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# Integration by Trigonometric Substitution vs Table of Integral Solution

I'm not sure how to phrase this question, but I find myself confused over the correctness of a particular solution in the table of integral, specifically: $$\int\frac{dx}{\sqrt{x^2-a^2}}=\ln|\sqrt{x^2-a^2}+x|+C$$

I find that if you try to solve it with trigonometric substitution, you get a different answer:

So, for the example, since the form $$\sqrt{x^2-a^2}$$ is present, we use the substitution $$x=asec \theta$$, and $$dx=asec\theta tan \theta d\theta$$. Doing so:

$$\int\frac{asec\theta tan \theta d\theta}{\sqrt{(asec \theta)^2-a^2}}$$ $$\int\frac{asec\theta tan \theta d\theta}{\sqrt{a^2sec^2 \theta-a^2}}$$ $$\int\frac{asec\theta tan \theta d\theta}{a\sqrt{sec^2 \theta-1}}$$ Using the trigonometric identity $$tan^2 \theta=sec^2 \theta -1$$ $$\int\frac{asec\theta tan \theta d\theta}{atan \theta}$$

Cancelling like terms, we get $$\int sec \theta$$, which finally integrates to $$ln|sec \theta + tan \theta|+C$$

Undoing the substitution, using this triangle:

https://i.stack.imgur.com/6tkR7.png

With $$sec \theta = \frac{x}{a}$$ and $$tan \theta = \frac{\sqrt{x^2-a^2}}{a}$$, the final answer then is:

$$ln\biggl|\frac{x}{a} + \frac{\sqrt{x^2-a^2}}{a}\biggl|+C$$

Which is evidently different from the one from the table of integrals. Is my solution wrong or something?

• looks the same to me, since $-\ln |a|$ can be absorbed into $C$? – Calvin Khor Feb 7 at 13:48
• Oh right, that can be done. Thanks for pointing that out – Xyzar Feb 7 at 13:52

$$\ln \left|\dfrac{x+\sqrt{x^2-a^2}}{a}\right| +c= \ln |x+\sqrt{x^2-a^2}| - \ln|a| +c = \ln(x+\sqrt{x^2-a^2}) +k$$
$$k = c-\ln |a| =$$ constant

format: plain


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# In the equation 2KClO_3 -> 2KCI + 3O_2, how many moles of oxygen are produced when 5.05 m of KClO_3 decompose completely?

##### 1 Answer
Nov 18, 2016

7.58 moles of ${O}_{2}$ are produced.

#### Explanation:

There are multiple lines of reasoning for solving this. Let me give you the most detailed:

You start with a plan: we will convert moles of $K C l {O}_{3}$ to moles of ${O}_{2}$ using the balanced chemical equation:

You will get 3 moles of ${O}_{2}$ for every 2 moles of $K C l {O}_{3}$ you decompose.

Now let's substitute the number of moles of $K C l {O}_{3}$ that you actually started with:

$5.05 \text{ moles "KClO_3 xx (3 " moles " O_2)/(2 " moles " KClO_3) = 7.575 " moles } {O}_{2}$

Finally, since we started with three significant figures, we have to end up with three significant figures since all we did was multiply and divide, so...

7.58 moles ${O}_{2}$


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# Highlights

• New network architecture (SDNet) that factorizes 2D medical images into spatial anatomical factors and non-spatial modality factors;
• Various experiments to show that the learned representation is well-suited to a variety of image analysis tasks, including:
• semi-supervised segmentation
• multi-task segmentation and regression (e.g. Left Ventricular Volume estimation)
• image-to-image synthesis

# Introduction

Learning a decomposition of data into a spatial content factor and a non-spatial style factor has been a focus of recent research in computer vision[…].

This focus can be explained by the advantages of such a representation:

• Meaningful representation of the anatomy that can be generalized to any modality

• Suitable format for pooling information from various imaging modalities

# Methods

## Spatial Decomposition Network (SDNet)

The SDNet can be seen as an autoencoder that learns multiple factors, namely:

• $$s = f_A (x)$$: a multi-channel output of binary maps, representing the anatomical components
• $$z = f_M (x)$$: the $$Q(z \vert X)$$ multivariate Gaussian as in a standard VAE, representing the modality components

Other sub-networks are added to provide feedback over multiple tasks, with varying degrees of supervision:

• $$g$$: a self-supervised decoder that tries to reconstruct the input image from its $$s$$ and $$a$$ decomposition;
• $$h$$: a segmentor network that predicts the cardiac segmentation from $$s$$.
• When a ground truth is available for the current image, $$h$$ is trained using a standard Dice loss;
• Otherwise, the training is semi-supervised through a GAN-like adversarial loss, where a discriminator network tries to predict whether a segmentation mask was predicted by $$h$$ or comes from a pool of groundtruth segmentations.

The overall loss function is the following weighted sum (determined empirically):

$L = \lambda_1 L_{KL} + \lambda_2 L_{segm} + \lambda_3 L_{adv} + \lambda_4 L_{rec} + \lambda_5 L_{z_{rec}}$

where $$L_{KL}$$ and $$L_{rec}$$ make up the autoencoder’s self-supervision, and $$L_{segm}$$ and $$L_{rec}$$ correspond to the aforementioned segmentation supervision or semi-supervision. The additional factor is $$L_{z_{rec}}$$, which is enforces a modality factor reconstruction:

$L_{z_{rec}} = \mathbb{E}_{z,y} [\|z - f_{modality}(y, f_{anatomy}(y))\| _{1}]$

where $$y$$ is an image produced using a random $$z$$ sample. This is done to avoid a posterior collapse, where the decoder would ignore parts or the totality of the modality factor.

# Data

## Semi-supervised segmentation

• Training subset of the ACDC dataset:
• 1920 images with manual segmentations (ED and ES) and 23,530 images with no segmentations
• Edinburgh Imaging Facility QMRI: 26 healthy volunteers with around 30 cardiac phases each, acquired on a 3T scanner
• 241 images with manual segmentations (ED) and 8353 images with no segmentations

## Multimodal segmentation and modality transformation

• Data from the 2017 Multi-Modal Whole Heart Segmentation (MM-WHS) Challenge: 20 cardiac CT/CT angiography (CTA) volumes and 20 cardiac MRI volumes
• 3626 MR and 2580 CT images, all with manual segmentations of seven heart structures: myocardium, left atrium, left ventricle, right atrium, right ventricle, ascending aorta and pulmonary artery

## Modality estimation

• cine-MR and CP-BOLD images of 10 canines[…]. Two almost identical sequences with the only difference that CP-BOLD modulates pixel intensity with the level of oxygenation present in the tissue.

• 129 cine-MR and 264 CP-BOLD images with manual segmentations from all cardiac phases

# Results

## Others

Other experiments were also conducted relative to modality type estimation (from modality factors) and modality factor traversal (named factor sizes in the paper).

# Remarks

• The authors present a lot of details about their design choices, so it seems possible to reproduce accurately their experiments.


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# 8.3 Nash inequalities

xxx For NOTES: Nash vs. Sobolev

A Nash inequality for a chain is an inequality of the form

 $\|g\|^{2+\frac{1}{D}}_{2}\leq C\,[\mbox{{\cal E}}(g,g)+{\textstyle\frac{1}{T% }}\|g\|^{2}_{2}]\,\|g\|^{1/D}_{1}$ (8.46)

that holds for some positive constants $C$, $D$, and $T$ and for all functions $g$. We connect Nash inequalities to mixing times in Section 8.3.1, and in Section 8.3.2 we discuss a comparison method for establishing such inequalities.

## 8.3.1 Nash inequalities and mixing times

A Nash inequality implies a useful bound on the quantity

 $N(t)=\|{\bf P}_{t}\|_{2\to\infty}$ (8.47)

appearing in the mixing time Lemma 8.13. This norm is continuous in $t$ and decreases to $\|{\bf E}\|_{2\to\infty}=1$ as $t\uparrow\infty$. Here is the main result:

###### Theorem 8.17

If the Nash inequality (8.46) holds for a continuous-time reversible chain, some $C,D,T>0$, and all $g$, then the norm $N(s)$ at (8.47) satisfies

 $0

Proof. First note $N(t)=\|{\bf P}_{t}\|_{1\to 2}$ by Lemma 8.12. Thus we seek a bound on $h(t):=\|{\bf P}_{t}g\|^{2}_{2}$ independent of $g$ satisfying $\|g\|_{1}=1$; the square root of such a bound will also bound $N(t)$.

Substituting ${\bf P}_{t}g$ for $g$ in (8.46) and utilizing the identity in Lemma 8.10(a) and the fact that ${\bf P}_{t}$ is contractive on $L^{1}$, we obtain the differential inequality

 $h(t)^{1+\frac{1}{2D}}\leq C\,[-{\textstyle\frac{1}{2}}h^{\prime}(t)+{% \textstyle\frac{1}{T}}h(t)],\ \ \ t\geq 0.$

Writing

 $H(t):=\left[{\textstyle\frac{1}{2}}Ch(t)e^{-2t/T}\right]^{-1/(2D)},$

the inequality can be equivalently rearranged to

 $H^{\prime}(t)\geq\left[2D(C/2)^{1+\frac{1}{2D}}\right]^{-1}e^{\frac{t}{DT}},\ % \ \ t\geq 0.$

Since $H(0)>0$, it follows that

 $H(t)\geq T\left[2(C/2)^{1+\frac{1}{2D}}\right]^{-1}\left(e^{\frac{t}{DT}}-1% \right),\ \ \ t\geq 0,$

or equivalently

 $h(t)\leq\left[\frac{T}{C}\left(1-e^{-t/(DT)}\right)\right]^{-2D},\ \ \ t\geq 0.$

But

 $0

so for these same values of $t$ we have

 $\displaystyle h(t)$ $\displaystyle\leq$ $\displaystyle\left[\frac{t}{C}\left(1-e^{-1/D}\right)\right]^{-2D}$ $\displaystyle=$ $\displaystyle e^{2}\left[\frac{t}{C}\left(e^{1/D}-1\right)\right]^{-2D}\leq[e(% DC/t)^{D}]^{2},$

as desired.

We now return to Lemma 8.13 and, for $t\geq T$, set $s=T$. (Indeed, using the bound on $N(s)$ in Theorem 8.17, this is the optimal choice of $s$ if $T.) This gives

xxx NOTE: In next theorem, only need conclusion of Theorem 8.17, not hypothesis!

###### Theorem 8.18

In Theorem 8.17, if $c\geq 1$ and

 $t\geq T+\tau_{2}\left(D\log\left(\frac{DC}{T}\right)+c\right),$

then $\sqrt{{\hat{d}}(2t)}\leq e^{1-c}$; in particular,

 ${\hat{\tau}}\leq T+\tau_{2}\left(D\log\left(\frac{DC}{T}\right)+2\right).$

The following converse of sorts to Theorem 8.17 will be very useful in conjunction with the comparison method.

###### Theorem 8.19

If a continuous time reversible chain satisfies

 $0

then it satisfies the Nash inequality

 $g$

with

 $C^{\prime}:=2\,(1+{\textstyle\frac{1}{2D}})\,[(1+2D)^{1/2}C]^{1/D}\leq 2^{2+% \frac{1}{2D}}\,C^{1/D}.$

xxx Detail in proof to be filled in (I have notes): Show $\mbox{{\cal E}}({\bf P}_{t}g,{\bf P}_{t}g)\downarrow$ as $t\uparrow$, or at least that it’s maximized at $t=0$. Stronger of two statements is equivalent to assertion that $\|{\bf P}_{t}f\|^{2}_{2}$ is convex in $t$.

