, 11, 15]
for nn in knn:
gclf = skm.GraphEigenvalueNearestNeighbors(n_neighbors=nn)
test_pred = []
for ix in range(0, len(train_graphs)):
train_loo_graphs, train_loo_labels, test_graph, test_label = leave_one_out_cv(ix, train_graphs, train_labels)
gclf.fit(train_loo_graphs, train_loo_labels)
test_pred.append(gclf.predict([test_graph])[0])
print("Accuracy on test for %s neighbors: %0.3f" % (nn, accuracy_score(train_labels, test_pred)))

Accuracy on test for 1 neighbors: 0.788
Accuracy on test for 3 neighbors: 0.798
Accuracy on test for 5 neighbors: 0.768
Accuracy on test for 7 neighbors: 0.747
Accuracy on test for 9 neighbors: 0.758
Accuracy on test for 11 neighbors: 0.788
Accuracy on test for 15 neighbors: 0.788

Clearly, 5 was suboptimal, and the best results (at least on these data) are coming from 3-NN classification. Almost 80%, in fact! This is great, but let's see what we can do with a more traditional feature engineering approach and a boosted classifier.

In [ ]:

In [ ]:


text:
# Mathematical principle for decomposing metric fluctuation

1. Feb 17, 2009

### aristurtle

What's the underlying principle to demonposite metrict fluctuation as scalar ,vector and tensor?
Is this decomposition complete?unique?
for scalar mode,
$$\delta g_{\mu \nu}=a^{2} \left( \begin{array}{cccc} 2\phi & -B,_{i} \\ -B,_{i} & 2(\psi \delta_{ij}-E,_{i,j}) \end{array}\right)$$
why should 00 term to be a "scalar"?but it is not a lorentz scalar?and why shoud 0i terms to look like a 3 vector? thank you.

2. Feb 18, 2009

### Chalnoth

The '0' index represents time. The 'i' index goes from 1-3 and represents the three spatial directions. Does that help?

3. Feb 18, 2009

### aristurtle

thank you! But I know that already ,of course.
00 term of metric fluctuation is a SO(3) scalar,but when a boost happens to change the whole metric tensor,00 term is not a scalar. when you infer a quantity or a set of quantitites as scalar , vector or tensor,you should specify a coordinate thansformation to make the definition valid.what kind of coordinate transformation here is used to define the "scalar","vector",and "tensor" fluctuations?

4. Feb 18, 2009

### Chalnoth

Well, these are just useful divisions for us to make sense of what's going on here.

In the stress-energy tensor, for instance, the (00) term is the energy density, the (0i) terms are the linear momentum density, and the (ii) terms are pressure and stress.

I don't think this is an actual coordinate transformation, just a grouping of components of the original tensor into components within a particular coordinate system that allow for easier interpretation of the results.

5. Feb 18, 2009

### cristo

Staff Emeritus
Quantities in the perturbed metric tensor are decomposed into scalar, vector and tensor quantities depending upon how they transform under a spatial coordinate transformation in the background spacetime. See, for example, Phys. Rev. D 22, 1882 - 1905 (1980). It is useful to decompose the perturbations in this way since, to linear order, the different types decouple.

6. Feb 19, 2009

### aristurtle

O I almost got there! Yes! it is SO(3) transformation! I was reading Brandenberger's lecture notes and run into the problem. I checked another textbook by Mukhanov and find out that it is SO(3), plus spatial translation!
Now the underlying principle is the cosmic principle to constrain metric to what it looks like according to the text book.The homegeneity requires the metric to be invarient under spatial translation, and isotropic requires metric to be ,wait, here is another problem.isotropic requires the metric function to be same after a rotation.but it is not! it just behaves like this: 0i terms behaves like SO(3) vector ,ij term SO(3) tensor.
yes you attacked the problem from another side of einstein equation ,the energy momentum tensor.I want to point out that energy momentum tensor is not necessarily stress energy tensor.Because only a sub set of observers in whose eyes the system is isotropic can call energy momentum tensor as stress energy tensor.In the eye of another observer who is moving with respect to the isotropic observer, the system is not isotropic and sterams of momentun (i,j terms) do not vanish.

7. Feb 19, 2009

### Chalnoth

Right, so the cosmological principle only applies for a specific subset of observers. So more accurately stated, it is the statement that there exist some observers for whom the universe appears homogeneous and isotropic. A general observer will certainly not see this.

format: plain


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Lemma 64.12.2. Let $S$ be a scheme. Let $j : R \to U \times _ S U$ be an étale equivalence relation. Let $X = U/R$ be the associated algebraic space (Spaces, Theorem 63.10.5). There is a canonical bijection

$R\text{-invariant locally closed subschemes }Z'\text{ of }U \leftrightarrow \text{locally closed subspaces }Z\text{ of }X$

Moreover, if $Z \to X$ is closed (resp. open) if and only if $Z' \to U$ is closed (resp. open).

Proof. Denote $\varphi : U \to X$ the canonical map. The bijection sends $Z \to X$ to $Z' = Z \times _ X U \to U$. It is immediate from the definition that $Z' \to U$ is an immersion, resp. closed immersion, resp. open immersion if $Z \to X$ is so. It is also clear that $Z'$ is $R$-invariant (see Groupoids, Definition 39.19.1).

Conversely, assume that $Z' \to U$ is an immersion which is $R$-invariant. Let $R'$ be the restriction of $R$ to $Z'$, see Groupoids, Definition 39.18.2. Since $R' = R \times _{s, U} Z' = Z' \times _{U, t} R$ in this case we see that $R'$ is an étale equivalence relation on $Z'$. By Spaces, Theorem 63.10.5 we see $Z = Z'/R'$ is an algebraic space. By construction we have $U \times _ X Z = Z'$, so $U \times _ X Z \to Z$ is an immersion. Note that the property “immersion” is preserved under base change and fppf local on the base (see Spaces, Section 63.4). Moreover, immersions are separated and locally quasi-finite (see Schemes, Lemma 26.23.8 and Morphisms, Lemma 29.20.16). Hence by More on Morphisms, Lemma 37.49.1 immersions satisfy descent for fppf covering. This means all the hypotheses of Spaces, Lemma 63.11.5 are satisfied for $Z \to X$, $\mathcal{P}=$“immersion”, and the étale surjective morphism $U \to X$. We conclude that $Z \to X$ is representable and an immersion, which is the definition of a subspace (see Spaces, Definition 63.12.1).

It is clear that these constructions are inverse to each other and we win. $\square$

There are also:

• 2 comment(s) on Section 64.12: Reduced spaces

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text:
# Tag Info

16

I assume that an edge $(u,v)$ means that $u$ has to be executed before $v$. If this is not the case, turn around all edges. I furthermore assume that you are less interested in paths (those are already given by the DAG) than in a good execution strategy given the dependencies. You can easily adapt the topological sort procedure: instead of appending, merge ...

9

In searching around for an online presence for one of the classics in this field (Coffman, Denning: Operating Systems Theory, Prentice Hall, 1983) I came upon what looks like a comprehensive textbook with a Google books preview Pinedo: Scheduling: Theory, Algorithms, and Systems, Springer 2008. The author's homepage also has pages devoted to each of his ...

9

Given a 3CNF with clauses $\phi_1,\ldots,\phi_k$ on variables $x_1,\ldots,x_n$. Suppose both $x_i$ and $\overline{x_i}$ appear in the formula for at most $k_i$ times respectively. We design a colored DAG $G$ whose vertices consists of three parts: "Assignment" vertices $v_i(j)$ and $\bar{v}_i(j)$, $1\leq i\leq n$, $1\leq j\leq k_i$. Color $v_i(j)$ with the ...

8

You need to determine at what time each job is completed. With a first-come-first-served scheduler, this is simple to calculate: each job starts as soon as the processor becomes free, and takes exactly its burst time to complete. You've already calculated the start and end times to calculate the wait times, so use that to obtain the turnaround time. For ...

8

As $m = k \times n$, we can look at this in terms of $k$ and $n$ instead of $n$ and $m$. Let's say $T_i$ is the time it takes the $i$-th processor to finish its work. As $n$ grows, the probability that $T_i$ = $5k$ (the processor was assigned only $T=5$ tasks) for some $i$ approaches $1$, so makespan being defined as $\mathrm{max}(T_i)$, $E[M]$ approaches $... 7 The graph of overlapping jobs is an interval graph. Interval graphs are perfect graphs. So what you are trying to do is find a maximum weight independent set (i.e., no two overlap) in a perfect graph. This can be solved in polynomial time. The algorithm is given in "Polynomial Algorithms for Perfect Graphs", by M. Grötschel, L. Lovász, and A. Schrijver. ... 7 There are two categories of solutions to this sort of problem that I'm aware of: biased lotteries and filtered/generated random sequences. First, let's dispense with easy but wrong solutions that keep no state. Any lottery-style solution that maintains no state will have the number of wins in a binomial distribution, which fails the "as many times" ... 6 Greedy algorithm can't help in that case. And it couldn't be compared with both fractional or 0-1 knapsack problems. The first could be resolved by greedy algorithm in O(n) and the second is NP. The problem you have could be brute-forced in O(2^n). But you could optimize it using dynamic programming. 1) Sort intervals by start time. 2) Initialize int[] ... 5 Update: For the new version where you try to minimize the makespan, your static schedule still has the optimal expected value. Let$M$be the random variable for the makespan. Let$F_i$be the time slave$i$is finished. We then have that$M = \max_i(X_i)$. Let$c_i$be the number of jobs allocated to slave$i$. Then we have that$X_i = \sum_{i=1}^{c_i} X =...

5

The question is proven NP-hard in "Minimizing Makespan in a Two-Machine Flow Shop with Delays and Unit-Time Operations is NP-Hard" by W. Yu, H. Hoogeveen, and J.K. Lenstra (2004). This is proven in section 9 of the paper: Theorem 24. The problem of minimizing the makespan on a single machine with two unit time operations per job with arbitrary ...

5

This looks like the so called master-slave scheduling model introduced by Sahni. In particular, your problem falls under the Single-Master Master-Slave Systems. You can distinguish several cases: 1) If you do not add any additional constraint on the order of job execution (as in your case), the problem is called unconstrained minimum finish time problem (...

5

I am slightly confused. If $k$ is a constant, then you can try all $O(n^k)$ possible tours. Hence the problem is in ${\sf P}$. However, if $k$ is part of the input, the decision problem is ${\sf NP}$-complete. This can be shown by reducing $\text{HAM-CIRCUIT}$ to the problem, by the following reduction. Assume that we want to determine if a $n$-vertex ...

5

One could implement this in O(nlogn) Steps: Sort the intervals based on end time define p(i) for each interval, giving the biggest end point which is smaller than the start point of i-th interval. Use binary search to obtain nlogn define d[i] = max(w(i) + d[p(i)], d[i-1]). initialize d[0] = 0 The result will be in d[n] n- the number of intervals. ...

