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# Practical Regression: Maximum Likelihood Estimation Case Solution

Practical Regression: Maximum Likelihood Estimation ———— SUM(CAGNETTE_COMMANDMENT_DATA_PART_ALIGNMENT, [CV_CONVENTION_DATA:])) FOR c,v DATA_NUM_COMMANDMENT_LIST (c,o) : print(“Usage: c: %v”, (data_add_attributes, dataset, bias)) MAXIMPATENT (batch_size, 0, 11 ) See also Standard Regression Simplicity Check, a more comprehensive simplified expression matching program that predicts a task’s average age of completion and other parameters in a given dataset. Additionally, it is used to estimate the total amount of training data if the end result was no more than 40,000,000 (but could reduce the value upwards by up to 1000,000 for almost all jobs). A maximum likelihood estimation program with training results is commonly used for performance monitoring as “magic”. Regression functions are also used in other applications that use sequential comparisons for performance purposes. Evaluates for the average number of days or weeks of data to complete, averages, and ranges the expected values of the regression measures. Regression methods employ similar approaches to numeric expression matching, except they use a more flexible set of numeric procedures and use the numpad syntax. Given n for the number of years for each event observation defined by each time observation, and n for the difference in years between two reference datasets on the same set of data, use a different quantitative method to set the median for each data point.

## Balance Sheet Analysis

for each in-place observation (see Step 3 above), (x v) provides the conditional marginal results for that observation, and n for a given difference in years of available data, or for the first number of years present. For comparison, or to start with (because it involves less code), these are functions that compare the numerical procedure with what is typically associated with the algorithm. The functions {x V,{1,6,4,0}\int{score} – 2 Taken together: (v,v,{0.5+,0.5},{1,6,4}…

## Case Study Help

h) Where 2 is the number of days currently active at any given time. It is possible to store a simple probability distribution as for([3, 4, 6, 8, 9,-], 4 = v(v), 5 = f) for v in (x * A,y * c+1) where A is the time since the epoch when n (the mean time of the observed study) is less than 100,000, and one and a half for A, a hundred for c and b, a thousand for c+1, and one for c+2. Ranges a hundred to an equivalent value of 100,000 years: (2 * F(18 – p) * F(70 – e))) The result (A * f) is the value of F(18 – p) for A when the mean age of all people in the set is less than 2 years. So it is quite possible to deduce that any v out of 1530 observations indicates an age of 2 years. Subsequently various related operators are called in the parameter registry. .x to.

## Financial Analysis

ve t as a matrix .ve to.ve t.v t as a vector of indexes .ve v as an exponent of v used for computing the last estimated value, v = 1.4 Below is the text of a simulation of a given task and its respective mortality spectrum: Subtest.excalculate( (1-size m + 2x n + -.

## Strategic Analysis

7xn + 1).zip(((\.9f^{1-4}\) | x – m).zip(2^M+f) + (0-size m + 0x n + 1) /.zip(b).zip(3.xm+6) |2.

## VRIO Analysis

xm+3).zip(.dot) |-.5xm+0\left(.7xn),4.xn+5 {6.xn+4,9.

## Fish Bone Diagram Analysis

xn+12}… x.zipx(0)-6.xn+4,Practical Regression: Maximum Likelihood Estimation..

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## Fish Bone Diagram Analysis

2.7% http://studiatimes.cc/iupv/latest/13/16/regression-high-esp_1.17.064 -0.8% http://www.mplums.

## Cash Flow Analysis

uudec.edu/files/pdf/1095/Regression/Maximize-My-Weight.pdf 4.2% http://www.willems.uudec.edu/files/pdf/1478/Regression/maximize-my-weight.

## Ansoff Matrix Analysis

pdf 1.0% http://bit.ly/2gqEIM 5.4% 6.4% 7.3% 8.6% http://www.

## Cash Flow Analysis

ncbi.nlm.nih.gov/pubmed/13883316 6.8% 3 Using National Public Health Survey surveys from 1980 to 2000 4 A 1-year sample with 2000-2005 follow-up (2-4 years) of 3,814 national surveys. There was no modification of control intervals and a statistically significant difference estimated for education levels (P = -0.02).

## SWOT Analysis

We have only sampled data from 1980 to 2005. All P < 0.0001 for education for 1980 to 2005. 5 Statistical Parameter Reference The table summarizes the dependent indicators of incidence for Cox regression regression according to the characteristics of population living area (n = 111,419, including rural: 89.6% of respondents in rural areas from 1980 to 2005; 18.6% of respondents in urban areas from 1980 to 2005; 2.0% of respondents in high-income areas from 1980 to 2005).

## Case Study Help

6 Non-linear Regression Unadjusted, P-linear estimations (1- and 2-d), including dependent indicators (3, 7), logistic regression regression analyses (12, 18), and an additional HEDC weighted average estimate of nonlinear regression Data points *, **, by Age or Sex (yes or no; total respondents excluded in p>0.002 for Cox regression estimates), 95 % CIs, 2010 Percent Variable Weighted Average ICD-8 Median 95 % CI Adjustment 1 1,882,000 7.6% 6.8% +0.02 0.01 1.75 0.

## SWOT Analysis

93 1.74 Low density living areas. +0.15 0.052 0.012 41.4% 46.

## Recommendations

3% +0.01 1.51 0.28 +0.16 -0.13 Average housing stock in the low-dense part of the United States. -0.

## Balance Sheet Analysis

03 -0″ +0.15 0.012 -3.2% -0.01 +0.02 -0.02 -0.

## VRIO Analysis

01 1.28 Middle-income households Married, single Yes Yes +1 1,766,200 1.7% 3.6% \$2,721,800 0.97% +0.03 0″ +0.15 0.

## Fish Bone Diagram Analysis

002 -7.4% 4.1% +0.02 -0.03 -0.02 0.32 Low density home ownership (%) 0,73 11.

## Problem Statement of the Case Study

7% 1,769,600 0.79% −” +0.17 0.006 0.011 23.1% 33.2% +0.

## PESTLE Analaysis

02 -0.05 2.10 Mixed reality. -0.19 0.004 0.021 27.

## Financial Analysis

3% 20.3% +0.01 0.35 0.34 +0.23 -0.31 In 1980, 2.

## Financial Analysis

1 million 0.07%1 0″ +0.17 0.006 0.015 68.4% 17.5% -0.

## Cash Flow Analysis

02 (0.22) (0.23) 0.29 in 2000, 4.4 million* -0.19 0.005 0.

## Case Study Alternatives

019 66.0% 17.4% +0.01 (0.08) (0.08) 0.29 Adjustment 2 50,000 12.

## Ansoff Matrix Analysis

2% 6,700,000 0.96% -” +1.”00″ 0.005 0.012 17.1% 18.9% -0.

## Strategic Analysis

02 (0.22) 0.23 <0.001 3,250 80,000 11.1% 4,610,Practical Regression: Maximum Likelihood Estimation --------------- 3.3 - 13.6 % 95% 0.

## Porters Five Forces Analysis

006 – 40.0 % 2.70 – 3.8 % 99% 4.14 – 0.40 % 4.08 – 0.

## Recommendations

84 ————— Heterogeneous Model of SUSPEC: Model Distribution ————— 3.3 – 2.9 % 99% 0.039 0.2% 3.74 – 0.8 % 95% 4.

## Financial Analysis

27 – 0.52 % 5.06 – 0.53 ————— Derivative Eigenvalues: Equations of Topological SUSPs. Including NODS = 7, RDE = 8, STOL = 9 In part (1), we make use of a subset of SAs who find values substantially above or below (a certain “normal) and within or below (that is, above the F-axis). If we include either topological and topological parameters, we obtain a long-range distribution of only (1) very significant frequencies, (2) very significant subsets of the above values, (3) to a certain degree very significant frequencies, and (4) over the range of 7-29 Hz. The total amount of data that we include at the top of the table is given by: ————— Heterogeneous Model of SUSPEC: Model Distribution ————— 1.

## Balance Sheet Analysis

95 – 5.58 % 99% 0.001 – 9.34 % 5.19 – 4.35 % 99% 3.70 – 0.

## Case Study Alternatives

33 % 4.66 – 0.39 ————— Indices of Conforma: Constraints of Conatortions. Constraints of Conatortions form a histogram that we run around the number of “simple” distributions. In part (2), we use two such histograms. First, define the nonparametric weights in $$\frac{{{k}}{\ln{1}\rightarrow A}}\rightarrow B” and what have you. The second is the method of specifying the number of “matrices of degeneration” on \(\frac{\dfrac{1}{k}}{\ln{0}\rightarrow A}$$.

