text:
Attempting to prove the Intermediate Value Theorem

Homework Statement:
Define a continuous function ##f:[a,b]\longrightarrow \mathbb{R}## such that ##f(b)>0## and ##f(a)<0##. Prove that there is a real number ##c\in (a,b)## such that ##f(c)=0##.
Relevant Equations:
A function ##f## is continuous at ##a\in \textrm{dom}(f)## if for all positive ##\epsilon##, there is a ##\delta>0## such that for ##x\in \textrm{dom}(f)##, if ##0<|x-a|<\delta## then ##|f(x)-f(a)|<\epsilon##
Proof goes like this:
(1) Prove the existence of open intervals centered around the end-points of the domain such that the image of the points in these intervals through ##f## has the same sign as the image of the end-point through ##f##. In other words, prove that there is a ##\delta>0## such that if ##b>x>b-\delta##, ##f(x)f(b)>0##; and similarly for ##a##.

(2) Conclude there is a point ##c\in (a,b)## such that for any ##\delta>0##, the image of ##(c-\delta,c+\delta)## contains both positive and negative points.

(3) By definition of continuity, there must be a real number ##L## that ##f(c)## converges to. Reason that ##L## cannot be negative or positive.

===(1)===
Suppose that there are no such intervals. Consider ##b##. Choose ##\epsilon\leq f(b)##. Hence, for any ##\delta>0##, if ##|b-x|<\delta##, then ##f(x)<0##.
\begin{align*}
|f(x)-f(b)|&=&f(b)+|f(x)|\\
&\geq&f(b)\\
&\geq&\epsilon
\end{align*}
This contradicts the fact that ##f## is continuous. Hence, there must be some ##\delta>0## such that ##f(x)>0## if ##b-\delta<x<b##.

===(2)===
Now choose ##c=b-\delta_b## where ##\delta_b>0## with the property that if ##x<b-\delta_b##, then ##f(x)\leq 0##.
Similarly, ##c=a+\delta_a## where ##\delta_a>0## with the property that if ##x>a+\delta_a##, then ##f(x)\geq 0##.
% Note: I feel like I should be invoking completeness here, but wasn't sure if it applies or if it does, how to invoke it.

Let ##\delta_c>0##.

If ##\delta_c+c>x>c=a+\delta_a##, then ##f(x)\geq 0##.
If ##c-\delta_c<x<c=b-\delta_b##, then ##f(x)\leq 0##.

===(3)===
Suppose ##f(c)>0## and choose ##\epsilon\leq f(c)##. Let ##\delta>0## and choose ##x\in (c-\delta,c)##.

Then ##f(x)\leq 0## and:
\begin{align*}
|f(c)-f(x)|&=&-f(x)+f(c)\\
&=&|f(x)|+f(c)\\
&\geq&f(c)\\
&\geq&\epsilon
\end{align*}
Hence, ##f## cannot be continuous at ##x=c## if ##f(c)>0##.

Now suppose ##f(c)<0## and choose ##\epsilon\leq -f(c)##. Let ##\delta>0## and choose ##x\in (c,c+\delta)##.

Then ##f(x)\geq 0##:
\begin{align*}
|f(x)-f(c)|&=&f(x)+(-f(c))\\
&=&|f(x)|-f(c)\\
&\geq&-f(c)\\
&\geq&\epsilon
\end{align*}
Hence, ##f## cannot be continuous at ##x=c## if ##f(c)<0##.

This leaves only the one possibility by the trichotomy of the real numbers.

Last edited:

Office_Shredder
Staff Emeritus
Gold Member
I don't understand step 2. Why can I pick a single c which satisfies both the condition for b and the condition for a? It feels like that's the whole problem.

docnet
pasmith
Homework Helper
Alternatively: Define $P \subset [a,b]$ such that $x \in P$ if and only if $f(y) < 0$ for all $y \in [a,x]$. Then show $f(\sup P) = 0$.

docnet
Why can I pick a single c which satisfies both the condition for b and the condition for a?

Define ##D_b=\{\delta>0:f(b)f(x)>0\textrm{ if }x\in(b-\delta,b)\}##.

We reason that this is bounded. Suppose ##\delta=|b-a|##. Then the interval ##(b-|b-a|,b)=(a,b)## contains points from the interval around ##a## such that ##f(x)<0##. Hence, this choice of ##\delta## cannot belong to ##D_b##. If ##\delta>|b-a|##, then for ##x\in(b-\delta,b)##, ##f(x)## is not even defined. Hence, ##D_b## is bounded from above by ##|b-a|##. By the completeness of real numbers, there must exist a ##\sup D_b##; denote it as ##\delta_b##.

If we define ##D_a## similarly, then we can make a symmetric argument.

I was going to reason that ##a+\delta_a=b-\delta_b##, then call it ##c##.

By definition, if ##a<x< a+\delta_a##, then ##f(x)<0##; if ##x\geq a+\delta_a##, then ##f(x)\geq 0##.
By definition, if ##b>x> b-\delta_b##, then ##f(x)>0##; if ##x\leq b-\delta_b##, then ##f(x)\leq 0##.

This is what I have so far; as for how to figuratively connect the dots, I do not know.

One thing that comes to mind is just repeating step three, replacing ##c## with either ##a+\delta_a## or ##b-\delta_b##.

Then show ##f(\sup P)=0##.

I feel like I'd be repeating step three with this, to be honest, if I replaced ##c## in my post with ##\sup P##.