Proof. As in the proof of Theorem 8.17, we note $N(t)=\|{\bf P}_{t}\|_{1\to 2}$. Hence, for any $g$ and any $0,

 $\displaystyle\|g\|^{2}_{2}$ $\displaystyle=$ $\displaystyle\|{\bf P}_{t}g\|^{2}_{2}-\int_{s=0}^{t}{\textstyle\frac{d}{ds}}\|% {\bf P}_{s}g\|^{2}_{2}\,ds$ $\displaystyle=$ $\displaystyle\|{\bf P}_{t}g\|^{2}_{2}+2\int_{s=0}^{t}\mbox{{\cal E}}({\bf P}% _{s}g,{\bf P}_{s}g)\,ds\mbox{\ \ \ by Lemma~{}\ref{basic-1}(a)}$ $\displaystyle\leq$ $\displaystyle\|{\bf P}_{t}g\|^{2}_{2}+2\mbox{{\cal E}}(g,g)t\mbox{\ \ \ xxx % see above}$ $\displaystyle\leq$ $\displaystyle 2\mbox{{\cal E}}(g,g)t+C^{2}t^{-2D}\|g\|^{2}_{1}.$

This gives

 $\|g\|^{2}_{2}\leq t[2\mbox{{\cal E}}(g,g)+{\textstyle\frac{1}{T}}\|g\|^{2}_{% 2}]+C^{2}t^{-2D}\|g\|^{2}_{1}$

for any $t>0$. The righthand side here is convex in $t$ and minimized (for $g\neq 0$) at

 $t=\left(\frac{2DC^{2}\|g\|^{2}_{1}}{2\mbox{{\cal E}}(g,g)+T^{-1}\|g\|^{2}_{2% }}\right)^{1/(2D+1)}.$

Plugging in this value, raising both sides to the power $1+{\textstyle\frac{1}{2D}}$ and simplifying yields the desired Nash inequality. The upper bound for $C^{\prime}$ is derived with a little bit of calculus.

## 8.3.2 The comparison method for bounding $N(\cdot)$

In Section 8.1 we compared relaxation times for two chains by comparing Dirichlet forms and variances. The point of this subsection is that the comparison can be extended to the norm function $N(\cdot)$ of (8.47) using Nash inequalities. Then results on $N(\cdot)$ like those in Section 8.2.3 can be used to bound mixing times for other chains on the same state space.

xxx For NOTES?: Can even use different spaces. New paragraph:

To see how this goes, suppose that a benchmark chain is known to satisfy

 $0 (8.48)

By Theorem 8.19, it then satisfies a Nash inequality. The $L^{1}$- and $L^{2}$-norms appearing in this inequality can be compared in the obvious fashion [cf. (8.18)] and the Dirichlet forms can be compared as in Theorem 8.1. This shows that the chain of interest also satisfies a Nash inequality. But then Theorem 8.17 gives a bound like (8.48) for the chain of interest, and Theorem 8.18 can then be used to bound the $L^{2}$ threshold time ${\hat{\tau}}$.

Here is the precise result; the details of the proof are left to the reader.

###### Theorem 8.20 (comparison of bounds on $N(\cdot)$)

If a reversible benchmark chain satisfies

 $0

for constants ${\tilde{C}},{\tilde{D}},{\tilde{T}}>0$, then any other reversible chain on the same state space satisfies

 $0

where, with $a$ and $A$ as defined in Corollary 8.2, and with

 $a^{\prime}:=\max_{i}({\tilde{\pi}}_{i}/\pi_{i}),$

we set

 $\displaystyle D$ $\displaystyle=$ $\displaystyle{\tilde{D}},$ $\displaystyle C$ $\displaystyle=$ $\displaystyle a^{-(2+\frac{1}{D})}{a^{\prime}}^{1/D}A\times 2(1+{\textstyle% \frac{1}{2D}})[(1+2D)^{1/2}{\tilde{C}}]^{1/D}$ $\displaystyle\leq$ $\displaystyle a^{-(2+\frac{1}{D})}{a^{\prime}}^{1/D}A\times 2^{2+\frac{1}{2D}}% {\tilde{C}}^{1/D},$ $\displaystyle T$ $\displaystyle=$ $\displaystyle\frac{2A}{{a^{\prime}}^{2}}{\tilde{T}}.$

xxx Must correct this slightly. Works for any $A$ such that ${\tilde{\cal E}}\leq A\mbox{{\cal E}}$, not just minimal one. This is important since we need a lower bound on $T$ but generally only have an upper bound on ${\tilde{\cal E}}/\mbox{{\cal E}}$. The same goes for $a^{\prime}$ (only need upper bound on ${\tilde{\pi}}_{i}/\pi_{i}$): we also need an upper bound on $T$.

###### Example 8.21

Random walk on a $d$-dimensional grid.

As in Example 8.5, consider the continuized walk on the $d$-dimensional grid $I=\{0,\ldots,m-1\}^{d}$. In Example 8.5 we compared the Dirichlet form and variance for this walk to the $d$-fold product of random walk on the $m$-path with end self-loops to obtain

 $\tau_{2}\leq{\tilde{\tau}}_{2}=d(1-\cos{\textstyle\frac{\pi}{m}})^{-1}\leq{% \textstyle\frac{1}{2}}dm^{2};$ (8.49)

using the simple bound $\sqrt{{\hat{d}}(2t)}\leq\pi^{-1/2}_{*}e^{-t/\tau_{2}}\leq(2n)^{-1/2}\exp\left(% -\frac{2t}{dm^{2}}\right)$ we then get

 ${\hat{\tau}}\leq{\textstyle\frac{1}{4}}m^{2}d[\log(2n)+2],$ (8.50)

which is of order $m^{2}d^{2}\log m$. Here we will see how comparing $N(\cdot)$, too, gives a bound of order $m^{2}d^{2}\log d$. In Example 8.43, we will bring log-Sobolev techniques to bear, too, to lower this bound to order $m^{2}d\log d$

xxx which is correct, at least for TV. New paragraph:

Recalling (8.45), we may apply Theorem 8.20 with

 ${\tilde{D}}=d/4,\ \ \ {\tilde{C}}=(4dm^{2})^{d/4},\ \ \ {\tilde{T}}=dm^{2}/16,$

and, from the considerations in Example 8.5,

 $a\geq{\textstyle\frac{1}{2}},\ \ \ A=1,\ \ \ a^{\prime}=2.$

xxx See xxx following Theorem 8.20. Same paragraph:

This gives

 $D=d/4,\ \ \ C\leq 2^{6+\frac{10}{d}}dm^{2}\leq 2^{16}dm^{2},\ \ \ T=dm^{2}/32.$

Plugging these into Theorem 8.18 yields

 ${\hat{\tau}}\leq{\textstyle\frac{1}{8}}m^{2}d^{2}\log d+({\textstyle\frac{19}{% 8}}\log 2)m^{2}d^{2}+{\textstyle\frac{33}{32}}m^{2}d,$ (8.51)

which is $\leq 3m^{2}d^{2}\log d$ for $d\geq 2$.

Other variants of the walk, including the thinned-grid walk of Example 8.6, can be handled in a similar fashion.

xxx Do moderate growth and local Poincaré? Probably not, to keep length manageable. Also, will need to rewrite intro a little, since not doing $\Delta^{2}$-stuff (in any detail).


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# Homotopy Type Theory function extensionality

## Idea

Remember that when two functions $f,g : A \to B$ are homotopic we have a witness to

$\prod_{x : A} f(x)=g(x)$

but as discussed in the homotopy article, it is not the case that $f=g$.

Function extensionality simply assumes this is the case as an axiom.

## Definition

The axiom of functional extensionality says that for all $A,B$, $f,g : A \to B$ there is a function

$funext : (f \sim g) \to (f = g)$

This allows homotopic functions to be equal.


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        If Santa Claus 🎅 distributes gifts to every single child in every nook and cranny of the world in a single night, (taking every unit surface area of the earth to contain a single child) and assuming he has the entire globe to distribute to, in only a matter of 8.5 hours, what is his speed? [ std. value of earth's radius is considered ]
4 years ago

Manas Baul
18 Points
							Radius of Earth= 6378.1 KmTotal Surface Area = 4/3$\pi r^3$ = 510,072,000 km2Every Unit Surface Area Has one child $\therefore$ Total Number of Childs = 510,072,000Total Time = 8.5 hrsHence, Speed = 510,072,000(2)(17)-1= 6.00084706 X 108 Km/hWhich is way greater then speed of light.

4 years ago
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## Number Theory

Suppose { $$a_{1}, a_{2}, ..a_{k}$$ } is a complete residue system modulo k where k is prime. Prove that for each integer n and each nonnegative integer s there exists a congruence of the form $$n = \sum_{j=0}^{s}b_{j}k^{j} (mod k^{s+1})$$ where $$b_{j}$$ is one of the $$a_{j}$$


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# Gamma of a gas mixture

1. May 10, 2014

### Jahromi24

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

An experiment you're designing needs a gas with γ = 1.49. You recall from your physics class that no individual gas has this value, but it occurs to you that you could produce a gas with γ = 1.49 by mixing together a monatomic gas and a diatomic gas.
What fraction of the molecules need to be monatomic?

2. Relevant equations

$\gamma$=5/3 for Monatomic
$\gamma$=7/5 for Diatomic

3. The attempt at a solution

$\gamma$=5x/3+7/5(1-x)=1.49
I solve for x and get 0.3375, but this is wrong.
I can't figure out what I'm doing wrong because everywhere I look it seems correct to me.

2. May 10, 2014

### Feodalherren

The 7/5 for the diatomic gas is a simplified version of reality where you're saying it has 5 degrees of freedom. x,y,z and then it can rotate around the x and z axis. In reality it also rotates around the y-axis but the number is very small. Just an idea.

3. May 10, 2014

### Staff: Mentor

Cp and Cv of a gas mixture are proportional to the number of moles of each gas, but the ratio of Cp to Cv for the mixture is not. You need to treat Cp and Cv separately, and also make use of the condition that Cp - Cv of each species is equal to R.

chet

4. Dec 12, 2016

### Hoang-Lan Jones

Could you please go through the steps? I still do not understand, and I have my final tomorrow.

5. Dec 12, 2016

### Mister T

$\gamma=\frac{C_P}{C_V}$

Can you use that relation to work out the value of $\gamma$ if the mixture is half monatomic and half diatomic?

6. Dec 12, 2016

### Hoang-Lan Jones

I think you're supposed to use that, but I'm not sure how.
Also, I found this, and I think it helps, but I'm not sure what to do with this either.

Last edited by a moderator: Apr 15, 2017
7. Dec 12, 2016

### Mister T

Can you answer my question? Yes or no?

8. Dec 12, 2016

### Hoang-Lan Jones

Okay, I have actually figured it out! Thank you for leading me to find the right path, Mister T!
My work is below! :) I was actually working with a gamma that was 1.52 for the problem, but the math should still be correct.