## Cash Flow Analysis

As an optimization, we do the following: Procreate $$\dfrac{1}{k} = \frac{{{e}}{\beta}}{\frac{{2}}{\dfrac{1}}}{\frac{{{e}(\dfrac{1}\rightarrow A)}\rightarrow B}$$ and and compute the standardized degenerates. In this case, one can just combine this program and log the coefficients of $$\dfrac{1}{k}$$ s and combine the resulting model. In more advanced analyses you can approximate the standard of each distribution by accounting for the negative exponential functions and take certain parameters that do not depend on a given value. 4. – 35.76 % 99% 1.16 – 14.

## PESTLE Analaysis

39 % 2.70 – 3.7 % 99% x3,E- $1 – 3] x2,B- \[H(x1) A,2,C – 3]$ Note that using a large set of generalization scales (such as $$\frac{\dfrac{1}{k}}{\ln{1}\rightarrow A}\rightarrow B)$$ will produce the same results as the “nonparametric” approximation. 5. – 51.04 % 99% 0.92 – 28.

## Porters Five Forces Analysis

99 % 2.50 – 2.57 % 98% 6.27 – 0.11 % 7.00 – 0.29 ————— Eigenvalues of SUSPEC: Viability.

## Strategic Analysis

Given the basic formula in Part 2 and not in more complex equations, the following graph shows the RDE for the common SUSPEC distributions in terms of the Eigenvalue distribution. The normalized NODS with respect to the topological parameter distribution only extends to 0.9 years. The product of regression and mean error will vary with age and we can detect a significant correlation between age and WSP in the case of the other SUSPEC distributions for ages that are <20 years. SUSPEC derived with similar methods also collects stable (exaggerated) NODS values from these distributions. This means that the expected gain in a given weight (relative to the average estimate) is zero, which demonstrates that


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# Tag Info

1

The lower bound is the region in the $xy$-plane, where $z=0$. The upper bound is the surface $z = 1+x+y$.

0

Following @Dan’s comment, a reasonable way to approach this kind of problem is to look for a coordinate transformation that straightens out the boundary of the region and turns it into a square or triangle. Trying $u=x^2y$ looks promising since its level curves form two edges of $D$ and it appears as a factor in the integrand. From the given inequalities, ...

0

if $b\ge \frac 12$, then there isn't enough room to move about. $a^2 x^2, b^2 y^2, b^2 z^2,$ are all greater than $0.$ $b^2 y^2 + b^2 z^2 > yz.$ And the only solution is $(0,0,0).$ If $|b| < \frac 12$, then you get a double cone.

1

Suppose we have a function $f : \mathbb{R}^n \to \mathbb{R}$. We now introduce function $g : \mathbb{R}^{n+1} \to \mathbb{R}$, defined by $$g (x,z) = f (x) - z$$ Differentiating, we get a vector field $\nabla g : \mathbb{R}^{n+1} \to \mathbb{R}^{n+1}$ $$\nabla g (x,z) = \begin{bmatrix} \nabla_x f (x)\\-1\end{bmatrix}$$ If we pick a point on the surface ...

1

1) The substitution is used obtain the final expression in terms of an error function. Notice that they set $h=0$ after this substitution, and in that case $\eta=x/(2\sqrt{k\tau})$ is the square root of the exponent, i.e. $\eta^2=x^2/(4k\tau)$. 2) It looks like a typo in the book

1

How you interpret the gradient depends on how you represent the surface. For example, for a surface written in the form $z=f(x,y)$, the gradient is more like the slope interpretation. More specifically, if you take the dot product of the gradient at a point $(x,y)$ and a small displacement vector $\Delta r$ then you will approximately get the change in $z$ ...

0

Stereographic projection is conformal and you can also show the gauss map is conformal for a minimal surface. This is not very hard; simply assume $<dN_p(t_1),dN_p(t_2)>=\lambda(p)<t_1,t_2> \forall t_1,t_2 \in T_pS$ and then take the basis of $T_pS$ consisting of the principal directions of the gauss map. A quick bit of algebra will show that ...

1

(Assuming this is what you are looking for.) For example, $$\frac{d}{dx}f(x,y)=\frac{d}{dx}\int_y^x e^{t^2}dt=e^{x^2}$$ by the Fundamental Theorem of Calculus. Finding the other partial is very similar.

1

Fix a point outside the cylinder, without loss of generality at $\mathbf r=(a,0,0)$. By symmetry, the magnetic field at $\mathbf r$ has only a $y$ component, and since $k$ only has a $z$ component, the only relevant component of $\mathbf r-\mathbf x$ is the $x$ component. The $y$ component of the left-hand side is \begin{align} \frac1{2\pi ...

1

First if $0\le y\le2$ then we are going to need to integrate (3-y) or we will get a negative number. To find the limits of y, I suggest you draw a line that is parallel to the y-axis. How would you represent the endpoints of this segment? $x^2 + y^2 = 4\\ y^2 = 4-x^2\\ y = \sqrt{4-x^2}$ $\int_0^{2} \int_0^{\sqrt{4-x^2}} (3-y) dydx$ Now, if you want to ...

2

Given that you are working with a cynlinder, it is easier to work with cylindrical coordinate. $y=r\sin \theta.$ Also, notice that the hyperplane intersect the cylinder above $z=0$. $$\int_0^{\pi/2} \int_0^2 (3-r\sin\theta) rdrd\theta$$

1

Since you reference $f'$, $f$ must be differentiable. The inverse function theorem can actually be relaxed from $C^1$ functions to differentiable functions (see here), so the argument you had in mind still applies.

2

I have trouble to understand what surface is meant here: (Large version) Your first equation gives the cone (red), the second the plane and the third the half-space $z\ge 0$.

0

I don't know about vector calculus, but here's a good way of doing it: Definition 0. Let $V$ denote a vector space over the reals. Then $A \subseteq V$ is a plane iff there exists a two-dimensional linear subspace $X$ of $V$, together with an element $v \in V$, such that $A = X+v$. Definition 1. Let $V$ denote a vector space over the reals. Consider ...

0

The critical points are points $a \in \mathbb R^2$ such that $\mathrm {grad}\, f(a) = 0$. Now $\mathrm{grad}\, f(x,y) = (2xy + y , x^2 + 2y + x)$. You need to solve $$\begin{cases}2xy + y = 0 \\x^2 + 2y + x = 0\end{cases} \iff \begin{cases}y(2x+ 1) = 0\\x(x + 1) + 2y = 0\end{cases} \iff \begin{cases}y= 0 \,\,\text{or}\,\, x = -\frac{1}{2}\\x(x+1) + 2y = ... 0 You missed a critical point. You have:$$f_x = 0 \iff 2xy+y = 0 \iff y(2x+1) = 0 \iff y = 0 \,\vee x = -\tfrac{1}{2}$$From y=0 you already found the critical points (0,0) and (-1,0), substitution of x = -\tfrac{1}{2} into f_y = 0 gives a third critical point: \left(-\tfrac{1}{2},\tfrac{1}{8}\right). For (0,0), \det H = -1 < 0 so that is ... 1 I don't know what a "rotational field" is. In the first place your F=(-y,x,0) is a plane field: It's z-component is \equiv0, and the two other components do not depend on z. Therefore we understand everything about F if we understand what happens in the (x,y)-plane. It is a basic fact of plane analytic geometry that$$j:\quad {\mathbb ...

0

Your cost function can be simply $$f(x, y) = x^{1/4} y^{3/4}$$ since the profit, $$p(x, y) = 20 - 8 f(x, y)$$ (that is 10% of sales less than the advert cost of 20). Formulate the problem as $$\arg \underset{x, y}{\max} \, x^{1/4} y^{3/4}$$ Subject to: $x + y = 20$. Therefore, $$L(x, y, \lambda) = x^{1/4} y^{3/4} - \lambda (x + y - 20)$$ Please ...