Last edited:
Office_Shredder
Staff Emeritus
Gold Member
Eclair, try applying your way of computing c to the function ##(x-1)(x-2)(x-3)## on the interval [0,4]. I think you will find you did not succeed at finding a single c. But you should consider what you did find, and if you actually care about those two things lining up. This basically just turns into what pasmith is suggesting.

try applying your way of computing c to the function ##(x-1)(x-2)(x-3)## on the interval [0,4].

Okay.

##a=0##
##b=4##

##D_0=\{\delta>0:f(x)f(0)>0\textrm{ if }x\in(0,\delta)\}##

Assert ##\sup D_0=1## and note that ##f(0)=(-1)(-2)(-3)=-6<0##.

Note that if ##0<x<1##, then:

##-1<x-1<0##
##-2<x-2<-1##
##-3<x-3<-2##

Hence, the function ##(x-1)(x-2)(x-3)## is a product of negative numbers if ##x\in (0,1)##; the function must be negative as a result.

If ##x\geq1##:

##x-1\geq 0##
##x-2\geq -1##
##x-3\geq -2##

And the last two numbers are of some concern. I could only guarantee that they would both be negative, thus guaranteeing that the function is positive, but only if ##x\leq 2##. So it wouldn't be the case that ##f(x)>0## if ##x\in (1-\delta,1+\delta)## for any positive ##\delta##.

But you should consider what you did find, and if you actually care about those two things lining up.

I think I found the value for ##\sup P##, which is ##a+\delta_a## and the value for ##\inf([a,b]\setminus P)##, which is ##b-\delta_b##, where ##P## was defined by Mr. pasmith. I mean, there would be some neighborhood around these two points that would contain points whose images are positive and negative. They'd be the same if the function was zero only at one point.

Office_Shredder
Staff Emeritus
Gold Member
Eclair, why are you evaluating the individual signs of that factored polynomial? f is a function, you can evaluate the function but shouldn't do anything else with it

I think you should try to find ##\delta_b## to see why your method doesn't work on this function.

Remember, your goal is to find a zero of f. When finding ##\delta_a##, you found the number 1. I am impressed by your ability to not just evaluate f(1) and wonder about it, but I recommend you do that now :)

##f(1)=(1-1)(1-2)(1-3)=0(-1)(-2)=0##

Office_Shredder
Staff Emeritus
Gold Member
Ok, so you're done for this function, right? You computed ##\delta_a##, and it found you a zero, which was your goal.

Is it a coincidence that this worked, or can you turn it into a proof?

turn it into a proof

No, it was not really a coincidence. There are a bunch of points that the function evaluates to positive values immediately after ##x=1##, and the function cannot just jump from a positive number to a negative number, since otherwise it would not be continuous.

Office_Shredder
Staff Emeritus
Gold Member
No, it was not really a coincidence. There are a bunch of points that the function evaluates to positive values immediately after ##x=1##, and the function cannot just jump from a positive number to a negative number, since otherwise it would not be continuous.

What if instead the function was ##(x-1)^2(x-2)##? What is ##\delta_a##? Is the function positive afterwards?

The value of ##x## does not affect the sign of ##(x-1)^2##. And ##x-2## is negative if ##0<x<2## and positive if ##4>x>2##. ##\delta_a## would be ##2## in this case; easy to see if we just subtract two from these inequalities.

Office_Shredder
Staff Emeritus
Gold Member

##D_0:=\{\delta>0:f(x)f(0)>0\textrm{ if }x\in(0,\delta)\}##

Set ##\delta=2## and choose ##x=1.5##. Note that ##1.5\in(0,2)##.

##f(x)=(x-1)^2(x-2)## on the interval ##[0,4]##.

##f(1.5)=0.5^2(-0.5)##.
##f(0)=-2##
##f(1.5)f(0)=0.25>0##

Hence, ##2\in D_0##.

I'm not even sure that ##1## is an upper bound for ##D_0##.

In any case, can I ask if all that I'd written up until post #4 would have been correct if ##f## was identically zero at just one point on its domain?

Office_Shredder
Staff Emeritus
Gold Member
What is f(1)? If ##\delta_a >1##, it needs to be negative the way you defined it.

Oh. ##f(1)=0##, so ##f(0)f(1)=0##, which means that if ##\delta>1##, then you can choose the point ##x=1## in ##(0,\delta)## such that ##f(0)f(1)## is not positive, which would mean that ##\delta\notin D_0##.

Office_Shredder
Staff Emeritus
Gold Member
Yes.

So how does that help you? You know that ##D_0## is bounded above by ##b-a##, ##D_0## is not empty, so it has a supremum. In both of the examples we looked at, the supremum had ##f(c)=0##. Is that always true?

Whenever ##x\in(a,a+\delta_a)##, ##f(x)<0##. If ##x\geq a+\delta_a##, then ##f(x)\geq0##. Let's say that ##x=a+\delta_a##. Then ##f(x)=0## or ##f(x)>0##. If the latter held, then the function would fail to be continuous at that point.

Denote ##L=f(a+\delta_a)## and suppose it is positive. Set ##\epsilon\leq L## and let ##\delta>0##. If ##x\in (a+\delta_a-\delta,a+\delta_a)##, then ##f(x)<0##. Hence,
\begin{align*}
|L-f(x)|&=&L+(-f(x))\\
&=&L+|f(x)|\\
&\geq&L\\
&\geq&\epsilon
\end{align*}

This leaves just the other possibility, I think.