5

This seems to be exercise 9 from Kleinberg's Algorithm Design, chapter six. I'll report how i tried to solve it. Let's define $OPT(i)$ as the maximum amount of work from day $1$ to day $i$, with $i \leq n$ For every $i$, there can be a reboot in any previous day $1\leq j < i$. If it was on day $j$, than you processed $\sum_{k=j+1}^i \min\left\{s_{k-j}, ... 4 Let$\{P_1,...,P_n\}$be the set of croissant's byers. Let${v_i}^k-1$be the amount spent by$P_i$on croissant up to the day$k$(it can be the number of times he buy croissant if they always spend the same money whatever the number of peoples present, which don't look clever enough for our croissants lover); the$-1$is for initialisation and to avoid ... 4 The problem minimizing maximal lateness is discussed in the algorithm book by Kleinberg and Tardos under greedy algorithms. The solution is like yours, ordering the tasks by their deadline. Indeed the length of the tasks is not relevant. The main ingredient of the proof is that any optimal solution$O$can be transformed into this greedy one$G$by swapping ... 4 Hint: Let$\sigma$be the optimal schedule. Suppose that jobs$k$and$k+1$are switched. How is the cost affected? What does the optimality of$\sigma$tell us about$w_k,w_{k+1},t_k,t_{k+1}$? 4 For the Round-Robin Scheduler the quantum time is to ensure that each process has a share to the CPU and we don't have starvation problems. It's known for being fair, such that each process shares the CPU equality. Quantum time is it's the amount of time spent on each process in the CPU until we context switch to the next process in the ready queue. For ... 4 This sounds very much like a Hamilton path problem. If you set the edge weights all to$1$and the deadline of the vertices all to$|V| - 1$, then each vertex can only be visited once, since there is not enough time to visit one twice before the deadline runs out. Finding a Hamilton path on the other hand is NP-hard. Thus, your problem is NP-hard as well. 3 Use a dynamic programming algorithm to compute an$n \times n$table$T$, where the entry$T(j,k)$answers the question: suppose you wish to schedule$j$out of the first$k$jobs. What is the earliest time you can complete processing these? How do we compute$T(j,k+1)$? Either the job$k+1$is included in the best set of$j$out of$k+1$jobs, so$$T(j,k+... 3 Here's a naive algorithm. It relies ultimately on brute force, but may perform okay sometimes. Each constraint$(\sigma_{m_i}, \sigma_{m_j}, \sigma_{m_k}) \in S \implies i < k \wedge \neg(i < j < k)$consists of two conjuncts; let's call them type-$A$,$i < k$, and type-$B$,$\neg(i < j < k)$. Each type-$B$constraint can be equivalently ... 3 This has always troubled me too. The problem is with the origins of the English word and the difference between ordinal numbers ("first", "second", "third", ...) and cardinal numbers (the numbers you use to count things). In schedulers we use the word priority in the sense of: the right to take precedence or to proceed before others[1] It came into ... 3 The shortest-task-first heuristic is a 1/2-approximation algorithm. For a tight example, image that a compatible set$\{T_i\mid 1\leq i\leq m\}$sorted by their finish time and the other shorter tasks$Y_i$,$1\leq i\leq m/2$, such that$Y_i$overlaps with$T_{2i-1}$and$T_{2i}$for each$i$. The shortest-first heuristic finds$m/2$tasks while the optimal ... 3 This is more a question of English than computer science. For almost any regular verb (and "schedule" is a perfect example), the [verb]er is the thing that does the [verb]ing. The scheduler is the software that schedules. 3 Since a customer must either be assigned all the requested hours or none at all, this problem is$NP$-complete. This can be seen by reduction from Exact Cover. A polynomial algorithm is unlikely to exist. Starting with biggest job first might be a reasonable heuristic for many cases, but there are no performance guarantees. 3 When an I/O is started, it will typically have an I/O request structure associated with it that includes items like the process ID that the I/O belonged to. When the I/O completes the device driver will typically fork into an OS level I/O subsystem in the OS which will queue the IO request packet notification to the process and move the process to a ... 3 The greedy scheduling with a first fit resource selection is actually not correct. Consider the case where the intervals are:${ (1, 3) , (2, 5) , (6, 7) , (4, 8) }$, for simplicity assume we only have 2 resources. The greedy algorithm will assign intervals 1, 2, 3 to resources 1, 2, 1 respectively and will fail to assign a resource for the last interval. ... 3 This scheduling problem is called non-preemptive job shop scheduling. The classic reference is Edward G Coffman, Jr, and Peter J Denning: Operating Systems Theory, Prentice-Hall, 1973. Job-shop scheduling is discussed in Section 3.7 and 3.8 (although they mainly focus on a restricted version of the problem called the flow shop problem where the$a_i$... 3 This solution has problems and will be deleted soon; see templatetypedef's comment. You can solve this in polynomial time using minimum-cost flow. In the following, all edges have unit capacity. Create a source vertex$s$, and a target vertex$t$. We will send$k$units of flow from$s$to$t$. Create$n+1$vertices$v_0, \dots, v_n$to represent the ... 3 The algorithms for$n$processors will do the job just fine for the 2-processor case. In this context,$n\$ does not refer to the number of tasks to be scheduled, which probably caused the confusion when you looked for algorithms. Then, there is a great website for searching for scheduling problems, which you can access at http://www-desir.lip6.fr/~durrc/...

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


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# Sine Double angle formula

## Formula

$\large \sin{2\theta} \,=\, 2\sin{\theta}\cos{\theta}$

### Other forms

Sine of double angle rule is also written in some standard forms.

Angle Double angle Formula
$x$ $2x$ $\sin{2x}$ $\,=\,$ $2\sin{x}\cos{x}$
$A$ $2A$ $\sin{2A}$ $\,=\,$ $2\sin{A}\cos{A}$
$\alpha$ $2\alpha$ $\sin{2\alpha}$ $\,=\,$ $2\sin{\alpha}\cos{\alpha}$

It expresses that the angle of the right angled triangle can be represented by any symbol but the double angle is twice to that symbol and the sine of double angle property is expanded in terms of sine and cosine of the angle in the same form.

### Proof

Take theta $(\theta)$ is an angle of a right angled triangle and its double angle is denoted by $2\theta$. Sine of double angle is written as $\sin{2\theta}$. The expansion of the sine of double angle can be derived in geometrical method.

01

### According to ΔICG

Consider $\Delta ICG$. The angle of this triangle is $2\theta$. So, sine of double angle is written as $\sin 2\theta$ and express it in terms of ratio of the lengths of the sides.

$\sin 2\theta \,=\, \dfrac{GI}{CG}$

The length of the side $\overline{GI}$ can be written as the sum of the length of the sides $\overline{GH}$ and $\overline{HI}$.

$\implies \sin 2\theta \,=\, \dfrac{GH+HI}{CG}$

$\implies \sin 2\theta \,=\, \dfrac{GH}{CG}+\dfrac{HI}{CG}$

The sides $\overline{HI}$ and $\overline{DF}$ are parallel lines and their lengths are also equal exactly. Hence, replace the length of the side $\overline{HI}$ by $\overline{DF}$.

$\implies \sin 2\theta \,=\, \dfrac{GH}{CG}+\dfrac{DF}{CG}$

02

### According to ΔHGF

Now, focus on right angled triangle $HGF$ to transform the length of the side $\overline{GH}$.

$\cos \theta \,=\, \dfrac{GH}{FG}$

$\implies GH = FG \cos \theta$

Now, the length of the side $\overline{GH}$ can be replaced by the product of the length of the side $\overline{FG}$ and cosine of angle theta.

$\therefore \,\,\, \sin 2 \theta = \dfrac{FG \cos \theta}{CG} + \dfrac{DF}{CG}$

03

### According to ΔDCF

Look at the $\Delta DCF$ to transform the length of the side $\overline{DF}$.

$\sin{\theta} \,=\, \dfrac{DF}{CF}$

$\implies DF = CF \sin{\theta}$

Now, express the length of the sides $\overline{DF}$ as the product of the length of the side $\overline{CF}$ and sine of angle theta.

$\therefore \,\,\, \sin 2 \theta = \dfrac{FG \cos \theta}{CG} + \dfrac{CF \sin{\theta}}{CG}$

04

### According to ΔFCG

$\implies \sin{2\theta} = \dfrac{FG}{CG} \times \cos{\theta} + \dfrac{CF}{CG} \times \sin{\theta}$

Now, it is time to replace the ratio of the lengths of the sides. The right angled triangle $FCG$ is the only triangle which is formed by the sides of $\overline{FG}$, $\overline{CF}$ and $\overline{CG}$.

$\sin{\theta} = \dfrac{FG}{CG}$

$\cos{\theta} = \dfrac{CF}{CG}$

Now, substitute these two ratios in terms of trigonometric functions in the equation to obtain the expansion of the sine of double angle.

$\implies \sin{2\theta}$ $\,=\,$ $\sin{\theta} \times \cos{\theta}$ $+$ $\cos{\theta} \times \sin{\theta}$

$\implies \sin{2\theta}$ $\,=\,$ $\sin{\theta} \cos{\theta}$ $+$ $\sin{\theta} \cos{\theta}$

$\therefore \,\,\,\,\,\, \sin{2\theta}$ $\,=\,$ $2\sin{\theta} \cos{\theta}$

It is an expansion of the sine of double angle in terms of sine and cosine of the angle. It is used as a trigonometric formula to expand sine of angle double as twice the product of sine and cosine of angle.

#### Verification

Take $\theta = 30^\circ$ and get values of both sides and compare the results of them.

Evaluate the left hand side expression for angle $30^\circ$.

$\sin{2(30^\circ)}$

$= \sin{60^\circ}$

$= \dfrac{\sqrt{3}}{2}$

Similarly, find the right hand side expression for the angle $30^\circ$.

$2\sin{(30^\circ)}\cos{(30^\circ)}$

$= 2 \times \dfrac{1}{2} \times \dfrac{\sqrt{3}}{2}$

$= \dfrac{2}{2} \times \dfrac{\sqrt{3}}{2}$

$\require{cancel} = \dfrac{\cancel{2}}{\cancel{2}} \times \dfrac{\sqrt{3}}{2}$

$= 1 \times \dfrac{\sqrt{3}}{2}$

$= \dfrac{\sqrt{3}}{2}$

Now, compare both results to verify the double angle formula of the sine function. They both are equal.

$\sin{2(30^\circ)}$ $\,=\,$ $2\sin{(30^\circ)}\cos{(30^\circ)}$ $\,=\,$ $\dfrac{\sqrt{3}}{2}$

The expansion of the sine of double angle is true for all the angles. Therefore, the expansion of the sine of double angle is also called as a trigonometric identity.

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# Prove that the inverse of an integer matrix is also an integer matrix

1. Jan 5, 2009

### bigevil

1. The problem statement, all variables and given/known data
A is an invertible integer matrix. Prove that if det A = 1 or det A = -1, then the inverse of A is also an integer matrix.

Also prove the reverse, if A-inverse is an integer matrix then its determinant is 1 or -1.

2. Relevant equations

I'm not too sure how to start here. My first instinct is to work backwards from some properties of determinants that I know of.

det A-inv = det A
det [A(A-inv)] = det(A)det(A-inv) = det(A) det(A) = det I = 1, so det A = 1 or -1.

Same can be proven for A-inv, det (A-inv) det (A-inv) = det I = 1, so det A-inv = 1 or -1.

3. The attempt at a solution

I'm not so sure how to approach this from here. Sorry if this is a stupid question, I am studying on my own, not in college and not many other resources to turn to.

2. Jan 5, 2009

### HallsofIvy

Staff Emeritus
One method of finding the inverse of a matrix is to set a-1ij[/b]= Aji/d where Aji is the determinant of the "minor" (matrix you get by removing jth row and ith column) and d is the determinant of A. If A is $\pm 1$, what does that tell you?