9. May 23, 2017

### SwapDogg

γ = n1 (f1+2) + n2(f2+2)
n1f1 + n2f2


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Show that $x \ln(ex) - \sqrt{x}\geq 0$ for all $x\geq 1$

How do I continue to prove this?

Show that $$x \ln(ex) - \sqrt{x}\geq 0$$ for all $$x \geq1$$

My try:

$$\begin{eqnarray*} \\ \ln(e^x) + \ln(x^x) &\geq& \sqrt{x} \\ \\ \ln(e^x) &\geq& \sqrt{x} - \ln(x^x) \\ \\ e^x &\geq & e^\sqrt{x} e^{-\ln(x^x)} \\ \\ e^x&\geq& e^\sqrt{x} (1/x^x) \\ \\ e^x - \frac{e^\sqrt{x}}{x^x}&\geq& 0 \end{eqnarray*}$$

I "see" that this is bigger than 0, but I think that there are more calculations to do.

-
Let $x$ be positive but less than $1/e$. Then our expression is clearly negative. So you will have to change the condition $x\ge 0$ to something else. –  André Nicolas Sep 27 '12 at 15:32
@AndréNicolas He actually wants $x\geq 1$. –  Pedro Tamaroff Sep 27 '12 at 15:33
You have an error: $e^{-\ln x^x}\ne -x^x$; but rather $e^{-\ln x^x}= e^{\ln (1/x^x)}=1/x^x$. –  David Mitra Sep 27 '12 at 15:33
@DavidMitra: Thanks. –  Julian Assange Sep 27 '12 at 15:34
Okay, but now I think it's obivous that this is bigger than 0? –  Julian Assange Sep 27 '12 at 15:41

Consider the function

$$f\left( x \right) = x\log \left( {ex} \right) - \sqrt x = x\log x + x - \sqrt x$$

$\log x$ is positive for $x>1$ and negative for $0<x<1$.

And $x>\sqrt x$ for $x>1$, and $x<\sqrt x$ for $0<x<1$. Thus

$$\begin{cases} f(x)>0 \text{ for } x>1\\ f(x)<0 \text{ for } 0<x<1\\f(x)=0 \text{ for }x=1\end{cases}$$

-
Sorry, but I meant x>=0. –  Julian Assange Sep 27 '12 at 15:51
@JulianAssange I'm showing you the inequality is not true for $x\geq 0$. –  Pedro Tamaroff Sep 27 '12 at 16:00
Sorry, I meant x >= 1. :) –  Julian Assange Sep 27 '12 at 16:03
@JulianAssange I showed you that is true, too. Did you read the answer carefully? –  Pedro Tamaroff Sep 27 '12 at 16:10
Yeah, I just wanted you to know. :=) –  Julian Assange Sep 27 '12 at 16:12

Substitute 1/e for x. Of course 1/e is greater than zero. But ln(e*1/e) equals zero, so your Left Hand Side is negative, equals to -1/sqrt{e}. Therefore, the inequality you are trying to prove is false.

-
You should probably edit this to say that it was for the original version of the problem, which claimed the inequality for all $x\ge 0$. –  Brian M. Scott Sep 27 '12 at 20:32

Let $f(x)=x\ln(ex)-\sqrt x$. Note that $f'(x)=\ln(ex)+1 -\tfrac{1}{2\sqrt x}>0 \;(*)$ for $x\geq1$, and so $f$ is increasing from $x=1$. But $f(1)=0$, and so $f(x)\geq 0$ for $x\geq 1$.

$(*)$ This can be seen since $\ln(ex)+1$ is increasing, and $\tfrac{1}{2\sqrt x}$ is decreasing, so $f'(x)$ must be increasing. Since $f'(1)>0$, we thus must have this inequality for all $x\geq1$.

-
The inequality is true for $x \geq 1$. Put $f(x)=x\ln({\rm e}x)-\sqrt{x}$ and use the derivative test to prove that.


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2 added 5 characters in body

The problem with respect to applications of the abelianization is that the abelian categories one produces are almost uniformly horrible. More precisely they are just too big to deal with. So using them to produce results which aren't formal statements about some class of triangulated categories seems like it would be very challenging.

As a concrete(ish) example suppose that $R$ is a discrete valuation ring and let $D(R)$ be the unbounded derived category of $R$-modules. Then the abelianization $A(D(R))$ is not well-copowered - the image of the stalk complex $R$ in degree zero in $A(D(R))$ has a proper class of quotient objects (the reference for this is Appendix C of Neeman's book Triangulated Categories). Since in this case $R$-Mod is hereditary the derived category is really pretty good - we understand the compacts $D^b(R-mod)$ very well and we know all of the localizing and smashing subcategories of $D(R)$. It even comes with a natural tensor product making it rigidly compactly generated and it has a DG-enhancement. On the other hand the abelianization is kind of crazy.

One can take approximations of the abelianization by abelian categories which are more managable. However, I still don't know of any specific applications that aren't just shadows of facts which work for all sufficiently nice triangulated categories.

1

The problem with respect to applications of the abelianization is that the abelian categories one produces are almost uniformly horrible. More precisely they are just too big to deal. So using them to produce results which aren't formal statements about some class of triangulated categories seems like it would be very challenging.

As a concrete(ish) example suppose that $R$ is a discrete valuation ring and let $D(R)$ be the unbounded derived category of $R$-modules. Then the abelianization $A(D(R))$ is not well-copowered - the image of the stalk complex $R$ in degree zero in $A(D(R))$ has a proper class of quotient objects (the reference for this is Appendix C of Neeman's book Triangulated Categories). Since in this case $R$-Mod is hereditary the derived category is really pretty good - we understand the compacts $D^b(R-mod)$ very well and we know all of the localizing and smashing subcategories of $D(R)$. It even comes with a natural tensor product making it rigidly compactly generated and it has a DG-enhancement. On the other hand the abelianization is kind of crazy.

One can take approximations of the abelianization by abelian categories which are more managable. However, I still don't know of any specific applications that aren't just shadows of facts which work for all sufficiently nice triangulated categories.


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# Newton's third law in terms of inertial position vectors for n-body system

 P: 30 Assuming $$\vec{r_{a}}$$ and $$\vec{r_{b}}$$ is calculated from an inertial frame of reference. then for any two objects (named a and b) in a system of more than two objects, Is this the newton's third law, $$\frac{d^{2}}{dt^{2}}m_{a}\vec{r_{a}}=-\frac{d^{2}}{dt^{2}}m_{b}\vec{r_{b}}$$ i think this cant be right because then this implies $$\frac{d^{2}}{dt^{2}}m_{i}\vec{r_{i}}=0$$ for every object in that system. so i think i have misunderstood the law, so my question is can anyone state the law in terms of above variables for n-body system ? Edit 1 (fix) fixed a embarrassing mistake d/dt -> d^2/dt^2 thank you
 P: 1,020 see if this helps https://docs.google.com/viewer?a=v&q...dUaHm8yd0n31kg
 P: 30 No it does not :( i think the problem i am facing is which frame of reference to use, for example 1. if i use a frame of reference such as its origin and the center of mass of the object from which the position vector is to be calculated, coinsides .so $$\vec{R_{AB}}$$ is position vector of object A from a frame of reference such as its origin and center of mass of object B coinsides. and $$\vec{R_{BA}}$$ is position vector of object B from a frame of reference such as its origin and center of mass of object A coinsides. and the law would be $$\frac{d^{2}}{dt^{2}}m_{A}\vec{R_{AB}}=-\frac{d^{2}}{dt^{2}}m_{B}\vec{R_{BA}}$$ but the problem what $$\theta_{AB},\theta_{BA}$$ to choose, certainly the above equation is not true for arbitrary $$\theta_{AB},\theta_{BA}$$ . 2. may be a inertial frame of reference is that frame and following eqaution is the law, $$\frac{d^{2}}{dt^{2}}m_{A}(\vec{r_{A}}-\vec{r_{B}})=-\frac{d^{2}}{dt^{2}}m_{B}(\vec{r_{B}}-\vec{r_{A}})$$ hence $$\frac{d^{2}}{dt^{2}}\left(m_{A}-m_{B}\right)\left(\vec{r_{A}}-\vec{r_{B}}\right)=0$$ where $$\vec{r_{A}},\vec{r_{B}}$$ is calculated from an inertial frame of reference . please tell me which one is correct ? if both not correct please state the law too in terms of $$\vec{r_{A}},\vec{r_{B}}$$ ? thank you
 P: 1,020 Newton's third law in terms of inertial position vectors for n-body system it just says F12=-F21,where F12=m2 dv2/dt(acting on second one).similarly for F21.Don't count others position vector into first one or so.
 Mentor P: 41,459 Newton's 3rd law simply says: $$\vec{F}_{ab} = - \vec{F}_{ba}$$ When you drag in the acceleration you are really talking about Newton's 2nd law, which involves the net force. No reason to think that the net force on particle a will be equal and opposite to the net force on particle b if other particles exist.
 P: 30 ok but since $$\vec{F}_{AB}$$ is a vector what is the frame of reference ? can $$\vec{F}_{AB}$$ be expressed in terms of any positional vectors or any other variables ?
 P: 1,020 ANY inertial reference frame will work.
 P: 30 i think i got it. $$\vec{F}_{AB}=\left|\vec{F}_{AB}\right|\hat{F}_{AB}$$ where $$\left|\vec{F}_{AB}\right|$$ is all those forces like gravitational etc . i did not know how to express $$\vec{F}_{AB}$$ . i thought it was same force as defined in 2nd law. thanks andrien, Doc Al .
Mentor
P: 41,459
 Quote by ato i think i got it. $$\vec{F}_{AB}=\left|\vec{F}_{AB}\right|\hat{F}_{AB}$$
Well, that would be true for any vector.
 where $$\left|\vec{F}_{AB}\right|$$ is all those forces like gravitational etc .
Not sure what you are saying here.

When I wrote Newton's 2nd law as $$\vec{F}_{ab} = - \vec{F}_{ba}$$
$\vec{F}_{ab}$ stood for the force on a exerted by b and $\vec{F}_{ba}$ stood for the force on b exerted by a.
P: 30
what i mean is that before tackling the third law, it is necessary that the notion $$\vec{F}_{ab}$$ be understood, means how to calculate or express it , which i did not. the point is $$\vec{F}_{ab}$$ is completely different than $$\vec{F}_{a}$$ defined in 2nd law. actually its confusing that both are referred as force.

i think the law is saying that when defining (or at least for all defined) inter-body forces (like gravitational,electric or magnetic), you only need (to define)
$$\vec{F}_{ab-interbody-force}$$
the third law would automatically define,
$$\vec{F}_{ba-interbody-force}$$

 Quote by Doc Al $$\vec{F}_{ab} = - \vec{F}_{ba}$$ $\vec{F}_{ab}$ stood for the force on a exerted by b and $\vec{F}_{ba}$ stood for the force on b exerted by a.
i suppose its for individual interbody force , then thats what i am saying . i am just trying to avoid any personification i can. but if its not for individual forces, then i dont know how to get $$\vec{F}_{ab}$$ if there are more than two forces involved.
my guess would be for two forces gravitational and electric,

$$\left(\vec{F}_{ab-gravitational}+\vec{F}_{ab-electrical}\right)=-\left(\vec{F}_{ba-gravitational}+\vec{F}_{ba-electrical}\right)$$

but then i think newton's laws would be insufficient to imply this
$$\vec{F}_{ab-gravitational}=-\vec{F}_{ba-gravitational}$$
and
$$\vec{F}_{ab-electrical}=-\vec{F}_{ba-electrical}$$

please tell me if i am wrong .
thanks
 Mentor P: 41,459 Newton's 3rd law concerns individual forces (interactions between two bodies) not the net force on a body.