Office_Shredder
Staff Emeritus
Gold Member
No, on
Whenever ##x\in(a,a+\delta_a)##, ##f(x)<0##. If ##x\geq a+\delta_a##, then ##f(x)\geq0##.

No, in the example we saw, ##f(x)<0 ## for ##x>a+\delta_a##. We know ##f## is continuous, and ##f(x)<0## for ##x<a+\delta_a##. Use that to show ##f(a+\delta_a)\leq 0##. Then suppose it's strictly less than zero- you can show that if that's true, ##\delta_a## is not actually the supremum, which is a contradiction.

Let ##\epsilon>0##. Then there is a ##\delta>0## such that if ##|x-(a+\delta_a)|<\delta##, then ##|f(x)-f(a+\delta_a)|<\epsilon##. This implies that:

##f(a+\delta_a)<f(x)+\epsilon<\epsilon##
##f(a+\delta_a)-\epsilon<f(x)<0##

This implies that ##f(a+\delta_a)## cannot be greater than some arbitrarily small ##\epsilon>0##. Hence, either ##f(a+\delta_a)=0##, or ##f(a+\delta_a)<0##. It cannot be positive for the reason in post #19.

Office_Shredder
Staff Emeritus
Gold Member
Why is ##f(x)+\epsilon < \epsilon##?

Because ##f(x)<0## if you choose ##x<a+\delta_a##. Adding ##\epsilon## to both sides of the former inequality gives that result.

Specifically:

##|f(a+\delta_a)-f(x)|<\epsilon##
##-\epsilon<f(a+\delta_a)-f(x)<\epsilon##
##f(a+\delta_a)<\epsilon+f(x)##

Office_Shredder
Staff Emeritus
Gold Member
But all you have is ##|x-(a+\delta_a)|< \delta##. So x could be as large as ##a+\delta_a+\delta##

You can also choose ##x## to be as small as ##(a+\delta_a)-\delta##, though.

And ##f(x)## is negative if ##(a+\delta_a)>x>(a+\delta_a)-\delta##, no matter how small ##\delta## is.

On the other hand, if ##f(x)\geq0## whenever ##(a+\delta_a)+\delta>x>(a+\delta_a)##:

##f(x)-f(a+\delta_a)<\epsilon##
##-f(a+\delta_a)<\epsilon-f(x)\leq\epsilon##
##-f(a+\delta_a)-\epsilon<-f(x)\leq0##
##f(a+\delta_a)+\epsilon>f(x)\geq0##

This implies that if ##f(a+\delta_a)<0## cannot exceed some arbitrarily small ##\epsilon## in magnitude.

This means that ##f(a+\delta_a)\geq0##. In addition, we also have another possibility for when ##x## is on the left-side of this interval: ##f(a+\delta_a)\leq0##. If ##f(a+\delta_a)\neq0##, then we have continuity issues as discussed previously. I think I covered this in the third step in the opening post of this topic?

Also, if ##f(x)<0## whenever ##(a+\delta_a)+\delta>x>(a+\delta_a)##, this would contradict yet again the fact that ##\delta_a## is the supremum for ##D_a##.

Last edited:


text:
Prove frac{cos(a)}{1+sin(a)}+(1+sin)

illusiia 2020-11-09 Answered
Prove $\frac{\mathrm{cos}\left(a\right)}{1+\mathrm{sin}\left(a\right)}+\left(1+\mathrm{sin}\right)$
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2k1enyvp
Answered 2020-11-10 Author has 94 answers

$\frac{\mathrm{cos}\left(a\right)}{1+\mathrm{sin}\left(a\right)}+\frac{1+\mathrm{sin}\left(a\right)}{\mathrm{cos}\left(a\right)}$
$=\frac{\left(1+\mathrm{sin}\left(a\right){\right)}^{2}+{\mathrm{cos}}^{2}\left(a\right)}{\left(1+\mathrm{sin}\left(a\right)\right)\mathrm{cos}\left(a\right)}$
$=\frac{\left(1+\mathrm{sin}\left(a\right){\right)}^{2}+1-{\mathrm{sin}}^{2}\left(a\right)}{\left(\left(1+\mathrm{sin}\left(a\right)\right)\mathrm{cos}\left(a\right)}$
$=\frac{2}{\mathrm{cos}\left(a\right)}=\frac{2}{\left(\frac{1}{\mathrm{sec}\left(a\right)}\right)=\frac{2}{\mathrm{sec}\left(a\right)}}$

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text:
## Facts

Words:
Earth - (noun) Saxon=Erda. Dirt or the planet on which we live.
Terra - (noun) Latin.
Terran - (adjective)
Moon - (noun) From the Latin word Mensis (month).
Luna - (noun) Latin.
Lunar - (adjective)
Terran equatorial diameter: 12,756km (<8000miles) - - Average radius = 6,371,000m
Lunar equatorial diameter: 3475km (~earth/4)
Earth mass:
Moon mass: (0.01 X Mass of Earth)
Earth's Average distance from Sol: 150,000,000km (93,000,000 miles)
Moon's Average distance from Earth: 384,400km (250,000 miles)
Age:
Orbits because of:
Earth orbits around Sun:
Moon orbits around Earth (Sidereal Period): 27 d 7 h 43.1 min
Moon orbits around Earth (Synodic Period): 29 d 12 h 44.0 min
Earth rotates:
Moon rotates:
Earth's Inclination: 23.5° to the plane of the Solar System
Moon's Inclination: 5.145° to the ecliptic
Earth Atmosphere:
Moon Atmosphere:
Gravity on Earth:
Gravity on Moon:

