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# linear_regression_simple

Simple model that learns W and b by minimizing mean squared errors via gradient descent.

library(tensorflow)

# Create 100 phony x, y data points, y = x * 0.1 + 0.3
x_data <- runif(100, min=0, max=1)
y_data <- x_data * 0.1 + 0.3

# Try to find values for W and b that compute y_data = W * x_data + b
# (We know that W should be 0.1 and b 0.3, but TensorFlow will
# figure that out for us.)
W <- tf$Variable(tf$random_uniform(shape(1L), -1.0, 1.0))
b <- tf$Variable(tf$zeros(shape(1L)))
y <- W * x_data + b

# Minimize the mean squared errors.
loss <- tf$reduce_mean((y - y_data) ^ 2) optimizer <- tf$train$GradientDescentOptimizer(0.5) train <- optimizer$minimize(loss)

# Launch the graph and initialize the variables.
sess = tf$Session() sess$run(tf$global_variables_initializer()) # Fit the line (Learns best fit is W: 0.1, b: 0.3) for (step in 1:201) { sess$run(train)
if (step %% 20 == 0)
cat(step, "-", sess$run(W), sess$run(b), "\n")
}


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# Homework Help: Differential Calculus - Question

1. Oct 17, 2011

### studentxlol

1. The problem statement, all variables and given/known data

I don't understand how this can be written like this:

$$\frac{1}{(2x+1)^{1/2}}-\frac{1}{(2x+1)^{3/2}}=\frac{2x+1-1}{(2x+1)^{3/2}}=\frac{2x}{(2x+1)^{3/2}}$$

What's the rule which makes this possible and explain please.

2. Oct 17, 2011

### HallsofIvy

The "rule" is one that you probably learned in the third grade: you add or subtract fractions by getting a "common denominator".

The first fraction has denominator $(2x+1)^{1/2}$. The second fraction has denominator $(2x+ 1)^{3/2}= (2x+1)^{1+ 1/2}= (2x+1)(2x+1)^{1/2}$.

In other words, the common denominator is $(2x+1)^{3/2}$ and you get it by multiplying the numerator and denominator of the first fraction by 2x+1.

format: plain


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# Order of element in a group

In Fraleigh's example he states:

I get pretty much everything that is happening, except I am confused by one fact. How does he know he know that $\mathbb Z_4 \times \mathbb Z_3$ has an element of order $4$ while $\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_3$ does not? I know the maximum possible order of an element in $\mathbb Z_4 \times \mathbb Z_3$ is $\text{lcm}(4,3)=12$ while the maximum possible order of element in $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3$ is $\text{lcm}(2,2,3)=6$. How do I specifically know just by looking at $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3$, that it does not have an element of order $4$. Is there a theorem which states the different possible orders of elements of a group such as $\mathbb Z_2\times \mathbb Z_2\times \mathbb Z_3$? Or would I just have to write down all the possible elements of each group and calculate the order by hand of each element? Thank you.

## 4 Answers

You're quite close in recognizing that least common multiples play a central role. The key fact is that if we have a product group $\displaystyle G = \displaystyle \prod_{k=1}^n G_k$, then the order of an element $(x_1, ..., x_n) \in G$ is precisely $\text{lcm}(|x_1|, ..., |x_n|)$ where $|x_k|$ denotes the order of $x_k$ in its respective group. So if we want to determine the possible orders of elements in $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3$, we just need to consider the possible orders of elements in $\mathbb{Z}_2$ and in $\mathbb{Z}_3$ and see if the least common multiple of any triplet of possibilities can be $4$.

If we were to take a more naive approach by trying to find the order of each element by hand, it would be perhaps easier to recognize the isomorphism$^\dagger$ $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \cong \mathbb{Z}_2 \times \mathbb{Z}_6$ before one begins (fewer coordinates to deal with). However, as one attempts to find the order of each element in the latter product manually, it quickly becomes apparent, even to a previously-uninformed student, that the rule discussed above holds!

$^\dagger$This comes from this fact.

• So is the possible of order of elements in Z2: 1 and 2 and the possible order of elements in Z3: 1,2,3? – rain Mar 20 '18 at 9:46
• I see, thanks, makes perfect sense. Thanks. – rain Mar 20 '18 at 9:50
• @rain, amidst my tiredness, I made an error. You are correct about the order of elements in Z2, but the order of an element in Z3 is going to be either $1$ (in the case of the identity) or $3$ since $1+1 = 2 \quad \rightarrow \quad 2+1 = 0$ and $2+2=1 \quad \rightarrow \quad 1+2 =0$. This doesn't change the fundamental argument though. – Kaj Hansen Mar 20 '18 at 11:19
• Thanks, makes sense. – rain Mar 21 '18 at 1:03

That's quite simple: a triple $(x,y,z)$ satisfies $4(x,y,z)=(4x,4y,4z)=0$ if and only if $$\begin{cases}4x=0\\4y=0\\4z=0\end{cases}.$$ However $4$ is a unit mod.3, so $4z=0$ implies $z=0$, and you element is $(x,y,0)$. But $x$ and $y$ have order $1$ or $2$, so $(x,y,0)$ has order $1$ or $2$.

Proof by direct calculation is quite simple here. Let $(a,b,c) \in Z_2 \times Z_2 \times Z_3$. Then

$4(a,b,c) = (0,0,0)$

$\Rightarrow (4a,4b,4c)=(0,0,0)$

$\Rightarrow 4c = 0 \mod 3$

$\Rightarrow c=0 \mod 3$

but any $(a,b,0) \in Z_2 \times Z_2 \times Z_3$ has an order of 1 or 2, not 4. So no element of $Z_2 \times Z_2 \times Z_3$ has an order of 4.

Not only the maximum possible order of an element in $\mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3$ is lcm $(2,2,3)=6$, but we actually have $6a=0$ for all $a \in \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_3$. Therefore, the order of every element divides $6$ and so cannot be $4$.


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# Changing the basis of a transformation

Given $T: \mathbb{R}^2 \to \mathbb{R}^2 : T\begin{bmatrix} x \\ y \end{bmatrix} \to \begin{bmatrix} 2x+y \\ x-3y \end{bmatrix}$ with standard basis $\mathcal{B}$ and basis $$\mathcal{B'}=\left\{\begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 2 \end{bmatrix} \right\}$$ Find the matrix $[T]_\mathcal{B'}^\mathcal{B'}$ representing the transformation $T$ with respect to the basis $\mathcal{B'}$ by making use of a change of basis matrix. Then calculate $[T]_\mathcal{B'}^\mathcal{B'}$ directly and check that you get the same answer.

Method 1

$T \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix} \qquad \quad T \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ -3 \end{bmatrix}$

$\Rightarrow [T]_\mathcal{B}^\mathcal{B} = \left[\begin{array}{rr} 2 & 1 \\ 1 & -3 \end{array}\right]$

To find the transition matrix $[I]_\mathcal{B}^\mathcal{B'}$ from $\mathcal{B}$ to $\mathcal{B'}$ I row reduced the augmented coefficient matrix $$\left[\begin{array}{cc|cc} 1 & 1 & 1 & 0 \\ 1 & 2 & 0 & 1 \end{array}\right] \rightarrow \left[\begin{array}{cc|rr} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & 1 \end{array}\right]^*$$ So $[I]_\mathcal{B}^\mathcal{B'}= \left[\begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array}\right]$ and $\left([I]_\mathcal{B}^\mathcal{B'}\right)^{-1}= \left[\begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array}\right]$

Now, $[T]_\mathcal{B'}^\mathcal{B'}= \left([I]_\mathcal{B}^\mathcal{B'}\right)^{-1} [T]_\mathcal{B}^\mathcal{B} [I]_\mathcal{B}^\mathcal{B'} = \left[\begin{array}{rr} 1 & 1 \\ 1 & 2 \end{array}\right] \left[\begin{array}{rr} 2 & 1 \\ 1 & -3 \end{array}\right] \left[\begin{array}{rr} 2 & -1 \\ -1 & 1 \end{array}\right] = \left[\begin{array}{rr} 8 & -5 \\ 13 & 9 \end{array}\right]$

Method 2

$T \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ -2 \end{bmatrix} \qquad \quad T \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 4 \\ -5 \end{bmatrix}$

$\Rightarrow [T]_\mathcal{B'}^\mathcal{B'}= \left[\begin{array}{rr} 3 & 4 \\ -2 & -5 \end{array}\right]$

Why are my solutions different?

• I'm not going to accept an answer to this question in the foreseeable future because I don't understand linear algebra well enough to understand the correct answers. – ahorn May 24 '16 at 11:21

$T(1,1)$ and $T(1,2)$ should be expressed in basis $\mathcal B'$, not $\mathcal B$.

Let $\{e_1, e_2\}$ be the canonical basis. Set $u_1=\begin{bmatrix}1\\1 \end{bmatrix}=e_1+e_2$, $\,u_2=\begin{bmatrix} 1\\2\end{bmatrix}=e_1+2e_2$. From these, you deduce: $$e_1=2u_1-u_2,\quad e_2=u_2-u_1.$$

You've proved $\,T(u_1)= 3e_1-2e_2$, $\,T(u_2)=4e_1-5e_2$, whence: $$T(u_1)=8u_1-5u_2,\quad T(u_2)=13u_1-9u_2$$ so that $$T_\mathcal B^\mathcal{B'}=\begin{bmatrix}8&13\\-5&-9 \end{bmatrix}.$$

Incidentally, we proved the change of basis matrix from $\mathcal B'$ to $\mathcal B$ is $\,\begin{bmatrix}2&-1\\-1&1 \end{bmatrix}$.

• Did I calculate my transition matrix in Method 1 correctly? I think that it is wrong, and I should have gotten the transition matrix simply by putting the basis vectors of $\mathcal{B'}$ together to form $\begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}$ – ahorn Apr 27 '15 at 10:18
• The transition matrix I got in Method 1, $[T]_\mathcal{B}^\mathcal{B'}$, is supposed to be used by multiplying it with a vector in $\mathcal{B}$ to get a vector in $\mathcal{B'}$. Thus, it is not the transition matrix but rather the inverse. – ahorn Apr 27 '15 at 10:37
• What exactly did I do wrong in method 2? – ahorn Apr 27 '15 at 10:42
• Nothing is wrong. There's one more step to do. – Bernard Apr 27 '15 at 11:05
• What does $\begin{bmatrix} 3 & 4 \\ -2 & -5 \end{bmatrix}$ represent? – ahorn Apr 27 '15 at 11:10


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# The space between two 15 cm long concentric cylinders is filled with glycerine (viscosity 0.5...