 Related Discussions General Math 9 Calculus & Beyond Homework 3 Introductory Physics Homework 4 Introductory Physics Homework 3 Advanced Physics Homework 0


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# Math Help - Maclaurin Series for a Differential Equation

1. ## Maclaurin Series for a Differential Equation

Consider the differential equation:
$\frac{dy}{dx} = y - x + 1$
The initial condition is y(0) = 1. Construct the Maclaurin series for y(x).

I know I'm supposed to differentiate it, but I don't remember how to use the initial condition to start me off...

2. The McLaurin series for y is...

$y= \sum_{n=0}^{\infty} a_{n}\cdot x^{n}$ (1)

... so that...

$y^{'}= \sum_{n=0}^{\infty} n\cdot a_{n-1}\cdot x^{n-1}$ (2)

The 'initial condition' $y(0)=1$ extablishes that $a_{0}=1$. If now write the DE in terms of (1) and (2) we obtain...

$\sum_{n=0}^{\infty} \{(n+1)\cdot a_{n+1} -a_{n}\}\cdot x^{n}= 1-x$ (3)

.. so that is...

$a_{1}=2$

$a_{2}=\frac{1}{2}$

$a_{n+1}= \frac{a_{n}}{n+1}, n>2$ (4)

Kind regards

3. see attachment


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$$\require{cancel}$$

# 5.2: Supported and Tethered Membranes

Model lipid bilayers or synthetic lipid bilayers are membranes that are created in vitro. In vivo studies of lipid bilayers and the proteins embedded within them can be difficult due to the complexity of the cellular dynamics of membranes. Model bilayers use biological membranes, along with artificial constraints, to investigate the structure and function of lipid bilayers. Supported and tethered membranes are two types of model bilayers frequently used in model studies. Importantly, these hybrid systems are not employed as vesicles, but, instead, maintain a non-curved structure placed against or physically attached to a non-fluid surface. This method of studying membrane biology is valuable in that it allows for the utilization of many techniques are not as applicable to “freestanding” membranes, since supported and tethered membranes are more stable [1]. Furthermore, the complexity of natural membranes makes them difficult to study, especially in vivo. Model bilayers provide an excellent model for studying the natural functions of lipids and membrane proteins in a simple manner. Moreover, many techniques require a physical interaction with the membrane, which would be difficult to perform on freestanding membranes that are capable of floating through fluid.

# Supported Membranes

Supported lipid bilayers (SLBs) are biological or synthetic membranes with one side of the membrane exposed to fluid environment, while the other side is lined against a flat solid surface. There is typically a very thin fluid layer separating the membrane from the surface (Figure $$\PageIndex{1}$$). This type of configuration allows SLBs to maintain their fluid integrity and to diffuse laterally in 2D space. However, this small distance between the membrane and its surface (1-2nm) could potentially lead to interactions/friction between the lipids and the surface [2]. Another type of model bilayer has been designed to relieve this issue.

Regardless, SLBs usually remain stable for long periods of time as well [3]. In addition to the new techniques developed from using SLBs (see Scientific Applications), established methods have been used with SLBs to further understand lipid bilayer dynamics and function [4]. In one of the first examples, researchers employed fluorescence microscopy to examine the diffusion of two different fluorescent labeled lipids (Figure $$\PageIndex{2}$$).

## Preparation of Supported Membranes

SLBs can be built in a number of different ways. One of the first methods for synthesizing SLBs used alkylated glass sides and applied only a monolayer of lipids to an air-water interface (Figure $$\PageIndex{3}$$) [5]. This first type of configuration was mostly used to study the lateral movement of monolayers via fluorescence microscopy. Since this approach was employed, many different surface types have been used, including glass, mercury, and silicon-oxide [5]. Silicon oxide is the most popular surface type [6]. Self-assembly of lipid bilayers is central for maintaining biological integrity. Self-assembly is also important for maintaining the biology of any embedded proteins. A more recent technique for forming supported membranes uses a previously constructed lipid monolayer that is supported on a surface as a template to fuse an additional monolayer from a vesicle [7]. In this technique, the monolayer from the vesicle fuses only with the already supported monolayer to form a new lipid bilayer. This prevents any preservation issues that might arise from the monolayer and proteins being exposed directly to a planar surface. For instance, exposing a membrane protein directly to the support surface can prevent any lateral movement of the protein. This issue would prevent charactering the native movements and functions of the membrane protein. An additional technique for preparing supported membranes is called vesicle deposition [8]. In this method, either unilamellar vesicles (SUVs) or giant unilamellar vesicles (GUVs) are fused with the solid support surface. This is achieved by increasing the temperature at which the reaction is taking to place to above the melting temperature of lipids. At the temperature, the vesicles with naturally fuse with the surface.

The number of surface-constricted membrane systems has greatly expanded. Solid-supported lipid bilayers, polymer-cushioned lipid bilayers, hybrid bilayers, tethered lipid bilayers, suspended lipid bilayers, and supported vesicular layers have all become well-established models [9].

## Tethered Membranes

To further increase the stability of hybrid membrane systems, some lipid bilayer are physically attached to a planar surface via a protein or a polymer. Tethered lipid bilayers membranes (tLBMs), also referred to as polymer-supported bilayers, employ a “tether” that is not part of the studied membrane to connect it to a surface (Figure $$\PageIndex{4}$$). tLBMs also require the use of membrane proteins for solid-surface support. tLBMs provide some advantages compared to SLBs. Because tLBMs are attached to their surface via a tether, the space between lipids and the surface is much greater than that of SLBs [2]. This feature is more likely to prevent friction between the membrane and the solid. Addtionally, SLBs can more easily lose contact with the contact surface or can result in only specific parts of the membrane interacting with the surface, as opposed to the entire membrane [3]. These issues can lead to a system the is no longer biologically relevant. The nature of this type of model lipid bilayer allows the membrane to be exposed to fluids on both sides, more like a naturally occurring membrane.

## Scientific Applications

One of the most useful applications for tethered and supported membranes is the study of peripheral types membrane proteins, . The stable nature of these bilayers causes the proteins embedded within the membrane to remain immobile and well oriented, making it easier to observe protein function. Specifically, studies of the immune and endocrine systems were able to examine lipid embedded T-cell receptors and hormone receptors (Figure $$\PageIndex{5}$$). One study used SLBs containing T-cell receptors to show that the actual T-cell, when bound to the receptor through the antigen and class II major histocompatibility complex, stabilizes that entire interaction of the complex [10]. Other studies have also employed tBLMs and SLBs in conjunction with antibodies to research cancer. Using a similar method to the T-cell receptor study, researchers used SLBs chemically linked to antibodies to capture and purify circulating tumor cells (CTCs) [11]. In another application, SLBs and tLBMs are subjected to atomic force microscopy (AFM) and fluorescent imaging as a way to investigate the effects of environmental stressors [12]. In this particular study, the authors used SLBs to show that yeast cells increase the amount of unsaturated lipid content and ergosterol as a means of ethanol tolerance. The authors employed AFM against SLBs to show the change in morphology.

Tethered and supported membranes have been used to investigate many other biological phenomenon including, vesicle/receptor force relationship and effect of undulation forces between a vesicle and a rigid wall [13].

## References

1. Naumann, R., Jonczyk, A., Kopp, R., van Esch, J., Ringsdorf, H., Knoll, W., & Gräber, P. (1995). Incorporation of Membrane Proteins in Solid‐Supported Lipid Layers. Angewandte Chemie International Edition in English, 34(18), 2056-2058.
2. Rebaud, S., Maniti, O., & Girard-Egrot, A. P. (2014). Tethered bilayer lipid membranes (tBLMs): Interest and applications for biological membrane investigations. Biochimie, 107, 135-142.
3. Sinner, E. K., & Knoll, W. (2001). Functional tethered membranes. Current opinion in chemical biology, 5(6), 705-711.
4. Groves, J. T., Ulman, N., Cremer, P. S., & Boxer, S. G. (1998). Substrate-membrane interactions: mechanisms for imposing patterns on a fluid bilayer membrane. Langmuir, 14(12), 3347-3350.
5. von Tscharner, V., & McConnell, H. M. (1981). Physical properties of lipid monolayers on alkylated planar glass surfaces. Biophysical journal, 36(2), 421.
6. http://www.cmns.leeds.ac.uk/SSbiomemb.htm
7. Kalb, E., Frey, S., & Tamm, L. K. (1992). Formation of supported planar bilayers by fusion of vesicles to supported phospholipid monolayers. Biochimica et Biophysica Acta (BBA)-Biomembranes, 1103(2), 307-316.
8. Morandat, S., Azouzi, S., Beauvais, E., Mastouri, A., & El Kirat, K. (2013). Atomic force microscopy of model lipid membranes. Analytical and bioanalytical chemistry, 405(5), 1445-1461.
9. Richter, R. P., Bérat, R., & Brisson, A. R. (2006). Formation of solid-supported lipid bilayers: an integrated view. Langmuir, 22(8), 3497-3505.
10. Watts, T. H., Gaub, H. E., & McConnell, H. M. (1986). T-cell-mediated association of peptide antigen and major histocompatibility complex protein detected by energy transfer in an evanescent wave-field.
11. Wu, J. C., Tseng, P. Y., Tsai, W. S., Liao, M. Y., Lu, S. H., Frank, C. W., ... & Chang, Y. C. (2013). Antibody conjugated supported lipid bilayer for capturing and purification of viable tumor cells in blood for subsequent cell culture.Biomaterials, 34(21), 5191-5199.
12. Vanegas, J. M., Contreras, M. F., Faller, R., & Longo, M. L. (2012). Role of unsaturated lipid and ergosterol in ethanol tolerance of model yeast biomembranes. Biophysical journal, 102(3), 507-516.
13. Sackmann, E. (1996). Supported membranes: scientific and practical applications. Science, 271(5245), 43-48.


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How to line-break a framed text box?

Is it possible to line-break a framed text box in LaTeX? If so, is it necessary to use some package to do so, and what is the name of that package?

EDIT: For greater clarity, let me give an example. Suppose I am typing a line of text and I put a long word in a framed box at the end of this line. Because by default LaTeX will not line-break such a framed box it will stick out of the right margin of text. My question concerns the possibility of line-breaking this framed text box over to the left side of the next line. In this way, both the text and the frame itself would be line-breaked from the right side of the given line to the left side of the next line.