## Gravity (Newton & Einstein)

1. Aristotle thought that heavier objects fell faster than lighter objects. He partially understood momentum, thinking that moving objects had a sort of moving energy. If a cannonball was flying through the air, it would continue to move in a straight line (he thought) until it ran out of energy. Then (he thought) it would fall straight down. Thus, according to Aristotle, projectiles always followed the path of a right triangle. The church taught these ideas until the end of the Renaissance.
2. Galileo discovered/proved that Objects on Earth all fall at the same rate. g = -9.81m/s2 - - - TeacherTube Video of Dave Scott
3. DesCartes & Kepler each had their own theories about planetary motion. Kepler figured out their true paths, but they were otherwise both wrong about gravity.
4. Newton wrote equation to calculate gravitational force between and 2 obj. He figured out that gravity and inertia keep them in orbit.
Newton's Law of Universal Gravitation (p263):
where G is the constant of universal gravitation,
the m's are the masses of the 2 objects, and r is the distance between them.
Force is measured in Newtons
Gravitational force is localized to the center of a spherical mass. (so to find your weight, we would need to know the radius and mass of the Earth.)
5. Einstein found the link between mass, gravity, velocity and time. Mass curves time-space, causing objects to fall into gravity wells.

## Tides

We travel through 2 zones of high tides a day, because there are 2 tidal bulges that follow the Moon as the Earth rotates below it.

1. Spring Tides-
2. Neap Tides-

## Phases

4 weeks, 8 phases position, West to East, orbit, Sun, Moon, Earth light, shadow, visible parts, reflection, rotation speed, ecliptic, The near side of the Moon is always turned toward the Earth. Position of Moon changes as goes around ---- in its orbit... causes different phases at different times. We only see part of the Moon that the Sun is shining on it.

The moon has 8 phase positions that re-occur every four weeks. The phases of the moon are caused by its revolution around the earth, and the angle of the Sun's light as seen from our point of view. During the moon's orbit, we see changing amounts of the 51% of the moon that is illuminated. The moon roughly follows the ecliptic, rising in the East and setting in the West. However, since it is moving around the Earth about 12 degrees per day, it slowly moves across the sky, seeming to go backward every evening, until it is back at its starting point and in the same starting phase. The near side of the moon is always facing the earth, which shows each phase.

## Eclipses

1. Solar Eclipses-
2. Lunar Eclipse-
	Why do they last so long?

Define Umbra, Penumbra, Partial eclipse

## Origins of the Moon

Theories concerning the Origin of Luna

1. Sister world formed at the same time of the same stuff (popular, but moon has no iron core)
2. Parted at birth (“)
3. Parted because of a collision much later (but could Earth have healed the wound?)
4. Captured (but moon has similar water isotope content)

## Why is the Moon so important to humans?

1. Romance / Religion / Our closest neighbor, it is somehow our connection to the cosmos
2. Science
3. Exploration
4. Launching platform / staging area / practice for exploring other places:
1. gives a 2300mph boost to speed when launching from moon as opposed to earth
2. lesser gravity means it's easier to escape
3. let's use moon materials instead of earth materials!!!


text:
## Fall 2015, problem 6

Let $S$ denote the unit sphere in $\mathbb{R}^3$, let $x_1,\dots,x_n \in S$, and let $d(\cdot,\cdot)$ denote the standard distance function on $\mathbb{R}^3$. Prove that $$\sum_{i,j} d(x_i,x_j) \leq n^2$$ and determine for which points $x_1,\dots,x_n$ the above sum is $n^2$.


text:
# Tag Info

## New answers tagged dictionaries

1 vote

### Dictionary with sets as keys where lookup can be set intersection

My understanding: You might have three values in the table with keys (a, b), (a, c), (b, c). When you lookup (d, e, c) you want the values for the keys (a, c) and (b, c) containing a “c”. First you ...
• 25.2k
1 vote

### Dictionary with sets as keys where lookup can be set intersection

Unfortunately, no, this isn't achievable (at least not if you want to handle arbitrary sets as keys). The reason is simple: if $N$ denotes the number of elements in a key set, then it takes $O(N)$ ...
• 141k

Top 50 recent answers are included


text:
# SN1 like reaction acid-cataylzed hydrolysis of epoxide

What would be the reaction mechanism for the reaction between epoxide and $\ce{H2SO4}$?

-
It would be better if you would propose the mechanism and someone can verify it for you and explain some of the underlying concepts. –  jonsca Nov 30 '12 at 2:32
Agreed. In order to give you a better answer, we need to know what about this mechanism you understand and what you don't. –  Ben Norris Nov 30 '12 at 11:47

Note that your question could do with some more data. Here is my answer based on what I think you mean:

You ought to get a glycol.

First, a proton from the acidic medium attacks the neucleophile, $\ce{O}$. This creates a positively charged trivalent oxygen. The electrons in one of the two $\ce{C-O}$ bonds transfer to the $\ce{O+}$, breaking the bond and effectively transferring the positive charge to the carbon.

At this point, note that the carbocation thus formed may rearrange.

Now, we have the nucleophilic attack. An oxygen atom (specifically the lone pair of an oxygen) from a nearby water molecule attacks the carbocation, forming another trivalent positively charged oxygen. This is deprotonated, giving the final glycol product.