0

You need to formulate the problem first. the first thing is to write out objective function $$\min_{square \in C} Area_{square}$$ where implicit constraint C is a set of all inscribed squares. For inscribed square, its edge length should satisfy: $$l\cos \theta + l \sin\theta=L$$ Thus, the problem becomes $$\min_{\theta \in (0,\pi/2)} ... 0 Hint. You may use polar coodinates, x=r \cos \theta, y=r \sin \theta, then you initial function writes$$ \frac{\sin (r^2)}{r^2} $$and consider r \to 0. 0 Hint: Change to polar coordinates and use a well-known limit. One does not even need to go to polar coordinates, but that is a useful move in general when the denominator is x^2+y^2. (Your solution is correct.) 0 As mentioned in other answer, curl\,\vec F=0\; , so we can try to find its potential \;U(x,y,z)\;:$$U_x=6xy+4xz\implies U=3x^2y+2x^2z+C(y,z)\;,\;\;C(y,z)\;\text{a constant in}\;x\implies3x^2+2yz=U_y=3x^2+C_y(y,z)\implies C(y,z)=y^2z+K(z)=\text{ constant in y}\implies2x^2+y^2=U_z=2x^2+y^2+K'(z)\implies K(z)=T=\text{ constant}\implies$$... 1 Indeed \text{curl}\, \vec F=0, so \vec F is conservative. Then you have$$ \int_\Gamma \vec F\cdot d\vec r=g(1,1,2)-g(0,0,0), $$where g is a potential function for \vec F. One such function is g(x,y,z)=3x^2y+y^2z+2x^2z. Then$$ \int_\Gamma \vec F\cdot d\vec r=g(1,1,2)-g(0,0,0)=3+2+4=9. $$6 By setting x=au and y=bv the problem boils down to computing$$ I(a,b) = ab\iint_{\mathbb{R}^2}\sqrt{u^2+v^2} e^{-(u^2+v^2)}\,du\,dv = 2\pi ab \int_{0}^{+\infty} \rho^2 e^{-\rho^2}\,d\rho = \pi a b\cdot\Gamma\left(\frac{1}{2}\right).$$1 Since \Bbb R^3 is simply connected, such \phi will exist if and only if \nabla \times F = 0 , so it is always good to check if this is the case, so you don't waste time. You want to solve$$ \begin{cases} \frac{\partial \phi}{\partial x}(x,y,z) = 2xy + 4xz \\ \frac{\partial \phi}{\partial y}(x,y,z) = x^2 + 6yz \\ \frac{\partial \phi}{\partial z}(x,y,z) ...

0

The task poses a system of partial derivatives: $$\partial_x \phi = 2xy + 4xz \\ \partial_y \phi = x^2 + 6yz \\ \partial_z \phi = 2x^2 + 3y^2$$ Integration gives $$\phi = x^2y + 2x^2z + f(y,z) \\ \phi = x^2y + g(x,z) + 3y^2z \\ \phi = h(x,y) + 2x^2z + 3y^2z$$ Comparison gives $$\phi = x^2 y + 2 x^2 z + 3 y^2 z + C$$ where $C$ is some constant.

2

Just follow the definition: $\nabla \phi = (\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z})$ Hence, you have to solve: $$\begin{cases} 2xy + 4xz =\frac{\partial \phi}{\partial x} \\ x^2 + 6yz = \frac{\partial \phi}{\partial y} \\ 2x^2 + 3y^2 = \frac{\partial \phi}{\partial z} \end{cases}$$ The first ...

0

0

If you want to do this rigorously, you need to show that $$|f(x_0+h,y_0+k) - f(x_0,y_0) - \partial_xf(x_0,y_0)h - \partial_yf(x_0,y_0)k|= o(||(h,k)||).$$ Using the triangle inequality, $$|f(x_0+h,y_0+k) - f(x_0,y_0) - \partial_xf(x_0,y_0)h - \partial_yf(x_0,y_0)k| \\ \leqslant |f(x_0+h,y_0+k) - f(x_0,y_0+k) - \partial_xf(x_0,y_0)h| + | f(x_0,y_0+k)- ... 1 The objective function is non-negative. If you take \{x_n\} and \{y_n\} to be orthogonal, it is clear that the minimum is equal to 0. Take a_n>0 such that \sum_{n=2}^\infty a_n^2=s<\infty and let$$ x_n=a_n,\quad y_n=-a_n,\quad n\ge2, $$and x_1, y_1 such that x_1y_1=s. Then \sum_{n=1}^\infty x_ny_n=s-\sum_{n=2}^\infty a_n^2=0. 1 Assuming your * is matrix multiplication$$J = \pmatrix{4cx+1 & 0\cr 0 & 4cy+1\cr}$$so this will be a contraction with the Euclidean norm in any convex region where |4cx+1|<1 and |4cy+1|<1. For this to be true in your rectangle (x,y) \in [0.93, 1.52] \times [0.41, 1] you need -25/76 < c < 0. 0 Of course both o(dx) and o(dy) are o(||(dx, dy)||). Then you have only to prove that  \epsilon(dx)dy is o(||(dx, dy)||). Now$$ \frac{|\epsilon(dx)dy|}{||(dx, dy)||}\leq\frac{|\epsilon(dx)||dy|}{|dy|}=|\epsilon(dx)|\rightarrow 0 $$1 If we parameterize x and y as both equal to t, we are not guaranteed that the derivative with respect to t equals the derivative with respect to x or y. Using the rule for total derivative:$$ \frac{df}{dt}=\frac{\partial f}{\partial x}\frac{d x}{dt}+ \frac{\partial f}{\partial y}\frac{d y}{dt}. $$2 The direction vector v is a unit vector:$$ v = (1/\sqrt{2}) \, (1,1)^t $$and as f has a total derivative we get the directional derivative$$ \partial f / \partial v = \DeclareMathOperator{grad}{grad} \grad f \cdot v = (1,1)^t \cdot (1/\sqrt{2}) \, (1,1)^t = (1 / \sqrt{2}) \, 2 = \sqrt{2}  Checking the change of $f$ for the unit step in ...

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# Calculating EFL of a Three Lens System

1. ### nek26

1
If I have three lenses, each one 50 mm in diameter and with focal lengths of 500mm, -125mm and 150 mm, What would be the EFL if I combined them all together (assuming zero distance between them).

I have been looking online for focal lengths of three lens systems but cannot find anything. Would I just use the 1/f = 1/f1 + 1/f2 equation twice and basically turn two of the lenses into one focal length then use that focal length again with the third? Thanks.

2. ### Bob S

Here is the general matrix solution to a three lens system with arbitrary drifts between the lenses. A drift of length d is represented by a 2 x 2 matrix

1 d
0 1

and a thin lens is represented by

1 0
k 1

where k=-1/f.

I show the matrix equations, and a solution for three 25-cm FL lenses separated by two 2-cm drifts. For your case with no spacing between the lenses, the focus is 150 cm beyond the last lens. If the spacing between the lenses is 1 cm, the focal length is reduced to 96 cm. If the spacing is 0.1 cm, the focal length is 141.5 cm.

Bob S

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# Factoring polynomials and the Frobenius

Following Scott’s post, here is another post about some basic number theory that I haven’t seen written well in one place. First, a list of puzzles. In these puzzles, “polynomial” always means “polynomial with integer coefficients”:

• Give an example of a polynomial which has a root modulo every prime, but does not have a rational root.
• Give an example of a polynomial which has a nontrivial factorization modulo every prime, but is irreducible over the integers.
• Prove that, if a polynomial has at least k linear factors for every prime p, then it has at least k irreducible factors over the integers. Furthermore, if the degree of the polynomial is greater than k, then this lower bound can be improved to k+1
• Give an example of two irreducible polynomials f and g so that, modulo every prime, f and g have the same number of factors of the same degrees, but such that $\mathbb{Q}[x]/f(x)$ and $\mathbb{Q}[x]/g(x)$ are not isomorphic as fields.
• Everyone of these sounds like a problem about fields but, with a little bit of knowledge, they are actually problems about finite group actions. By the way, although every problem is solvable as written, they become easier if you allow yourself a finite number of exceptions instead requiring them to actually work for all primes, and I will only explain how to solve them with this proviso. I do assume that you know the basics of Galois theory.

Here is the main idea. Let f be a polynomial with integer coefficients and without repeated roots. To make my life simpler, I’ll assume that f has leading coefficient one, so that the roots of f are algebraic integers, although actually all I need to do in the more general case is to exclude those primes that divide the leading term of f. Let K be the splitting field of f, that is to say, the smallest extension of $\mathbb{Q}$ in which f has all of its roots. Let the roots of f be $\theta_1$, …, $\theta_n$. Define R to be the subring of K generated (over $\mathbf{Z}$) by the roots of f. Let G be the Galois group of K over $\mathbb{Q}$. Then G permutes the $\theta_i$ and hence acts on R. (Note for experts: the standard thing to do is to take the integral closure of $\mathbb{Z}$ inside K, rather than use R. This is undoubtedly the right thing to do, but I can avoid a lot of technical explanation by avoiding it.)

By definition, an $\overline{\mathbb{F}_p}$-point of $\mathrm{Spec} R$ is a ring homomorphism from R to the algebraic closure of $\mathbb{F}_p:=\mathbb{Z}/p$. (Yes, we are fixing an algebraic closure of $\mathbb{F}_p$.) I think that this is far too complicated a name for this idea so, in this blog post, I will call such a homomorphism a $p$-widget. If I really need to be explicit, I’ll call it a $p$-widget of f.