• No, you do not need a package, e.g. \documentclass{article} \begin{document} \fbox{\begin{minipage}{3cm} This is a rather long sentence with no meaning. \end{minipage}} \end{document} works, but you may enjoy the nice options tcolorbox offers. – Schrödinger's cat Nov 4 '19 at 3:58
• Perhaps I'm not being clear: I want the frame itself to break over a line. – The Riddler Nov 4 '19 at 4:00
• Sorry, can you try to rephrase? Or are you looking for tex.stackexchange.com/q/411339 ? – Schrödinger's cat Nov 4 '19 at 4:01
• I think that's what I'm looking for, although it looks awfully complicated :( – The Riddler Nov 4 '19 at 4:08
• Can you post MWE? and also provide more clear picture on your requirement.. – MadyYuvi Nov 4 '19 at 5:11

I know only a trick, which works only manually:

\documentclass[12pt,a5paper]{article}
\usepackage{fbox}
\begin{document}

We look inside the alternate reality of the president’s
Twitter account, where he absorbs and amplifies a
noxious stream of disinformation.

We look inside the alternate reality of the \fbox[lbt]{presi-} \fbox[rbt]{dent’s}
Twitter account, where he absorbs and amplifies a
noxious stream of disinformation.

\end{document}

• Where can I find documentation on the optional [lbt] and [rbt] arguments to the \fbox command? – The Riddler Nov 4 '19 at 13:15
• I suppose in the documentation of the package ... ;-) – user187802 Nov 4 '19 at 13:18
• Ah, didn't realize fbox was also a package since one can use the \fbox command without loading any packages. Thanks. – The Riddler Nov 4 '19 at 13:19

Easy way to create a framed box is:

\usepackage{framed}
\begin{framed}
blah blah...
\end{framed}

This allows line break, page break etc.

Framed boxes with many custom styles then tcolorbox and tikz packages are preferable.

mdframed package also one of nice package which supports framed box with breaking option...


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LYRICS.AZ APPLICATION

# The Beta Band - Unknown lyrics

[Verse 1]
Something
Why did you point at me?
Something
Why did you laugh at me?
Something
What did you say to me?
Something
What did it mean to you?

[Chorus]
I suppose I'm the same but in a different way
I suppose I'm the same but in a different way

I suppose I'm the same but in a different way
[Lyrics from: https:/lyrics.az/the-beta-band/hot-shots-ii/unknown.html]
I suppose I'm the same but in a different way

[Verse 2]
Something


text:
Industrial Utility Efficiency

# Metals

Since completion of the system upgrade in the fall of 2020, PC Forge is on track to save an average of 1.9 million kWh and $266,000.00 per year in energy costs – and increase the production capability of its forging operation by 40%. The project also achieved a one-year payback with a \$245,000 incentive from Government of Ontario’s utility Independent Electricity System Operator (IESO).


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This paper describes the characteristics of the heat transfer and flow of a swirling laminar impinging jet in a comparatively narrow space with a confined wall. Air is impinged on a flat surface with constant wall temperature. The heat transfer and flow field were analyzed numerically by solving three-dimensional governing equations. Heat transfer experiment and flow visualization were also performed. Numerical heat transfer was compared with experimental results. Temperature distribution and velocity vectors in the space were obtained for various swirl numbers at Reynolds number $Re=2000$. The numerical and experimental results show that the swirling jet enhances or depresses the local heat transfer, and the average Nusselt number ratio with and without swirl takes a peak at a certain swirl number.

1.
Martin
,
H.
, 1977, “
Heat and Mass Transfer Between Impinging Gas Jets and Solid Surface
,”
, Vol.
13
,
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Hartnett
and
T. F.
Irvine
, eds.,
,
New York
.
2.
Viskanta
,
R.
, 1993, “
Heat Transfer to Impinging Isothermal Gas and Flame Jets
,”
Exp. Therm. Fluid Sci.
0894-1777,
6
, pp.
111
134
.
3.
Ichimiya
,
K.
, 2003, “
Heat Transfer Characteristics of an Annular Turbulent Impinging Jet With a Confined Wall Measured by Thermosensitive Liquid Crystal
,”
Heat Mass Transfer
0947-7411,
39
, pp.
545
551
.
4.
Ichimiya
,
K.
, 1995, “
Heat Transfer and Flow Characteristics of an Oblique Turbulent Impinging Jet Within Confined Walls
,”
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, pp.
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.
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, and
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,
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,
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, 1978, “
Heat Transfer Under a Laminar Swirling Impinging Jet—A numerical Study
,”
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, pp.
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.
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Ward
,
J.
, and
Mahmood
,
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, 1982, “
Heat Transfer From a Turbulent, Swirling, Impinging Jet
,”
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, pp.
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.
8.
Wicker
,
R. B.
, and
Eaton
,
J. K.
, 1999, “
Effect of Injected Longitudinal Vorticity on Particle Dispersion in a Swirling, Coaxial Jet
,”
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0022-1481,
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, pp.
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772
.
9.
Wen
,
M. Y.
, and
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,
K. J.
, 2003, “
An Impingement Cooling on a Flat Surface by Using Circular Jet With Longitudinal Swirling Strips
,”
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, pp.
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4667
.
10.
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,
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, and
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,
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, 2006, “
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.
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,
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, 1988,
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,
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,
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.


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# Asymptotic Expansion for Exponential Integral Function

## Theorem

$\displaystyle \map \Ei x \sim \frac {e^{-x} } x \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {n!} {x^n}$

where:

$\Ei$ is the exponential integral function
$\sim$ denotes asymptotic equivalence as $x \to \infty$.


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Discussion and comments : strong versus weak significance tests and the role of meta-analytic procedures
Journal de la société française de statistique, Volume 145 (2004) no. 4, p. 47-53
@article{JSFS_2004__145_4_47_0,
author = {Kwan, Ernest and Friendly, Michael},
title = {Discussion and comments : strong versus weak significance tests and the role of meta-analytic procedures},
journal = {Journal de la soci\'et\'e fran\c caise de statistique},
publisher = {Soci\'et\'e fran\c caise de statistique},
volume = {145},
number = {4},
year = {2004},
pages = {47-53},
language = {en},
url = {http://www.numdam.org/item/JSFS_2004__145_4_47_0}
}

Kwan, Ernest; Friendly, Michael. Discussion and comments : strong versus weak significance tests and the role of meta-analytic procedures. Journal de la société française de statistique, Volume 145 (2004) no. 4, pp. 47-53. http://www.numdam.org/item/JSFS_2004__145_4_47_0/

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text:
## Introduction

When new drugs or vaccines are brought to market, it is important to continue to monitor their safety. The phase 2 and 3 trials preceding marketing may uncover most common adverse events, but rare adverse events may go undetected due to limited sample size of the trials. Other adverse events may go undetected if they only affect a subpopulation that was excluded from the clinical trials. To catch these types of adverse events requires post-marketing drug and vaccine safety surveillance, for example using routinely collected healthcare data, such as administrative claims and electronic health records. To detect adverse events as quickly as possible, data should be analyzed as it accrues, for example once every month analyzing all data up to that point in time. It is important to realize that such sequential testing for a safety signal is a form of multiple testing that needs to be adjusted for if one aims to maintain the prespecified type 1 error (i.e. if one aims to keep the probability of rejecting the null when the null is true at the specified alpha level). A common approach to adjusting for sequential testing is maximum sequential probability ratio testing (MaxSPRT), used by the U.S Food and Drug Administration (FDA) in its safety surveillance.

However, there is another important reason why the true type 1 error might deviate from the prespecified alpha, and that is systematic error due to the observational nature of the study. Systematic error can manifest from multiple sources, including confounding, selection bias, and measurement error. For example, in a study comparing vaccinated to unvaccinated, the vaccinated group may differ from the unvaccinated one in terms of age and fragility that impact the baseline probability of having the outcome, and failing to adjust for these differences will lead to biased effect size estimates.

This vignette demonstrates combining adjusting for sequential testing through MaxSPRT with adjusting for systematic error using empirical calibration. We use a simulated dataset of survival data for several negative controls and a single positive control.

## Simulated data

For this vignette we will use simulated data that can be generated using the simulateMaxSprtData function

maxSprtSimulationData <- simulateMaxSprtData(
n = 10000,
pExposure = 0.5,
backgroundHazard = 0.001,
tar = 10,
nullMu = 0.2,
nullSigma = 0.2,
maxT = 100,
looks = 10,
numberOfNegativeControls = 50,
numberOfPositiveControls = 1,
positiveControlEffectSize = 4
)
head(maxSprtSimulationData)
##         time outcome exposure lookTime outcomeId
## 6  5.4443501       0     TRUE       10         1
## 18 5.7940466       0     TRUE       10         1
## 35 7.5386315       0     TRUE       10         1
## 51 5.4168833       0     TRUE       10         1
## 62 0.5159339       0    FALSE       10         1
## 74 9.9375227       0    FALSE       10         1

This is simulated survival data for 50 negative controls (outcome IDs 1-50), and 1 positive control (outcome ID 51). The data provides 10 looks at data accruing over time (at times t = 10, 20, 30, …, 100), with each look having more data than the one before. We simulated moderate systematic error (mean = 0.2, SD = 0,2), and the positive control has a true hazard ratio of 4.

## MaxSPRT

MaxSPRT (Maximum Sequential Probability Ratio Testing) is a well-known procedure for adjusting for sequential testing. It requires computing the log likelihood ratio (LLR), and comparing it against a prespecified critical value to determine whether there is a signal (suggesting an effect).

### Log likelihood ratio

In this case, because we have survival data we will use a Cox proportional hazards model. For any outcome, at any look, we can fit a Cox model using the Cyclops package. In this example, we use outcome 51, and the look at t = 50:

library(Cyclops)
library(survival)

dataOutcome51 <- maxSprtSimulationData[maxSprtSimulationData$outcomeId == 51, ] dataOutcome51LookT50 <- dataOutcome51[dataOutcome51$lookTime == 50, ]

cyclopsData <- createCyclopsData(
Surv(time, outcome) ~ exposure ,
modelType = "cox",
data = dataOutcome51LookT50
)
fit <- fitCyclopsModel(cyclopsData)
coef(fit)
## exposureTRUE
##     1.829709

For MaxSPRT, we’re not interested in the point estimate of the hazard ratio, but rather the LLR. This is the log of the ratio between the maximum likelihood, and the likelihood of the null hypothesis. The maximum likelihood is already available from the fit object:

# Maximum log likelihood:
fit$log_likelihood ## [1] -1167.396 We can use Cyclops to compute the log likelihood of parameter value 0 (hazard ratio = 1): llNull <- getCyclopsProfileLogLikelihood( object = fit, parm = "exposureTRUE", x = 0 )$value
llNull
## [1] -1210.618

And we can compute the LLR by subtracting the two. Note that we use a single-sided test, so we will set LLR to 0 if the point estimate (the maximum likelihood estimate) is below 0. It might also be the case that no maximum likelihood estimate exists, for example because there are zero counts, in which case we also set the LLR to 0:

if (fit$return_flag == "ILLCONDITIONED" || coef(fit) < 0) { llr <- 0 } else { llr <- fit$log_likelihood - llNull
}
llr
## [1] 43.22212

### Critical value

The critical value for the LLR is computed based on the total number of looks that are expected to be made, and the expected sample size at each look. Because we have already simulated all looks, we will use the actual sample sizes instead of expected ones. For Cox models it is recommended to use the binomial critical value computation:

outcomesPerLook <- aggregate(outcome ~ lookTime, dataOutcome51, sum)
# Need incremental outcomes per look:
outcomesPerLook <- outcomesPerLook$outcome[order(outcomesPerLook$lookTime)]
outcomesPerLook[2:10] <- outcomesPerLook[2:10] - outcomesPerLook[1:9]

exposedTime <- sum(dataOutcome51$time[dataOutcome51$exposure == TRUE &
dataOutcome51$lookTime == 100]) unexposedTime <- sum(dataOutcome51$time[dataOutcome51$exposure == FALSE & dataOutcome51$lookTime == 100])
cv <- computeCvBinomial(
groupSizes = outcomesPerLook,
z = unexposedTime / exposedTime,
minimumEvents = 1,
alpha = 0.05
)
## Selected alpha: 0.046 (least conservative value below 0.05)
cv
## [1] 2.689342
## attr(,"alpha")
## [1] 0.04597

By comparing the LLR to the critical value we can determine whether there is a signal:

llr > cv
## [1] TRUE

## MaxSPRT and empirical calibration

### Likelihood profiles

For empirical calibration, we’ll need more than just the likelihood at the maximum and at the null hypothesis. Instead, we’ll need information on the entire likelihood distribution. A convenient way to approximate the likelihood function is the likelihood profile. This profile provides, at various values the hazard ratio, the computed likelihood. The Cyclops package has a convenient function for computing likelihood profiles:

llProfileOutcome51LookT50 <- getCyclopsProfileLogLikelihood(
object = fit,
parm = "exposureTRUE",
bounds = log(c(0.1, 10))
)
head(llProfileOutcome51LookT50)
## # A tibble: 6 x 2
##   point  value
##   <dbl>  <dbl>
## 1 -2.30 -1411.
## 2 -2.29 -1410.
## 3 -2.29 -1410.
## 4 -2.28 -1409.
## 5 -2.27 -1408.
## 6 -2.26 -1407.