Here's the same mechanism, showing movement of electrons (pairs):

-


text:
# Inverse Decoupled Transform (Six-Phase)

Implement dqz1z2o1o2 to abc and xyz transform

• Library:
• Simscape / Electrical / Control / Mathematical Transforms

## Description

The Inverse Decoupled Transform (Six-Phase) block converts the direct, quadrature, z1, z2, o1, and o2 components in a rotating reference frame to the components of a six-phase system in the abc and xyz reference frame. The three-phase systems abc and xyz are spatially shifted by 30 electrical degrees.

You can configure the block to align the phase a-axis of the five-phase system to either the d-axis or q-axis of the rotating reference frame at time t = 0.

θ = ωt + Offset is the angle between the a and q axes for the q-axis alignment or the angle between the a and d axes for the d-axis alignment, where:

• ω is the rotational speed of the rotational reference frame.

• t is the time, in s, from the initial alignment.

• Offset=\begin{array}{c}0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}d-axis\text{\hspace{0.17em}}alignment\\ -\frac{\pi }{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}q-axis\text{\hspace{0.17em}}alignment\end{array}

### Equations

The Inverse Decoupled Transform (Six-Phase) block implements the transform as

`$\left[\begin{array}{c}a\\ b\\ c\\ x\\ y\\ z\end{array}\right]=\left[\begin{array}{cccccc}\mathrm{cos}\theta & -\mathrm{sin}\theta & 1& 0& 1& 0\\ \mathrm{cos}\left(\theta -\frac{2\pi }{3}\right)& -\mathrm{sin}\left(\theta -\frac{2\pi }{3}\right)& -\frac{1}{2}& -\frac{\sqrt{3}}{2}& 1& 0\\ \mathrm{cos}\left(\theta +\frac{2\pi }{3}\right)& -\mathrm{sin}\left(\theta +\frac{2\pi }{3}\right)& -\frac{1}{2}& \frac{\sqrt{3}}{2}& 1& 0\\ \mathrm{cos}\left(\theta -\frac{\pi }{6}\right)& -\mathrm{sin}\left(\theta -\frac{\pi }{6}\right)& -\frac{\sqrt{3}}{2}& \frac{1}{2}& 0& 1\\ \mathrm{cos}\left(\theta -\frac{5\pi }{6}\right)& -\mathrm{sin}\left(\theta -\frac{5\pi }{6}\right)& \frac{\sqrt{3}}{2}& \frac{1}{2}& 0& 1\\ \mathrm{cos}\left(\theta +\frac{\pi }{2}\right)& -\mathrm{sin}\left(\theta +\frac{\pi }{2}\right)& 0& -1& 0& 1\end{array}\right]\left[\begin{array}{c}d\\ q\\ {z}_{1}\\ {z}_{2}\\ {o}_{1}\\ {o}_{2}\end{array}\right]$`

The z1z2 and o1o2 pairs do not form a rotating magnetic potential. They are generalized zero-sequence components.

## Ports

### Input

expand all

Direct, quadrature, z1-axis, z2-axis, o1-axis, and o2-axis components of the system in the rotating reference frame, specified as a vector.

Data Types: `single` | `double`

Angular position of the rotating reference frame, specified as a scalar.

Data Types: `single` | `double`

### Output

expand all

Components of the three-phase system in an abc reference frame, returned as a vector.

Data Types: `single` | `double`

Components of the three-phase system in an xyz reference frame, returned as a vector.

Data Types: `single` | `double`

## Parameters

expand all

Whether to align the a-phase vector of the abc reference frame to the d-axis or q-axis of the rotating reference frame.


text:
#### DOCUMENTA MATHEMATICA, Vol. 20 (2015), 37-64

Hubert Kalf, Takashi Okaji, and Osanobu Yamada

The Dirac Operator with Mass m_0 >= 0 : Non-Existence of Zero Modes and of Threshold Eigenvalues

A simple global condition on the potential is given which excludes zero modes of the massless Dirac operator. As far as local conditions at infinity are concerned, it is shown that at energy zero the Dirac equation without mass term has no non-trivial $L^2$-solutions at infinity for potentials which are either very slowly varying or decaying at most like $r^{-s}$ with $s \in (0,1)$. When a mass term is present, it is proved that at the thresholds there are again no such solutions when the potential decays at most like $r^{-s}$ with $s \in (0,2)$. In both situations the decay rate is optimal.

2010 Mathematics Subject Classification: Primary 35P15; Secondary 81Q10.

Keywords and Phrases: Dirac operators, virial theorem, threshold eigenvalue, zero mode.

Full text: dvi.gz 41 k, dvi 107 k, ps.gz 302 k, pdf 221 k.

Home Page of DOCUMENTA MATHEMATICA


text:
Mapping to custom reference minimap2/IGV
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0
Entering edit mode
3.1 years ago
sendhelp • 0

I'm trying to map nanopore reads to a mtDNA reference using minimap2 for the purpose of determining how many mitochondrial reads are present in the nanopore data.

The minimap command I'm using is;

~/apps/minimap2/minimap2 -ax map-ont mtref.fa reads.fq  > aln.sam

I know there are mtDNA reads present as I can align them to the reference, however when I look at the resulting indexed and sorted sam file in IGV there's no coverage?