Now, $\Delta:=\prod_{i \neq j} (\theta_i - \theta_j)$ is an integer (since it is Galois invariant) and is nonzero (since we assumed that f doesn’t have repeated roots). There are only finitely many primes which divide $\Delta$; for any prime p which does not divide $\Delta$ we know that f continues to have distinct roots after reduction modulo p. I’ll call the primes dividing $\Delta$ the bad primes, and call all the others the good primes. For those trying to make the translation to standard terminology, “good” implies, but is not implied by, “unramified”.

What’s good about the good primes? Well, for one thing, let $\pi$ be any p-widget. If $\pi$ is good then $\pi(\theta_i) \neq \pi(\theta_j)$ for $i \neq j$. Why is this? Well, $\pi$ is a ring homomorphism, so $\pi(\Delta)=\prod_{i \neq j} (\pi(\theta_i) - \pi(\theta_j))$. But the right hand side is just the image in $\overline{F_p}$ of an integer which we assumed not divisible by p, so every factor on the right must be nonzero. Therefore, if p is a good prime, then G acts faithfully on $p$-widgets. It turns out, although I don’t see an easy proof, that G also acts transitively. I believe that this latter fact is due to Frobenius.

Now, there is also another action on $p$-widgets. Instead of acting on the ring R, we can act on the field $\overline{\mathbb{F}_p}$. Specifically, there is a map $x \mapsto x^p$ from $\overline{\mathbb{F}_p}$ to itself, called the Frobenius and we can turn one $p$-widget into another by post composing with the Frobenius. We’ll denote the Frobenius by $F$. If p is a good prime, and $\pi$ is a $p$-widget, then $F \circ \pi=\pi \circ g$ for some unique g in G, as we saw that G acts transitively and faithfully on the $p$-widgets. If, instead of $\pi$, we chose some other $p$-widget $\pi \circ h$, this would change the resulting value of $g$ to $h^{-1} g h$. In other words, the conjugacy class of g is determined by uniquely by p, and is known as the Frobenius class of p and denoted $Frob(p)$.

Now, here are the parts that I think could be laid out more explicitly in the books I’ve read. As those of you who are familiar with Galois theory know, the irreducible factors of f (over the rationals) correspond (in number and size) to the G-orbits in the set $\{ \theta_1, \ldots, \theta_n \}$. Now, similarly, F generates the Galois group of any finite extension of $\mathbb{F}_p$. Thus, the irreducible factors of f over $\mathbb{F}_p$ correspond to the orbits of $Frob(p)$ on the same set. Note that, although the actual orbits depend on the choice of a representative for $Frob(p)$, their number and size do not.

It shouldn’t be too hard to convince yourself of the following point, as well. If K is both the splitting field for f and for another polynomial g, and p is good with respect to both f and g, then $Frob(p)$ is the same element of G whether it is computed using f or g. (I will concede that this is one thing which is easier when one uses the standard approach in terms of integral closure.)

You now know enough to solve puzzles (1), (2) and (4). For example, puzzle (1) asks us to find a group G acting on a finite set S such that every element of G fixes a point of S, but no point of S is fixed by all of G. The simplest example is $G=\mathbb{Z}/2 \times \mathbb{Z}/2$ where S has three orbits, each stabilized by a different 2-element subgroup of G. Technically, to solve the puzzle we must actually find a field extension with this Galois group — the easiest solution is to take $f(x)=(x^2-2)(x^2-3)(x^2-6)$ so $K=\mathbb{Q}(\sqrt{2}, \sqrt{3})$, which has this Galois group. Puzzle (2) is only a little harder. For puzzle (4), it will help you to have the following hint: f and g should have the same splitting field.

To get puzzle (3), you need some elementary group theory (which I leave to you) plus a result that says that every conjugacy class of G occurs as $Frob(p)$ for some p. This is
Chebotarev’s Density Theorem : Let C be a conjugacy class in G. Then the density of primes p such that $Frob(p)=C$ is $|C|/|G|$.

Finally, let me make one more connection to standard terminology. The ring $R[\Delta^{-1}]$ is a so-called Dedekind domain, so every ideal factors uniquely as a product of prime ideals. (Again, I am inverting $\Delta$ in order to smear over my failure to take integral closures.) For a good prime p, the prime ideal factors of p correspond to the orbits of the $p$-widgets under the action of Frobenius. If $\pi$ is a $p$-widget, the corresponding prime ideal is the kernel of $\pi$. If I had chosen R correctly to begin with, then R itself would be a Dedekind domain and I could define the following standard objects for you: if $\mathfrak{p}$ is the kernel of a $p$-widget $\pi$, then the decomposition group of $\mathfrak{p}$ is the stabilizer of $\mathfrak{p}$ and the inertia group is the stabilizer of $\pi$.

## 12 thoughts on “Factoring polynomials and the Frobenius”

1. There is another way of seeing that $(x^2-2)(x^2-3)(x^2-6)$ has a factor modulo every (good) prime. Namely, at least one of 2,3, or 6 must be a square mod $p$ by quadratic reciprocity. (To avoid problems with $p=2, q=3$ one can instead choose a pair of primes whic are squares modulo each other.)

The group theory related to question 3 is interesting to me. Cannot one use it to say something more specialized: that if an irreducible polynomial has a root modulo every prime, then it has a rational root?

2. Cannot one use it to say something more specialized: that if an irreducible polynomial has a root modulo every prime, then it has a rational root?

By which you mean: any irreducible polynomial with roots mod p for all p is linear?

That’s right, because of an old lemma of Burnside’s, which says that the number of orbits of a G-set is the average over the group of the number of fixed points of each group element on that set. If X is a G-set on which each group element has at least one fixed point, then this average must be strictly larger than 1 (since the identity fixes more than one point, as long as X isn’t a single point).

3. Yes. What you said is equivalent to what I said.

Thanks for reminding me it was Burnside’s Lemma. According to Dummit and Foote (page 843, exercise 8) the result originated with Frobenius–which ties back nicely with the thread!

4. That should be an 8, followed by a parenthesis. Not a smiley face.

5. alex says:

Give an example of a polynomial which has a nontrivial factorization modulo every prime, but is irreducible over the integers.

======================================
I believe that irreducible polynomial

x^9-72*x^8-1056*x^7+13888*x^6+419328*x^5+64512*x^4 +53960704*x^3-190070784*x^2+746913792*x-4954783744

has a nontrivial factorization modulo every prime.

In fact, when prime p==2 mod 3
then factorization pattern is

1+1+1+2+2+2 if p==8 or 20 mod 21,

3+6 , otherwise

6. Jaime Montuerto says:

Dear David,

Your question about polynomials that has roots or factors modulo every prime is somewhat similar vein on my question to you about every prime numbers divides a Mersenne number. The difference is that every Mersenne numbers cannot be expressed by single polynomial. My take on your question is that there are no polynomials of those properties. A polynomial can have an infinite set of primes that divides it but not every primes in that set. The relationship of Mersennes are recurrence M_i+1 = M_i * 2 + 1.

My observation is that every prime numbers has distinct period/cyclic, in fact all integers have period.

Here’s the algorithm on calculating the period of integer N:

1. Start from any integer n < N.
2. double n, then modulo N, take note of the result.
3. If the result is not in the set yet then do process 2,ie double the result modulo N.

Obviously this won’t run forever, the limit will be the totient function of
N (Euler function) = N-1. If N is prime, but not all prime behave this way. Some integers examples: period of 23 is 11, 11 is 10, 15 is 4, 7 is 3.

4. the number of processes or doubling will be the period/cycle of N.

About a^2 + b^2, you asked what form are their factors. It seems that sum of squares are of 1 mod 4 forms and it seems to me that the factors are of the same forms, 1 mod 4, essentially a Proth like primes. What is your observation?

Regards

7. “… is a so-called Dirichlet domain”. That’s a new one for me. I guess the “so-called” allows you to get away with anything, but probably you meant “Dedekind domain”, right?

8. David Speyer says:

Oops! Thanks.

9. One substitution of “Dedekind” for “Dirichlet” down, one to go. (And I remembered what Dirichlet domains are: a certain kind of fundamental domain for the action of a discrete group on a symmetric space.)