3. Jan 6, 2009

### bigevil

Halls, I'm not sure if I'm familiar with that equation. Do you mean the adjoint matrix of A, ie, A-1 = Adj (A) / det A? Also not very sure about your question, what do you mean by A is +1 or -1?, the determinant?

If det A = 1 or -1, then the adjoint = inverse.

I have thought of something, not sure if this is the track you're pushing me along:

If det A = 1 or -1, then

A det A = I = A A-1. Given A is an integer matrix, and already given that det A is an integer, then A-1 is also an integer matrix. Does that fit together right?

4. Jan 6, 2009

### Dick

Yes, Hall's is talking about Adj(A). It's the transposed matrix of cofactors. If A is integer, then Adj(A) is definitely integer. For the second part, det(I)=det(A)*det(A^(-1)). If A and A^(-1) are integer matrices, then what about det(A) and det(A^(-1))? det(I)=1. Hence?

5. Jan 7, 2009

### bigevil

Ok, Dick:

If A and A-inverse are integer matrices then det A and det (A-inverse) are integers because of the Laplace expansion. det (I) = 1, so det (A A-inv) =1

det (A-inv) det (A-inv) = 1, and since both are equal, det (A-inv) must be 1 or -1.

Is this the right reasoning? Thanx Dick.

6. Jan 7, 2009

Indeed, the question itself provides the answer. When you say, integer, it means if you have f as a function of integers, then it can be easily manipulated into Drichilet functions and/or Dirac-Delta Functions. And, from here, you can use a lot of identities or transforms to achieve what is desired. Another method is to use induction. Another is to make a 2x2 matrix, where a ij element of I. Then find its inverse. And then prove every element of A inverse, is also an integer. Quite tedious, but has some charming results in the end.

7. Jan 7, 2009

### Dick

No. Something seems to be evading you. There is NO reason why det(A-inv)*det(A-inv)=1. What you have is det(A)*det(A-inv)=1. det(A) and det(A-inv) are both integers. How many ways are there to multiply two integers and get 1?

8. Jan 7, 2009

### bigevil

Oh... sorry Dick I got it mixed up, was doing a question on determinant of transposes before this.

det A = det (A-inv) = 1 or
det A = det A(-inv) = -1, hence. Thanks Dick...

Madhawk, I'm not much of a mathematician...


text:
20 added 13 characters in body

now we see under what conditions,$(a+bI)^N=a-bI\pmod q$, we can conclude N dosent have $4k+1$ factors, suppose there is a factor $q=4k+1$ for $N$and$N=q*d$ , then $(a+bI)^N=((a+bI)^q)^d=(a-bI)^d=a^d+b^d(-I)^{4k+3}=a^d+b^d*I\pmod q$. so N dosent have $4k+1$ factors. and There are 3,or 5 or 7 etc.factors of form 4k+3 for N. now to the 3 factors. Suppose N exists. $(a+bI)^{q_1q_2q_3}=a-bI\pmod {q_1q_2q_3}$
$((a+bI)^{q_1})^{q_2q_3}=((a-bI)^{q_2q_3}=a-bI\pmod {q_1}$
$((a+bI)^{q_2})^{q_1q_3}=((a-bI)^{q_1q_3}=a-bI\pmod {q_2}$
$((a+bI)^{q_3})^{q_1q_2}=((a-bI)^{q_1q_2}=a-bI\pmod {q_3}$
as$((a-bI)^{(q_1+1)})^{(q_1-1)k_1}=(a^2+b^2)^{(q_1-1)k_1}=1\pmod {q_1}$
multiply (a-bI) to the both sides.
$(a-bI)((a-bI)^{(q_1+1)})^{(q_1-1)*k_1}=(a-bI)^{(q_1^2-1)*k_1+1}\pmod {q_1}$
$(a-bI)((a-bI)^{(q_2+1)})^{(q_2-1)*k_2}=(a-bI)^{(q_2^2-1)*k_2+1}\pmod {q_2}$
$(a-bI)((a-bI)^{(q_3+1)})^{(q_3-1)*k_3}=(a-bI)^{(q_3^2-1)*k_3+1}\pmod {q_3}$
so we have
$q_2q_3=(q_1^2-1)*k_1+1$ (1)
$q_1q_3=(q_2^2-1)*k_2+1$ (2)
$q_1q_2=(q_3^2-1)*k_3+1$ (3)
(1)/(2)
$\frac{q_2}{q_1}=\frac{(q_1^2-1)*k_1+1}{(q_2^2-1)*k_2+1}$
introduce variable $t$ and it must be a integer.
$(q_1^2-1)k_1+1=q_2t$ (4)
$(q_2^2-1)k_2+1=q_1t$ (5)
tanspose
$(q_1^2-1)k_1=q_2t-1$ (6)
$(q_2^2-1)k_2=q_1t-1$ (7)
(6)*(7)
$q_1q_2t^2-(q_1+q_2)t+1-(q_1^2-1)(q_2^2-1)k_1k_2=0$
$\Delta=(q_1+q_2)^2-4q_1q_2(1-(q_1^2-1)(q_2^2-1)k_1k_2)$
$=(q_1-q_2)^2+4q_1q_2(q_1^2-1)(q_2^2-1)k_1k_2$
$t=\frac{(q_1+q_2)\pm\sqrt{(q_1-q_2)^2+4q_1q_2(q_1^2-1)(q_2^2-1)k_1k_2}}{2q_1q_2}$
$(q_1+q_2)\pm\sqrt{\Delta}=(q_1+q_2)\pm(q_1-q_2)\ne 0\pmod {q_1q_2}$
so $t$ does not exist. therefore the Compsite Number for $N=4k+3$ dose not exist.

19 added 1 characters in body

now we see under what conditions,$(a+bI)^N=a-bI\pmod q$, we can conclude N dosent have $4k+1$ factors, suppose there is a factor $q=4k+1$ for $N$and$N=q*d$ , then $(a+bI)^N=((a+bI)^q)^d=(a-bI)^d=a^d+b^d(-I)^{4k+3}=a^d+b^d*I\pmod q$. so N dosent have $4k+1$ factors. and There are 3,or 5 or 7 etc.factors of form 4k+3 for N. now to the 3 factors. Suppose N exists. $(a+bI)^{q_1q_2q_3}=a-bI\pmod {q_1q_2q_3}$
$((a+bI)^{q_1})^{q_2q_3}=((a-bI)^{q_2q_3}=a-bI\pmod {q_1}$
$((a+bI)^{q_2})^{q_1q_3}=((a-bI)^{q_1q_3}=a-bI\pmod {q_2}$
$((a+bI)^{q_3})^{q_1q_2}=((a-bI)^{q_1q_2}=a-bI\pmod {q_3}$
as$((a-bI)^{(q_1+1)})^{(q_1-1)k_1}=(a^2+b^2)^{(q_1-1)k_1}=1\pmod {q_1}$
multiply (a-bI) to the both sides.
$(a-bI)((a-bI)^{(q_1+1)})^{(q_1-1)*k_1}=(a-bI)^{(q_1^2-1)*k_1+1}\pmod {q_1}$
$(a-bI)((a-bI)^{(q_2+1)})^{(q_2-1)*k_2}=(a-bI)^{(q_2^2-1)*k_2+1}\pmod {q_2}$
$(a-bI)((a-bI)^{(q_3+1)})^{(q_3-1)*k_3}=(a-bI)^{(q_3^2-1)*k_3+1}\pmod {q_3}$
so we have
$q_2q_3=(q_1^2-1)*k_1+1$ (1)
$q_1q_3=(q_2^2-1)*k_2+1$ (2)
$q_1q_2=(q_3^2-1)*k_3+1$ (3)
(1)/(2)
$\frac{q_2}{q_1}=\frac{(q_1^2-1)*k_1+1}{(q_2^2-1)*k_2+1}$
introduce variable $t$ and it must be a integer.
$(q_1^2-1)k_1+1=q_2t$ (4)
$(q_2^2-1)k_2+1=q_1t$ (5)
tanspose
$(q_1^2-1)k_1=q_2t-1$ (6)
$(q_2^2-1)k_2=q_1t-1$ (7)
(6)*(7)
$q_1q_2t^2-(q_1+q_2)t+1-(q_1^2-1)(q_2^2-1)k_1k_2=0$
$\Delta=(q_1+q_2)^2-4q_1q_2(1-(q_1^2-1)(q_2^2-1)k_1k_2$ \Delta=(q_1+q_2)^2-4q_1q_2(1-(q_1^2-1)(q_2^2-1)k_1k_2)=(q_1-q_2)^2+4q_1q_2(q_1^2-1)(q_2^2-1)k_1k_2t=\frac{(q_1+q_2)\pm\sqrt{(q_1-q_2)^2+4q_1q_2(q_1^2-1)(q_2^2-1)k_1k_2}}{2q_1q_2}(q_1+q_2)\pm\sqrt{\Delta}=(q_1+q_2)\pm(q_1-q_2)\ne 0\pmod {q_1q_2}$so$t$does not exist. therefore the Compsite Number dose not exist. 18 added 2 characters in body; deleted 12 characters in body now we see under what conditions,$(a+bI)^N=a-bI\pmod q$, we can conclude N dosent have$4k+1$factors, suppose there is a factor$q=4k+1$for$N$and$N=q*d$, then$(a+bI)^N=((a+bI)^q)^d=(a-bI)^d=a^d+b^d(-I)^{4k+3}=a^d+b^d*I\pmod q$. so N dosent have$4k+1$factors. and There are 3,or 5 or 7 etc.factors of form 4k+3 for N. now to the 3 factors. Suppose N exists.$(a+bI)^{q_1q_2q_3}=a-bI\pmod {q_1q_2q_3}((a+bI)^{q_1})^{q_2q_3}=((a-bI)^{q_2q_3}=a-bI\pmod {q_1}((a+bI)^{q_2})^{q_1q_3}=((a-bI)^{q_1q_3}=a-bI\pmod {q_2}((a+bI)^{q_3})^{q_1q_2}=((a-bI)^{q_1q_2}=a-bI\pmod {q_3}$as$((a-bI)^{(q_1+1)})^{(q_1-1)k_1}=(a^2+b^2)^{(q_1-1)k_1}=1\pmod {q_1}$multiply (a-bI) to the both sides.$(a-bI)((a-bI)^{(q_1+1)})^{(q_1-1)*k_1}=(a-bI)^{(q_1^2-1)*k_1+1}\pmod {q_1}(a-bI)((a-bI)^{(q_2+1)})^{(q_2-1)*k_2}=(a-bI)^{(q_2^2-1)*k_2+1}\pmod {q_2}(a-bI)((a-bI)^{(q_3+1)})^{(q_3-1)*k_3}=(a-bI)^{(q_3^2-1)*k_3+1}\pmod {q_3}$so we have$q_2q_3=(q_1^2-1)*k_1+1$(1)$q_1q_3=(q_2^2-1)*k_2+1$(2)$q_1q_2=(q_3^2-1)*k_3+1$(3) (1)/(2)$\frac{q_2}{q_1}=\frac{(q_1^2-1)*k_1+1}{(q_2^2-1)*k_2+1}$introduce variable$t$and it must be a integer.$q_1(q_1^2-1)k_1+1=q_1t$(q_1^2-1)k_1+1=q_2t$ (4)
$q_2(q_2^2-1)k_2+1=q_2t$ (q_2^2-1)k_2+1=q_1t$(5) tanspose$q_1(q_1^2-1)k_1=q_1t-1$(q_1^2-1)k_1=q_2t-1$ (6)
$q_2(q_2^2-1)k_2=q_2t-1$ (q_2^2-1)k_2=q_1t-1$(7) (6)*(7)$q_1q_2t^2-(q_1+q_2)t+1-(q_1^2-1)(q_2^2-1)k_1k_2$q_1q_2t^2-(q_1+q_2)t+1-(q_1^2-1)(q_2^2-1)k_1k_2=0$
$\Delta=(q_1+q_2)^2-4q_1q_2(1-(q_1^2-1)(q_2^2-1)k_1k_2$
$=(q_1-q_2)^2+4q_1q_2(q_1^2-1)(q_2^2-1)k_1k_2$
$t=\frac{(q_1+q_2)\pm\sqrt{(q_1-q_2)^2+4q_1q_2(q_1^2-1)(q_2^2-1)k_1k_2}}{2q_1q_2}$
$(q_1+q_2)\pm\sqrt{\Delta}=(q_1+q_2)\pm(q_1-q_2)\ne 0\pmod {q_1q_2}$
so $t$ does not exist. therefore the Compsite Number dose not exist.