MEC3102-2019-S1- assignment 1 (due date 18 April 2019, Weighting 15%)
Q1: The space between two 15 cm long concentric cylinders is filled with glycerine
(viscosity 0.5 N.s/m2
), figure 1. The inner cylinder has a radius R = 7 cm and the gap
width between the cylinders a =b= 0.25 cm, Determine the torque and the power required
to rotate the inner cylinder at ?=180 rev/min. The outer cylinder is fixed. Assume the
velocity distribution in the gap to be linear.
(50 points)
Figure 1
Q2: The rigid gate OAB of figure 2 is hinged at O and rests against a rigid support at B.
What minimum horizontal force, P, is required to hold the gate closed if its width is 3 m?
Neglect the weight of the gate and friction in the hinge. The back of the gate is exposed
to the atmosphere.
(50 points)
Figure 2
Q3: Water at 20° C is flowing in the pipe and out to the atmosphere as in figure 3. If friction can be
neglected, calculate the flow rate.
Note that the elevation given in the figure is in meters and the Mercury (Hg) have specific gravity of
13.55.
(50 points)
Figure 3
Document Preview:

MEC3102-2019-S1- assignment 1 (due date 18 April 2019, Weighting 15%) Q1: The space between two 15 cm long concentric cylinders is filled with glycerine 2 (viscosity 0.5 N.s/m ), figure 1. The inner cylinder has a radius R = 7 cm and the gap width between the cylinders a =b= 0.25 cm, Determine the torque and the power required to rotate the inner cylinder at ?=180 rev/min. The outer cylinder is fixed. Assume the velocity distribution in the gap to be linear. (50 points) Figure 1Q2: The rigid gate OAB of figure 2 is hinged at O and rests against a rigid support at B. What minimum horizontal force, P, is required to hold the gate closed if its width is 3 m? Neglect the weight of the gate and friction in the hinge. The back of the gate is exposed to the atmosphere. (50 points) Figure 2Q3: Water at 20° C is flowing in the pipe and out to the atmosphere as in figure 3. If friction can be neglected, calculate the flow rate. Note that the elevation given in the figure is in meters and the Mercury (Hg) have specific gravity of 13.55. (50 points) Figure 3

Attachments:


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# A Potential Energy

Tags:
1. Jul 13, 2016

### observer1

The form of the Lagrangian is: L = K - U

When cast in terms of generalized coordinates, the kinetic energy (K) can be a function of the rates of generalized coordinates AND the coordinates themselves (velocity and position); a case would be a double pendulum.

However, the potential energy (U) is ONLY a function of the generalized coordinates and NOT the rates: position, NOT velocity.

Yes, I understand that the potential energy is the potential function whose negative gradient delivers the force (from the path independence of the work done) and I can anticipate this will only be function of position and not velocity.

However, could anyone extend a few descriptive words - not equations: I got those - on why it is natural to expect the potential energy NOT to depend on velocities. Because... I am not really content with my explanation and am looking for something... just a few more words so I can rest more assured.

2. Jul 13, 2016

### wrobel

Your question is about the definitions. Actually there is also a generalized potential $V=V(q,\dot q)$, it is determined as follows $Q_j=\frac{d}{dt}\frac{\partial V}{\partial \dot q^j}-\frac{\partial V}{\partial q^j}$. In classical mech. $V$ depends on $\dot q$ linearly: $V=v_i(q)\dot q^i+w(q)$. For example the Lorentz force possesses the generalized potential.

3. Jul 13, 2016

### observer1

OK, this is new to me. If, as you say (and from experience on your responses, I have no reason to doubt you) in classical mechanics, the potential can be a function of velocity, can you give me an example? Springs? dashpots? gravity? mechanisms? I just don't see how. I am a bit shocked by your response because it was not what I was expecting. For if, in classical mechanics, the force must be the negative of the gradient of the potential function, I cannot see how the potential function can be a function of velocities.

4. Jul 13, 2016

### wrobel

I have brought an example, another example is the Coriolis force if you are writing the equations in a non-inertial frame

5. Jul 13, 2016

### observer1

Wow... a flash of new insight... Can I say it in my words and ask you to assess my words?

We stand on a rotating merry go round and fire a ball outward and observe the motion of the ball from the rotating frame. It will exhibit unexpected motion due to the Coriolis effect. We relegate this as force and call it the Coriolis Force. Now if we solve the equations of motion of this system using variational methods (Hamilton's principle), then the potential energy is a function of the velocity.

If those words are correct, I am afraid I still do not see it. How can the potential energy be a function of the velocity of the ball if we are observing it in the rotating frame?

6. Jul 13, 2016

### wrobel

Assume we have a particle of mass $m$ which is undergone to the force $\boldsymbol F=\boldsymbol B\times\boldsymbol v,\quad \boldsymbol v=(\dot x,\dot y,\dot z)$. For simplicity suppose that $\boldsymbol B=(B_x,B_y,B_z)=const$.
Then the equations of motion of this particle are the Lagrange equations with the Lagrangian $L=\frac{m}{2}(\dot x^2+\dot y^2+\dot z^2)-V,$ where
$$-2V=(-yB_z+zB_y)\dot x+(-zB_x+xB_z)\dot y+(yB_x-xB_y)\dot z.$$

7. Jul 13, 2016

### observer1

Oh!

Thank you!

8. Jul 13, 2016

### vanhees71

This is not always correct. The form $L=K-U$, where $K$ is the kinetic energy and $U$ a potential is only true when the kinetic-energy term is quadratic in the (generalized) velocities. In any case you have to determine the form of energy from Noether's theorem applied to time-translation invariance, leading to the Hamiltonian
$$H=p_k \dot{q}^k-L, \quad p_k=\frac{\partial L}{\partial \dot{q}^k}.$$
In relativistic physics usually the interaction part also depends on $\dot{q}^k$, e.g., the Lagrangian for the motion of a relativsitic charged particle in an external em. field in Cartesian coordinates reads
$$L=-m c^2 \sqrt{1-\vec{v}^2/c^2}-\frac{q}{c} \dot{x}^{\mu} A_{\mu}, \quad (x^{\mu})=(ct,\vec{x}).$$
As you can easily check, in this case $L \neq K-U$.


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testOutliers {DHARMa} R Documentation

## Test for outliers

### Description

This function tests if the number of observations outside the simulatio envelope are larger or smaller than expected

### Usage

testOutliers(simulationOutput, alternative = c("two.sided", "greater",
"less"), margin = c("both", "upper", "lower"), type = c("default",
"bootstrap", "binomial"), nBoot = 100, plot = T)

### Arguments

 simulationOutput an object of class DHARMa, either created via simulateResiduals for supported models or by createDHARMa for simulations created outside DHARMa, or a supported model. Providing a supported model directly is discouraged, because simulation settings cannot be changed in this case. alternative a character string specifying whether the test should test if observations are "greater", "less" or "two.sided" (default) compared to the simulated null hypothesis margin whether to test for outliers only at the lower, only at the upper, or both sides (default) of the simulated data distribution type either default, bootstrap or binomial. See details nBoot number of boostrap replicates. Only used ot type = "bootstrap" plot if T, the function will create an additional plot

### Details

DHARMa residuals are created by simulating from the fitted model, and comparing the simulated values to the observed data. It can occur that all simulated values are higher or smaller than the observed data, in which case they get the residual value of 0 and 1, respectively. I refer to these values as simulation outliers, or simply outliers.

Because no data was simulated in the range of the observed value, we don't know "how strongly" these values deviate from the model expectation, so the term "outlier" should be used with a grain of salt. It is not a judgment about the magnitude of the residual deviation, but simply a dichotomous sign that we are outside the simulated range. Moreover, the number of outliers will decrease as we increase the number of simulations.

To test if the outliers are a concern, testOutliers implements 2 options (bootstrap, binomial), which can be chosen via the parameter "type". The third option (default) chooses bootstrap for integer-valued distribubtions with nObs < 500, and else binomial.

The binomial test considers that under the null hypothesis that the model is correct, and for continuous distributions (i.e. data and the model distribution are identical and continous), the probability that a given observation is higher than all simulations is 1/(nSim +1), and binomial distributed. The testOutlier function can test this null hypothesis via type = "binomial". In principle, it would be nice if we could extend this idea to integer-valued distributions, which are randomized via the PIT procedure (see simulateResiduals), the rate of "true" outliers is more difficult to calculate, and in general not 1/(nSim +1). The testOutlier function implements a small tweak that calculates the rate of residuals that are closer than 1/(nSim+1) to the 0/1 border, which roughly occur at a rate of nData /(nSim +1). This approximate value, however, is generally not exact, and may be particularly off non-bounded integer-valued distributions (such as Poisson or neg binom).

For this reason, the testOutlier function implements an alternative procedure that uses the bootstrap to generate a simulation-based expectation for the outliers. It is recommended to use the bootstrap for integer-valued distributions (and integer-valued only, because it has no advantage for continuous distributions, ideally with reasonably high values of nSim and nBoot (I recommend at least 1000 for both). Because of the high runtime, however, this option is switched off for type = default when nObs > 500.

Both binomial or bootstrap generate a null expectation, and then test for an excess or lack of outliers. Per default, testOutliers() looks for both, so if you get a significant p-value, you have to check if you have to many or too few outliers. An excess of outliers is to be interpreted as too many values outside the simulation envelope. This could be caused by overdispersion, or by what we classically call outliers. A lack of outliers would be caused, for example, by underdispersion.

### Author(s)

Florian Hartig

testResiduals, testUniformity, testOutliers, testDispersion, testZeroInflation, testGeneric, testTemporalAutocorrelation, testSpatialAutocorrelation, testQuantiles, testCategorical


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# Showing that a sequence is exact

Proof: We need to show that $0 \rightarrow \ker f \rightarrow M$ and $N \rightarrow \coker f \rightarrow 0$ are both exact. Firstly, let $\iota: \ker f \rightarrow M$ be the inclusion. Then $0 \rightarrow \ker f \rightarrow M$ is injective since $\ker f$ is exact, i.e. \ker f = \img j = ${0}$ where $j$ sends $0$ to the $\ker f$.Thus exactness follows. Next let $\pi: N \rightarrow \coker f$ be the projection onto the quotient $N/\img (f)$. Then $\coker f$ sends everything in the $\img(f)$ to $0$. So $\img(f) \rightarrow 0$. Thus $N/\img(f) = N/0 \cong N$.

This is where I'm stuck and confused. I'm reading from Rotman's Introduction to Homological Algebra and he doesn't have many problems showing how to deal with these type of problems and in terms of proving, I'm not entirely sure how to approach these problems.

In terms of the above problem I'm confused with whether or not I'm approaching it correctly and if it makes any sense.

Any help or advice would be greatly appreciated.

• This is correct but you also need to show that $\ker(f) \to M \to N$ and $M \to N \to \text{coker}(f)$ is exact. May 18 '18 at 16:30
• I originally thought about doing exactly that but wasn't sure May 18 '18 at 16:32

To solve the problem, go back to the very definition of an exact sequence. Do as you said: you consider the inclusion map $\iota:\ker f\to M$, the map $f:M\to N$, the quotient map $\pi:N\to\text{coker} f$ and the zero map $\text{coker}f\to0$. Then, in this order, prove that the image of one is the kernel of the next.


text:
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### Course: 6th grade (Eureka Math/EngageNY)>Unit 1

Lesson 1: Topic A: Representing and reasoning about ratios

# Basic ratios

## Problem

The table shows how many hours Sara spent at several activities one Saturday.
ActivityHours
Soccer practice2
Birthday party3
Science project1
What is the ratio of hours spent at soccer practice to hours spent at a birthday party?


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• anonymous
If a merchant has two types of tea, one work $2.70 per kilogram and the other worth$3.00 per kilogram, how many kilograms of each type should the merchant use in order to produce 30 kilograms of a blend that is worth \$2.95 per kilogram?
Mathematics

Looking for something else?

Not the answer you are looking for? Search for more explanations.


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# How does one prove $(1+x/n)^n\to e^x$ pointwise through series expansion?

We have $$e^x=\sum_{k=0}^\infty \frac{x^k}{k!}$$ and $$\left(1+\frac{x}{n}\right)^n=\sum_{k=0}^n \frac1{n^k}{n\choose k}x^k$$ therefore $$\left\lvert e^x-\left(1+\frac{x}{n}\right)^n\right\rvert\le\sum_{k=n+1}^\infty \frac{\lvert x\rvert^k}{k!}+\sum_{k=0}^n \lvert x\rvert^k\left\lvert\frac1{n^k}{n\choose k}-\frac1{k!}\right\rvert.$$

Question: As $n\to\infty$, the first sum trivially goes to $0$, but how to properly bound the second one?