10. Pingback: Polynomials ~


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# Browse Abstracts & Presentations - 2021 International Symposium on Molecular Spectroscopy by Subject "Atmospheric science"

• (International Symposium on Molecular Spectroscopy, 2021-06-24)
The Great Oxygenation Event, the introduction of O$_{2}$ into the Earth’s atmosphere approximately 2.5 billion years ago, is a critical stage in the development of life on Earth. The exact timing of this event is thought ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-25)
New particle formation (NPF) plays a crucial role in the formation of secondary aerosol particles, which in turn actively influence the climate radiative forcing balance. The magnitude of this impact is currently unclear, ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-24)
The accuracy of atmospheric trace gas retrievals depends directly on the accuracy of the molecular absorption model used within the retrieval algorithm. For remote sensing of well-mixed gases, such as carbon dioxide (CO$_2$), ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Several experimental studies have shown that 1,4-hydroxycarbonyls can undergo sequential reactions involving cyclization followed by dehydration to form dihydrofurans.\footnote{Atkinson, R. et al. Atmos. Environ. 2008, 42, ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-24)
\begin{wrapfigure}{r}{0pt} \includegraphics[scale=0.2]{Fig_nu4.eps} \end{wrapfigure} Silicon tetrafluoride (SiF$_4$) should be a normal trace component of volcanic gases. However, a better knowledge of spectroscopic ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-25)
The fundamentals of XH stretching modes typically lie in the 2,500 – 3,700 cm$^{-1}$ (2.7 - 4.0 $\mu$m) range. Continuous-wave optical parametric oscillators (CW OPOs) enable high-resolution spectroscopy in this region. A ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-25)
We used two previously described [1,2] cavity ring-down spectroscopy systems to accurately measure line intensities in the following three $^{12}$C$^{16}$O$_{2}$ rovibrational bands near 1.6 $\mu$m: (30012) $\leftarrow$ ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Greenhouse gases were monitored on an open-path over kilometer length with calibrated wavelength modulation spectroscopy (WMS). A movable platform was developed capable of detecting atmospheric variations of the concentrations ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-25)
Ozonolysis of alkenes is an important oxidation pathway of alkenes in the troposphere because it is involved in the production of organic aerosol and OH radicals. The mechanism of ozonolysis of alkenes involves the formation ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-24)
Vegetation is a major source of emission of Biogenic Volatile Organic Compounds (BVOCs), which play an important role in atmospheric chemistry. Apart from methane and isoprene, monoterpenes (\chem{C_{10}H_{16}}) are emitted ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-25)
We present measurements of molecular transition frequencies based on the comb-locked cavity ring-down spectroscopy technique [1], reporting vacuum transition frequencies of Doppler-broadened $^{12}$C$^{16}$O$_{2}$ in the ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-25)
For the first time, THz spectroscopy was used for a direct monitoring of molecular species in the Dunkirk atmospheric simulation chamber called CHARME for Chamber for the Atmospheric Reactivity and the Metrology of the ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-24)
The \~{B} - \~{X} laser-induced fluorescence (LIF) and dispersed fluorescence (DF) spectra of the atmospherically important $\beta$-monofluoro ethoxy (MFEO), $\beta$,$\beta$-difluoro ethoxy (DFEO), and $\beta$,$\beta$,$\ ... application/pdf PDF (20kB) • (International Symposium on Molecular Spectroscopy, 2021-06-24) Isoprene (C$_{5}$H$_{8}$) is an important molecular target in trace gas sensing due to its presence in Earth's atmosphere and human breath. Isoprene is the most abundant of the so-called biogenic volatile organic compounds ... application/pdf PDF (18kB) • (International Symposium on Molecular Spectroscopy, 2021-06-24) The version 4.0 dataset from the Atmospheric Chemistry Experiment – Fourier Transform Spectrometer (ACE-FTS) on SCISAT, released in March of 2019, has sulfur dioxide (SO$_{2}$) volume mixing ratio (VMR) profiles as a routine ... application/pdf PDF (18kB) • (International Symposium on Molecular Spectroscopy, 2021-06-25) New particle formation is a process in which trace vapors in the atmosphere cluster and grow into larger particles. Recently, we have employed a multi-trap gas-phase ion preparation technique and cryogenic ion vibrational ... application/pdf PDF (13kB) • (International Symposium on Molecular Spectroscopy, 2021-06-24) The pure rotational spectrum of 2-chloroethyl radical, CH$_{2}$ClCH$_{2}\$, has been observed for the first time using a cavity-based Fourier-transform microwave spectrometer in the frequency region of 10 GHz – 34 GHz. The ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-24)
After almost 18 years in low-Earth orbit, the ACE satellite is making near-real time measurements of numerous trace gases, thin clouds, aerosols and temperature by solar occultation. The primary instrument is a high-resolution ...

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• (International Symposium on Molecular Spectroscopy, 2021-06-24)
Absorption of solar radiation by water vapor in the near-UV region is a poorly-understood but important issue in atmospheric science. To better understand water vapor near-UV absorption, we constructed a cavity ring-down ...

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# How do you find the uncertainty in a function of two variables?

1. Sep 30, 2012

### JDoolin

I'm trying to make up an automatic spreadsheet for my Physics students so they can enter the data and see everything calculated for them, but I ran into a snag.

I have two random variables, h and t with uncertainties Δh, and Δt, respectively.

The uncertainty in h is simply estimated, based on the precision and technique used to get the data.

The uncertainty in t is taken as the standard deviation of 4 trials of an experiment.

With these two numbers, h and t, I'm generating the function $a=2h/t^2$. (Derived from $h=\frac{1}{2} a t^2$.

Here's my question: What would be the standard method for figuring the uncertainty in a?

I'm thinking the uncertainty in $t^2$ is $2 t \Delta t$ and the uncertainty in $h$ is $\Delta h$, and then you'd multiply these together to get the uncertainty in $a$.

Then to get an idea of uncertainty, percentage-wise, you just take that and divide by the average value of $a$ which is $2 h / \bar t^2$.

Does that sound like the right approach? I'm troubled because I feel like there's an implied factor, $\Delta h/h$ in the final result is going to always decrease the uncertainty. If this is done correctly, the uncertainty in the height should increase the total uncertainty.

2. Sep 30, 2012

### JDoolin

Another thought:

Using the product rule from calculus, $\Delta (2 h t^{-2})= (2 h) (-2 t^{-3}) \Delta t + 2 (\Delta h) (t^{-2})$, right?

3. Sep 30, 2012

### chiro

Hey JDoolin.

When you want to look at what a function of a random variable is and then talk about it's uncertainty, I'm assuming you are referring to either its variance or some kind of probability interval.

In the case that you are talking about variance, then what you want to calculate is Var[t^2] which is given as E[t^4] - E[t^2]^2 and you can calculate E[g(t)] in the standard way even if you have a sample of data with no known distribution as opposed to one with a distribution of an analytic form.

If this is truly a random variable, then this is the only real way you can quantify uncertainty with a single number (and it's typically what is done when quoting variance or standard deviation which is the square root of variance).

4. Sep 30, 2012

### chiro

You can't do this and it's one of the reasons why stochastic calculus needs all the higher mathematics tools and why all the "quants" for lack of a better word, need to know all this math (or at least be able to learn it quick enough to know what the theory says).

In the stochastic calculus model, you have a measure that is with respect to a random variable. In normal calculus your measure (if it's Riemannian) is the infinitesimal one and if it's Lebesgue it's the Lebesgue one.

The thing about the measure for random variables (the starting point is typically the Weiner process which you can think of as a Normal distribution) is that not only is this measure random, but it can also be negative as well as positive.

Because of these attributes, the chain rule that you take for granted doesn't work because these measures are completely different (the dWt is not the same as dt) and because of this, people needed to come up with a way to differentiate functions with respect to these random variables.

The big step for this after Weiner introduced his process was with Ito who showed a formula for derivatives of processes based on Weiner processes and this became useful for mathematical finance.

Now you can use approximation methods to get approximate values of stochastic integrals in the exact same way you use the trapazoidal method or simpsons rule to approximate a normal integral, but instead of the dt you use the difference in the values of the random numbers generated which means you will get all kinds of values (and this can help illustrate why you can't use calculus in the same way).

5. Sep 30, 2012

### JDoolin

Well, let's see, I think I've broken down the uncertainties into

(1) precision uncertainty--introduced by either limitations in the measurement device, or just a lack of precision in measurement on the part of the experimenter. Don't have any data on this one; the experimenter just has to just be honest about how closely he measured it, although with a meter-stick, you can probably put a minimum of around 1 millimeter.
(2) random uncertainty--introduced by limitations on the ability of the experimenter to start a timer as an object is released, and stop the timer when it gets to a certain mark.
(3) systematic uncertainty--introduced by many different things... For instance, (a) if the experimenter's eye isn't lined up with the mark, so the object actually doesn't fall all the way to the mark, (b) if the experimenter consistently starts the timer late or early because of the way the object is dropped.

I'm trying to get an estimate of uncertainty based on (1) and (2) and figure out how uncertainty in the two variables (height, and time) translate into the uncertainty in acceleration.