17 added 113 characters in body; edited body
16 added 7 characters in body
15 deleted 4 characters in body; deleted 3 characters in body; added 1 characters in body


text:
# [dev-context] removing .pgf file with --purgeall

Hans Hagen pragma at wxs.nl
Wed Jun 10 11:27:52 CEST 2009

Mojca Miklavec wrote:
> On Tue, Jun 9, 2009 at 17:48, Hans Hagen wrote:
>> Mojca Miklavec wrote:
>>> Hello Hans,
>>>
>>> may I request removing \jobname.pgf when doing context/ctxtools
>>> --purgeall (in both mkii/mkiv)? That's a file that TikZ creates by
>>> default (similar to \jobname-mpgraph*, only serving a different
>>> purpose) and is only needed when a document is compiled on computer
>>> without tikz installed.
>> is this name configureable? can we turn it into something \jobname-temp.pgf
>> or so? (as we do with the mp temp files)
>> or maybe \jobname-pgfgraphic.pgf or so.
>
> I don't believe that it's configurable (though it could be possible to
> send a request to Till Tantau). The code is just:
>
> \def\pgfutil at aux@read at hook{
>   \pgfutil at IfFileExists{\jobname.pgf}{\input \jobname.pgf\relax}{}
>   \csname newwrite\endcsname\pgfutil at auxout
>   \csname openout\endcsname\pgfutil at auxout\jobname.pgf
> }
>
> Is the main concern being afraid of removing some file with the same ending?

indeed; it's why i have the -mpgraph

it would be handy if there was

\def\pfgfilename{\jobname.pgf}

or so, but for the moment i've just added the suffix to the remove list
and we leave it to the user not to use the pgf suffix otherwise

Hans

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tel: 038 477 53 69 | fax: 038 477 53 74 | www.pragma-ade.com
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text:
# Category Archives: Computer Vision

## Configure OpenCV-Python development environment in Windows

In older days, it is relative complex to configure a OpenCV development environment. You have to install an IDE, e.g. Mircosoft Visual Studio, and have to set many configurations of the project, such as the include folder, the header folder and where the IDE can find the third-party OpenCV library. But after Python becomes one of the main streams, things has changed. Python has proven to be a better and easier way to program and test computer vision programs, compared with C++. In this article, I will show you the simple steps that could help you to configure an OpenCV-Python ready environment in Windows.

1. First, you have to download and install Python 2.7 and Numpy. Install them to their default locations, in order to avoid any further issues. The reason to use Python 2.7 instead of Python 3.x is that Python 2.7 is best supported by most third-party libraries, while Python 3 is too new for all the scientific libraries to be fully supported.

2. Download the latest OpenCV release from Sourceforge and extract it into a system folder, e.g., C:\Program Files\opencv.

3. Goto $(opencv folder)\build\python\2.7\x86 or \x64 based on your machine type. Copy cv2.pyd to$(python folder)\lib\site-packages.

That’s all you have to do. Really easy and will only cost you 10 minutes. Now let’s open the Python IDLE and test if the environment is working:

import numpy
import cv2

print cv2.__version__

Run the code and if you got the version printed and no error message, then congratulations: you have succeed. But if you did get error message, then you should check the version of both the Python and the OpenCV is correct. If you have multiple versions of Python, then check if you have run the right version of IDLE.

Before we try more complicated vision algorithms, let’s simply connect a USB camera first and try to get the video stream from the camera and display it on the screen:

import cv2

cap = cv2.VideoCapture(0)

while(True):
cv2.imshow('camera', frameInput)

if cv2.waitKey(33) == 27:
break

cv2.destroyAllWindows()
cap.release()

What you should got is a pop-up window displaying the current image captured from the camera. You don’t have to understand this code at the moment, but the thing is once you got the live video stream from the USB camera, further steps can be done to process, modify and save it. More details and examples will be discussed later in my further blogs.

References
[1] OpenCV-Python Tutorials, Install OpenCV-Python in Windows, available at: http://opencv-python-tutroals.readthedocs.org/en/latest/py_tutorials/py_setup/py_setup_in_windows/py_setup_in_windows.html

## AR.Drone Position Servoing and Visual Tracking

A demonstration of my Master’s Thesis: Visual-Based Localization and Tracking of a UGV with a Quadcopter. In this project, a visual tracking framework is designed to track the UGV with an AR.Drone quadcopter from Parrot. The system utilizes a centralized control by a ground station which is running ROS and Ubuntu 12.04 LTS.

The first two experiments were taken with the support of a global vision system which was designed using a low cost web camera. While in the last experiment, the quadcopter simply uses IMU data for navigation. The image was captured from the bottom camera of the AR.Drone and processed with OpenCV. Four PID controllers were designed to control the motion of the quadcopter to make it hold at a position or track a trajectory.

The next step is to use such a robot system for factory and infrastructure inspection. But since I have to return my quadcopter to the department, it is really problematic for me to imply this idea. Hope I can find the chance to get another AR.Drone soon.

（国内用户请访问优酷：http://v.youku.com/v_show/id_XNzczOTg0MDY0.html


text:
Share

# A Die is Tossed Twice. a 'Success' is Getting an Even Number on a Toss. Find the Variance of Number of Successes. - CBSE (Science) Class 12 - Mathematics

ConceptBernoulli Trials and Binomial Distribution

#### Question

A die is tossed twice. A 'success' is getting an even number on a toss. Find the variance of number of successes.

#### Solution

$\text{ We have,}$
$p =\text{ probability of getting an even number on a toss } = \frac{3}{6} = \frac{1}{2} and$
$q = 1 - p = 1 - \frac{1}{2} = \frac{1}{2}$
$\text{ Let X denote a success of getting an even number on a toss . Then, }$
$\text{ X follows binomial distribution with parameters n = 2 and } p = \frac{1}{2}$
$\therefore Var\left( X \right) = \text{ npq } = 2 \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{2}$

Is there an error in this question or solution?

#### Video TutorialsVIEW ALL [1]

Solution A Die is Tossed Twice. a 'Success' is Getting an Even Number on a Toss. Find the Variance of Number of Successes. Concept: Bernoulli Trials and Binomial Distribution.
S


text:
# A command for table entry by columns

This question is in followup to this question; an excellent response there has allowed me to very easily make tables for examples in the grammar book I am working on which have one language on one row, and the corresponding translation on the following row. For example, the following example is produced by the code underneath it:

\begin{ExampleRow}
\dakota{čhaŋté + oyúze,   čhaŋtóyuze,    čhaŋté + ohnáka,  čhaŋtóhnaka}
\english{heart + condition of,  disposition,    heart + to place inside,    to cherish}
\gap
\dakota{ȟe + o + púza, ȟópuza,    napé + ožáža  ,napóžaža}
\english{mountain+ place + dry, desert, hand + washing place,   sink}
\end{ExampleRow}

Such a method of quickly producing tables which are uniform in style has been very helpful; and I have been trying to take what I have learned since asking the last question and apply it to make a similar command for producing tables which function similarly (i.e. you need not specify the number of rows or columns beforehand) but instead are transposed, with words of the same language / font appearing in columns instead of rows.

More specifically, I am looking to create a command ExampleCol which, when given the following input,

\begin{ExampleCol}
\dakota{čhaŋté + oyúze,   čhaŋtóyuze,    čhaŋté + ohnáka,  čhaŋtóhnaka}
\english{heart + condition of,  disposition,    heart + to place inside,    to cherish}
\end{ExampleCol}

Would produce the following table:

I'll try my best to summarize what I have so far: here is a minimal working example for the first table (which inputs by rows)

\documentclass{memoir}
\usepackage{fontspec}
\usepackage{booktabs}
\usepackage{etoolbox}

\newfontfamily\dak{Linux Biolinum O}
\newfontfamily\eng{JosefinSans-SemiBold}

\renewcommand{\arraystretch}{1.2}
\newcommand{\gap}{\\[-1.2ex]}

\newcommand{\doenglish}[1]{\appto\temp{&\eng{#1}}}
\newcommand{\dodakota}[1]{\appto\temp{&\dak{#1}}}

\newcommand{\english}[1]{%
\def\temp{}% initialize to empty
\appto\temp{\\}% end the row
\temp % deliver contents
}

\newcommand{\dakota}[1]{%
\def\temp{}% initialize to empty
\appto\temp{\\}% end the row
\temp % deliver contents
}

\newenvironment{ExampleRow}
{\begin{center}
\begin{tabular}{ c@{} *{\maxcolumns}{c} } \toprule
{\bottomrule\end{tabular}
\end{center}}

\def\maxcolumns{20}

\begin{document}
\begin{ExampleRow}
\dakota{čhaŋté + oyúze,   čhaŋtóyuze,    čhaŋté + ohnáka,  čhaŋtóhnaka}
\english{heart + condition of,  disposition,    heart + to place inside,    to cherish}
\gap
\dakota{ȟe + o + púza, ȟópuza,    napé + ožáža  ,napóžaža}
\english{mountain+ place + dry, desert, hand + washing place,   sink}
\end{ExampleRow}
\end{document}

After looking through previous questions on this site, I tried to modify one of the solutions given in this question to suit my needs. Here's my attempt (which is add-able to the minimal example file given above to run)

\newcount\rowc
\makeatletter

\def\ExampleCol{
\centering
\hbox\bgroup
\let\\\cr
\def\ruleb{
\ifnum\rowc=1\hrule height 1pt \else
\fi}
\valign\bgroup
\global\rowc\@ne
\hbox to 6em{\strut \hfill##\hfill}%
\ruleb
&&%
\hbox to 6em{\strut\hfill##\hfill}%
\ruleb
\cr}
\def\endExampleCol{%
\crcr\egroup\egroup}

Using this, the following code produces the accompanying picture:

\begin{ExampleCol}
\dakota{wáŋča, núŋpa, yámni, tópa, záptaŋ,sákpe}
\english{one, two, three, four, five, six}
\end{ExampleCol}

Which is looking great (with the exception that I would like a horizontal bar at the bottom of the table as well but don't want to specify its height beforehand, and the top bar is a bit too close to the text)

However, this solution doesn't work for me as the cells seem to be of a fixed size, if I try and enter longer words in the table neighboring cells overlap. For example, the following code

\begin{ExampleCol}
\dakota{čhaŋté + oyúze,   čhaŋtóyuze,    čhaŋté + ohnáka,  čhaŋtóhnaka}
\english{heart + condition of,  disposition,    heart + to place inside,    to cherish}
\end{ExampleCol}

produces this table:

Any suggestions on a good way to make an environment for entering tables in columns like this that doesn't have the same problems would be greatly appreciated!