Thanks!

mapping minimap2 igv nanopore • 2.2k views
0
Entering edit mode

Have you made a custom genome in IGV using the same reference? By default IGV shows you the entire reference so you may need to zoom in before you start seeing the reads.

0
Entering edit mode

I have. I solved the initial problem by actually creating a sorted bam file and using that in IGV rather than just the same file. Thanks!

1
Entering edit mode
3.1 years ago
ATpoint 64k

You can simply check the full alignment using samtools idxstats for the number of reads mapping to chrM.No need to align against a specific subset (which is also not recommended).

0
Entering edit mode

Thank you! Can I ask why it isn't recommended to align against a subset (I think you're referring to aligning genomic reads to a single chromosome reference?)

1
Entering edit mode

The aligner will always try to find the best match for each read in the library. If your library contains reads from the entire genome but you include only one chromosome (or one region) with the true origin of the read missing the aligner might false-positively put it somewhere else. I never quantified what the extend of false-positives is but this opinion is pretty much what I read here at Biostars from many of the more experienced folks.

0
Entering edit mode

Ah, makes sense! There isn't a reference genome available for me currently so I guess this is the only method?

1
Entering edit mode

Well I am spoiled by working with mouse and human only (and I often forget that when writing answers :-D ) but yeah if you have no reference I would include as much of the draft genome or the contigs you have and then filter for the regions/contigs/chromosomes you need.


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# Weightless at the center of mass of two merging black holes?

When $$r = 0$$ the Christoffel symbol $$\Gamma_{tt}^r$$ is zero and that means the radial four-acceleration is zero and that means you're weightless.

This example is for a spherically symmetric system, but it is basically saying that because the gravitational attraction cancels out from all directions, you would feel weightless. As an analogy, if we could assume that the mass is distributed so along the binary BH system, that there is a point (center of mass) where the gravitational attraction cancels out from both sides, then at this center of mass point we could calculate the same Christoffel symbol.

Now at the center of mass of the black hole binary system, based on this, if this calculation showed 0 (I have not found such calculation), you could feel weightless. So far so good.

What gives me doubt, is that in a merging binary system, initially the center of mass point could be outside the event horizons.

But eventually as the event horizons join, the center of mass point is incorporated by the joining event horizons, and thus the center of mass point moves inside the event horizon (of the now joint system). At that moment, in the process, there are two singularities, and if I understand correctly, being inside the horizon means having the spatial and temporal dimensions oddly interchanging roles.

One of the consequences of this is a inevitable move of any object toward the singularity, the singularity becomes the future. Could this in any way change the fact of weightlessness at the center of mass of the binary system?

As per the comments, I am asking about whether we could feel weightless at the COM of the BH binary (not moving through that point but being held there with no relative motion to the COM).

Question:

1. Could you feel weightless (is $$\Gamma_{tt}^r$$=0) at the center of mass of the BH binary?
• What is the centre of mass? I mean what is the definition?
– MBN
Jan 6, 2020 at 10:12
• @MBN arxiv.org/pdf/1807.11489.pdf this defines COM for BH binaries. Jan 6, 2020 at 17:19
• Can you say which page/paragraph?
– MBN
Jan 7, 2020 at 9:42
• @MBN 2.1 paragraph Jan 7, 2020 at 15:46
• @safesphere thank you I edited to make it clear, do you think it is OK now? Jan 7, 2020 at 17:57

• thank you do you think that at the center of mass of the binary $\Gamma_{tt}^r$=0? Jan 5, 2020 at 23:42


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# A three phase star connected alternator develops an rms voltage of V1 per phase at the fundamental frequency. It also generates a third harmonic component whose rms value is V3 . What is the line voltage of the alternator?

This question was previously asked in
DSSSB JE EE 2019 Official Paper (Held on 5 Nov 2019)
View all DSSSB JE Papers >
1. $$\sqrt {3(V_1^2 + V_3^2)}$$
2. $$\sqrt {(V_1^2 + V_3^2)}$$
3. $$\dfrac{\sqrt {V_1^2 + V_3^2}}{\sqrt 3}$$
4. $$\sqrt {2(V_1^2 + V_3^2)}$$

Option 1 : $$\sqrt {3(V_1^2 + V_3^2)}$$

## Detailed Solution

Concept:

The RMS of an alternating quantity having different harmonic components is given by

$${V_{rms}} = \sqrt {V_1^2 + V_2^2 + \ldots + V_n^2}$$

Where Vn represents the nth harmonic component.

V1 = Fundamental component of voltage

Calculation:

Given harmonic component phase voltages of a three phase star connected alternator are fundamental and third.

As three phase star connected alternator can possibly have Line voltage of third harmonics, Hence third harmonic phase voltage is considered in calculation of total Line voltage.

Total phase voltage  $${V_{ph\ rms}} = \sqrt {V_1^2 + V_3^2}$$

Total line voltage VL = $$\sqrt3$$ Vph rms

VL = $$\sqrt3( \sqrt {V_1^2 + V_3^2})$$

= $$\sqrt {3(V_1^2 + V_3^2})$$

Important Points

If the three phase delta connected alternator is present, then there is no existance of third harmonic component of line voltage implies no consideration of third harmonic component of phase voltage is taken for the calculation.


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# moments¶

dragon.vm.tensorflow.nn.moments(
x,
axes=None,
keepdims=False,
name=None
)[source]

Compute the mean and variance of input along the given axes.