My end goal is to calculate the acceleration due to gravity. If the actual acceleration of gravity lies within the uncertainty, then there are probably not any significant systematic errors. But if the actual acceleration of gravity lies well outside the uncertainty, then the students need to explain what might have happened during their measurements to create systematic errors.

6. Sep 30, 2012

### chiro

Well precision uncertainty is usually something bounded (i.e. a constant) with a plus/minus value. You could give this a distribution if you wanted to (like a uniform distribution for example) but you don't necessarily have to.

In terms of the random error, a lot of this kind of error uses a normal distribution and for the kind of error you are describing, a normal distribution is a good model not only mathematically but physically as well (since you expect most values to cluster around a mean and get other values that decay as things get further). It may not be exact, but it is a good model to use (Remember: all models aren't correct but some are more useful than others).

In terms of 3 and the final error, you introduce random variables for each "error" and the total error becomes the sum of those errors. You can figure out the mean and variance of the sum of these errors using standard expectation and variance operator identities.

You can actually correlate these errors if you want to and make them dependent: it's up to you. Also you can obtain confidence intervals for the final random variable (remember they are just sums of individual ones) or get variance/standard-deviations and use that as well.

7. Sep 30, 2012

### JDoolin

Alright, let me tell you another way I thought of doing it.

Instead of dealing with the uncertainty in the time, directly, I calculate all four accelerations from each individual time. Then at the end, I calculate the standard-deviation in the acceleration. However, this does not include the uncertainty in the height.

In order to fix that, I take the uncertainty in the height as a fraction and add 1. $1+\frac{\Delta h}{h}$. Then I multiply this height uncertainty factor by the uncertainty in acceleration found earlier.

Just for clarification though, I'm not sure this is a stochastic process... The true quantity that the students are trying to measure is not random at all. However, the experimental results they have; the numbers that are written on the paper--that may be a random variable, after all. So I'm basically trying to get an estimate of the true number from a set of random numbers generated from that true number. If the experiment were done perfectly, I would expect the time to be exactly the same for each trial, within the precision capabilities of the measurement devices. Almost all of the randomness in the experiment comes from a lack of precision and accuracy in the measurements of the heights and times.

The experiment itself might be a stochastic process, but the number we're trying to estimate from it is a constant, if that helps.

8. Sep 30, 2012

### JDoolin

Let's quantify that a little more... Of course I can't directly add the uncertainty in h to the uncertainty in t or $t^2$, because they don't have the same units.

But the factor of Δh/h as a percentage uncertainty in the acceleration can be added on to the uncertainty in the acceleration due to the timing as a percentage increase.

That's what I was doing in my last post, by multiplying by $1+\frac{\Delta h}{h}$.

9. Sep 30, 2012

### chiro

I'll comment on the other points later, but this problem of a true parameter being a constant but being embedded in a sample is exactly what estimation in statistics is all about.

The one that most people are familiar with is estimating a population mean: the population mean is something that is fixed and doesn't change at all yet with a limited sample, we can't actually determine this with certainty so we need to construct an estimator that is the best estimator that we can use and this is exactly what has happened with t-tests, ANOVA's, chi-squared tests, F-tests, and all the results related to estimation and hypothesis testing (which is quite a lot).

So if for example your process was modelled by a normal distribution with the true value of the experiment being the mean, and the actual realizations that account for error being the possible values in the distribution, then given this assumption, you can take your sample and estimate either of these values or you could even assume that one of these values takes a specific value (for example if you know theoretically what the value should be for the real result you can set the population mean to that value and it becomes a known quantity).

In the case of the above, the only thing now is to estimate the variance (I'm still assuming normality, but this doesn't have to be the case: it's easy for introducing the methods and thinking).

So you have a known mean, you assume an underlying Normal distribution: you then check if you meet the assumptions for a chi-square test in which you can get a distribution involving the population variance and the sample variance which is a function of your observations.

This is an extremely simple model to account for random errors and is used in many statistical models as well as in error analysis (which is what physicists and scientists use to understand errors).

So point being is that you can take a sample and you can assume some distribution constraint (like normal). If you do you should check it meets the criteria by looking at the Histogram and getting some fit-test parameters (like Shapiro-Wilk, but also use other tests like the Jacque-Vera, Kolmogorov-Smirnov and anything to give indicators to support evidence for normality).

You can then assume a population parameter being equal to something (you might do this from a theoretical estimate using physics calculations) or you can take your sample and make an inference on what that parameter should be and then given some significance level, see what the values of this parameter are given a specific interval corresponding to a specific probability and then make up your mind on whether this makes sense physically.

The parameters for a normal are mean and variance and you can estimate both from the sample, or set any of them to a fixed value (as explained above) but from what you are saying, you probably will not fix the variance and you will want to estimate this even if you do fix the mean by using a theoretical physical calculation.

What I'd recommend though for the mean is that you do inference on the mean by doing your theoretical calculation and then testing the hypothesis that the population mean equals the value of the theoretical calculation and see if you fail to reject that.

If you do reject that, it means that the data supports the notion that the theory is wrong (you can't conclude this, but it provides evidence for this) if all things have been setup and done according to the right protocol.


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# Need help modifying a motion detector circuit

I'm working with my son on a 4-H project, and we have great ambitions, but very limited knowledge.

We're looking to use a motion detector to switch off the current to electromagnets that will hold the doggie door closed.

We found a Motion Triggered Room Light Kit at Jameco, and it looks like it would do the trick for us except that it's designed to turn power on when motion is detected - we need it to cut the power to the magnets so the door will swing.

The other thing we'd like to be able to do is modify it to use 2 motion detectors instead of one - one on each side of the door so it unlocks no matter which way the dog is coming through.

We'd also like to eliminate the light sensor portion, as we want it to work day and night. It looks like replacing the CDS photo resistor with a regular resister of the appropriate resistance (it looks like 100KΩ should do the trick, but I'm not 100% sure). We can simply leave the light sensor inside the box so it always thinks it's night if need be.

Here is the schematic provided with the kit:

Sorry it's not particularly good quality, I simply pulled the image from the PDF of the instructions I downloaded from Jameco

We can do the soldering and follow the very detailed instructions in the kit, but neither of us have enough electronics knowledge to be able to modify the circuit to do what we need.

This modification consists of two parts: an easy part and a slightly harder part :) You'll need a:

• 74HC32 OR gate IC in a DIP package
• Another PIR sensor, identical to that in the kit
• A scrap of perfboard/veroboard/...
• More of the phone wire called for by the kit
• And a 200kΩ, 1/4W resistor (Radded in the schematic below)

The first step is to enable the circuit to work with an ORed set of PIR sensors, while disabling the light sensor function. For this, you'll need parts listed above.

Make up two PIR sensors according to the kit instructions, but do not include the CdS photoresistor or R1, and do not connect them to the kit. You can then make up the OR gate perfboard according to the following schematic, with R1 being the 10kΩ resistor the kit has you install in the sensor enclosure:

simulate this circuit – Schematic created using CircuitLab

Once this is made up, the +5V, OPTO, MOTION, and GND leads from the perfboard can be wired to the mainboard in place of the original sensor, while the two modified sensors can be connected to +5V1/MOTION1/GND1 and +5V2/MOTION2/GND2 respectively. This hardwires the system in "dark" mode, and means that either or both motion sensors active will activate the relay.

Now, to the easy part of the modification, and that's changing it to turn OFF the device it's controlling when it's triggered. That's simply a matter of taking the wire from the relay to the HOT side of the output and wiring it to the normally closed contact of the relay (the currently-unused pin on the 2-pin end of the relay) instead of the normally open contact of the relay (the center pin on the 3 pin end of the relay).

• This looks fantastic, thank you! Parts are already in the shopping cart. Quick question - the new resistor is 200kΩ and you have it labeled Radded. What does that mean? Is it a specific type of resistor, or is that just the label you gave it? – FreeMan Jun 26 '15 at 3:56
• Also, can you confirm that this OR gate IC will work? And will this or this or this work for the resistor? If so, which would be preferable? – FreeMan Jun 26 '15 at 4:15
• Instead of using an OR IC, you could simply construct the two kits with the photosensor modification, and wire the normally closed contacts of the two relays in series. When either relay energises, the circuit is broken. – Nick Johnson Jun 26 '15 at 6:53
• Thanks for the suggestion @NickJohnson. As I said in OP: we have great ambitions, but very limited knowledge. Would you make an answer out of that comment and include a schematic? I can follow instructions, but I have very little idea of how to get there myself. – FreeMan Jun 26 '15 at 13:43
• @FreeMan -- use an OR gate IC in DIP instead, and you want a fixed resistor, not a variable one. (The 220kΩ resistor will do in a pinch -- Jameco only sells 200kΩ resistors in packs of 100 for some bloody reason...) – ThreePhaseEel Jun 26 '15 at 22:13

An even simpler solution than ThreePhaseEel's is to build two of the kits, with the CDS substituted with a resistor as you suggest. Additionally, modify the connection to the relay, so that power flows via the normally closed contact rather than the normally open contact.