Here we’ve specified that the true hazard ratio is almost certainly in the [0.1, 10] range, so the profile will be restricted to this region. This function uses an adaptive grid, using more grid points if the likelihood shows higher curvature to avoid loss of information. If we want, we could also plot the profile:

library(ggplot2)
ggplot(llProfileOutcome51LookT50, aes(x = point, y = value)) +
geom_line()

Note that the x-axis is the log of the hazard ratio, and the y-axis represents the log of the likelihood.

### Fitting a null distribution using likelihood profiles

We can fit our null distribution as usual, using the point estimates and standard errors of the negative controls as described in the other vignettes, but when counts are low this may lead to biased estimates of the null distribution parameters. Especially for the earlier looks, we can expect counts to be low, and therefore it is better to take the full likelihood distribution of each negative control into account. Again the likelihood profiles are a convenient way of capturing the full likelihood function. Here we compute the likelihood profile for all negative controls (having outcome IDs 1-50) at time t = 50:

negativeControlProfilesLookT50 <- list()
dataLookT50 <- maxSprtSimulationData[maxSprtSimulationData$lookTime == 50, ] for (i in 1:50) { dataOutcomeIlookT50 <- dataLookT50[dataLookT50$outcomeId == i, ]
cyclopsData <- createCyclopsData(
Surv(time, outcome) ~ exposure ,
modelType = "cox",
data = dataOutcomeIlookT50
)
fit <- fitCyclopsModel(cyclopsData)
llProfile <- getCyclopsProfileLogLikelihood(
object = fit,
parm = "exposureTRUE",
bounds = log(c(0.1, 10))
)
negativeControlProfilesLookT50[[i]] <- llProfile
}

We can use these profiles to fit the null distribution:

nullLookT50 <- fitNullNonNormalLl(negativeControlProfilesLookT50)
## Detected data following grid distribution
nullLookT50
## Estimated null distribution
##
##      Estimate
## Mean   0.3145
## SD     0.1860

This shows substantial systematic error in the negative controls, with mean and standard deviation away from 0.

### Computing a calibrated critical value

Using the fitted null distribution, we can compute an empirically calibrated CV:

calibratedCv <- computeCvBinomial(
groupSizes = outcomesPerLook,
z = unexposedTime / exposedTime,
minimumEvents = 1,
alpha = 0.05,
nullMean = nullLookT50[1],
nullSd = nullLookT50[2]
)
## Selected alpha: 0.050 (least conservative value below 0.05)
calibratedCv
## [1] 16.71318
## attr(,"alpha")
## [1] 0.049571

Which is substantially higher than the uncalibrated CV we computed earlier (CV = 2.6893419). We can compare the LLR computed earlier to the calibrated critical value to determine whether there is a signal:

llr > calibratedCv
## [1] TRUE

## Demonstrating type 1 error with and without calibration

It is always a good idea to compute the type 1 error based on the negative controls, to understand the operating characteristics of your study. Unfortunately, when doing sequential testing we cannot evaluate the overall performance of our surveillance until our final look at the data. Here we compute type 1 error both with and without calibration, comparing it to the nominal type 1 error, to demonstrate why empirical calibration can be important.

### Computing log likelihood ratios and (uncalibrated and calibrated) critical values

We re-use the example code provided earlier to compute the log likelihood ratios, uncalibrated critical values and calibrated critical values for each negative control at each look. Note that this code will take some time to complete.

allCvsAndLlrs <- data.frame()
for (t in unique(maxSprtSimulationData$lookTime)) { # Compute likelihood profiles and LLR for all negative controls: negativeControlProfilesLookTt <- list() llrsLookTt <- c() dataLookTt <- maxSprtSimulationData[maxSprtSimulationData$lookTime == t, ]
for (i in 1:50) {
dataOutcomeIlookTt <- dataLookTt[dataLookTt$outcomeId == i, ] cyclopsData <- createCyclopsData(Surv(time, outcome) ~ exposure, modelType = "cox", data = dataOutcomeIlookTt ) fit <- fitCyclopsModel(cyclopsData) # likelihood profile: llProfile <- getCyclopsProfileLogLikelihood( object = fit, parm = "exposureTRUE", bounds = log(c(0.1, 10)) ) negativeControlProfilesLookTt[[i]] <- llProfile # LLR: llNull <- getCyclopsProfileLogLikelihood( object = fit, parm = "exposureTRUE", x = 0 )$value
if (fit$return_flag == "ILLCONDITIONED" || coef(fit) < 0) { llr <- 0 } else { llr <- fit$log_likelihood - llNull
}
llrsLookTt[i] <- llr
}

# Fit null distribution:
nullLookTt <- fitNullNonNormalLl(negativeControlProfilesLookTt)

# Compute calibrated and uncalibrated CV for all negative controls:
cvs <- c()
calibratedCvsLookT <- c()
for (i in 1:50) {
dataOutcomeI <- maxSprtSimulationData[maxSprtSimulationData$outcomeId == i, ] outcomesPerLook <- aggregate(outcome ~ lookTime, dataOutcomeI, sum) # Need incremental outcomes per look: outcomesPerLook <- outcomesPerLook$outcome[order(outcomesPerLook$lookTime)] outcomesPerLook[2:10] <- outcomesPerLook[2:10] - outcomesPerLook[1:9] exposedTime <- sum(dataOutcomeI$time[dataOutcomeI$exposure == TRUE & dataOutcomeI$lookTime == 100])
unexposedTime <- sum(dataOutcomeI$time[dataOutcomeI$exposure == FALSE &
dataOutcomeI$lookTime == 100]) # Note: uncalibrated CV will be same for every t, but computing in loop # over t for clarity of code: cv <- computeCvBinomial( groupSizes = outcomesPerLook, z = unexposedTime / exposedTime, minimumEvents = 1, alpha = 0.05 ) cvs[i] <- cv calibratedCv <- computeCvBinomial( groupSizes = outcomesPerLook, z = unexposedTime / exposedTime, minimumEvents = 1, alpha = 0.05, nullMean = nullLookTt[1], nullSd = nullLookTt[2] ) calibratedCvsLookT[i] <- calibratedCv } # Store in a data frame: allCvsAndLlrs <- rbind( allCvsAndLlrs, data.frame( t = t, outcomeId = 1:50, llr = llrsLookTt, cv = cvs, calibrateCv = calibratedCvsLookT ) ) } ### Computing overall type 1 error We compare the computed LLRs to the computed critical values to determine how often a signal is generated for the negative controls (where the null hypothesis is true): signals <- c() calibratedSignals <- c() for (i in 1:50) { idx <- allCvsAndLlrs$outcomeId == i
signals[i] <- any(allCvsAndLlrs$llr[idx] > allCvsAndLlrs$cv[idx])
calibratedSignals[i] <-  any(allCvsAndLlrs$llr[idx] > allCvsAndLlrs$calibrateCv[idx])
}
# Type 1 error when not calibrated (should be 0.05):
mean(signals)
## [1] 0.34
# Type 2 error when calibrated (should be 0.05):
mean(calibratedSignals)
## [1] 0.04`

This shows that type 1 error is much closer to nominal when using empirical calibration.


text:
# Tag Info

33

Blast can definitely kill you, although it is only lethal at much shorter ranges compared to shrapnel. A building can be destroyed by 5psi overpressure while a Human can withstand up to 45psi and live. Some data here: A 5 psi blast overpressure will rupture eardrums in about 1% of subjects, and a 45 psi overpressure will cause eardrum rupture in ...

5

This is a fun question and has about 4 or 5 different factors at play: Types of forces involved: Pressure vs. Inertia Types of resistance involved: Rigidity vs. Plasticity Objects involved: Soldiers vs. Buildings Scenario: Damage from a bomb blast (energy wave) vs. shrapnel impact/penetration (kinetic collision). How damage is applied with: Blast - the ...

4

If you add mass keeping the volume constant, the pressure would change. When a source says that pressure is independent of mass, it means that pressure is independent of mass if you keep the density constant, not the volume. The idea is that if you take two identical copies of a system, then combine them to make a system twice as large, extensive ...

4

You don't want to use compression ignition in gasoline engines. If you attempt to compress air and gasoline vapors together to the point where they ignite, the ignition point will not be "controllable". Ideally, the ignition should occur 10-15 degrees before "top dead center", meaning that the ignition starts when the crank shaft is "under rotated" 10-15 ...

3

When the thermos is closed, the air at the top of the thermos will be warm relative to the ice-cold water below it. As time proceeds, the two materials that are in contact with each other at the surface of the water will exchange heat due to their different temperatures. This will cool the air. When gasses are cooled they move more slowly and they don't ...

3

the air [pressure] is lowered by heating the air inside the bottle. I think you need to watch the video again. The air pressure is raised by heating the air, not lowered. This increase in pressure forces air out of the bottle, which is visible by the egg vibrating on the rim. The egg acts like a one-way valve. When the pressure inside is greater, it ...

3

The simple answer, which is what I think you're hoping for is the following: At constant volume, the system (by definition) is not able to do work on the surroundings because work involves a change in volume. All the heat you put in is spent raising the temperature (internal energy). At constant pressure, some of the energy you put in goes into raising the ...

3

The blast overpressure of the explosion is a very strong shock wave which can kill humans. There are a number of ways an explosion without shrapnel can do harm to people: Rupturing of the hollow organs due to rapid compression and expansion by the shock wave. The body can get thrown through the air if a strong detonation occurs nearby. Impact of the body ...

2

Only the symmetric stress tensor is physical, since thermodynamics demands a symmetric stress tensor. Since the symmetric stress tensor is unique, your option 2 is the correct one. (Canonical versions may be simpler but need not be physical; cf. the canonical momentum, which is often different from the physical momenetum.) This is completely unrelated to ...