$\begin{cases} \text{mean} = \frac{1}{n}\sum(\text{input}) \\ \text{variance} = \frac{1}{n}\sum(\text{input} - \text{mean})^{2} \end{cases}$

The argument axis could be negative or None:

x = tf.constant([[1, 2, 3], [4, 5, 6]])

# A negative axis is the last-k axis
print(tf.nn.moments(x, 1))
print(tf.nn.moments(x, -1))  # Equivalent

# If axes is None, the vector-style reduction
# will be applied to return a scalar result
print(tf.nn.moments(x))  # Mean is 3.5, Var is 2.916667

# Also, axes could be a sequence of integers
print(tf.nn.moments(x, [0, 1]))  # Mean is 3.5, Var is 2.916667

Parameters:
• x (dragon.Tensor) – The input tensor.
• axes (Union[int, Sequence[int]], optional) – The axis to reduce.
• keepdims (bool, optional, default=False) – Keep the reduced dimensions or not.
• name (str, optional) – The operation name.
Returns:

• dragon.Tensor – The mean tensor.
• dragon.Tensor – The variance tensor.


text:
# Katherine Whitaker, Ph.D.

Assistant Professor, Department of Physics | University of Connecticut

Storrs, CT, United-States

Katherine Whitaker, Ph.D., is an expert in star and galaxy formation.

## Biography

Katherine Whitaker is an Assistant Professor in the Department of Physics at the University of Connecticut.

## Areas of Expertise (3)

Stars Physics Galaxies

## Affiliations (1)

• American Astronomical Society, Member

## Accomplishments (3)

Yale University

#### FAMOUS Travel Grant

American Astronomical Society

## Media Appearances (2)

#### Hubble Finds Patches of Newborn Stars in Galaxy 11 Billion Light-Years from Earth

Science News  online

2017-07-07

Astronomers using the NASA/ESA Hubble Space Telescope have captured a detailed image of an extremely distant, gravitationally lensed galaxy they believe contains two dozen patches of newborn stars.

view more

#### Free talk about the mysteries of the galaxies

The Weston Forum  online

2017-07-11

"UCONN astrophysicist Dr. Katherine E. Whitaker is giving a talk about the mysteries of galaxy formation and evolution on Tuesday, July 18 at 8 p.m., at the Westport Astronomical Society." (...)

view more

## Articles (5)

The Mass, Color, and Structural Evolution of Today's Massive Galaxies Since z ~ 5 The Astrophysical Journal

Allison R. Hill, Adam Muzzin, Marijn Franx, Bart Clauwens, Corentin Schreiber, Danilo Marchesini, Mauro Stefanon, Ivo Labbe, Gabriel Brammer, Karina Caputi, Johan Fynbo, Bo Milvang-Jensen, Rosalind E. Skelton, Pieter van Dokkum, and Katherine E. Whitaker

2017

In this paper, we use stacking analysis to trace the mass growth, color evolution, and structural evolution of present-day massive galaxies ($\mathrm{log}({M}_{* }/{M}_{\odot })=11.5$) out to z = 5. We utilize the exceptional depth and area of the latest UltraVISTA data release, combined with the depth and unparalleled seeing of CANDELS to gather a large, mass-selected sample of galaxies in the NIR (rest-frame optical to UV). Progenitors of present-day massive galaxies are identified via an evolving cumulative number density selection, which accounts for the effects of merging to correct for the systematic biases introduced using a fixed cumulative number density selection, and find progenitors grow in stellar mass by $\approx 1.5\,\mathrm{dex}$ since z = 5. Using stacking, we analyze the structural parameters of the progenitors and find that most of the stellar mass content in the central regions was in place by $z\sim 2$, and while galaxies continue to assemble mass at all radii, the outskirts experience the largest fractional increase in stellar mass. However, we find evidence of significant stellar mass build-up at $r\lt 3\,\mathrm{kpc}$ beyond $z\gt 4$ probing an era of significant mass assembly in the interiors of present-day massive galaxies. We also compare mass assembly from progenitors in this study to the EAGLE simulation and find qualitatively similar assembly with z at $r\lt 3\,\mathrm{kpc}$. We identify $z\sim 1.5$ as a distinct epoch in the evolution of massive galaxies where progenitors transitioned from growing in mass and size primarily through in situ star formation in disks to a period of efficient growth in r e consistent with the minor merger scenario.

Predicting Quiescence: The Dependence of Specific Star Formation Rate on Galaxy Size and Central Density at 0.5 < z < 2.5 The Astrophysical Journal

Katherine E. Whitaker, Rachel Bezanson, Pieter G. van Dokkum, Marijn Franx, Arjen van der Wel, Gabriel Brammer, Natascha M. Förster-Schreiber, Mauro Giavalisco, Ivo Labbé, Ivelina G. Momcheva, Erica J. Nelson, and Rosalind Skelton

2017

In this paper, we investigate the relationship between star formation and structure, using a mass-complete sample of 27,893 galaxies at 0.5 < z < 2.5 selected from 3D-HST. We confirm that star-forming galaxies are larger than quiescent galaxies at fixed stellar mass (${M}_{\star }$). However, in contrast with some simulations, there is only a weak relation between star formation rate (SFR) and size within the star-forming population: when dividing into quartiles based on residual offsets in SFR, we find that the sizes of star-forming galaxies in the lowest quartile are 0.27 ± 0.06 dex smaller than the highest quartile. We show that 50% of star formation in galaxies at fixed ${M}_{\star }$ takes place within a narrow range of sizes (0.26 dex). Taken together, these results suggest that there is an abrupt cessation of star formation after galaxies attain particular structural properties. Confirming earlier results, we find that central stellar density within a 1 kpc fixed physical radius is the key parameter connecting galaxy morphology and star formation histories: galaxies with high central densities are red and have increasingly lower SFR/${M}_{\star }$, whereas galaxies with low central densities are blue and have a roughly constant (higher) SFR/${M}_{\star }$ at a given redshift. We find remarkably little scatter in the average trends and a strong evolution of >0.5 dex in the central density threshold correlated with quiescence from z ~ 0.7–2.0. Neither a compact size nor high-n are sufficient to assess the likelihood of quiescence for the average galaxy; instead, the combination of these two parameters together with ${M}_{\star }$ results in a unique quenching threshold in central density/velocity.