• I think, it would be better to store the English/Dakota word pairs together instead of using two lists. Splitting can be done later on. – user31729 Jul 10 '15 at 12:30
• Is there any feedback on this? – user31729 Mar 30 '18 at 14:54

This is perhaps not the best way to achieve this, but the traversing of a list downwards within a table is tedious, since you have to store which row is currently processed, due to cell grouping.

This is perhaps a very ineffective solution -- it processes the list every time a row counter is incremented.

\documentclass{article}

\usepackage{etoolbox}

\newcounter{rowcounter}
\newcounter{temprowcounter}

\newcommand{\processlist}[2]{%
\typeout{#1}%
\stepcounter{temprowcounter}%
\ifnumless{\value{temprowcounter}}{#1}{%
}{%
\xdef\@@@@@localtemp{#2}%
\listbreak%
}%
}

\newcommand{\showrows}[2]{%
#1 & #2 \tabularnewline
}

\newcommand{\processrows}[2]{%
\stepcounter{rowcounter}%
\setcounter{temprowcounter}{0}%
\forlistloop{\processlist{\number\value{rowcounter}}}{#1}%
\showrows{#2}{\@@@@@localtemp}%
}
\newcommand{\followrows}[2]{%
\setcounter{rowcounter}{0}%
\forlistloop{\processrows{#2}}{#1}%
}

\begin{document}

\begin{tabular}{lr}
\followrows{\Englishlist}{\Dakotalist}
\tabularnewline
\end{tabular}

And now reversed...

\begin{tabular}{lr}
\tabularnewline
\followrows{\Dakotalist}{\Englishlist}
\end{tabular}

\end{document}


text:
# That’s nitpicking, isn’t it?

I keep Monty Python and the Holy Grail and This is Spın̈al Tap1 on my iPhone at all times. They’re among my favorite movies and bear up under repeated (and repeated and repeated) viewings, not just because they’re funny, but because there’s a depth to them. The Pythons knew a lot about Arthur and medieval England, and the Tap crew knew a lot about rock and roll.

So it was with some trepidation that I started listening to The Incomparable podcast’s recent episode on Spın̈al Tap, called, with a certain inevitability, “These Go to Eleven.” Jason Snell hosted the show and the guests were Andy Ihnatko and Ben Boychuk. I’ve enjoyed listening to all of them on The Incomparable,2 and I’ve always liked Jason and Andy’s writing,3 but I don’t think of any of them as being particularly steeped in rock and roll culture. Andy’s always recommending musicals on AMC, not usually the sign of a rocker. And then there’s the matter of age—Spın̈al Tap came out in 1983 and is parodying people and events that took place in the 60s and, especially, the 70s. A lot of the humor comes from understanding what was going on back them, and I didn’t think any of the three were old enough to have that understanding.

Despite my doubts, the show was a lot of fun. Jason, Andy, and Ben obviously have a deep affection for the movie, and they had good insights into what makes it work. Jason in particular, I thought, was spot on in his discussion of the “these go to eleven” scene, which works so well, both because of the absurdity of the markings on the amps (a brilliant idea) and because Rob Reiner and Christopher Guest play it absolutely straight. It’s Marty’s sincerity and Nigel’s obtuseness that turn a good joke into one that’s lasted 30 years.

Still, some topics were missed that I think are essential to any discussion of Spın̈al Tap.

First on my list is All You Need is Cash, Eric Idle’s wonderfully askew retelling of the Beatles’ story through a retrospective documentary about the fictional Rutles, the band whose legend will last a lunchtime. I saw Cash only once, when it first ran on NBC in the late 70s, but there are bits of it that I simply cannot forget. And the music! Tap’s music was, as the Incomparable guys said, both catchy and funny, but Neil Innes’s songs for the Rutles were absolutely uncanny in how they evoked Lennon and McCartney with only a little outright theft.

The big difference between the two films is that while Cash focuses exclusively on the Beatles, Tap wanders all over the rock scene for stories to satirize. There are certainly Beatles references—like their outfits in the “Gimme Some Money” clip and the song “All The Way Home,” a pretty obvious nod to “The One After 909”—but they mine a much bigger and richer vein:

• That Spın̈al Tap was “England’s loudest band” was taken directly from Deep Purple, who were designated the world’s loudest band by the Guinness Book of World Records.
• The story of their second drummer’s death—“you can’t dust for vomit”—was just a slight adjustment to the cause of death for both Jimi Hendrix and John Bonham. (I was under the impression Bon Scott had also choked on his own vomit, but apparently he died of alcohol poisoning. And let me warn you, you can get lost for hours on this Wikipedia page on rock and roll deaths.)
• The scene in which Nigel and David yell at each other during a recording session recalls the legendary fuck-filled argument among the Troggs when one of the guys couldn’t play his part.4

• Many bands played at festivals on the Isle of Wight. There’s a great Who concert film of its appearance there in 1970. Spın̈al Tap played a jazz-blues fest (or was it a blues-jazz fest?) on the Isle of Lucy.

• Nigel Tufnel is, of course, the spitting image of Jeff Beck, but his performances are more reminiscent of Beck’s Yardbirds bandmate, Jimmy Page. There’s a great scene in The Song Remains the Same in which Page plays his guitar with a violin bow.

Not to be outdone, Nigel plays his guitar with a violin.

• I’ve never been able to make a rational connection, but there’s something about the interviews with the pipe-smoking Derek Smalls that reminds me of this scene in The Kids Are Alright in which the country gentleman John Entwistle goes out for a little shooting practice in his wellies.

I could draw more parallels (Duke Fame = Marc Bolan), but you get the idea. There’s a lot more to Spın̈al Tap than foil-wrapped cucumbers.

1. The character n̈, called the n-diaeresis or n-umlaut, doesn’t have it’s own Unicode code point (although it does have its own Wikipedia page, in which Spın̈al Tap features prominently). The way to get it is through the combining character for the umlaut itself, U+0308. Since 308 in hex translates to 776 in decimal, we can generate the combining umlaut with ̈. Thus, n̈ = n̈. This works in both HTML and Markdown. (I’m using a dotless i, ı, just before the n̈ because it keeps the middle of the word from being overcrowded with dots and because that’s what the band’s logo uses.)

2. We’ll set aside Andy’s poorly formed opinions on The Avengers and how they derailed that particular show

3. I haven’t read Ben’s work, which seems to be largely political. Given that he’s affiliated with the Heartland Institute, I suspect his writings would infuriate me, but that shouldn’t affect his views on pop culture.

4. How wonderful is the internet? When I was young, I’d read about this Troggs tape but never got a chance to hear it. Now I can type “Troggs fuck yelling” into my browser and Google’s first hit led me to this clip. Quite exciting, this computer magic!


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# Number of divisors of $20^{20}$ with exactly $20$ divisors [closed]

How many positive integers $$x$$ with $$x\mid 20^{20}$$ have exactly $$20$$ divisors ?

## closed as off-topic by Gregory J. Puleo, Eevee Trainer, Alexander Gruber♦Apr 29 at 23:47

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Gregory J. Puleo, Eevee Trainer, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.

• Hint : The number is equal to the number of the divisors of $20$ (Try to find out why!) – Peter Apr 13 at 11:48
• You might say that's a "perfect" hint. – Oscar Lanzi Apr 13 at 11:52

Since $$20^{20}=2^{40}5^{20}$$, the general form of such a factor is $$2^a5^b$$ for non-negative integers $$a,\,b$$ with $$(a+1)(b+1)=20$$. Note in particular the ordered pair $$(a,\,b)$$ is what matters, not the unordered pair $$\{a,\,b\}$$. There are exactly as many of these as there are factors of $$20=2^25$$, i.e. $$3\times2=6$$.


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Address 2761 Wawiorka Rd, Krakow, WI 54137 (920) 899-3493

# error propagation through ln Lena, Wisconsin

Not the answer you're looking for? The uncertainty u can be expressed in a number of ways. With only 1 variable this is not even a bad idea, but you get troubles when you have a function f(x,y,...) of more input, which is why the method presented in This is $Revision: 1.18$, $Date: 2011/09/10 18:34:46$ (year/month/day) UTC.

Determinate errors have determinable sign and constant size. Is there any alternative to sed -i command in Solaris? When propagating error through an operation, the maximum error in a result is found by determining how much change occurs in the result when the maximum errors in the data combine Rule 2 If: or: then: In this case also the errors are combined in quadrature, but this time it is the fractional errors, i.e.

Propagation of Error http://webche.ent.ohiou.edu/che408/S...lculations.ppt (accessed Nov 20, 2009). Note that even though the errors on x may be uncorrelated, the errors on f are in general correlated; in other words, even if Σ x {\displaystyle \mathrm {\Sigma ^ Ïƒ p.2. Retrieved 2013-01-18. ^ a b Harris, Daniel C. (2003), Quantitative chemical analysis (6th ed.), Macmillan, p.56, ISBN0-7167-4464-3 ^ "Error Propagation tutorial" (PDF).

with ΔR, Δx, Δy, etc. Also averaging df = (df_up + df_down)/2 could come to your mind. First, the measurement errors may be correlated. doi:10.6028/jres.070c.025.

Now make all negative terms positive, and the resulting equuation is the correct indeterminate error equation. Now a repeated run of the cart would be expected to give a result between 36.1 and 39.7 cm/s. the error in the quantity divided by the value of the quantity, that are combined. See Ku (1966) for guidance on what constitutes sufficient data2.

Since $$\frac{\text{d}\ln(x)}{\text{d}x} = \frac{1}{x}$$ the error would be $$\Delta \ln(x) \approx \frac{\Delta x}{x}$$ For arbitraty logarithms we can use the change of the logarithm base:  \log_b The final result for velocity would be v = 37.9 + 1.7 cm/s. Retrieved 22 April 2016. ^ a b Goodman, Leo (1960). "On the Exact Variance of Products". For example, the bias on the error calculated for logx increases as x increases, since the expansion to 1+x is a good approximation only when x is small.

Calculate (1.23 ± 0.03) × . When the variables are the values of experimental measurements they have uncertainties due to measurement limitations (e.g., instrument precision) which propagate to the combination of variables in the function. Guidance on when this is acceptable practice is given below: If the measurements of a and b are independent, the associated covariance term is zero. In a probabilistic approach, the function f must usually be linearized by approximation to a first-order Taylor series expansion, though in some cases, exact formulas can be derived that do not

A piece of music that is almost identical to another is called? Note: Where Δt appears, it must be expressed in radians. This mathematical procedure, also used in Pythagoras' theorem about right triangles, is called quadrature. Pearson: Boston, 2011,2004,2000.

If you know that there is some specific probability of $x$ being in the interval $[x-\Delta x,x+\Delta x]$, then obviously $y$ will be in $[y_-,y_+]$ with that same probability. In such cases there are often established methods to deal with specific situations, but you should watch your step and consult your resident statistician when in doubt. f = ∑ i n a i x i : f = a x {\displaystyle f=\sum _ Ïƒ 4^ Ïƒ 3a_ Ïƒ 2x_ Ïƒ 1:f=\mathrm Ïƒ 0 \,} σ f 2 The determinate error equations may be found by differentiating R, then replading dR, dx, dy, etc.