2

Well, the temperature profile of an object on the macroscopic scale is governed by, $${{dT} \over {dt}}=\kappa \cdot \nabla^2 T$$ Where T is the temperature, t is time, and $\kappa$ is the thermal diffusivity. It's the constant that matters here. The larger the constant, the faster the temperature changes. The equation itself is called the Heat Equation. ...

2

First law of thermodynamics $$\delta Q=\delta E+p\delta V$$ $$\therefore\frac {\delta Q}{\delta T} = \frac{\delta E}{\delta T}+nR$$ As $p\delta V$ can be written as $nR\delta T$ When volume is constant $$\delta Q= \delta E$$ $$\therefore\frac {\delta Q}{\delta T} = \frac{\delta E}{\delta T}$$ $\frac {\delta Q}{n\delta T}$ is Molar heat capacity. ...

2

are you filling the bottles with CO2? or regular air. If your bottles are air tight, you could add a small bit of dry ice to each bottle then seal it tight and then refrigerate it, as CO2 absorbs better in colder water. If you plan on drinking this, I'd do more research cause I'm not sure dry ice is safe for consumption - so check that out if this is a ...

2

I think it is partially a perceptional problem. The pressure, i.e. force per area is depdendend on the depth of water in this case, and it doesn't matter much, if you are naked or if you wear waders. But if you wear waders, the pressure is implied on the surface of your clothes, not on your skin. Your clothes will now dent to give in to the pressure, until ...

2

Take a 1-cm square tube and place it vertically in the container from top to bottom, touching the bottom so that the bottom of the container is the bottom of the tube. The pressure at the bottom of the tube is nothing but the weight of water it is supporting - the water in the tube. Supporting means to keep from falling. (Forget the air pressure - that's ...

2

For most substances, increasing the pressure when a system is in equilibrium between liquid and solid phases will increase the phase transition temperature. Water is one of a few special substances for which the pressure lowers the temperature of transition. The basic reason is that water actually expands when it goes from the liquid to solid phase. In ...

1

I like BowlofRed's answer, but I think there's more going on here than that. I also don't believe that straight forward cooling would suck in the egg that fast. The egg gets sucked in very quickly. Certainly as the burning starts, air is escaping. It's impossible to see precisely when the egg stops vibrating, as some vibration might be quite small and ...

1

Acid Jazz is quite right. Generally any instability in air medium can result in sound waves radiation. Vortices causes local pressure instabilities and the medium is "able to propagate the image of it" (well, the Wave equation). Kármán vortex street is nice (and most of all solvable!) example of vortex shedding. But in certain domain of the Reynolds number ...

1

First you half fill the pipette with hot water. So at that point it's filled half with hot water in the lower part, half filled with air at room temperature in the upper part. Then you block the top end of the pipette with the tip of your finger and turn the pipette upside down, so the hot water starts flowing down to the blocked side. However, the presence ...

1

Short answer: The major difficulty lies with the definitions themselves, and none of the possibilities given has a real physical meaning which can be univoquely related to stress in non extensive systems in its conventional mechanical original meaning. The long one: What kind of systems does this apply to? This is not answered by referring to systems with ...

1

A force does not require a constant input of energy to exist. Energy is only required to perform work, which is exerting a force over a distance. $$W = \mathbf F \centerdot \Delta \mathbf x$$ That distance is key. In your example, if the size of the container does not change, no energy is expended no matter how long the force lasts. If the force is used ...

1

As a simplification, you can consider that you have a 2D viscous flow between two boundaries that approach each other. Assuming that the flow is symmetrical about the line (with the line along the Y direction), you can simplify this further to "no flow at x=0". What you are left with is a pressure distribution $p(x,t)$ whose integral in $x$ should equal the ...

1

Ok, here's one possible explanation. First, if you're graphing pressure vs payload, you should switch your axes--I was confused! Anyway, the payload obviously is going to increase the pressure in your tires. However, your tires are also elastic. So, they can "stretch" a little, to increase their volume in response to a change in pressure. However, most ...

1

Force due to pressure over a given area is $F=AP$. A bubble will have three forces to consider. 1. Air pressure from air inside the bubble pushing the wall of the bubble outward (trying to make the bubble bigger) 2. Air pressure from air outside the bubble pushing the wall of the bubble inward (trying to make the bubble smaller), and 3. a surface tension of ...

1

I think this picture covers the gist of it. You need rubber or plastic, a mailable material that flattens when pressed against the object, but that wants to return to it's original shape when released. Think of, as part of measuring the weight of an object, you measure the pressure all around it. If you have more air pressure on the bottom than the ...

1

If you remove the bottom you no longer have a container. The pressure is atmospheric and you just have gravity at work.

1

The other answers already mention pressure and heat. A bomb sets nearby bodies in motion with a speed depending on the strength of the explosion, the distance to the body, and how much surface area of the body was facing the bomb. While - as explained in the other answers - being set in motion is rarely lethal, being smashed against a wall can easily lead ...

1

By the way, it's customary to use primes or some other indicator inside the integral to be slightly less confusing. I.e. $$W(t) = \int_{t_0}^{t} P(t')Q(t')dt'$$ Be careful about how you think about Q(t). You've described it as a "flow rate", but really, you are saying $Q(t) = \frac{dV}{dt}$, or the rate of change of the volume with time. Now, the term ...

1

Here is an interesting article that shows how water was frozen at high temperature under pressure http://www.azom.com/news.aspx?newsID=8016 Here is an extract Sandia Convert Water to Ice in Nanoseconds Published on March 19, 2007 at 1:15 AM Sandia’s huge Z machine, which generates temperatures hotter than the sun, has turned water to ice in nanoseconds. ...

1

The variable $h$ in this equation always refers to a vertical distance from the point where you want to know the pressure up to the fluid surface in contact with the air. In the tilted cup you would still measure straight up to the fluid surface. The bit of tea in the middle of the cup doesn't "know" anything about the walls of the cup. Their orientation ...

1

When high pressure is applying on a gas, it gets converted into liquid form. Similarly, if more pressure is applied to the liquid, force of attraction increases so that the liquid is converts into solid state. As the pressure increases the rate of crystallization also increases. i.e., the freezing point also increases.

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


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# Answer this question. Im getting confused very much

Eshan
3 years ago
Dear student,

The electric field inside a conductor is always zero. When there is an external electric field, electric field is induced in the direction opposite to that of external electric field and of same magnitude so that net electric field inside it is zero.

The electric field at the center of sphere due to the charge is

Hence the induced electric field at that point is

However potential at that point due to the charge is simply given by


text:
# Tag Info

1

In the standard model the mass for the quarks and other elementary particles comes from from the Higgs mechanism. In the case of the quarks, the masses given in the table of the link are calculated using convoluted theoretical models and data input from scattering experiments.4 At the moment chiral symmetry breaking does not contribute to the quark masses ...

2

I think its easiest to understand this if one has a minimal understanding of QFT. I'm not sure about your background knowledge but hopefully this isn't gibberish to you. The QCD Lagrangian for massless quarks is given by, {\cal L} = - g \sum_i \bar{\psi} _i A _\mu \gamma ^\mu \psi _i - \frac{1}{4} F _{ \mu \nu } F ^{ \mu \nu } ...

3

Confinement, i.e. the fact that bound states in QCD are colour singlets (or in other words: there are no free quarks or gluons) is experimentally well-established. There is no doubt that it is not just a prediction of lattice simulations, but a feature of strongly interacting systems at low energies, as no free quarks (or gluons) have yet been found within ...

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text:
# Homework Help: Concavity Of The Determinant

1. Jul 11, 2012

### Combinatorics

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

How can I prove that given two $nXn$ positive semi-definite matrices $A,B$, then the following inequality holds:

$det(A+B)^\frac{1}{n} \geq det(A) ^\frac{1}{n} + det(B)^\frac{1}{n}$

2. Relevant equations

Brunn-Minkowski Inequality:
http://en.wikipedia.org/wiki/Brunn–Minkowski_theorem

3. The attempt at a solution
I've tried proving that these kind of determinants represent volumes of n-dimensional compact bodies, but without any success.. Is there any algebraic/computational way of doing it?

Thanks in advance !

2. Jul 11, 2012

### Latrace

Maybe you can use that because $A$ and $B$ are positive semi-definite, they can be written as the Gram-matrix of certain vectors $a_1, a_2, ..., a_n$ and $b_1, b_2, ..., b_n$ respectively, and the determinant of a Gram-matrix is the content (or volume) squared of the $n$-dimensional box spanned by the vectors of which it is composed? Then maybe the Brunn-Minkowski inequality? This seems intuitively right to me, but it's not a rigorous argument!

3. Jul 12, 2012

### chiro

Hey Combinatorics.


text:
WeBWorK Main Forum

by Joel Trussell -
Number of replies: 3
A collage in ECE would like to ask the students to derive the differential equation for a simple circuit. The usual answer for this is something like that in the attached file. Is there a way the students can enter derivatives or indefinite integrals in their answer - such a way that it can be checked?

Re: answer contains derivatives and integrals

by Michael Gage -
You could easily check the last question (with the values in the box).  We've had good luck checking purely symbolic (but more math related) questions which could probably handle questions b -- d.  (see image attached)

Possibly even question (a) could be handled but I would suggest that might be better suited to an essay question to be graded by hand.

Re: answer contains derivatives and integrals

by Joel Trussell -
I could handle all the symbolic variables except the operators. image attached. How do I insert images into the message window?

the code is
DOCUMENT();

"PGstandard.pl",
"MathObjects.pl",
"parserFunction.pl",
"PGcourse.pl",
"PG.pl",
);

TEXT(beginproblem());

#############################
#  Setup

## required to use step function
step => {
class => 'Parser::Legacy::Numeric',
perl => 'Parser::Legacy::Numeric::do_step'
},
);
t=>"Real", C=>"Real", L=>"Real",  R=>"Real",  I=>"Real",  Vs=>"Real", w=>"Real",
V0=>"Real"
);

# Change tolerance to account for difference in Matlab and Webwork computations
# I don't know the problem yet

Context()->flags->set(
tolerance => 0.001,
tolType => "absolute",
);
parserFunction("u(t)" => "step(t)");

$answer[0] = Formula("L")->reduce;$answer[1] = Formula("1/C")->reduce;

$answer[2] = Formula("R")->reduce;$answer[3] = Formula("R*I + j*w*L*I - j*I/(w*C)")->reduce;
$answer[4] = Formula("V0*exp(j*pi/3)")->reduce;$answer[5] = Formula("V0*exp(j*pi/3)/(R +  j*w*L - j/(w*C))")->reduce;

#############################
#  Main text

Context()->texStrings;
BEGIN_TEXT
This is a test problem
$PAR \{image("RLC_Series_Circuit.png",height=>140, width=>270)\}$BR
Problem is related to Problem x.xx  in the text (for ECE303).
$PAR$PAR For the above circuit, with $$v_s(t) = V_0 cos(\omega t + \pi/3)$$ Volts, write the voltage loop equation in terms of the current $$i(t)$$, using the symbolic values $$R$$, $$L$$, $$C$$, $$v_s(t)$$.