I can show convergence in different ways but I'm interested in this specific approach.

• Interesting question! To avoid getting other methods, I would specifically write down: **Question **: how can one show that the sum (the second sum in your question) goes to zero? (does it even go to zero?). Because there are many things one can do before the triangle inequality step to prove the claim, but that boils down doing the classic proof. That is what I think at least. Dec 12 '17 at 10:23
• This is one of the standard ways of proving the equality of the exponential limit and the exponential series and if you search enough you will find this approach in detail on this website. Dec 12 '17 at 12:41
• @Shashi I've followed your advice, thanks Dec 12 '17 at 18:10
• See this answer math.stackexchange.com/a/1898375/72031 Dec 16 '17 at 7:04

\begin{align}\sum_{k=0}^n \lvert x\rvert^k\left\lvert\frac1{n^k}{n\choose k}-\frac1{k!}\right\rvert &\le\sum_{k=1}^{\infty}\frac{\lvert x\rvert^k}{k!} \underbrace{\left\lvert \big(\tfrac{n\cdot(n-1)\cdot\ldots\cdot(n-(k-1))}{n^k }- 1\big) \right\rvert}_{:=a_k(n)} \end{align}

Now let $\epsilon>0$ be arbitrary.

On the one hand $0\le a_k(n)\le 1$ for all $k$, hence the above series is bounded by $\exp(|x|)-1$. In particular there exists some $N$ such that $\sum_{k=N}^\infty a_k(n)\frac{|x|^k}{k!} \le \sum_{k=N}^\infty \frac{|x|^k}{k!} < \epsilon$.

On the other hand, since $a_k(n) \to 0$ for all $k$ we can then also find some $n_\epsilon$ such that $a_k(n_\epsilon) < \epsilon$ for all $k<N$. Putting the two together we conclude

\begin{align} \sum_{k=1}^{\infty}a_k(n_\epsilon)\frac{|x|^k}{k!} \le \sum_{k=1}^{N-1}\epsilon\frac{|x|^k}{k!} + \sum_{k=N}^{\infty}\frac{|x|^k}{k!} \le \epsilon(\exp(|x|)-1) + \epsilon = \exp(|x|)\epsilon \end{align}

Which demonstrates point-wise convergence.

• Thanks! +1, I understand and appreciate your justification. I was wondering, since the infinite sum at the beginning converges for all $n$ and doesn't depend on $n$, couldn't we also directly take the limit as $n\to\infty$ and get $0$ as $a_k(n)\to 0$ for all $k$? Dec 12 '17 at 13:52
• @Richard I did that in an earlier answer that I deleted; user Shashi rightly pointed out that the swapping of the limit with the sum needs some justification. This could be done by a uniform convergence argument, but I then rather chose to give the elementary approach that doesn't need any more machinery. Dec 12 '17 at 15:36
• Didn't you do that when the sum still depended on $n$? I mean, the infinite sum does not, so taking the limit there should be safe, right? Dec 12 '17 at 17:24
• To me it doesn't look different from $\sum_{k=0}^\infty \frac{x^k }{(n+k)!} \to 0$ as $n\to\infty$ Dec 12 '17 at 17:32
• Sorry to bother you, could you please consider my doubt? Dec 13 '17 at 18:13

All the coefficients in the 2nd sum are non-positive. Indeed,

\begin{multline*}\frac{1}{n^k}\binom{n}{k}-\frac{1}{k!}=\frac{1}{k!}\left(\frac{n!}{n^k(n-k)!}-1\right)=\frac{1}{k!}\left(\frac{(n-k+1)(n-k+2)\cdot\ldots\cdot n}{n^k}-1\right)\\\le \frac{1}{k!}\left(\frac{n^k}{n^k}-1\right) = 0.\end{multline*}

Performing your computation without absolute value and observing that for $x>0$ we have the difference on the left is positive, we have done for $x>0$. It seems to me that for $x<0$ the similar argument will work (with the reverted inequalities). Please check it. For $x=0$ the checking is direct.


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# Upper and Lower Fence Calculator

Created by Rijk de Wet
Reviewed by Wojciech Sas, PhD and Steven Wooding
Last updated: Dec 12, 2022

Welcome to the upper and lower fence calculator, where we'll discuss the use of fences in statistics and show you how to find the lower fence and the upper fence of a dataset. These fences are vital to finding that pesky outlier in your dataset.

## What is the upper and lower fence?

Before we can make sense of our upper and lower fence calculator, we must define what fences in statistics mean. The upper and lower fences of a dataset are the thresholds, outside of which values can be considered outliers. Outliers, therefore, are any values that fall below the lower fence or above the upper fence.

## The use of fences in statistics

Besides helping us find outliers, fences can be a suitable replacement for the minimum and maximum in descriptive statistics. In most cases, box plots (a helpful tool to visualize a dataset) use the minimum and maximum as the box's whiskers. It's much more insightful, though, to use the upper and lower fences for the whiskers and then indicate outlying values with distinct points. Take a look at the figures below:

## How to calculate the upper and lower fence

Now that we know what the fences are, we'd love to know exactly how to find the upper and lower fence. There are a few steps involved, so let's get started!

Before we can get to the upper and lower fence formulas, we need two basic but important values: the dataset's quartiles. We denote the first quartile with $Q_1$ and the third quartile with $Q_3$. We can then calculate the interquartile range (IQR) as $\text{IQR} = Q_3 - Q_1$.

Now that we have the building blocks, how do we find the lower fence, and how do we find the upper fence?

• The formula for the upper fence is $\text{Upper fence} = Q_3 + 1.5\times\text{IQR}$.

• The formula for the lower fence is $\text{Lower fence} = Q_1 - 1.5\times\text{IQR}$.

Some select sources replace the $1.5$ in our formulas with other values, like $2$ or sometimes even $3$. The choice of the multiplier is problem-specific and depends on the dataset distribution. An outlier in a dataset of children's heights would still be close to the rest of the dataset, so we'd choose a smaller multiplier like $1.5$. A millionaire's salary in a dataset of salaries would be much more outlying, so we'd pick a bigger multiplier, like $3$ or more. If you'd like to use a different multiplier for your dataset, you can enter the Advanced mode and change it from the default of $1.5$.

So, there you have it! We've found the formula for the upper fence and the lower fence, and now we can find outliers in any dataset we want.

## An example of how to calculate the upper and lower fence

Let's use our upper and lower fence formula in a practical example. Suppose we have a dataset of each year's rainfall volumes for January from 2010 to 2021 in New York, and it looks like this:

$1.33$, $1.96$, $3.12$, $2.20$, $1.58$, $2.04$, $1.80$, $6.32$, $1.90$, $3.84$, $2.93$, $2.34$.

Can you already guess which value is going to be an outlier?
Let's back up that guess with some math.

1. We first have to sort the dataset in increasing order:

$1.33$, $1.58$, $1.80$, $1.90$, $1.96$, $2.04$, $2.20$, $2.34$, $2.93$, $3.12$, $3.84$, $6.32$.

2. We can determine that the dataset's first and third quartiles are $Q\_1= 1.85$ and $Q\_3 = 3.025$.

3. Using $Q\_1$ and $Q\_3$, we can calculate the interquartile range with

$\text{IQR} = Q\_3 - Q\_1 = 3.025 - 1.85 = 1.175$

4. Finally, we can use the upper fence formula to get

$\text{Upper fence} = Q\_3 + 1.5 \times \text{IQR} = 3.025 + 1.5 \times 1.175 = 4.7875$

5. …and we can use the lower fence formula to get

$\text{Lower fence} = Q\_1 - 1.5 \times \text{IQR} = 1.85 - 1.5 \times 1.175 = 0.0875$

We can look at our data with these upper and lower fences and see that 2017's rainfall of $6.32$ is an outlier. Was your initial guess correct?

## How to find the upper and lower fence with our calculator

Our upper and lower fence calculator takes all these steps for you and gives you the fences in the blink of an eye so that you can get to find outliers in your dataset.

1. Enter your dataset's individual values in the rows. You can input up to 50 values.
2. Optionally, change the multiplier used in the fence formulas in the Advanced mode.
3. The calculator will determine the fences and display them at the bottom of the list of values, along with your dataset's outliers and the steps taken to calculate them. Happy outlier hunting!

## FAQ

### What is an outlier?

Outliers are values in a dataset that differ significantly from other values. The presence of outliers can be a problem, although it depends on what task you're using the data for. Outliers can be legitimate data, like a CEO's salary in a salary dataset. Outliers can also be invalid or due to mistakes; this could be a poorly calibrated sensor, or a typing error made when copying handwritten data over to a spreadsheet.

### How do I find outliers?

To find outliers in your dataset, you need to calculate the upper and lower fences of the dataset. You'd then see which of the dataset's values fall outside of the fences — those values are all outliers.

### How do I calculate the upper and lower fences?

Multiply your dataset's interquartile range with 1.5, then add and subtract that from your dataset's first and third quartiles, respectively. Those are your upper and lower fences.

### What is the upper fence formula?

You can calculate the upper fence with Q₃ + 1.5 × IQR, where Q₃ is your third quartile and IQR is your interquartile range. Any value in your dataset above the upper fence is an outlier.

### What is the lower fence formula?

You can calculate the lower fence with Q₁ − 1.5 × IQR, where Q₁ is your first quartile and IQR is your interquartile range. Any value in your dataset below the lower fence is an outlier.

Rijk de Wet
Values (enter up to 50 numbers)
#1
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Results
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### Coffee kick

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text:
## CryptoDB

### Paper: OptORAMa: Optimal Oblivious RAM

Authors: Gilad Asharov , Bar-Ilan University Ilan Komargodski , NTT Research Wei-Kai Lin , Cornell University Kartik Nayak , Duke University Enoch Peserico , Università degli Studi di Padova Elaine Shi , Cornell University DOI: 10.1007/978-3-030-45724-2_14 (login may be required) Search ePrint Search Google Slides EUROCRYPT 2020 Oblivious RAM (ORAM), first introduced in the ground-breaking work of Goldreich and Ostrovsky (STOC '87 and J. ACM '96) is a technique for provably obfuscating programs' access patterns, such that the access patterns leak no information about the programs' secret inputs. To compile a general program to an oblivious counterpart, it is well-known that $\Omega(\log N)$ amortized blowup is necessary, where $N$ is the size of the logical memory. This was shown in Goldreich and Ostrovksy's original ORAM work for statistical security and in a somewhat restricted model (the so called \emph{balls-and-bins} model), and recently by Larsen and Nielsen (CRYPTO '18) for computational security. A long standing open question is whether there exists an optimal ORAM construction that matches the aforementioned logarithmic lower bounds (without making large memory word assumptions, and assuming a constant number of CPU registers). In this paper, we resolve this problem and present the first secure ORAM with $O(\log N)$ amortized blowup, assuming one-way functions. Our result is inspired by and non-trivially improves on the recent beautiful work of Patel et al. (FOCS '18) who gave a construction with $O(\log N\cdot \log\log N)$ amortized blowup, assuming one-way functions. One of our building blocks of independent interest is a linear-time deterministic oblivious algorithm for tight compaction: Given an array of $n$ elements where some elements are marked, we permute the elements in the array so that all marked elements end up in the front of the array. Our $O(n)$ algorithm improves the previously best known deterministic or randomized algorithms whose running time is $O(n \cdot\log n)$ or $O(n \cdot\log \log n)$, respectively.
##### BibTeX
@inproceedings{eurocrypt-2020-30214,
title={OptORAMa: Optimal Oblivious RAM},
booktitle={39th Annual International Conference on the Theory and Applications of Cryptographic Techniques, Zagreb, Croatia, May 10–14, 2020, Proceedings},
series={Lecture Notes in Computer Science},
publisher={Springer},
keywords={oblivious RAM;tight compaction;randomized algorithms},
volume={12105},
doi={10.1007/978-3-030-45724-2_14},
author={Gilad Asharov and Ilan Komargodski and Wei-Kai Lin and Kartik Nayak and Enoch Peserico and Elaine Shi},
year=2020
}


text:
Chapter 8

After studying this chapter, you should be able to:

• Describe the scope and overall purposes of the study of enzyme kinetics.
• Indicate whether ΔG, the overall change in free energy for a reaction, is dependent on reaction mechanism.
• Indicate whether ΔG is a function of the rates of reactions.
• Explain the relationship between Keq, concentrations of substrates and products at equilibrium, and the ratio of the rate constants k1/k−1.
• Outline how temperature and the concentration of hydrogen ions, enzyme, and substrate affect the rate of an enzyme-catalyzed reaction.
• Indicate why laboratory measurement of the rate of an enzyme-catalyzed reaction typically employs initial rate conditions.
• Describe the application of linear forms of the Michaelis–Menten equation to the determination of Km and Vmax.
• Give one reason why a linear form of the Hill equation is used to evaluate the substrate-binding kinetics exhibited by some multimeric enzymes.
• Contrast the effects of an increasing concentration of substrate on the kinetics of simple competitive and noncompetitive inhibition.
• Describe the ways in which substrates add to, and products depart from, an enzyme that follows a ping–pong mechanism and do the same for an enzyme that follows a rapid-equilibrium mechanism.
• Illustrate the utility of enzyme kinetics in ascertaining the mode of action of drugs.

Enzyme kinetics is the field of biochemistry concerned with the quantitative measurement of the rates of enzyme-catalyzed reactions and the systematic study of factors that affect these rates. Kinetic analysis can reveal the number and order of the individual steps by which enzymes transform substrates into products. Together with site-directed mutagenesis and other techniques that probe the protein structure, kinetic analyses can reveal details of the catalytic mechanism of a given enzyme.

A complete, balanced set of enzyme activities is of fundamental importance for maintaining homeostasis. An understanding of enzyme kinetics thus is important to understanding how physiologic stresses such as anoxia, metabolic acidosis or alkalosis, toxins, and pharmacologic agents affect that balance. The involvement of enzymes in virtually all physiologic processes makes them the targets of choice for drugs that cure or ameliorate human disease. Applied enzyme kinetics represents the principal tool by which scientists identify and characterize therapeutic agents that selectively inhibit the rates of specific enzyme-catalyzed processes. Enzyme kinetics thus plays a central and critical role in drug discovery and comparative pharmacodynamics, as well as in elucidating the mode of action of drugs.

A balanced chemical equation lists the initial chemical species (substrates) present and the new chemical species (products) formed for a particular chemical reaction, all in their correct proportions or stoichiometry. For example, balanced equation (1) describes the reaction of one molecule each of substrates A and B to form one molecule each of products P and Q:

The double arrows indicate reversibility, an intrinsic property of all chemical reactions. Thus, for reaction (1), if ...

Sign in to your MyAccess profile while you are actively authenticated on this site via your institution (you will be able to verify this by looking at the top right corner of the screen - if you see your institution's name, you are authenticated). Once logged in to your MyAccess profile, you will be able to access your institution's subscription for 90 days from any location. You must be logged in while authenticated at least once every 90 days to maintain this remote access.

Ok

## Subscription Options

### AccessPharmacy Full Site: One-Year Subscription

Connect to the full suite of AccessPharmacy content and resources including 30+ textbooks such as Pharmacotherapy: A Pathophysiologic Approach and Goodman & Gilman's The Pharmacological Basis of Therapeutics, high-quality videos, images, and animations, interactive board review, drug and herb/supplements databases, and more.


text:
• Professor
• APS Fellow
• Associate Chair
• Physics & Astronomy
• Condensed Matter Physics (Theory)
Blazer Dining, 343 Martin Luther King Blvd., BL373
859-257-4729
Research Interests:
Education

Ph.D., Yale University (1987)

Biography

PhD, Yale University, 1987

Faculty member at Boston University, 1990-1999

Research

Condensed Matter Theory

Much of my recent work is in the area of strongly correlated electron systems'', in which the interactions between electrons play a dominant role in selecting the ground state and low-energy excitations of the system. Usually Nature reorganizes the strongly interacting electrons into weakly interacting quasiparticles'', which are collective excitations of the electrons. Examples are Landau's Fermi liquid theory, where the quasiparticles have the same quantum numbers as the electron, or the Bardeen-Cooper-Schreiffer theory of superconductivity, where the gound state can roughly be described as a collection of Bose-condensed Cooper pairs (pairs of electrons of opposite spin and momentum). The collective excitations of this system are fermionic quasiparticles and bosonic phase fluctuations. More exotic reorganizations occur in fractional quantum Hall states, where the quasiparticles have a charge which is a fraction of the electron's charge. The fundamental problem of strongly correlated systems may be phrased as that of finding the correct weakly interacting quasiparticles, and the theory that describes their dynamics. Usually, because of the strong interactions, this connection between the original electrons and the final quasiparticles is nonperturbative (that is, the quasiparticle cannot be understood as a small deviation from a noninteracting electron). This makes the problem mathematically quite difficult.

The introduction of disorder, ubiquitous in real life in the form of lattice imperfections or impurity atoms, makes the problem much more challenging and interesting. Many symmetries of the clean system, such as translation and rotation, are now absent, and this makes the analysis of disordered systems much harder. More importantly, qualitatively new phenomena which are not present in systems without disorder can arise in systems with disorder, even when the electrons are noninteracting. Of course, real systems are both disordered and interacting, potentially leading to even more interesting new phenomena. I believe the interplay of interactions and disorder will be one of the dominant themes of condensed matter theory in the years to come, and I have started some investigations into this subfield. In particular, with my co-workers Herb Fertig, Harsh Mathur, R. Shankar, and Damir Herman, I have been able to obtain { nonperturbative} solutions to this problem in certain experimentally interesting regimes.

Specific topics I am working on are (i) The role of disorder in the bilayer quantum Hall systems, (ii) States with strong correlations and strong quantum fluctuations in mesoscopic systems, and (iii) The effect of disorder on gauge theories.

Education

• B.Tech., Physics, Indian Institute of Technology, Madras (1982)
• Ph.D., Theoretical Condensed Matter Physics), Yale University (1987)
Selected Publications:
• G. Murthy, Phys. Rev. Lett. 84, 350 (2000): A Composite Fermion Hofstader Problem: Partially Polarized Density Wave States in the FQHE''.
• G. Murthy, Phys. Rev. Lett. 85, 1954 (2000): Hall Crystal States at \nu=2 and Moderate Landau Level Mixing''.
• G. Murthy, Jour. Phys. Cond. Mat. 12, 10543 (2000) : Finite Temperature Magnetism in Fractional Quantum Hall Systems: Composite Fermion Hartree-Fock and Beyond''.
• G. Murthy, Phys. Rev. B 64, 195310 (2001): Hamiltonian theory of the fractional quantum Hall effect: Conserving approximation for in compressible fractions''.
• R. Narevich, G. Murthy, and H. A. Fertig, Phys. Rev. B 64, 245326 (2001): Hamiltonian Theory of the Composite Fermion Wigner Crystal''.
• G. Murthy, Phys. Rev. Lett. 87, 179701 (2001): Comment on Half-Polarized Quantum Hall States''''.
• G. Murthy, Phys. Rev. B 64, 241309 (2001): Effects of Disorder on the \nu=1 Quantum Hall State''.
• G. Murthy and R. Shankar, Phys. Rev. B 65, 245309 (2002): Hamiltonian Theory of the Fractional Quantum Hall Effect: Effect of Landau Level Mixing''.
• G. Murthy and H. Mathur, Phys. Rev. Lett. 89, 126804 (2002): Interactions and Disorder in Quantum Dots; Instabilities of the Universal Hamiltonian''.
• G. Murthy and R. Shankar, Phys. Rev. Lett. 90, 066801 (2003): Quantum Dots with Disorder and Interactions: A Solvable Large-g Limit''.
• G. Murthy and R. Shankar, Rev. Mod. Phys. 75, 1101 (2003): Hamiltonian Theory of the Fractional Quantum Hall Effects''.
• Y. N. Joglekar, H. K. Nguyen, and G. Murthy, Phys. Rev. B 68, 035332 (2003): "Edge Reconstructions in Fractional Quantum Hall Systems"
• I. F. Herbut, B. H. Seradjeh, S. Sachdev, and G. Murthy, Phys. Rev. B 68, 195110 (2003): Absence of U(1) Spin Liquids in Two Dimensions''.
• D. Herman, H. Mathur, and G. Murthy, cond-mat/0305364, Phys. Rev. B 69, 041301 (2004): Diamagnetic Persistent Currents and Spontaneous Time-Reversal Symmetry Breaking in Mesoscopic Structures''.
• G. Murthy, R. Shankar, D. Herman, and H. Mathur, cond-mat/0306529, Phys. Rev. B 69, 075321 (2004): A Solvable Regime of Disorder and Interactions in Ballistic Nanostructures, Part I: Consequences for Coulomb Blockade''.
• H. K. Nguyen, Y. N. Joglekar, and G. Murthy, Phys. Rev. B 70, 035324 (2004): Collective edge modes in fractional quantum Hall systems''.
• G. Murthy, Phys. Rev. B 70, 153304 (2004): Random matrix crossovers and quantum critical crossovers for interacting electrons in quantum dots''.
• G. Murthy, Phys. Rev. Lett. 94, 126803 (2005): Interplay between the mesoscopic Stoner and Kondo effects in quantum dots''.
• G. Murthy, R. Shankar, and H. Mathur, Phys. Rev. B 72, 075364 (2005) : Ballistic Quantum Dots with Disorder and Interactions: A numerical study on the Robnik-Berry billiard''.
• I. Rozhkov and G. Murthy, Phys. Rev. B 72, 193311 (2005): A nearly closed ballistic billiard with random boundary transmission''
• H. A. Fertig and G. Murthy, Phys. Rev. Lett. 95, 156802 (2005): A Coherence Network in the Quantum Hall Bilayer''
• I. Rozhkov and G. Murthy, Jour. Phys. A 38, 10843-10857 (2005): Ballistic dynamics of a convex smooth-wall billiard with finite escape rate along the boundary''.
• R. Shankar and G. Murthy, Phys. Rev. B 72, 224414 (2005): Deconfinement in d=1: A closer look''.

For the latest ones see the cond-mat arXiv

X


text:
# Lifting house with balloons [closed]

I have an assignment about the movie Up. Here is the question:

Critique the brief floating scene. Did the animators use enough balloons during this scene? What important variables, (which affect gases) did the animators seemingly fail to consider during the floating scenes as Carl's house is slowly rising after initial takeoff? What important variables, (which affect gases) did the animators fail to consider during the floating scenes as Carl traveled from the US to Venezuela? Explain why these factors should be considered.