Finally, plug the first kit into the power, and the second kit into the first kit. The output socket of the second kit will be on except when either kit detects motion.

• Thanks, Nick, that makes a lot of sense. Unfortunately, it's a bit pricey to use 2 complete kits, so I'm going to run with ThreePhaseEel's suggestion to just add one sensor. – FreeMan Jun 26 '15 at 23:11
• Fair enough! Different people have different tolerances for complexity and cost. You'll learn more with ThreePhaseEel's solution, too. – Nick Johnson Jun 27 '15 at 7:40


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# Why do some cations and anions interact with water to form corrresponding acids and bases(hydrolysis)? [duplicate]

I have seen in my textbook that on cations/anions formed on ionization of salts either exist as hydrated salts in aqueous solutions or interact with water to reform corresponding acids/bases depending upon nature of salts(Hydrolysis). It affects the pH of the solution.

For eg: $$\ce{CH3COONa -> CH3COO-_{(aq)} + Na+_{(aq)}}$$ Acetate ion formed undergoes hydrolysis in water to give acetic acid and $$OH^-$$ ions. $$\ce{CH3COO-_{(aq)} + H2O <-> CH3COOH_{(aq)} + OH-_{(aq)}}$$

It seems that this topic is oversimplified and needs some further explanation.

Why do some cations/anions interact with water to form corresponding acids and bases instead of existing independently in ionic form? (i.e. Why not acetate ion exist independently instead of reacting with water to form acetic acid?)

Question about the reaction

In the above example, why only the $$\ce{CH3COO-}$$ ions are getting hydrolysed when there are $${Na^+}$$ ions also?

I have seen the post Acidic and alkaline salt solutions: why do some salt form acid/bases while others don't? but it talks about the mechanism of the reaction and doesn't talk about the explaination for hydrolysis so I asked another question.


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# Finite State Automata for recognising consecutive characters

I'm currently working on this question as part of some homework, it has me stumped.

I'm familiar with finite state automata (FSA), I know how they work and I've read everything I can find on Google, but nothing's helped me come any closer to a solution.

If I don't know the length of the input string, or I'm not searching for a particular pattern, how can I design a machine that will always land on the final state?

I've tried drawing some, but that always end up being a little off.

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A.Schulz's answer gives you pretty much everything, but from the conceptual level, it may be worth recognising that you are looking for a particular pattern - $bb$. The difference is that you want to reject when you find it, rather than accept (and if you haven't found it, then everything is still okay). – Luke Mathieson Nov 17 '12 at 9:05
Thank you for your help, I got it! – Eogcloud Nov 17 '12 at 23:16

• $q_a$ represents: last character was not a b, and no bb yet
• $q_b$ represents: last character was a b, and no bb yet
• $q_{bb}$ represents: there was already a bb
Clearly, you start with $q_a$ and you accept in $q_a$ and $q_b$. Now try to figure out how to connect the states.


text:
## [TOPIC 3] LINEAR REGRESSION (3) – SOLVING NON-LINEAR PROBLEMS

Hi everyone! It has been a long time no see. I got extremely busy preparing my application for the deadline of Spring 2018 semester admission at universities to study at a higher level of education and do my further researches on the field of Machine Learning and Human-Computer Interface. It really was a difficult process, occupied most my time for travelling around and around the city for preparing necessary paperworks, writing statments like statment of purpose or study plan and sending emails to my professors to ask for recommendation letters.

## Introduction:

OK, just a little talk, to begin this topic, let’s remember, I introduced you how to implement a Linear Neuron Model to solve a linear problem with the aid of famous library Tensorflow. However, remember that how frequently do you face with the linear problem? Honestly, it rarely happens, even the “Materialism” problem related to the number of girl friend and monthly salary we dealt with in the first and second topic of “Linear Regression” is also not a pure linear problem. Let me show you what is a problem you get when using linear model for this. You might say that it was very good at trying to guess the number of my may-be-have girl friends (GF) in the first topic. But how about trying to use this linear model to predict the number of girl friends of the richest man in the world – Bill Gates (OK, this is just a joke, I mean no offence when I said that). From [1], we know Bill’s monthly income is $\216,666,666.00$, infer from the model, we get the number of GF he may have is $0.45753154 + 0.00087198*216,666,666.00 = 188,929.457$. How could it be real? Theoretically, may be. Practically, it’s infeasible, even Casanova couldn’t reach this number. Now, you see the problem here, if the monthly income of a man increase to infinite, the number of GF he has will increase to infinte too while the limit on the number of population worldwide or others such as morality, religion, etc. make it impossible. To deal with this, we need a restriction on the upper bound.

## Polynomial regression model

### Polynomial function

For our dataset, the polynomial model has form:

$\hat{y} = f(x, \omega) = \omega_{0} + x_{1} \omega_{1} + x_{1}^{2} \omega_{2} + ... + x_{1}^{n} \omega_{n} \indent (1)$

Or we can rewrite it into:

$\hat{y} = f(x, \omega) = \sum_{i=0}^{n}{x_{1}^{i} \omega_{i}}$

Or by matrix form:

$\hat{y} = f(x, \omega) = \bold{X} \omega \indent \indent \indent \indent (2)$

Let’s look at the equation (1), $f(x, \omega)$ is a nonlinear function of x though, it is a linear function of the coefficient $\omega$. Because of that, we can still implement the polynomial model as the same way as we do it with the linear one in the first topic. But now, we have the form of the matrix $\bold{X}$ is:

$\bold{X} = \begin{bmatrix} 1&x_{1}&x_{1}^{2}&...&x_{1}^{n}\\ 1&x_{2}&x_{2}^{2}&...&x_{2}^{n}\\..\\1&x_{k}&x_{k}^{2}&...&x_{k}^{n} \end{bmatrix}$

### Implementing polynomial model using TensorFlow

To form input vector after feeding it into the model, I wrote the function below.

Forming X data:

# Creating the input matrix X
def create_x(order):
global mans_monthly_income

# Creating element vector
X = np.ones((len(mans_monthly_income)), 1)
for i in range(1, order+1):
temp = np.power(mans_monthly_income, i)
x_temp = np.array(temp, dtype=float)
x_temp = np.transpose(np.matrix(x_temp))

X = np.column_stack((X, x_temp))

return X

Forming y data:

# Creating the target vector y
def create_y():
global total_number_of_gfs

y = np.array(total_number_of_gfs, dtype=float)
y = np.transpose(np.matrix(y))

return y

The results are showed below when using polynomial function with order of 3, 7 and 9, respectively:

### Testing the model

The question now is that how could we know which order is the best? To answer this, I made another dataset (red points in Fig. 5) for the purpose of testing our model by evaluating the root-mean-square (RMS) error (showed as Eq. (3)) each order from 0 to 10.

$E_{RMS} = \sqrt{\frac{1} {N} \sum_{i=1}^{N} {(\hat{y} - y)^{2}}} \indent \indent \indent \indent (3)$

##### Fig. 6 The relation between the value of RMS error and the order of polynomial function

From the Fig. 6, we can see that the RMS value of models with order higher than 8 begin increasing. In that case, we say our model was overfitted. By which, the model couldn’t work well with other test datasets (even it may work wrong), though it have done well with the training. To deal with this problem, there is some techniques can be used.

## Overfitting

### Increasing the size of the dataset

The first is to collect more and more data. This depends on the data acquisition step, sometimes we can, but sometimes not. There is a ‘lex non scripta’ supported by many is that the number of data points should be larger than some multiple (from 5 to 10 is OK) of the number of parameters in the model.

### Regularization

However, what happens if you are in the situation that you cannot acquire more data. In the case, I would like to introduce you a simple technique of regularization by adding $\frac{1} {2} \lambda \omega^{2}$ to the loss function $\mathcal{L} (\omega)$.  Now, the loss function become:

$\mathcal{L}(\omega) = \frac{1} {2} \sum_{i=1}^{N} {(y_{i} - \bold{x}_{i})^{2} \omega} + \frac{1} {2} \lambda \omega^{2} \indent \indent \indent \indent (4)$

The lambda in the adding element help us penalize the loss when the model become overfitting and keep the parameters small.