Example: An angle is measured to be 30Â°: Â±0.5Â°. University of California. However, if the variables are correlated rather than independent, the cross term may not cancel out. Students who are taking calculus will notice that these rules are entirely unnecessary.

Claudia Neuhauser. Please try the request again. The exact covariance of two ratios with a pair of different poles p 1 {\displaystyle p_{1}} and p 2 {\displaystyle p_{2}} is similarly available.[10] The case of the inverse of a Let's say we measure the radius of a very small object.

A. (1973). The end result desired is $$x$$, so that $$x$$ is dependent on a, b, and c. The extent of this bias depends on the nature of the function. more hot questions question feed about us tour help blog chat data legal privacy policy work here advertising info mobile contact us feedback Technology Life / Arts Culture / Recreation Science

The sine of 30Â° is 0.5; the sine of 30.5Â° is 0.508; the sine of 29.5Â° is 0.492. Since at least two of the variables have an uncertainty based on the equipment used, a propagation of error formula must be applied to measure a more exact uncertainty of the Sometimes, these terms are omitted from the formula. Journal of Sound and Vibrations. 332 (11).

We know the value of uncertainty for∆r/r to be 5%, or 0.05. Authority control GND: 4479158-6 Retrieved from "https://en.wikipedia.org/w/index.php?title=Propagation_of_uncertainty&oldid=742325047" Categories: Algebra of random variablesNumerical analysisStatistical approximationsUncertainty of numbersStatistical deviation and dispersionHidden categories: Wikipedia articles needing page number citations from October 2012Wikipedia articles needing The results of each instrument are given as: a, b, c, d... (For simplification purposes, only the variables a, b, and c will be used throughout this derivation). R., 1997: An Introduction to Error Analysis: The Study of Uncertainties in Physical Measurements. 2nd ed.

Uncertainty in logarithms to other bases (such as common logs logarithms to base 10, written as log10 or simply log) is this absolute uncertainty adjusted by a factor (divided by 2.3 Would you feel Centrifugal Force without Friction? Is there any job that can't be automated? Error Propagation in Trig Functions Rules have been given for addition, subtraction, multiplication, and division.

The derivative, dv/dt = -x/t2. In both cases, the variance is a simple function of the mean.[9] Therefore, the variance has to be considered in a principal value sense if p − μ {\displaystyle p-\mu } Generated Fri, 14 Oct 2016 15:16:55 GMT by s_wx1131 (squid/3.5.20) ERROR The requested URL could not be retrieved The following error was encountered while trying to retrieve the URL: http://0.0.0.10/ Connection It is important to note that this formula is based on the linear characteristics of the gradient of f {\displaystyle f} and therefore it is a good estimation for the standard


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• ### A lattice calculation of the hadronic vacuum polarization contribution to $(g-2)_\mu$(1710.10072)

Oct. 27, 2017 hep-ph, hep-lat
We present results of calculations of the hadronic vacuum polarisation contribution to the muon anomalous magnetic moment. Specifically, we focus on controlling the infrared regime of the vacuum polarisation function. Our results are corrected for finite-size effects by combining the Gounaris-Sakurai parameterisation of the timelike pion form factor with the L\"uscher formalism. The impact of quark-disconnected diagrams and the precision of the scale determination is discussed and included in our final result in two-flavour QCD, which carries an overall uncertainty of 6%. We present preliminary results computed on ensembles with $N_f=2+1$ dynamical flavours and discuss how the long-distance contribution can be accurately constrained by a dedicated spectrum calculation in the iso-vector channel.
• ### The hadronic vacuum polarization contribution to the muon $g-2$ from lattice QCD(1705.01775)

Oct. 10, 2017 hep-ph, hep-lat
We present a calculation of the hadronic vacuum polarization contribution to the muon anomalous magnetic moment, $a_\mu^{\mathrm hvp}$, in lattice QCD employing dynamical up and down quarks. We focus on controlling the infrared regime of the vacuum polarization function. To this end we employ several complementary approaches, including Pad\'e fits, time moments and the time-momentum representation. We correct our results for finite-volume effects by combining the Gounaris-Sakurai parameterization of the timelike pion form factor with the L\"uscher formalism. On a subset of our ensembles we have derived an upper bound on the magnitude of quark-disconnected diagrams and found that they decrease the estimate for $a_\mu^{\mathrm hvp}$ by at most 2%. Our final result is $a_\mu^{\mathrm hvp}=(654\pm32\,{}^{+21}_{-23})\cdot 10^{-10}$, where the first error is statistical, and the second denotes the combined systematic uncertainty. Based on our findings we discuss the prospects for determining $a_\mu^{\mathrm hvp}$ with sub-percent precision.
• ### Anomalous magnetic moment of the muon, a hybrid approach(1707.07715)

Sept. 26, 2017 hep-ph, hep-lat
A new QCD sum rule determination of the leading order hadronic vacuum polarization contribution to the anomalous magnetic moment of the muon, $a_{\mu}^{\rm hvp}$, is proposed. This approach combines data on $e^{+}e^{-}$ annihilation into hadrons, perturbative QCD and lattice QCD results for the first derivative of the electromagnetic current correlator at zero momentum transfer, $\Pi_{\rm EM}^\prime(0)$. The idea is based on the observation that, in the relevant kinematic domain, the integration kernel $K(s)$, entering the formula relating $a_{\mu}^{\rm hvp}$ to $e^{+}e^{-}$ annihilation data, behaves like $1/s$ times a very smooth function of $s$, the squared energy. We find an expression for $a_{\mu}$ in terms of $\Pi_{\rm EM}^\prime(0)$, which can be calculated in lattice QCD. Using recent lattice results we find a good approximation for $a_{\mu}^{\rm hvp}$, but the precision is not yet sufficient to resolve the discrepancy between the $R(s)$ data-based results and the experimentally measured value.


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Why is the function must be normalize ? what is the meaning of

1. Jun 9, 2012

sciboudy

why is the function must be normalize ? what is the meaning of normalization
I know it's ∫ψψ*=1 but what it mean ?

2. Jun 9, 2012

oli4

Re: Normalization

It means the total probability is 1

3. Jun 9, 2012

Simon Bridge

Re: Normalization

It means that the probability of finding the particle someplace in the Universe must be exactly 1.
 Oli4 just beat me to it.

We need to do this because it is often easier to work out the shape of the wave-function without worrying about keeping the probabilities in line at every step.

4. Jun 9, 2012

sciboudy

Re: Normalization

Okay thank you simon and oli4


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# Cannot upload .blend through Blend Exchange

I am trying to upload a .blend via Blend exchange. But after uploading, instead of getting a url to copy and paste, I am getting this. Any fixes?

• Do note that blend-exchange is run by a member of our community and it is a private site separate from the SE network. Thus this is not the best place for bug reports about that site. On the bottom of the help page it says to contact the site owner (GiantCowFilms), or open a github issue. – David Oct 2 '16 at 11:59
• It was probably caused by the filesize. 10 MB shouldn't be a problem, but I recently reconfigured PHP, and the upload limit might not be correct. I will look into fixing this now – GiantCowFilms Oct 2 '16 at 16:42

Thank you so much for pointing this out, this issue affected a large number of other users.

The exact error, verbatim from the log:

POST Content-Length of 12143120 bytes exceeds the limit of 8388608 bytes in Unknown on line 0

What was happening (to ALOT of users it seems, based on my error log), is that the file, which is sent in the HTTP post request, was to big for the post request limit set by PHP in the PHP configuration.

I swear I had expressly set the value to be big enough, but apparently it didn't work.

It has now been fixed.

The learning takeaway for me has been to actually test if config values have been updated, since sometimes something goes wrong in writing the new config.

Sorry for any inconvenience this may have caused.

• Did I thank you for your AWESOME .blend upload thing? [LOUD AND CLEAR] THANK YOU GiantCowFilms FOR WHAT YOU'RE GIVING TO US !!!! and thanks a lot also for the HDRs :) – Bithur Nov 25 '16 at 20:31
• @Bithur Thanks for the kind words :) – GiantCowFilms Nov 25 '16 at 21:03
• I just can't say how much you deserve my (our) thanks. – Bithur Nov 25 '16 at 21:16
• @GiantCowFilms - there seem to be errors during upload again, regardless of the file size. Again, many thanks for this awesome service! – metaphor_set Jun 17 at 5:50
• @metaphor_set Rewrite coming soon. I'm hoping that we will a) not have this many issues, and b) I will know sooner (I actually have some kind of error reporting). Anyway, I think the issue was caused by either a mysql or a php upgrade that I did in preparation to launch the rewrite, and did not fully test. Now Fixed. – GiantCowFilms Jun 17 at 6:03


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# 7 Effective energy landscape of an octahedral nanoparticle

In this section we analyze the energy landscape of a magnetic nanoparticle with the so-called truncated octahedral shape. The Co particles are usually reported as having this structure with fcc underlying lattice, see for example Ref. [20,14,8] and image in Figs. 1.1(c-d).

Regular truncated octahedrons having six squares and eight hexagons on the surface have been constructed cutting the ideal fcc lattice in an octahedral (two equal mutually perpendicular pyramids with square bases parallel to XY plane) and subsequent truncation. Equal surface densities in all hexagons and squares can be obtained if the fcc lattice is initially rotated in the XY plane, i.e. when the X axis is taken parallel to the direction and the Z axis to the direction. We perform the same calculations as before for a multi-spin particle cut from an fcc lattice, with cubic single-site anisotropy in the core and NSA.

In Fig. 2.11 we presents the values of the effective uniaxial and cubic anisotropy constants as a function of surface anisotropy value . The results show that the effective uniaxial contribution in this system is practically zero. We can also observe that the effective cubic anisotropy is modified with respect to the core by surface effects. It can be seen that, similarly to the results discussed above for spherical particles, the effective cubic anisotropy constant is again proportional to for small . This is mainly due to the two contributions, one coming from the initial core cubic anisotropy and the other from the surface contribution as in Eq. (2.14). It is interesting to note that the surface contribution can change the sign of the effective cubic anisotropy constant from the initially negative cubic core anisotropy. We can also observe an asymmetric behavior of the effective anisotropy constants with respect to the change of the sign of the surface anisotropy which we found, in general, in all particles with fcc underlying lattice.

If the fcc lattice is initially orientated with crystallographic lattice axes parallel to those of the system of coordinates, then different atomic densities are created on different surfaces. This way the surface density along the XY circumference is different from that along XZ one. In Fig. 2.12 we plot the dependence of the effective cubic constant as a function of for and . It can be seen that, similarly to the results discussed above, the effective cubic constant is again proportional to but now its increase with is slower. The surface contribution can again change the sign of the initially negative cubic core anisotropy. Besides, we clearly see that the multi-spin particle develops a negative uniaxial anisotropy contribution, induced by the surface in the presence of core anisotropy, according to Eq. (2.20).

Rocio Yanes


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# Confidence interval for two proportions examples

Confidence interval for two proportions examples

## Confidence interval for two proportions examples

In this tutorial we will discuss some examples on confidence interval for difference between two population proportions.

## Example 1

A random sample of 85 parts manufactured by machine A yields 10 defective and a random sample of 110 parts manufactured by machine B shows 28 defective.

Compute 95% confidence interval for difference between the proportions of defectives?