For reference, we are supposed to assume that Carl's house was 125,000 pounds. Could someone please help me with this question? Thanks!

I'm thinking for the important variables about pressure, volume, and number of moles.

-

## closed as too localized by Chris White, Waffle's Crazy Peanut, David Z♦May 7 '13 at 5:58

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

Hi Someone - this is a site for conceptual questions about physics, not general homework help. If you can edit your question to ask about the specific physics concept that is giving you trouble, I'll be happy to reopen it. See our FAQ and homework policy for more information. –  David Z May 7 '13 at 5:59

You need to roughly estimate the volume of one balloon first. I would say it's about 4 liters, (that's how many bottles of milk I think I could fit in there) so 4 $dm^3$ which ammounts to $4\cdot 10^{-3} m^3$

The buoyancy force upward extered on one balloon is

$F_b = \rho_{air}V_{balloon}g$

where $\rho$ is the density of displaced liquid (here air) and $V$ is the volume of liquid the balloon displaces - so its own volume. I assume the balloon is inflated with helium. So, the gravitational pull downward on the balloon is

$F_g = \rho_{helium}V_{Balloon}g$

The net force pushing up one balloon is

$F_{up} = (\rho_{air}-\rho_{helium})V_{balloon}g$

Now the densities are repsectively $1.18 \frac{kg}{m^{3}}$ and $0.18 \frac{kg}{m^{3}}$ for air and helium. So

$F_{up} \simeq 1\cdot V_{balloon}g = 3.9 \cdot 10^{-2} N$

This net force up is the force the balloon actually pulls on the house with. The weight of the house is

$P = M_{house}g = 56250*9.81 N = 5.52\cdot 10^{5} N$

Where I have translated 125'000 pounds into kilograms. You would then roughly need 10 million balloons to lift up the house. You can easily replace the volume of the balloon with what you see fit. However, I think the number of balloons is not enough :)

For what is being neglected : air density changes as a function of height. Moreover, temperature changes will affect the volume of each balloon due to perfect gas law. The pressure of the air is also something that changes with height.

-
Oh I'm so sorry, the balloon is supposed to be 3 feet in diameter, which would make its volume about 14.1 ft^3 –  Someone May 5 '13 at 15:45
I don't like to calculate with feet. 3 feet is about 0.92 meters. So the balloon volume is (spherical) $0.4 m^3$ which is a hundred times greater than what I had assumed. Since the force is linear in the volume, this means in the end you should need 100 times less baloons than I had calculated, so about 100 thousand. It's still a lot. –  Mathusalem May 5 '13 at 16:20
Yup. But what about the other parts of the question, such as the variables? –  Someone May 5 '13 at 16:24
@Someone: It's better if you put some thought into it. This site is to help people learn from each other, and that doesn't work if your brain doesn't kick in. –  Mike Dunlavey May 5 '13 at 19:39


text:
Online Calculator Resource

Loan Repayment Calculator

Loan Repayment Calculator
Loan shortened by 1 year 1 month,
savings $455.66 in interest, with new payments of$750.00.
Loan Change Totals

Current
New
Change
Months
48
35
-13

4yr 0mo
2yr 11mo
-1yr 1mo
Payment
$550.00$750.00
$+200.00 Interest$1,642.01
$1,186.35$-455.66
Actual Amortization has rounding differences
Create an Amortization Schedule

Calculator Use

Use this calculator to determine 1) how extra payments can change the term of your loan or 2) how much additional you must pay each month if you want to reduce your loan term by a certain amount of time in months. Try different loan scenarios for affordability or payoff. Create amortization schedules for the new term and payments.

Make Extra Payments
Calculate how much your loan term and interest will change by applying extra money to your payments each month
Reduce Term (Months)
Calculate how much extra you need to pay each month in order to pay off your loan early
Current Loan Balance
the original amount on a new loan or principal outstanding if you are calculating a current loan
Interest Rate
the annual interest rate (stated rate) on the loan
Remaining Term (Months)
number of months which coincides with the number of payments to repay the loan.  How much time is left on this loan.
Current Monthly Loan Payment
the amount currently to be paid on this loan on a monthly basis toward principal and interest only. You can likely look at your last statement to find the amounts applied to principal and interest and add these 2 numbers together. (payment = principal + interest)
Monthly Extra
the extra amount you plan to add to your monthly payments on this loan to be applied to principal

This calculator will provide good results but you may want to also talk to your loan provider to get a calculation from them.

Term Calculation

When investigating different payment amounts (loans with extra payments) you can use the following formula to calculate what your corresponding number of months on the loan will be:

$$n=\dfrac{log\left[\frac{\frac{PMT}{i}}{\frac{PMT}{i}-PV}\right]}{log(1+i)}$$

where n = number of months, PMT = monthly payment, i = monthly interest rate as a decimal (annual rate divided by 100 divided by 12), and PV = loan balance (present value). log is the logarithm base 10.

Payment Calculation

When investigating different terms (months) you can use the following formula to calculate what your corresponding monthly payment amounts will be:

$$PMT=\dfrac{PVi(1+i)^n}{(1+i)^n-1}$$

where n = number of months, PMT = monthly payment, i = monthly interest rate as a decimal (annual rate divided by 100 divided by 12), and PV = loan balance (present value).

Cite this content, page or calculator as:

Furey, Edward "Loan Repayment Calculator"; from https://www.calculatorsoup.com - Online Calculator Resource.


text:
# How to insert a vertically centered tilde into a URL?

I need to insert a vertically centered tilde into a URL in text mode (not inside the \url or \href commands). I've tried the solution proposed here but I can't get it to work. Any ideas?

Here is my code

\documentclass[]{beamer}

\usepackage{textcomp}
\newcommand{\textapprox}{\raisebox{0.5ex}{\texttildelow}}

\begin{document}

\begin{frame}
\frametitle{MWE}
https://foo.com/\textapprox{}user/
\end{frame}

\end{document}

This is the error I'm getting

! LaTeX Error: Command \textapprox already defined.
Or name \end... illegal, see p.192 of the manual.

See the LaTeX manual or LaTeX Companion for explanation.
Type  H <return>  for immediate help.
...

l.4 ...extapprox}{\raisebox{0.5ex}{\texttildelow}}

?

• well as the error says the command name already exist. So use another one or redefine the existing command (if you are sure that you don't need it). Jun 1 at 12:17
• Thank you! I feel so dumb. If you convert your comment into an answer I will be pleased to accept it. Jun 1 at 12:21


text:
Deleted:Difficulty returning appropriate info. with esearch and/or efetch
0
0
Entering edit mode
7 weeks ago
Vincent Laufer ★ 2.4k

Hello,

A few days ago I asked some questions regarding how to find associated genomic data easily.

GenoMax recommended efetch and esearch, as can be seen here: Extract nucleotide sequence from a RefSeq Transcript ID

Since this time I have been using these tools; for some reason, though, I am hitting a snag when pulling genetic variants within a given gene ... I have a couple working versions; for instance, e

esearch -db snp -query "FGFR2" | efetch -format docsum

gives me approximately the right information, but no rsIDs are provided. Other queries also get me close, but for instance without variant annotation information (i.e. this results in a missense).

is there a simple way to emend the above query so it gives me rsID, chr, pos, functional info, and (ideally) more?

TYVM

esearch dbsnp efetch • 113 views


text:
Uh Oh! It seems you’re using an Ad blocker!

We always struggled to serve you with the best online calculations, thus, there's a humble request to either disable the AD blocker or go with premium plans to use the AD-Free version for calculators.

Or

# Free Fall Calculator

Enter the required parameters and the tool will readily calculate the free fall distance, time, and height of the object.

To Calculate:

Gravitational (g)

Initial velocity (v₀)

Height (h)

Time of fall (t)

Velocity (V)

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Get the Widget!

Add this calculator to your site and lets users to perform easy calculations.

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How easy was it to use our calculator? Did you face any problem, tell us!

An online free fall calculator helps you to determine the speed, time, and height of a freely falling object under the influence of the gravitational force. But before you start using this calculator, you must go through the following context for a better understanding of the concept.

Stay in touch!

## What Is Free Fall?

In the context of physics:

“An object is said to be in free fall motion if it falls on the ground with acceleration due to gravity and having no external force acting on it”

### Important:

If we define the ideal case of the free fall, then it must be considered that there is no air resistance acting upon the object and the gravity is acting constantly on it. The value of the gravity is nearly $$9.8\frac{m}{s^{2}}$$ or $$32.17\frac{ft}{s^{2}}$$. But when it comes to practically, the velocity of the falling object is controlled by terminal velocity which is defined below:

### Terminal Velocity:

“It is the velocity of the body or any object falling from the rest and either through air or fluid”

Under the influence of terminal velocity, the body stops accelerating due to the influence of the drag force in the upward direction and falls toward the ground at the constant speed.

$$F_{d} = F_{g}$$

For example:

The terminal velocity of a parachutist is about 150km/hr. Similarly the terminal velocity of the rain drops is about 9m/s. As terminal velocity is achieved when both the gravitational and drag force become equal, the free acceleration due to gravity calculator helps you to find the acceleration of objects experiencing such impacts.

### Free Fall Equation:

Following are the free fall formulas to determine the various parameters of freely falling object:

#### Free Fall Speed:

Basically, the actual definition of the velocity gives us the free fall speed of the free fall object and is as below:

$$V = V_{o} + g* t$$

Where:

$$V_{o}$$ = Initial Velocity that is measured in either m/s or ft/s
t = FreeFall time which is expressed in seconds
g = Gravitational Acceleration measured in either $$\frac{m}{s^{2}}$$ or $$\frac{ft}{s^{2}}$$

#### Free Fall Height:

Here arise two cases:

When Initial Displacement And Velocity Are Zero:

In this case, the distance of the falling body is determined using the expression below:

$$S = \frac{1}{2} * g * t^{2}$$

Object With Initial Velocity:

The free fall equation physics for a body or any object falling with the initial velocity is as under:

$$S = V_{o} * t \frac{1}{2} * g * t^{2}$$

Here the free  initial velocity calculator also makes the use of this formula to calculate the distance of the falling object.

#### Free Fall Time:

You can make use of the formula below to determine the time of the free fall object:

$$t = \sqrt{\frac{2 * h}{g}}$$

### How To Examine The Free Falling Body?

No doubt the free fall calculator is the most quick and precise way of determining various parameters that are associated with the free fall motion of the object. But solving for them by hand is also very important. This is why we will be resolving a couple of examples to clarify your idea in more depth. Just stay focused!

Example # 01:

An object has fallen with the initial velocity of about $$55\frac{m}{s}$$. It takes a time span of about 30 sec to hit the ground. How to determine its height?

Solution:

As free fall acceleration formula is given as:

$$S = V_{o} * t + \frac{1}{2} * g * t^{2}$$
$$S = 0 * 30 + \frac{1}{2} * 9.81 * \left(30\right)^{2}$$
$$S = 1650 + 0.5 * 9.81 * 900$$
$$S = 1797.15 m$$

Apart from the manual calculations, try using a free fall calculator to speed up your calculations.

Example # 02:

Find velocity of a falling object which has an initial velocity of about $$21\frac{m}{s}$$. Keep in mind that the time it takes to hit the ground is exactly 10sec.

Solution:

As we know that:

$$V = V_{o} + g* t$$
$$V = 21 + 9.8 * 10$$
$$V = 21 + 98$$
$$V = 119\frac{m}{s}$$

### How Free Fall Calculator Works?

Make use of this free falling object calculator to reduce the time of the calculations. Let us find how!