$BR $$\Large{ v_s(t) = }$$ \{ ans_rule(10) \}$$\Large{ \frac{di(t)}{dt} }$$ + \{ ans_rule(10) \}$$\Large{ \int i(t) dt }$$ + \{ ans_rule(10) \}$$\Large{ i(t) }$$ \{ AnswerFormatHelp("formulas") \}$PAR Write the corresponding phasor-domain equation. Use upper case I for the phasor current $$\tilde{I}$$ and $$w$$ for the frequency $$\omega$$.

$BR $$Vs =$$ \{ ans_rule(40) \} \{ AnswerFormatHelp("formulas") \}$PAR Write the phasor representation of the voltage source $$v_s(t) = V_0 cos(\omega t + \pi/3)$$ Use Vs for the phasor $$\tilde{V}_s$$

$BR $$Vs =$$ \{ ans_rule(40) \} \{ AnswerFormatHelp("formulas") \}$PAR Using the phasor representation of the voltage source, solve the phasor equation to obtain an expression for the phasor current, I.

$BR $$I =$$ \{ ans_rule(40) \} \{ AnswerFormatHelp("formulas") \} END_TEXT Context()->normalStrings; ############################ # Answers$showPartialCorrectAnswers = 1;

ANS( $answer[0]->cmp() ); ANS($answer[1]->cmp() );
ANS( $answer[2]->cmp() ); ANS($answer[3]->cmp() );
ANS( $answer[4]->cmp() ); ANS($answer[5]->cmp() );

COMMENT("MathObject version.");

SOLUTION(EV3(<<'END_SOLUTION'));

END_SOLUTION

ENDDOCUMENT();


text:
Under the auspices of the Computational Complexity Foundation (CCF)

REPORTS > DETAIL:

### Revision(s):

Revision #2 to TR16-076 | 9th February 2017 12:32

#### Characterization and Lower Bounds for Branching Program Size using Projective Dimension

Revision #2
Authors: Krishnamoorthy Dinesh, Sajin Koroth, Jayalal Sarma
Accepted on: 9th February 2017 12:32
Keywords:

Abstract:

We study projective dimension, a graph parameter (denoted by ${pd}(G)$ for a graph $G$), introduced by Pudlak and Rodl (1992). For a Boolean function $f$ (on $n$ bits), Pudlak and
Rodl associated a bipartite graph $G_f$ and showed that size of the optimal branching program computing $f$ (denoted by ${bpsize}(f)$) is at least ${pd}(G_f)$ (also denoted by
${pd}(f)$). Hence, proving lower bounds for ${pd}(f)$ imply lower bounds for ${bpsize}(f)$. Despite several attempts (Pudlak and Rodl (1992), Ronyai et.al, (2000)), proving
super-linear lower bounds for projective dimension of explicit families of graphs has remained elusive.

We observe that there exist a Boolean function $f$ for which the gap between the ${pd}(f)$ and ${bpsize}(f)$) is $2^{\Omega(n)}$. Motivated by the argument in Pudlak and Rodl
(1992), we define two variants of projective dimension - projective dimension with intersection dimension 1 (denoted by ${upd}(f)$) and bitwise decomposable projective
dimension (denoted by ${bpdim}(f)$). We show the following results :

1 We observe that there exist a Boolean function $f$ for which the gap between ${upd}(f)$ and ${bpsize}(f)$ is $2^{\Omega(n)}$. In contrast, we also show that the bitwise decomposable
projective dimension characterizes size of the branching program up to a polynomial factor. That is, there exists a constant $c>0$ and for any function $f$, ${bpdim}(f)/6 \le {bpsize}(f) \le ({bpdim}(f))^c$.

2. We introduce a new candidate function family $f$ for showing super-polynomial lower bounds for ${bpdim}(f)$. As our main result, we demonstrate gaps between ${pd}(f)$ and the above two
new measures for $f$ : ${pd}(f) = O(\sqrt{n})$, ${upd}(f) = \Omega(n)$, ${bpdim}(f) = \Omega\left(\frac{n^{1.5}}{\log n}\right)$

3. Although not related to branching program lower bounds, we derive exponential lower bounds for two restricted variants of ${pd}(f)$ and ${upd}(f)$ respectively by observing that they
are exactly equal to well-studied graph parameters - bipartite clique cover number and bipartite partition number respectively.

Changes to previous version:

Revised version. Updated to fix display errors in abstract.

Revision #1 to TR16-076 | 9th February 2017 11:41

#### Characterization and Lower Bounds for Branching Program Size using Projective Dimension

Revision #1
Authors: Krishnamoorthy Dinesh, Sajin Koroth, Jayalal Sarma
Accepted on: 9th February 2017 11:41
Keywords:

Abstract:

We study projective dimension, a graph parameter (denoted by $\mathsf{pd}(G)$ for a graph $G$), introduced by Pudl\'ak and R\"odl (1992). For a Boolean function $f$(on $n$ bits), Pudl\'ak and R\"odl associated a bipartite graph $G_f$ and showed that size of the optimal branching program computing $f$ (denoted by $\mathsf{bpsize}(f)$) is at least $\mathsf{pd}(G_f)$ (also denoted by $\mathsf{pd}(f)$). Hence, proving lower bounds for $\mathsf{pd}(f)$ imply lower bounds for $\mathsf{bpsize}(f)$. Despite several attempts (Pudl\'ak and R\"odl (1992), R\'onyai et.al, (2000)), proving super-linear lower bounds for projective dimension of explicit families of graphs has remained elusive.

We observe that there exist a Boolean function $f$ for which the gap between the $\mathsf{pd}(f)$ and {\sf bpsize}$(f)$) is $2^{\Omega(n)}$. Motivated by the argument in (Pudlak, Rodl 1992), we define two variants of projective dimension - projective dimension with intersection dimension 1 (denoted by $upd(G)$) and bitwise decomposable projective dimension (denoted by $bitpdim(G)$). We show the following results :

1. We observe that there exist a Boolean function $f$ for which the gap between $\mathsf{upd}(f)$ and {\sf bpsize}$(f)$ is $2^{\Omega(n)}$. In contrast, we also show that the bitwise decomposable projective dimension characterizes size of the branching program up to a polynomial factor. That is, there exists a constant $c>0$ and for any function $f$, $\mathsf{bpdim}(f)/6 \le \textrm{\sf bpsize}(f) \le (\mathsf{bpdim}(f))^c$

2. We introduce a new candidate function family $f$ for showing super-polynomial lower bounds for $\mathsf{bpdim}(f)$. As our main result, we demonstrate gaps between $\mathsf{pd}(f)$ and the above two new measures for $f$ :

$\textrm{$\mathsf{pd}(f) = O(\sqrt{n})$,$\mathsf{upd}(f) = \Omega(n)$,$\mathsf{bpdim}(f) = \Omega\left(\frac{n^{1.5}}{\log n}\right)$}$

3. Although not related to branching program lower bounds, we derive exponential lower bounds for two restricted variants of $\mathsf{pd}(f)$ and $\mathsf{upd}(f)$ respectively by observing that they are exactly equal to well-studied graph parameters - bipartite clique cover number and bipartite partition number respectively.

### Paper:

TR16-076 | 27th April 2016 09:00

#### Characterization and Lower Bounds for Branching Program Size using Projective Dimension

TR16-076
Authors: Krishnamoorthy Dinesh, Sajin Koroth, Jayalal Sarma
Publication: 13th May 2016 13:00
Keywords:

Abstract:

We study projective dimension, a graph parameter (denoted by $pd(G)$ for a graph $G$), introduced by (Pudlak, Rodl 1992), who showed that proving lower bounds for $pd(G_f)$ for bipartite graphs $G_f$ associated with a Boolean function $f$ imply size lower bounds for branching programs computing $f$. Despite several attempts (Pudlak, Rodl 1992 ; Babai, Ronyai, Ganapathy 2000), proving super-linear lower bounds for projective dimension of explicit families of graphs has remained elusive.

We show that there exist a Boolean function $f$ (on $n$ bits) for which the gap between the projective dimension and size of the optimal branching program computing $f$ (denoted by bpsize$(f)$), is $2^{\Omega(n)}$. Motivated by the argument in (Pudlak, Rodl 1992), we define two variants of projective dimension - projective dimension with intersection dimension 1 (denoted by $upd(G)$) and bitwise decomposable projective dimension (denoted by $bitpdim(G)$). We show the following results about these parameters:

1. There is an explicit family of graphs on $N = 2^n$ vertices such that the projective dimension is $O(\sqrt{n})$, the projective dimension with intersection dimension $1$ is $\Omega(n)$ and the bitwise decomposable projective dimension is $\Omega(\frac{n^{1.5}}{\log n})$.

2. We show that there exist a Boolean function $f$ (on $n$ bits) for which the gap between $upd(G_f)$ and bpsize$(f)$ is $2^{\Omega(n)}$. In contrast, we also show that the bitwise decomposable projective dimension characterizes size of the branching program up to a polynomial factor. That is, there exists a large constant $c>0$ and for any function $f$, $bitpdim(G_f)/6 \le bpsize(f) \le (bitpdim(G_f))^c$.

3. We also study two other variants of projective dimension and show that they are exactly equal to well-studied graph parameters - bipartite clique cover number and bipartite partition number respectively. This immediately yields exponential lower bounds for these measures.

ISSN 1433-8092 | Imprint


text:
# fftshift

Version:

Computes the FFT shift spectrum.

## Syntax

c = fftshift(a)
c = fftshift(a, b)

## a

Vector or array of any dimension.

## b

Dimension for swapping. b is an integer. If you do not specify b, the function swaps each dimension sequentially.

## c

Shifted version of a. If a is a vector, fftshift(a) swaps the left and right parts of a. If a is a matrix, c = fftshift(a) swaps quadrants 1 and 3 and quadrants 2 and 4. . If a is a vector, c = fftshift(a) swaps each dimension sequentially. c is a vector or array of any dimension.

A = [1, 2, 3, 4; 4, 3, 2, 1]
C = fftshift(A)

Where This Node Can Run:

Desktop OS: Windows

FPGA: This product does not support FPGA devices


text:
Nigerian Scholars » » A wave has an amplitude equal to 4.0m, angular speed $$\frac{1}{3} \pi$$ rads-1 ...

# A wave has an amplitude equal to 4.0m, angular speed $$\frac{1}{3} \pi$$ rads-1 ...

### Question

A wave has an amplitude equal to 4.0m, angular speed $$\frac{1}{3} \pi$$ rads-1 and phase angle $$\frac{2}{3} \pi$$ rad. The displacement y of the wave particle is given as

### Options

A) y = 4 sin $$\frac{\pi}{3}$$(t + 2)

B) y = 4 sin $$\frac{\pi}{3}$$(t + $$\frac{2}{\pi}$$

C) y = 4 sin $$\frac{\pi}{3} (2t + 1)$$

D) y = 4 sin \(\frac{\pi}{3} (t + 2)


text:
# Neutral pions

by dEdt
Tags: neutral, pions
 P: 282 So apparently a neutral pion is a superposition of $u\bar{u}$ and $d\bar{d}$. I'm having trouble understanding what this means. I have no problem understanding how the decay products of some scattering experiment could be a superposition of these two states, but how can we treat this superposition as a single particle? And how would you draw the feynman diagram for some process involving a neutral pion?