Are the two machine differ significantly with respect to the proportion of defectives?

#### Solution

Given information

. Machine A Machine B
Sample size $n_1=85$ $n_2=110$
Observed no. of defectives $X_1=10$ $X_2=28$

We wish to determine $95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$.

#### Step 1 Specify the confidence level

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

#### Step 2 Given information

Given that $X_1 = 10$, $X_2 = 28$, $n_1 = 85$, $n_2 = 110$.

The estimate of the population proportions $p_1$ is $\hat{p}_1 =\frac{X_1}{n_1} =\frac{10}{85}=0.1176$ and
the estimate of the population proportion $p_2$ is $\hat{p}_2 =\frac{X_2}{n_2} =\frac{28}{110}=0.2545$.

#### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is

 \begin{aligned} (\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E. \end{aligned}
where $E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

#### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.025} = 1.96$.

#### Step 5 Compute the margin of error

The margin of error for the difference $(p_1-p_2)$ is

 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}\\ & = 1.96 \sqrt{\frac{0.1176*(1-0.1176)}{85}+\frac{0.2545*(1-0.2545)}{110}}\\ &= 0.1064. \end{aligned}

#### Step 6 Determine the confidence interval

$95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is

 \begin{aligned} (\hat{p}_1-\hat{p}_2) - E &\leq (p_1-p_2) \leq (\hat{p}_1-\hat{p}_2) + E\\ (0.1176-0.2545) - 0.1064 & \leq (p_1-p_2) \leq (0.1176-0.2545) + 0.1064\\ -0.2433 & \leq (p_1-p_2) \leq -0.0305 \end{aligned}

Thus, $95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is $(-0.2433,-0.0305)$.

#### Interpretation

We can be $95$% confident that the difference between two population proportions $(p_1-p_2)$ is between $-0.2433$ and $-0.0305$.

Because the $95$% confidence interval does not include the point zero, we conclude that at $0.05$ level of significance the two machine do not differ significantly with respect to the proportion of defectives.

## Example 2

In a recent survey of randomly selected adults 65 or older, 411 of 1012 men and 525 of 1062 women say they suffer from some form of arthritis.

a. Construct a 98% confidence interval for the difference between senior men and women who suffer from arthritis.

b. Does there appear to be a difference between senior men and women as far as suffering from arthritis?

#### Solution

Given information

. Men Women
Sample size $n_1=1012$ $n_2=1062$
Suffering from arthritis $X_1=411$ $X_2=525$

We want to determine $98$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$.

#### Step 1 Specify the confidence level

Confidence level is $1-\alpha = 0.98$. Thus, the level of significance is $\alpha = 0.02$.

#### Step 2 Given information

Given that $X_1 = 411$, $X_2 = 525$, $n_1 = 1012$, $n_2 = 1062$.

The estimate of the population proportions $p_1$ is $\hat{p}_1 =\frac{X_1}{n_1} =\frac{411}{1012}=0.4061$ and
the estimate of the population proportion $p_2$ is $\hat{p}_2 =\frac{X_2}{n_2} =\frac{525}{1062}=0.4944$.

#### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is

 \begin{aligned} (\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E. \end{aligned}

where $E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

#### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.01} = 2.33$.

#### Step 5 Compute the margin of error

The margin of error for the difference $(p_1-p_2)$ is

 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}\\ & = 2.33 \sqrt{\frac{0.4061*(1-0.4061)}{1012}+\frac{0.4944*(1-0.4944)}{1062}}\\ &= 0.0507. \end{aligned}

#### Step 6 Determine the confidence interval

$98$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is

 \begin{aligned} (\hat{p}_1-\hat{p}_2) - E &\leq (p_1-p_2) \leq (\hat{p}_1-\hat{p}_2) + E\\ (0.4061-0.4944) - 0.0507 & \leq (p_1-p_2) \leq (0.4061-0.4944) + 0.0507\\ -0.1389 & \leq (p_1-p_2) \leq -0.0375 \end{aligned}

Thus, $98$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is $(-0.1389,-0.0375)$.

#### Interpretation

We can be $98$% confident that the difference between two population proportions $(p_1-p_2)$ is between $-0.1389$ and $-0.0375$.

Because the $98$% confidence interval does not include the point zero, we conclude that at $0.02$ level of significance there appears to be no significant difference between the proportion of senior men and proportion of women who suffer from arthritis.

## Example 3

Two machines used in the same operation are to be compared. A random sample of 80 parts from the first machine yields 6 non-conforming ones. A random sample of 120 parts from the second machine shows 14 non-conforming ones.

Find a 95% confidence interval for difference in the proportion of non-conforming parts between the two machines.

#### Solution

Given information

. First Machine Second Machine
Sample size $n_1=80$ $n_2=200$
no.of non-confirming $X_1=6$ $X_2=14$

We want to determine $95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$.

#### Step 1 Specify the confidence level

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

#### Step 2 Given information

Given that $X_1 = 6$, $X_2 = 14$, $n_1 = 80$, $n_2 = 200$.

The estimate of the population proportions $p_1$ is $\hat{p}_1 =\frac{X_1}{n_1} =\frac{6}{80}=0.075$ and the estimate of the population proportion $p_2$ is $\hat{p}_2 =\frac{X_2}{n_2} =\frac{14}{200}=0.07$.

#### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is

 \begin{aligned} (\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E. \end{aligned}

where $E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}$ and $Z_{\alpha/2}$ is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

#### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus $Z_{\alpha/2} = Z_{0.025} = 1.96$.

#### Step 5 Compute the margin of error

The margin of error for the difference $(p_1-p_2)$ is

 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}\\ & = 1.96 \sqrt{\frac{0.075*(1-0.075)}{80}+\frac{0.07*(1-0.07)}{200}}\\ &= 0.0677. \end{aligned}

#### Step 6 Determine the confidence interval

$95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is

 \begin{aligned} (\hat{p}_1-\hat{p}_2) - E &\leq (p_1-p_2) \leq (\hat{p}_1-\hat{p}_2) + E\\ (0.075-0.07) - 0.0677 & \leq (p_1-p_2) \leq (0.075-0.07) + 0.0677\\ -0.0627 & \leq (p_1-p_2) \leq 0.0727 \end{aligned}

Thus, $95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is $(-0.0627,0.0727)$.

#### Interpretation

We can be $95$% confident that the difference between two population proportions $(p_1-p_2)$ is between $-0.0627$ and $0.0727$.

Because the $95$% confidence interval include the point zero, we conclude that at $0.05$ level of significance there is a significant difference between the proportion of non-confirming parts between the two machines.

Do read my step by step tutorial on Confidence interval for two proportions, tutorial will help you to understand Step by step procedure to estimate the confidence interval for difference between two population proportions is as follows:

Let me know in the comments if you have any questions on Confidence interval for two proportions examples and your thought on this article.

VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.


text:
# Hopping Frogs puzzle variant

There is an old puzzle which has various names such as Toads and Frogs, Jumping Frogs, Hopping Frogs, Leap Frog, etc., and which has been asked here before. I'd like to share a variant of this puzzle that I came up and which I haven't seen anywhere else.

There is a straight row of 9 squares (or lily pads if you prefer), each large enough to contain at most one frog. The middle square is empty, and there are 8 frogs on the other squares. The four frogs that start on the left can only move to the right, and the frogs that start on the right can only move to the left. The aim is for the two sets of frogs to pass each other so that they swap places.

In the original version of the puzzle, a frog can either walk forward one square or jump forward two squares, provided of course that the destination square is the empty one. So they start as:

AAAA.BBBB

The first few moves are:

AAA.ABBBB
AAABA.BBB
AAABAB.BB

and eventually, if you do things correctly, they end up as:

BBBB.AAAA

In my new variant, a frog can only jump forward two or three squares (i.e. jump over one or two other frogs to the empty square) - they cannot move forward just one square.

Question 1:
How can the two sets of four frogs pass each other using only forward jumps of two or three squares?

Question 2:
The same question, but now with a row of 13 squares and two sets of six frogs.

Further info:
I used a computer to search for solutions with other numbers of frogs. Whereas the original version can be solved with any number of frogs on the left and any number on the right, my variant seems to be unsolvable if the left and right numbers are different. When they are equal, it can be solved for 2+2, 4+4, 6+6, 8+8, 9+9, 10+10, 11+11, and 12+12 frogs, but I have not searched further. Though I have not yet examined the optimal solutions very closely, at first glance there is no obvious pattern to them so I do not know if a general optimal solution is possible. There may well be a general solution that is not optimal in all cases.
I expected such an obvious variant to have been analysed before, but if so, I haven't found it.

Edit::
It turns out that my computer program was buggy. The puzzle can be solved when the number of frogs on each side differs, except for a few cases. I re-analysed the cases with up to 12 frogs on either side, and the only ones that have no solution are: 1+0, 1+1, 3+1, 3+3, 4+1, 4+3, 5+4, 5+5, 6+1, 6+3, 7+4, 7+7, 9+1, and 9+4.
There is a general solution for even numbers of frogs. Thanks to astralfenix for the observation that led me to it. For 2r+2s frogs it uses r+s+3rs moves, which is not quite optimal in all cases.

• Is this the very same person that runs jaapsch.net? If so, I'd like to say that your website is extremely interesting and informative - have been following it for a while :) Thanks for running such a unique set of analyses. – TheGreatEscaper Dec 29 '16 at 9:30
• @TheGreatEscaper: Yes, jaapsch.net is my site. On it there is one page about the standard version of the Hopping Frogs puzzle. – Jaap Scherphuis Dec 29 '16 at 9:43

here's a way to do it in 33 moves for the 6 frog case. Interestingly, this involves putting the frogs in an alternating doubles pattern, 11221122 etc. The solution to the original version of the puzzle involves use of an alternating singles pattern (121212 etc).

• "In my new variant, a frog can only jump forward two or three squares (i.e. jump over one or two other frogs to the empty square)" is noted, so you cannot move forward I guess... – Oray Dec 29 '16 at 9:34
• Yes, moving one step forward is not allowed in my variant. – Jaap Scherphuis Dec 29 '16 at 9:39
• Good observation about the 11221122 doubles pattern. I think that gives rise to a general solution for n+n frogs with n even. – Jaap Scherphuis Dec 29 '16 at 10:16

Question 1

Initially AAAA.BBBB:

1. AA.AABBBB
2. AABAA.BBB
3. AAB.AABBB
4. AABBAA.BB
5. AABBAABB.
6. AABBA.BBA
7. AABBABB.A
8. AABB.BBAA
9. A.BBABBAA
10. ABB.ABBAA
11. .BBAABBAA
12. BB.AABBAA
13. BBBAA.BAA
14. BBB.AABAA
15. BBBBAA.AA
16. BBBB.AAAA

So total 16 moves for the first try :)

33 moves for 6+6.

• Well done. There could well be a general solution for even n that is of length n*n. However, the optimal solution that my computer found for 6+6 frogs is 33 moves. Maybe I should also search for non-optimal solutions if I want to find a general solution. – Jaap Scherphuis Dec 29 '16 at 9:51
• @JaapScherphuis I will let u know when I put this into my computer too :) – Oray Dec 29 '16 at 10:00


text:
# Math Help - Double-Angle Formulas

1. ## Double-Angle Formulas

Find the exact values of sin 2u, cos 2u, and tan 2u using the double angle formulas.:
cos u = -2/3, pi/2 < u < pi

So I know that the formulas are:
sin 2u = 2 sin u cos u
cos 2u = cos^2 u - sin^2 u
tan 2u = (2 tan u) / (1 - tan^2 u)

Now I am having trouble figuring out what sin u is. I believe that it is
sqrt(3), but I am not sure.
If I am right, the first formula would be:
sin 2u = (2) (sqrt3) (-2/3)
but I'm not sure how to multiply a square root with a fraction.