Input:

• First of all, select whether you want to determine either “Time & Velocity”, “Height & Velocity”, or “Time & Height”
• After making a certain selection, go for writing the parameters against it
• Also, select the units against each of the parameters
• Tap the calculate button

Output:

The free fall speed calculator determines:

• Free fall velocity
• Height of the object
• Time of free fall

## FAQ’s:

### Why does free fall motion occur?

When a light object is allowed to fall, it takes more time as it moves slowly towards the ground. But in case if we replace the object with a heavier one, it will take a less amount of time due to increased acceleration. This happens only when forces other than that of gravity are also at work. But if the object is considered to fall under vacuum, where no forces act on it other than just gravity, then it is said to be in free fall motion.

### What type of motion is the free fall?

As the acceleration increases at a uniform rate with respect to the time during free fall motion, this kind of motion is termed as uniform acceleration motion and could swiftly be determined by using free fall calculator.

### What do you mean by the term weightlessness?

A particular condition in which the object is in free fall motion is referred to as the weightlessness. This is because the weight of the object is described in terms of its gravitational acceleration.

### How does speed continue to increase during free fall motion?

The following table gives an idea the constant increase in the speed of the free fall object with respect to the time:

 Seconds After Object Is Fallen Speed$$\frac{m}{s}$$ 1 9.8 2 19.6 3 29.4 4 39.2

### What is the actual difference between free fall and weightlessness?

During free fall motion, the only force that acts on the object is the free fall motion. While in case of weightlessness, not any force acts on the object.

## Conclusion:

Free fall motion keeps the object in a certain trajectory. Moreover, it allows the object to accelerate at the constant speed during motion. All the planets revolving around the sun are in free fall motion under the influence of gravity caused by the Sun. students and professionals make a vast use of the free fall calculator to analyse the free motion of various objects.

## References:

From the source of wikipedia: Free fall, Free fall in Newtonian mechanics, Inverse-square law gravitational field, Free fall in general relativity

From the source of khan academy: Freefall review, Projectile height, Impact velocity

From the source pof lumen learning: Free-Falling Objects


text:
# When does a VW make sense?

### Help Support HomeBuiltAirplanes.com:

#### JohnR

##### Member
I'm building a full-size J-3 replica, and the numbers sure point to a VW conversion. However, I spoke with Great Plains, and the only way it would work well on a plane like this (lots of drag) is with a reduction drive.

So there go any weight savings over a Continental A65, and the complexity of the redrive is an added headache. The reason a redrive is needed is to get enough torque for the prop to convert to thrust.

I've run the numbers on a few engines, and nothing in the 65 HP family produces as much torque as a Cont. A65. The prices are good, it's simple and proven, and the weight is reasonable.

My plane was designed around the Rotax 582, but I'm scared of 2-strokes. I looked hard at the Hirth engines, but they're expensive, and they're still 2-strokes (and still don't put out the torque of an A65).

##### Well-Known Member
Personally, I'd go with an A65 or a Jab 2200. The Jab is lighter, more power, and more price.

#### JohnR

##### Member
I wish I could afford the Jab. I also like the 2300 RPM of the A65 vs. something like 3600 for the Jab. My ears will appreciate it!

##### Well-Known Member
3300 for the Jab; but the Jab gives at least as much power if you cap it at 2500 rpm as the A65, at a lot less weight... I recall the A65 being around 200 lbs, no? That's the same weight as an IO-233 at close to twice the power.

#### PTAirco

If you can find a cheap 65 hp Continental, I would go for it. I like VW's but once the conversion runs past $5000 or so, you might as well buy a used A-65 with maybe half its life remaining. Although when it comes to TBO, a VW would cost peanuts to re build compared to the Contintental. #### djschwartz ##### Well-Known Member Another factor to consider: the A series Continentals have no provisions for any electrical system. If you want a starter or alternator you would need a C series Continental (C-75, C-85, C90, or O-200). The original equipment starters and generators that these came with were very heavy. Excellent lightweight starters and alternators are available for these engines. I have a Skytec starter on my O-200 powered Luscombe and am very happy with it. I also have a certified 35 amp alternator since it is a standard category aircraft. There are yet lighter and less expensive units available for experimentals. Also, the dry weight of the A-65 is 170 lbs. This does not include exhaust system. The dry weight of the O-200 is 215 lbs which includes the old, heavy accessories but not the exhaust system. With the lightweight electricals it should weigh a bit less than 200 lbs depending on which ones you choose. And for that you get 100 HP at an RPM that will result in good propellor efficiency at the speeds you will operate a J-3 replica. #### BBerson ##### Light Plane Philosopher HBA Supporter A direct drive VW might work with a large low pitch prop. My Grob G109 has a direct drive 2000cc Limbach (VW) with a two position prop. In low pitch it will cruise about 70kts and 90kts in high pitch. If you are happy with 70kts, a large, direct drive fixed pitch prop might work if the plane is very light. (and the pilot also) VW is smoother than an A65 but less total thrust, I think. BB #### jumpinjan ##### Well-Known Member If you put that VW & PSRU on your J-3 replica, it will leap off the ground. Our club member Gwen, put a Great Plains VW & reduction gear on her Fokker D8 and never regretted it. (I have heard the same response from other WWI replica builders that upgraded to the re-drive from a direct drive engine. Forget the A65 and everything else recommended) http://wrightpatrol.org/Gwen/4_Aug_07_ 018.jpg Jan #### wally ##### Well-Known Member Hi, Looks like an engine and their reduction drive from Great Planes might be the way to go. At least I like it. You can buy a complete engine kit (you assemble) and a re-drive, all new stuff for about$4500. Or build your own engine up from whatever special piece parts GP sells, like a crank w/prop flange already on it, etc. and buy their re-drive. You might be under $3000, ready to fly with a little luck and work. A friend of mine just had his 85 Cont. overhauled on his Clipped Wing Cub. He said the final price was$12,000.!!!
Granted about all they saved was his case but still...Dang!
Best wishses,
Wally

#### Topaz

##### Super Moderator
Staff member
Log Member
For me, it depends upon how long you intend to keep this airplane and what you expect your financial capabilities to be in the future. The VW is a solid, well-performing engine, as is the Conti. Don't be afraid of the redrive on the VW. GP's redrive seems to have a good record of service.

The only technical reason to go with the VW that I can see is what DJSchwartz said: the VW will have full accessories for the price, but the Conti may not. Otherwise these are both very good engines.

In the long run, though, if you intend to keep the airplane through at least one TBO cycle, the VW will save you thousands of dollars. Even on a top-end, the VW will be far less to have work done, and you'll be able to get most of the parts at any auto supplier.

#### JohnR

##### Member
Lots of good info here, thanks all.

This idea sounds interesting:
A direct drive VW might work with a large low pitch prop. My Grob G109 has a direct drive 2000cc Limbach (VW) with a two position prop. In low pitch it will cruise about 70kts and 90kts in high pitch. If you are happy with 70kts, a large, direct drive fixed pitch prop might work if the plane is very light. (and the pilot also)
VW is smoother than an A65 but less total thrust, I think.
BB
That's about J-3 speed anyway, and I don't want an electrical system. As for a "large low-pitch prop" maybe a three blade Warp Drive or Ivoprop in the wider blade models?

#### JohnR

##### Member
Ok, the answer is no.

From Great Planes' website:

[SIZE=+1]Great Plains engines (when driven off the pulley end) are for use with wooden propellers only. Metal or composite propellers are not allowed.[/SIZE]

#### Grimace

##### Well-Known Member
Oh just go with a corvair motor and be done with it... ;-)

#### gschuld

##### Well-Known Member
:roll: As I was reading the thread, I was thinking the exact same thing!

#### JohnR

##### Member
At 220#, I don't think so.

#### Joe Kidd

##### Well-Known Member
You do have options for an engine aside from those mentioned. The newer BMW engines with redrive are lightweight powerhouses. Four cylinder Geo engines with a redrive are heavier but do a good job. Also if your looking at a 65 to 75 HP engine HKS has a beauty that comes with a redrive.
I'm weighing my options based on an ever tightening budget so the two most cost effective options I've found to build are the 2180 VW and a 2700 Corvair. But, I'm still researching and looking around the area.

#### macosxuser

##### Well-Known Member
An old C-75 or C-85 would probably be the easiest to find, and cheap. There was an O-200 core kicking around for $500 the other week in socal. Having RACED with these engines are Reno, I can tell you they are completely bulletproof, parts are easy to find still, and overhauls should run in the range of$5000-8000 depending on your level on involvement and whats wrong with them. The bottom ends usually last for oh, 4000 hours or 50 years, and the top end needs a refresh every 1500 hours or so.

We spin them up to 4200+ RPM and 250°+ of oil temp for 15-25 hours (one season), then pull the jugs, look at it... hey looks good, put it back together and run another season...

#### Topaz

##### Super Moderator
Staff member
Log Member
...and overhauls should run in the range of $5000-8000 depending on your level on involvement and whats wrong with them... While I would never argue against a certified engine for someone who wants one, part of the issue here is that$5,000-8,000 would buy you an complete brand-new VW-conversion engine and all-new accessories, instead of just an overhaul. The overhaul on the VW, even a major, will only be a fraction of the top-end overhaul price on the certified engine.

Total-cost-over-life for a certified engine like this is orders of magnitude higher than a solid VW conversion that's likely to be just as reliable in actual service. Certified engines make sense for the higher-performance, more mission-critical end of light aircraft, but for the smaller, lighter, slower, simpler end of things, they really don't, IMHO.

#### JohnR

##### Member
While I would never argue against a certified engine for someone who wants one, part of the issue here is that $5,000-8,000 would buy you an complete brand-new VW-conversion engine and all-new accessories, instead of just an overhaul. The overhaul on the VW, even a major, will only be a fraction of the top-end overhaul price on the certified engine. Total-cost-over-life for a certified engine like this is orders of magnitude higher than a solid VW conversion that's likely to be just as reliable in actual service. Certified engines make sense for the higher-performance, more mission-critical end of light aircraft, but for the smaller, lighter, slower, simpler end of things, they really don't, IMHO. Another aspect is availability - There always seem to be a handful of Continentals for sale, but you'll probably waste a bunch of time picking over duds before finding a good one. I've seen lots of A65s for sale between$1,800 and $3,000, and some for$6-7,000. A VW conversion is just a phone call away.

#### orion

##### R.I.P.
The bottom ends usually last for oh, 4000 hours or 50 years, and the top end needs a refresh every 1500 hours or so.
This is actually a rather important point and one that's often missed by those looking to do an auto conversion, and especially one that's based on the VW. While that Lyc or Cont will last you for quite some time, the VW conversions have on average a much shorter life. True, there are folks out there who get extended usage, some by quite a bit, but last I heard the average life of these small engines is in the range of about 200 hours. So you may have to figure the increased frequency of rebuild or maintenance into your cost numbers when comparing the auto alternatives and the certified engines.


text:
# cos(3x)-sin(3x)=0

## Simple and best practice solution for cos(3x)-sin(3x)=0 equation. Check how easy it is, and learn it for the future. Our solution is simple, and easy to understand, so dont hesitate to use it as a solution of your homework.

If it's not what You are looking for type in the equation solver your own equation and let us solve it.