Turn Eq. (4) to the matrix form:

$\mathcal{L}(\omega) = \frac{1} {2} (\bold{y} - \bold{X} \omega)^{2} + \frac{1} {2} \lambda \omega^{2} \indent \indent \indent \indent (5)$

Calculating the derivation of Eq. (5) and set it equal 0 to find roots of the model:

$\frac{\partial{\mathcal{L} (\omega)}} {\partial{\omega}} = -\bold{X}^{T} (\bold{y} - \bold{X} \omega) + \lambda \omega = 0$

Similarly,

$(\bold{X}^{T} \bold{X} + \lambda \bold{I}) \omega = \bold{X}^{T} \bold{y}$

Assume that $(\bold{X}^{T} \bold{X} + \lambda \bold{I})$ is a nonsingular matrix, we have parameters of model is the root of the equation:

$\omega = (\bold{X}^{T} \bold{X} + \lambda \bold{I})^{-1} \bold{X}^{T} \bold{y}$

#### Implementing polynomial with regularization using Scikit-Learn lib

The part of code to implement the model with regularization using Scikit-Learn is showed below:

# Creating the model and feeding data to it
model = linear_model.Ridge(alpha=lamda)
model.fit(X, y)

# Getting the coefficients
coef = model.coef_
params = coef[0]
print params

#### Implementting polynomial with regularization using TensorFlow lib

The part of code to implement the model with regularization using TensorFlow is showed below:

# Creating constant opt to store data
X_tensor = tf.constant(X)
y_tensor = tf.constant(y)

''' Finding the root of the model '''
# Calculating Xt*X
first_step_1 = tf.matmul(tf.transpose(X_tensor), X_tensor)
# Calculating lamda.I
temp_step = (order+1)*lamda*np.identity(order+1)
first_step_2 = tf.constant(temp_step)
# Calculating Xt*X + lamda.I
# Calculating (Xt*X + lamda.I)^(-1)
second_step = tf.matrix_inverse(first_step)
# Calculating (Xt*X + lamda.I)^(-1)*Xt
third_step = tf.matmul(second_step, tf.transpose(X_tensor))
# Calculating (Xt*X + 2.lamda.I)^(-1)*Xt*y
root = tf.matmul(third_step, y_tensor)

''' Getting parameters '''
with tf.Session() as sess:
params = sess.run(root)

print params

## Sigmoid function

Another way to confront with nonlinear problems is to use functions having features suiting dataset features. This is based on your experience much.

Look at our dataset, on the one hand, we can see points distributing in a curve that is partly look like a function called “sigmoid” (as showed as Fig. 7).

##### Fig. 7 The graph of sigmoid function

On the other hand, we have a restriction on the upper bound, so it can be easily recognised that these features suit sigmoid’s features well. For the reason, now, we will use the model showed in Eq. (6) instead of the polynomial model as before.

$\hat{y} = f(x, \omega) = \omega_{0} + \frac{1} {1 + e^{-x}} \omega_{1} \indent \indent \indent \indent (6)$

However, note that the input x is very large in our dataset (up to 20,000), so $e^{-x}$ is very small and can be set as 0. For that reason, we need to add a trick to the Eq. (6) by dividing x by a constant number k. In this case, I use k = 4000, the Eq. (6) turn into:

$\hat{y} = f(x, \omega) = \omega_{0} + \frac{1} {1 + e^{-x/4000}} \omega_{1} \indent \indent \indent \indent (7)$

The new model is illustrated in Fig. 8.

##### Fig. 8 The sigmoid model

Apart from polynomial or sigmoid function, we can also apply others like trigonometic or logaric function for our regression model.

So, that’s all I want to talk about how to use linear model for regression tasks. In the next topic, I would introduce the way to use linear model, but for classification tasks.

Hope you enjoy it, goodbye, and see ya,

Curious Chick

## References

[2] Simon Rogers, Mark Girolami – A first course in Machine Learning

[3] Christopher M. Bishop – Pattern Recognition and Machine Learning

## Introduction

Hi everybody,

In the last post, I depicted how to implement linear regression based on inverse matrix approach or normal equation using both Scikit-learn and TensorFlow libs. Continue reading “[TOPIC 2] LINEAR REGRESSION (2) – NEURAL NETWORK APPROACH”

## [TOPIC 1] LINEAR REGRESSION (1) – MATRIX INVERSE APPROACH

Hi everybody,

Today, I would like to talk about the linear model for regression task in this post. While you have seperate labels (which may be (0 and 1) or (-1 and 1)) in classification tasks, the outputs in regression are continuous (i.e. they are arbitrary numbers lie in particular ranges). Continue reading “[TOPIC 1] LINEAR REGRESSION (1) – MATRIX INVERSE APPROACH”

## MACHINE LEARNING – WHAT IS THE DATA TELLING YOU?

Hi everyone,

Once I mention the concept of Machine Learning (ML) to my friends, who doesn’t work in the discipline of technology or data science, at least one of them think that it may be related to some characters on sci-fi like The Terminator, Wall-E, or recently, Ultron. In fact, it is partly true, that ultimately, the final purpose of ML is to make a robot that can be able to think and do like human beings, but now, it seem to be used mainly for analyzing data and getting information from it.

Nowadays, lots of technical companies have been bringing researches to the life (Natural Language Processing (NLP), Pattern Recognition, etc.), many businesses also used ML to analyze their customer and stock stock, and it can be said that ML become one of the sexist field in the world. Basically, there is two types of ML: supervised and unsupervised learning (see Fig. 1). For supervised learning, you need a datasets including input and output to train your algorithms, it is used for regression or classification tasks. By contrast, unsupervised learning don’t need output to train and it is used for dimensionality reduction or cluster tasks. For more details about ML, you can refer to these references [1], [2].

##### Fig.1 Two types of Machine Learning

Actually, learning ML is not as tough as you can think and if you just stay at using it, it is no more than a tool. The hardest works is that how to collect and preprocess the data before you feed it into ML algorithms. Each discipline have their own way to do that. For example, speaking from personal experience in building Brain – Computer interface systems (BCIs), the data in this case is EEG signal which is acquired from the scalp. Firstly, for data collection stage, what you need to do before collecting signal is to ask yourself some question to consider the purpose of collecting data, kind of what is the sampling rate you want to design the suitable hardware? Which position is chose to collect signal on your scalp to select the number of electrodes? etc. Secondly, for the preprocessing stage, of course, the raw data often have a lot of noise, so you need to use filters to denoise your datasets after that, then extract features into a feature input vector and feed it to ML algorithms. I can say that feature extraction is one of the most creative work when you try to analyze the data, and it require several knowledge about the field you work in. For example, come back to the field of BCIs, you may use some common descriptive statistics such as mean, median, standard deviation, etc. as features of data, but if you know a little bit about how your brain works, like the magnitude of alpha will increase and that of beta decrease when you are relax or in meditation, blah blah, and instead of statistical features, you now use the energy of alpha and beta from every electrode to make feature vector, it will be more efficient (not only decrease the size of feature vector but also increase the classification results). OK, you see that a lot of things to do before using ML. However, these works are beyond of this topic, so I will not discuss more.

In this series, I just focus on some popular algorithms in both supervised and unsupervised learning to help us solve some basic problems. And as I mentioned before, even your work is out of the field of technology, but if you are interested in ML, you can also learn and discuss it.

On the other hand, all codes in this series will be implemented using Scikit-learn and TensorFlow. A little bit about TensorFlow, it is low-level library which is more complicated than Scikit-learn to be used to implement ML algorithm. However, I would like to use it in some of my real projects because it is more efficient. The reasons of this is that TensorFlow sources was built by using C. Moreover, a computational graph is needed to drawn before running the ML algorithms. These help TensorFlow run much faster than Scikit-learn. Actually, there is other libs like Theano using the same ideas as TensorFlow but I would rather using it. It depends the taste.

Another thing is how to install TensorFlow? If you use Ubuntu or MAC OS, it wouldn’t be problem, prefer to [3] and [4] to get how to install it. However, there is a problem when you try to install TensorFlow in your Windows computer. The problem is that, TensorFlow is only support for Python 3.5.* on Windows. I tried it and it seemed not to work as expectation. For example, I wrote a simple code that print ‘Hello World’ to the screen using TensorFlow, but the line that is printed on the screen respond for byte instead of string type.

##### Fig. 2 TensorFlow on Python 3.5

On the other hand, as I mentioned in the first tutorial in the series of Basic Python Tutorial, there are some important libraries that cannot be used on Python 3.5.* such as Scipy. Thus, to solve this problem, I use Ubuntu on Virtual Machine. The instruction to install Virtual Machine is described on [5].

##### How to install Ubuntu on VirtualBox in Windows [5]

OK, so have you set up all need of software on your computer for ML? In the next tutorial, we will familiar to the simplest ML algorithms called Linear Regression.

Hope you enjoy it,

Curious Chick