Any help is appreciated.

2. Originally Posted by iluvmathbutitshard
Find the exact values of sin 2u, cos 2u, and tan 2u using the double angle formulas.:
cos u = -2/3, pi/2 < u < pi

So I know that the formulas are:
sin 2u = 2 sin u cos u
cos 2u = cos^2 u - sin^2 u
tan 2u = (2 tan u) / (1 - tan^2 u)

Now I am having trouble figuring out what sin u is. I believe that it is
sqrt(3), but I am not sure.
If I am right, the first formula would be:
sin 2u = (2) (sqrt3) (-2/3)
but I'm not sure how to multiply a square root with a fraction.

Any help is appreciated.
nope, that's not the right value for sin(u). there are several ways you can get to sin(u). Two conventional ways are:

Use the fact that $\sin^2 u + \cos^2 u = 1$

Solve for $\sin u$. Note that we are in the second quadrant so you will need to take the positive square root. As sine is positive in that quadrant.

Another way: Draw a right-triangle with an acute angle $u$. The adjacent side to $u$ should be labeled $2$ while the hypotenuse should be labeled $3$. Use Pythagoras' Theorem to find the opposite side and use sine = opposite/hypotenuse

3. So sin u would be sqrt5/3?

If so, how would I multiply this by -2/3?

4. Originally Posted by iluvmathbutitshard
So sin u would be sqrt5/3?
yes, $\sin u = \frac {\sqrt 5}3$

If so, how would I multiply this by -2/3?

the same way you multiply any two fractions... multiply the numerators together to get the new numerator and the denominators together to get the new denominator. $\frac {\sqrt 5}3 \cdot \frac {-2}3 = \frac {-2 \sqrt 5}9$

5. Ok, great

SO then for cos 2u:
cos^2 u - sin^2 u
(4/9) - (5/9) = (-1/9)

Is this correct?

6. Originally Posted by iluvmathbutitshard
Ok, great

SO then for cos 2u:
cos^2 u - sin^2 u
(4/9) - (5/9) = (-1/9)

Is this correct?
yup

7. My answer for sin u was
(-4sqrt5) / (9)

And one last thing, is my tan correct:
(2 tan u) / (1 - tan^2 u)
(2 (-sqrt5/2)) / (1 - (5/4))

So this would simplify, I think, to
[(-2 sqrt5) / (2)] / [-1/4]
Is this correct?

8. Originally Posted by iluvmathbutitshard
My answer for sin u was
(-4sqrt5) / (9)
ok, good.

And one last thing, is my tan correct:
(2 tan u) / (1 - tan^2 u)
(2 (-sqrt5/2)) / (1 - (5/4))

So this would simplify, I think, to
[(-2 sqrt5) / (2)] / [-1/4]
Is this correct?
that's not simplified

for this last question, I would use the fact that $\tan 2 u = \frac {\sin 2u}{\cos 2u}$

just divide the first two answers you got...

9. So the answer would be 4 sqrt5?

10. Originally Posted by iluvmathbutitshard
So the answer would be 4 sqrt5?
yup

11. Thank you so much for your help.


text:
# About the logarithmic derivative of the Riemann zeta function

Let $\rho=\beta+i\gamma$ a non-trivial zeros of the Riemann zeta function and $s=\sigma+it$ a complex number. It is possible to prove that $$\frac{\zeta'}{\zeta}\left(s\right)=\sum_{\left|t-\gamma\right|\leq1}\frac{1}{s-\rho}+O\left(\log\left(t\right)\right) \tag{1}$$ uniformly for $-1\leq\sigma\leq2$ (see for example Titchmarsh, “The theory of the Riemann zeta function”, second ed., page $217$). So in particular if we take $\sigma=0$ holds $$\frac{\zeta'}{\zeta}\left(it\right)=\sum_{\left|t-\gamma\right|\leq1}\frac{1}{it-\beta-i\gamma}+O\left(\log\left(t\right)\right).$$ Question: is it possible to prove that $$\sum_{\left|t-\gamma\right|\leq1}\frac{1}{it-\beta-i\gamma}\ll\log\left(t\right)?$$ The problem is that it could be some zeros with real part very close to $0$ and so for $t=\gamma$ the sum is very "big". Thank you.

Addendum: My final goal it's to prove that $$\frac{\zeta'}{\zeta}\left(it\right)=O\left(\log\left(t\right)\right)$$ so also another proof of it (if exists) without the use of $(1)$ is welcome.

• Note that $\beta\gg\frac{1}{\log t}$ by the classical zero-free region and the symmetry $s\leftrightarrow 1-s$ for the zeros, so each term in the sum is $\ll\log t$. In fact the whole sum is $\ll\log t$, as shown by my response below. – GH from MO Mar 3 '16 at 14:57

I think your final goal follows by taking the logarithmic derivative of the functional equation: $$\frac{\zeta'}{\zeta}(s)+\frac{\zeta'}{\zeta}(1-s)=\log\pi-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{s}{2}\right)-\frac{1}{2}\frac{\Gamma'}{\Gamma}\left(\frac{1-s}{2}\right).$$ Applying this with $s=it$ and using the familiar asymptotic expansion of $\Gamma'/\Gamma$, we get $$\frac{\zeta'}{\zeta}(it)+\frac{\zeta'}{\zeta}(1-it)=O(\log t),\qquad t>2.$$ In fact we get $$\frac{\zeta'}{\zeta}(it)+\frac{\zeta'}{\zeta}(1-it)=-\log t+\log(2\pi)-\frac{i}{2t}+O(t^{-2}),\qquad t>2,$$ but observing $O(\log t)$ is sufficient on the right hand side. Indeed, the second term on the left hand side is $O(\log t)$, hence the same holds for the first term.
P.S. Of course this means that your sum is indeed $O(\log t)$. See also my comment below the original post.
• @Wolfgang: Yes, this is a classical result. See, for example, Theorem 3.5 in Titchmarsh: The theory of the Riemann zeta-function. The improved bound $O((\log t)^{2/3})$ is also known, see Section 6.19 of the same book. – GH from MO Mar 5 '16 at 16:26
• @Wolfgang: Let me correct my previous comment as I confused $\zeta'/\zeta$ with $\zeta$. Yes, this is a classical result. See, for example, Theorem 5.17 or Theorem 4.11 in Titchmarsh: The theory of the Riemann zeta-function (or Theorem 6.7 in Montgomery-Vaughan: Multiplicative number theory I). The improved bound $O((\log t)^{2/3}(\log\log t)^{1/3})$ is also known, see Section 6.19 of the same book. – GH from MO Mar 5 '16 at 16:50


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text:
A criterion on rationality of infinite series.

While I was studiyng I found some criterions to prove that an infinite convergent serie has an irrational sum.

That got me thinking. There are any criterions that prove the rationality of an infinite convergent serie sum?

• Potential duplicate: math.stackexchange.com/questions/66518/… – symplectomorphic Mar 26 '18 at 15:26
• There are particular cases. For example for $n\in \Bbb N$ we have $n!e=n!\sum_{k=0}^{\infty}\frac {1}{k!}=A(n)+B(n)$ where $A(n)\in \Bbb N$ and $B(n)=\sum_{k=n+1}^{\infty}\frac {n!}{k!}\in (0,1).$ Therefore $n!e\not \in \Bbb N$ for all $n\in \Bbb N$, so $e$ is irrational. – DanielWainfleet Mar 27 '18 at 1:45


text:
# How do you factor the trinomial v^2 + 20v + 100?

Then teach the underlying concepts
Don't copy without citing sources
preview
?

Write a one sentence answer...

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

Describe your changes (optional) 200

1
Mar 6, 2018

${\left(v + 10\right)}^{2}$

#### Explanation:

We must think of two numbers that have a product of 100 and a sum of 10. Thus, we can deduce that these two numbers are 10 and 10:

10+10=20
10*10=100

$\left(v + 10\right) \left(v + 10\right)$

Which can be summarised as:
${\left(v + 10\right)}^{2}$

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text:
20 terms

# MAFS.8.EE.1.1 & MAFS.8.EE.1.2

#### Terms in this set (...)

Multiplying powers
When multiplying powers with the same base, add the exponents.
Dividing powers
When dividing powers with the same base, subtract the exponents.
Positive exponents
Positive exponents represent how many times a number is being multiplied by itself.
Negative exponents
In a negative exponent, the base moves to the denominator in a fraction that has 1 for the numerator.
Rational (fractional) exponents
Rational exponents represent powers and roots at the same time.
exponent
The number that indicates how many times a base is used as a factor. Repeated multiplication.
base
When a number is raised to a power, the number that is used as a factor is the base.
power
An expression produced by raising a base to an exponent.
Zero Exponent Property
Any number, except for zero, taken to the zero power is equal to 1.
Negative Exponent Property
Take the multiplication inverse or reciprocal of the power. Repeated division.
Product of Powers Property
When multiplying powers with the same base, add the exponents.
Quotient of Powers Property
When dividing powers with the same base, subtract the exponents.
Power of a Power Property
Multiply the power outside the parenthesis by the power of each base inside the parenthesis.
Inverse Operations
The undoing of an opposite operation
Algebraic Equation
A mathematical sentence that contains an equals sign.
Reciprocal
Two numbers whose product is 1
Distributive Property
a(b + c) = ab + ac
When one of the factors of a product is a sum, multiplying each addend before adding does not change the product.
Combine like terms
terms that contain the same variable
Integer exponent
an exponent that is an integer
Integer
All whole numbers (both positive and negative) and zero.


text:
The maximum difference between the number of elements in the two sets of equal length of consecutive numbers that divisible by some prime numbers

Suppose that $$p_1,...,p_k$$ are distinct prime numbers.

Let $$f(n,l)$$ be equal to the number of elements from set $$\{n+1,n+2,...,n+l\}$$ that are divisible by some $$p_1,...,p_k$$. Is it true that $$\sup_{n,m,l \in \mathbb{N}} |f(n,l)-f(m,l)| = O(k).$$

For example, Rankin's construction of large prime gaps boils down to the fact that if $$p_1, \ldots, p_k$$ denote all prime numbers below $$x$$, then there exists some $$n$$ with $$f\left(n, \frac{x\log x\log\log\log x}{(\log\log x)^2}\right)=0$$.
If $$n=x, \ell=x^2-x$$, then $$f(n, \ell)=\pi(x^2)+\mathcal{O}(x)\sim\frac{x^2}{2\log x}$$, but the average taken over all $$n$$ is $$\ell \prod_{p\leq x}\left(1-\frac{1}{p}\right)\sim \frac{e^{-\gamma}x^2}{\log x}$$.
• Thanks for your answer. What do you think about the correct order? Is it $O(k^2 \log k)$? – Martin Hicks Jun 12 at 19:35
• A bound involving $k^2$ might be achievable, but I don't think anybody has done it yet. For the upper bound Selberg's sieve might suffice. The lower bound is probably more difficult. A good starting point might be Iwaniec's work on Jacobsthal function. – Jan-Christoph Schlage-Puchta Jun 13 at 8:52