## Solution for cos(3x)-sin(3x)=0 equation:

Simplifying
cos(3x) + -1sin(3x) = 0
Remove parenthesis around (3x)
cos * 3x + -1sin(3x) = 0
Reorder the terms for easier multiplication:
3cos * x + -1sin(3x) = 0
Multiply cos * x
3cosx + -1sin(3x) = 0
Remove parenthesis around (3x)
3cosx + -1ins * 3x = 0
Reorder the terms for easier multiplication:
3cosx + -1 * 3ins * x = 0
Multiply -1 * 3
3cosx + -3ins * x = 0
Multiply ins * x
3cosx + -3insx = 0
Solving
3cosx + -3insx = 0
Solving for variable 'c'.
Move all terms containing c to the left, all other terms to the right.
Add '3insx' to each side of the equation.
3cosx + -3insx + 3insx = 0 + 3insx
Combine like terms: -3insx + 3insx = 0
3cosx + 0 = 0 + 3insx
3cosx = 0 + 3insx
Remove the zero:
3cosx = 3insx
Divide each side by '3osx'.
c = ino-1
Simplifying
c = ino-1`


text:
# zbMATH — the first resource for mathematics

A conversation with Robert E. Kass. (English) Zbl 1420.62007
##### MSC:
 62-03 History of statistics 01A70 Biographies, obituaries, personalia, bibliographies 62C10 Bayesian problems; characterization of Bayes procedures
Kass, Robert E.
Full Text:
##### References:
 [1] Berger, J. O. and Delampady, M. (1987). Testing precise hypotheses. Statist. Sci.3 317-352. · Zbl 0955.62545 [2] Berger, J. O. and Selke, T. (1987). Testing a point null hypothesis: irreconcilability of P values and evidence. With comments and a rejoinder by the authors. J. Amer. Statist. Assoc.82 112-139. [3] Cramér, H. (1973). The Elements of Probability Theory and Some of Its Applications. Wiley, New York. [4] DeGroot, M. H. (1986). A conversation with David Blackwell. Statist. Sci.1 40-53. · Zbl 0955.01510 [5] Feller, W. (1950). An Introduction to Probability Theory and Its Applications. Vol. I. Wiley, New York, NY. · Zbl 0039.13201 [6] Feller, W. (1957). An Introduction to Probability Theory and Its Applications. Vol. II, 2nd ed. Wiley, New York. · Zbl 0077.12201 [7] Ferguson, T. S. (1967). Mathematical Statistics: A Decision Theoretic Approach. Probability and Mathematical Statistics, Vol. 1. Academic Press, New York. · Zbl 0153.47602 [8] Jarosiewicz, B. Chase, S. M. Fraser, G. W. Velliste, M. Kass, R. E. and Schwartz, A. B. (2008). Functional network reorganization during learning in a brain-machine interface paradigm. Proc. Natl. Acad. Sci. USA105 19486-19491. [9] Kass, R. E. (1989). The geometry of asymptotic inference. Statist. Sci.4 188-234. With comments and a rejoinder by the author. · Zbl 0955.62513 [10] Kass, R. E., Eden, U. T. and Brown, E. N. (2014). Analysis of Neural Data. Springer Series in Statistics. Springer, New York. · Zbl 1404.62002 [11] Kass, R. E., Kelly, R. C. and Loh, W.-L. (2011). Assessment of synchrony in multiple neural spike trains using loglinear point process models. Ann. Appl. Stat.5 1262-1292. · Zbl 1223.62159 [12] Kass, R. E. and Raftery, A. E. (1995). Bayes factors. J. Amer. Statist. Assoc.90 773-795. · Zbl 0846.62028 [13] Kass, R. E. and Steffey, D. (1989). Approximate Bayesian inference in conditionally independent hierarchical models (parametric empirical Bayes models). J. Amer. Statist. Assoc.84 717-726. [14] Kass, R., Ventura, V. and Brown, E. N. (2005). Statistical issues in the analysis of neuronal data. J. Neurophysiol.1 8-25. [15] Kass, R. E. and Vos, P. W. (1997). Geometrical Foundations of Asymptotic Inference. Wiley Series in Probability and Statistics: Probability and Statistics. Wiley, New York. · Zbl 0880.62005 [16] Kass, R. E. and Wasserman, L. A. (1996). The selection of prior distributions by formal rules. J. Amer. Statist. Assoc.91 1343-1370. · Zbl 0884.62007 [17] Kass, R. E., Caffo, B. S., Davidian, M., Meng, X.-L., Yu, B. and Reid, N. (2016). Ten simple rules for effective statistical practice. PLoS Comput. Biol.12 e1004961. [18] Kendall, M. and Stuart, A. (1977). The Advanced Theory of Statistics: Distribution Theory. Vol. 1, 4th ed. Macmillan Publishing, New York. · Zbl 0353.62013 [19] Mosteller, F. and Tukey, J. (1968). Data analysis, including statistics. In Handbook of Social Psychology, 2nd ed. (G. Lindzey and E. Aronson, eds.) 2. Wiley, New York. [20] Mosteller, F. and Wallace, D. L. (1964). Inference and Disputed Authorship: The Federalist. Addison-Wesley, Reading, MA. · Zbl 0122.14106 [21] Raftery, A. (2001). Statistics in the Twenty First Century, 1st ed. CRC Press, New York. [22] Rao, C. R. (1973). Linear Statistical Inference and Its Applications, 2nd ed. Wiley, New York. · Zbl 0256.62002 [23] Sellke, T., Bayarri, M. J. and Berger, J. O. (2001). Calibration of $$p$$ values for testing precise null hypotheses. Amer. Statist.55 62-71. · Zbl 1182.62053 [24] Tierney, L. (1994). Markov chains for exploring posterior distributions. Ann. Statist.22 1701-1762. · Zbl 0829.62080
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
## Cegah Corona

Stay at Home, Pakai Masker, Jaga Jarak, Cuci Tangan Selalu

## My Blog List

### How to Study Strategies for Every Learning Style

For example, visual learners sometimes struggle during essay exams because they can't recall test material that was presented orally during class. However, if the visual learner uses a visual aid when studying, like a colorful outline of test materials, he or she may retain more information. Read on to discover the characteristics of and ideal learning strategies for each learning style.

## Visual Learner Characteristics

Visual learners are those who learn through seeing. Visual learners typically share the following characteristics:

• Good at spelling, but forgets first names
• Find quiet study time beneficial
• Enjoy colors and fashion
• Dream in color
• Understand visual elements and charts
• Able to learn sign language easily

Learning Suggestions for Visual Learners

• Draw a map of events in history or draw scientific process.
• Make outlines of reading assignments.
• Copy what's on the board.
• Diagram sentences.
• Use flashcards.
• Take notes and make lists.
• Watch videos.
• Mark up your notes with highlighters, underlines, and color coding.

Test Types for Visual Learners

• Best test type: Diagram activities, map reading, essay tests, demonstrating a process.
• Worst test type: Listening tests

## Auditory Learner Characteristics

Auditory learners are those who learn best through hearing. They typically share the following characteristics:

• Like to read out loud
• Unafraid to speak up in class
• Good at giving explanations and oral reports
• Remember names
• Notice sound effects in movies
• Enjoy music
• Able to follow spoken directions
• Struggle to stay quiet for long periods of time
• Focused in study groups

Learning Suggestions for Auditory Learners

• Use word association to remember facts
• Record lectures
• Watch videos
• Repeat facts with eyes closed
• Participate in group discussions
• Use audio for language practice

Test Types for Auditory Learners

• Best test type: Oral exams and writing responses to lectures.
• Worst test type: Reading passages and writing answers in a timed test.

## Kinesthetic Learner Characteristics

Kinesthetic learners are those who learn through hands-on experience. Kinesthetic learners typically share the following characteristics:

• Good at sports
• Can't sit still for long
• May have sloppy handwriting.
• Learn well through lab and modeling activities
• Study with loud music
• Enjoy adventure books and movies
• Fidgety during lectures

Learning Suggestions for Kinesthetic Learners

• Study in short blocks
• Take lab-based classes
• Act out your study notes
• Take field trips to reinforce knowledge
• Study in groups
• Use flashcards and memory games

Test Types for Auditory Learners

• Best test type: Short definitions, fill-in-the-blank questions, and multiple choice.
• Worst test type: Essay tests and any overly long tests
Source : https://www.thoughtco.com

### Why Mathematics is Very Difficult For Some Students

In 2005, Gallup conducted a poll that asked students to name the school subject that they considered to be the most difficult. Not surprisingly, mathematics came out on top of the difficulty chart. So what is it about math that makes it difficult? Have you ever wondered?

Dictionary.com defines the word difficult as:

“...not easily or readily done; requiring much labor, skill, or planning to be performed successfully.”

This definition gets to the crux of the problem when it comes to math, specifically the statement that a difficult task is one that is not “readily” done. The thing that makes math difficult for many students is that it takes patience and persistence. For many students, math is not something that comes intuitively or automatically - it takes plenty of effort. It is a subject that sometimes requires students to devote lots and lots of time and energy.

This means, for many, the problem has little to do with brainpower; it is mostly a matter of staying power. And since students don't make their own timelines when it comes to "getting it," they can run out of time as the teacher moves on to the next topic.

## Math and Brain Types

But there is also an element of brain-style in the big picture, according to many scientists. There will always be opposing views on any topic, and the process of human learning is subject to ongoing debate, just like any other topic. But many theorists believe that people are wired with different math comprehension skills.

According to some brain science scholars, logical, left-brain thinkers tend to understand things in sequential bits, while artistic, intuitive, right-brainers are more global. They take in a lot of information at one time and let it "sink in." So left-brain dominant students may grasp concepts quickly while right-brain dominant students don’t. To the right brain dominant student, that time-lapse can make them feel confused and behind.

## Math as a Cumulative Discipline

Math know-how is cumulative, which means it works much like a stack of building blocks. You have to gain understanding in one area before you can effectively go on to “build upon” another area. Our first mathematical building blocks are established in primary school when we learn rules for addition and multiplication, and those first concepts comprise our foundation.

The next building blocks come in middle school when students first learn about formulas and operations. This information has to sink in and become “firm” before students can move on to enlarge this framework of knowledge.

The big problem starts to appear sometime between middle school and high school because students very often move on to a new grade or new subject before they’re really ready. Students who earn a “C” in middle school have absorbed and understood about half of what they should, but they move on anyway. They move on or are moved on, because

1. They think a C is good enough.
2. Parents don’t realize that moving on without a full understanding poses a big problem for high school and college.
3. Teachers don’t have time and energy enough to ensure that every single student understands every single concept.

So students move to the next level with a really shaky foundation. The outcome of any shaky foundation is that there will be a serious limitation when it comes to building and real potential for complete failure at some point.

The lesson here? Any student who receives a C in a math class should review heavily to make sure to pick up concepts they'll need later. In fact, it is smart to hire a tutor to help you review any time you find that you've struggled in a math class!

## Making Math Less Difficult

We have established a few things when it comes to math and difficulty:

• Math seems difficult because it takes time and energy.
• Many people don't experience sufficient time to "get" math lessons, and they fall behind as the teacher moves on.
• Many move on to study more complex concepts with a shaky foundation.
• We often end up with a weak structure that is doomed to collapse at some point.

Although this may sound like bad news, it is really good news. The fix is pretty easy if we’re patient enough!

No matter where you are in your math studies, you can excel if you backtrack far enough to reinforce your foundation. You must fill in the holes with a deep understanding of the basic concepts you encountered in middle school math.

• If you’re in middle school right now, do not attempt to move on until you understand pre-algebra concepts fully. Get a tutor if necessary.
• If you’re in high school and struggling with math, download a middle school math syllabus or hire a tutor. Make sure you understand every single concept and activity that is covered in middle grades.
• If you’re in college, backtrack all the way to basic math and work forward. This won’t take as long as it sounds. You can work forward through years of math in a week or two.

No matter where you start and where you struggle, you must make sure you acknowledge any weak spots in your foundation and fill the holes with practice and understanding!

Source : https://www.thoughtco.com/

### Perkalian Bentuk Aljabar Soal dan Pembahasan Penting (Kelas 7 SMP)

Kali ini kita belajar mengalikan operasi bentuk aljabar dengan mengikuti hukum distributif.

Contoh 1

Kalikan x3(x4 + 5a)

Jawab

Kita perluas dengan hukum distributif

x3(x4 + 5a)

x7 + 5ax3

Kita sudah tidak dapat menyederhanakan lagi.

## Contoh 2

Kalikan (x + 5)(a − 6)

Jawab

Uraikan hasilnya sebagai berikut :

(x + 5)(a − 6)

x(− 6) + 5(a − 6)

= ax − 6x + 5a − 30

Sudah tidak dapat disederhanakan kembali. Selesai sampai di situ.

## Contoh 3

Kalikan (2x + 3)(x2 − x − 5)

Jawab

Kita kalikan dengan memecah

(2x + 3)(x2− x − 5)

= (2x)(x2 − x − 5) + (3)(x2 − x − 5)

= (2x3 − 2x2 − 10x+ (3x2 − 3x − 15)

= 2x3 + x2 − 13x − 15

Kita jumlahkan dengan melepas kurung.

−2x2 + 3x2 = x2

dan

−10− 3x = −13x

## Contoh 4

Kalikan $\displaystyle{\left({x}-{3}\right)}^{2}$

Jawab

(x − 3)2

= (x − 3)(x − 3)

x(x − 3) − 3(x − 3)

x2 − 3x − 3x + 9

x2 − 6x + 9

## Catatan penting:

$\displaystyle{\left({x}-{3}\right)}^{2}$

tidak sama dengan

$\displaystyle{x}^{2}-{9}$

Contoh 5

Sedrerhanakan 5x[−4 + 10(x − y)] + 7x

Jawab

5x[−4 + 10(x − y)] + 7x

= 5x[−4 + 10x − 10y] + 7x

= −20x + 50x2 − 50xy + 7x

= −13x + 50x2 − 50xy

## Contoh 6

Sederhanakan (p − 1)(c − 1) + 2

Jawab

p(c − 1) − 1(c − 1) + 2

pc − p − c + 1 + 2

pc − p − c + 3


text:
Version: 0.6

# Authentication

##### info

The IOTA DIDComm Specification is in the RFC phase and may undergo changes. Suggestions are welcome at GitHub #464.

• Version: 0.1
• Status: IN-PROGRESS
• Last Updated: 2021-10-29

## Overview

This protocol allows two parties to mutually authenticate by disclosing and verifying the DID of each other. On successful completion of this protocol, it is expected that sender authenticated encryption may be used between the parties for continuous authentication.

### Example Use-Cases

• A connected sensor wants to make sure only valid well-known parties connect to it, before allowing access.
• A customer wants to make sure they are actually connecting to their bank, before presenting information.
• An organisation wants to verify the DID of the employer before issuing access credentials.

### Roles

• Requester: initiates the authentication.
• Responder: responds to the authentication request.

### Interaction

For guidance on diagrams see the corresponding section in the overview.

## Messages

### 1. authentication-request

• Type: iota/authentication/0.1/authentication-request
• Role: requester

Sent to initiate the authentication process. This MUST be a signed DIDComm message to provide some level of trust to the responder. However, even when signed it is possible to replay an authentication-request, so this message alone is insufficient to prove the DID of the requester. In addition to a unique requesterChallenge, the created_time and expires_time DIDComm message headers SHOULD be used to mitigate such replay attacks. Note that even a successful replay would only reveal the DID of the responder, authentication of a malicious requester will still fail without access to the original requester's private keys due to the use of challenges.

#### Structure

{  "did": DID,                   // REQUIRED  "requesterChallenge": string, // REQUIRED  "upgradeEncryption": string,  // REQUIRED}
FieldDescriptionRequired
didDID of the requester.1
requesterChallengeA random string unique per authentication-request by a requester to help mitigate replay attacks.
upgradeEncryptionA string indicating whether sender authenticated encryption should be used in the following messages. One of ["required", "optional", "unsupported"].2

1 The signing key used for the signed DIDComm envelope wrapping this message MUST be an authentication method in the DID document corresponding to did, as per the DIDComm specification.

2 The upgradeEncryption field allows negotiation of whether or not to use sender authenticated encryption for the authentication protocol and for all messages that follow it. It is RECOMMENDED to specify "required" as it offers various guarantees of continuous authentication and payload integrity for every message. The available options are:

#### Examples

{  "did": "did:iota:9rK6DPF46MCEzgfLD8AHFsaTuMqvmRo6kbXfjqQJPJmC",                     "requesterChallenge": "81285532-b72a-4a99-a9bd-b470475bc24f",                       "upgradeEncryption": "required",}

### 2. authentication-response

• Type: iota/authentication/0.1/authentication-response
• Role: responder

Sent in response to an authentication-request, proving the DID of the responder. Optionally establishes sender authenticated encryption based on the value of upgradeEncryption in the preceding authentication-request. If upgradeEncryption was "required" and this message is not encrypted, or "unsupported" and this message is encrypted, the requester MUST issue a problem-report and abort the authentication.

This message MUST be a signed DIDComm message, even if sender authenticated encryption is used. This is to ensure an authentication key is used to sign the challenge, in accordance with the DID specification, and because there may be increased security controls or guarantees compared to the keyAgreement keys used for sender authenticated encryption.

#### Structure

{  "did": DID,                   // REQUIRED  "requesterChallenge": string, // REQUIRED  "responderChallenge": string, // REQUIRED}
FieldDescriptionRequired
didDID of the responder.1
requesterChallengeMust match the requesterChallenge in the preceding authentication-request.
responderChallengeA random string unique per authentication-response by a responder to help mitigate replay attacks.

1 The signing key used for the signed DIDComm envelope wrapping this message MUST be an authentication method in the DID document corresponding to the did, as per the DIDComm specification.

#### Examples

1. Responder presenting their DID and offering a challenge to the requester:
{  "did": "did:iota:8cU6DPF56MDEugfLF8AHFaaTuMQvmRo6kbxfjqQJpJmC",  "requesterChallenge": "81285532-b72a-4a99-a9bd-b470475bc24f",  "responderChallenge": "b1f0dc02-85a3-4438-8786-b625f11f1be4",}

### 3. authentication-result

• Type: iota/authentication/0.1/authentication-result
• Role: requester

This message finalises the mutual authentication, proving control over the DID of the requester to the responder. Similar to authentication-response, this message MUST be a signed DIDComm message.

This MUST or MUST NOT use sender authenticated encryption depending on the outcome of the upgradeEncryption negotiation in the preceding authentication-response message, otherwise resulting in a problem-report and failure of the authentication protocol. For example, if upgradeEncryption was optional and the authentication-response used sender authenticated encryption, then the authentication-result MUST be encrypted to be valid.

#### Structure

{  "responderChallenge": string // REQUIRED}
FieldDescriptionRequired
responderChallengeMust match the challenge in the preceding authentication-response.1

1 The signing key used for the signed DIDComm envelope wrapping this message MUST be an authentication method in the DID document corresponding to the did of the requester in the authentication-request, as per the DIDComm specification.

#### Examples

1. Requester responding with the responder's challenge from the previous message:
{  "responderChallenge": "0768e82d-f498-4f38-8686-918325f9560d"}

### Problem Reports

The following problem-report codes may be raised in the course of this protocol and are expected to be recognised and handled in addition to any general problem-reports. Implementers may also introduce their own application-specific problem-reports.

For guidance on problem-reports and a list of general codes see problem reports.

CodeMessageDescription
e.p.msg.iota.authentication.reject-authenticationauthentication-request, authentication-response, authentication-resultThe party rejects an authentication request/response/result for any reason.
e.p.msg.iota.authentication.reject-authentication.missing-keysauthentication-request, authentication-response, authentication-resultThe party rejects an authentication request/response due to the other party lacking a supported keyAgreement section in the DID document.
e.p.msg.iota.authentication.reject-authentication.untrusted-identityauthentication-request, authentication-responseThe party rejects an authentication request/response due to the claimed DID of the other party.
e.p.msg.iota.authentication.reject-authentication.encyption-requiredauthentication-request, authentication-response, authentication-resultThe party rejects an authentication request/response/result due to the lack of sender authenticated encryption.
e.p.msg.iota.authentication.reject-authentication.encyption-unsupportedauthentication-request, authentication-response, authentication-resultThe party rejects an authentication request/response/result because it does not support sender authenticated encryption.

## Considerations

This section is non-normative.

• Trust: this authentication protocol only verifies that both parties have access to the necessary private keys (which could become compromised) associated with their DID documents. Verifying whether a DID document is bound to a physical identity may require additional interactions. Verifying whether a DID can be trusted can be achieved by, for instance:
• requesting a verifiable presentation of credentials issued by a trusted third party, such as a government,
• using the Well Known DID Configuration or embedding the DID in a DNS record to tie an identity to a website or domain,
• using an allowlist of trusted DIDs,
• exchanging DIDs out-of-band in a secure manner (note that some connection invitations could be altered by malicious parties depending on the medium).
• Authorisation: the permissions and capabilities of either party may still need to be established after authentication, either by verifiable presentation as mentioned above or other methods such as JWT tokens
• Privacy: the responder may be subject to probing whereby their DID may be revealed even with the use of sender authenticated encryption, as the skid message header is linked to their DID. This is possible if the responder chooses to accept the authentication-request of an unknown requester, or the requester successfully replays an authentication-request from a DID the requester trusts.

## Unresolved Questions

• Enforce signed DIDComm messages on top of sender-authenticated encryption or keep them optional? Usually unnecessary and DIDComm recommends against this since it's redundant and due to non-repudiation may decrease security and privacy by allowing participants to prove to third parties that authentication occurred.

• How to protect the DID of the responder (skid field in sender-authenticated message) to prevent probing identities even with anonymous encryption?

• Add examples of full signed and sender-authenticated messages with headers for better illustration?


text:
# How do you determine if Sigma (7*5^n)/6^n from n=[0,oo) converge and find the sums when they exist?

Feb 8, 2017

${\sum}_{n = 0}^{\infty} \frac{7 \cdot {5}^{n}}{6} ^ n = 42$

#### Explanation:

Write the series as:

${\sum}_{n = 0}^{\infty} \frac{7 \cdot {5}^{n}}{6} ^ n = 7 {\sum}_{n = 0}^{\infty} {\left(\frac{5}{6}\right)}^{n}$

so this is a geometric series of ratio $r = \frac{5}{6}$.

As $\left\mid r \right\mid < 1$ the series is convergent and its sum is:

$7 {\sum}_{n = 0}^{\infty} {\left(\frac{5}{6}\right)}^{n} = \frac{7}{1 - \frac{5}{6}} = \frac{7}{\frac{1}{6}} = 42$


text:
# How do you find the zeros, real and imaginary, of y= 3x^2-6x+2  using the quadratic formula?

$1 \pm \frac{\sqrt{3}}{3}$
$D = {d}^{2} = {b}^{2} - 4 a c = 36 - 24 = 12 - \to d = \pm 2 \sqrt{3}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = \frac{6}{6} \pm \frac{2 \sqrt{3}}{6} = 1 \pm \frac{\sqrt{3}}{3}$