Near-infrared Spectroscopy of Five Ultra-massive Galaxies at 1.7 < z < 2.7 The Astrophysical Journal

Erin Kado-Fong, Danilo Marchesini, Z. Cemile Marsan, Adam Muzzin, Ryan Quadri, Gabriel Brammer, Rachel Bezanson, Ivo Labbé, Britt Lundgren, Gregory Rudnick, Mauro Stefanon, Tomer Tal, David Wake, Rik Williams, Katherine Whitaker, and Pieter van Dokkum

2017

We present the results of a pilot near-infrared spectroscopic campaign of five very massive galaxies ($\mathrm{log}({M}_{\star }/{M}_{\odot })\gt 11.45$) in the range of $1.7\lt z\lt 2.7$. We measure an absorption feature redshift for one galaxy at ${z}_{\mathrm{spec}}=2.000\pm 0.006$. For the remaining galaxies, we combine the photometry with the continuum from the spectra to estimate continuum redshifts and stellar population properties. We define a continuum redshift (${z}_{\mathrm{cont}}$ ) as one in which the redshift is estimated probabilistically from the combination of catalog photometry and the observed spectrum using EAZY. We derive the uncertainties on the stellar population synthesis properties using a Monte Carlo simulation and examine the correlations between the parameters with and without the use of the spectrum in the modeling of the spectral energy distributions. The spectroscopic constraints confirm the extreme stellar masses of the galaxies in our sample. We find that three out of five galaxies are quiescent (star-formation rate of $\lesssim 1{M}_{\odot }\,{\mathrm{yr}}^{-1}$) with low levels of dust obscuration (${A}_{{\rm{V}}}\lt 1$) , that one galaxy displays both high levels of star formation and dust obscuration ($\mathrm{SFR}\approx 300{M}_{\odot }\,{\mathrm{yr}}^{-1}$, ${A}_{{\rm{V}}}\approx 1.7$ mag), and that the remaining galaxy has properties that are intermediate between the quiescent and star-forming populations.

Hierarchical Star Formation in Turbulent Media: Evidence from Young Star Clusters The Astrophysical Journal

K. Grasha, B. G. Elmegreen, D. Calzetti, A. Adamo, A. Aloisi, S. N. Bright, D. O. Cook, D. A. Dale, M. Fumagalli, J. S. Gallagher, D. A. Gouliermis, E. K. Grebel, L. Kahre, H. Kim, M. R. Krumholz, J. C. Lee, M. Messa, J. E. Ryon, and L. Ubeda

2017

We present an analysis of the positions and ages of young star clusters in eight local galaxies to investigate the connection between the age difference and separation of cluster pairs. We find that star clusters do not form uniformly but instead are distributed so that the age difference increases with the cluster pair separation to the 0.25–0.6 power, and that the maximum size over which star formation is physically correlated ranges from ~200 pc to ~1 kpc. The observed trends between age difference and separation suggest that cluster formation is hierarchical both in space and time: clusters that are close to each other are more similar in age than clusters born further apart. The temporal correlations between stellar aggregates have slopes that are consistent with predictions of turbulence acting as the primary driver of star formation. The velocity associated with the maximum size is proportional to the galaxy's shear, suggesting that the galactic environment influences the maximum size of the star-forming structures.

Star Formation at z = 2.481 in the Lensed Galaxy SDSS J1110 = 6459. I. Lens Modeling and Source Reconstruction* The Astrophysical Journal

Traci L. Johnson, Keren Sharon, Michael D. Gladders, Jane R. Rigby, Matthew B. Bayliss, Eva Wuyts, Katherine E. Whitaker, Michael Florian, and Katherine T. Murray

2017

Using the combined resolving power of the Hubble Space Telescope and gravitational lensing, we resolve star-forming structures in a $z\sim 2.5$ galaxy on scales much smaller than the usual kiloparsec diffraction limit of HST. SGAS J111020.0+645950.8 is a clumpy, star-forming galaxy lensed by the galaxy cluster SDSS J1110+6459 at $z=0.659$, with a total magnification $\sim 30\times$ across the entire arc. We use a hybrid parametric/non-parametric strong lensing mass model to compute the deflection and magnification of this giant arc, reconstruct the light distribution of the lensed galaxy in the source plane, and resolve the star formation into two dozen clumps. We develop a forward-modeling technique to model each clump in the source plane. We ray-trace the model to the image plane, convolve with the instrumental point-spread function (PSF), and compare with the GALFIT model of the clumps in the image plane, which decomposes clump structure from more extended emission. This technique has the advantage, over ray-tracing, of accounting for the asymmetric lensing shear of the galaxy in the image plane and the instrument PSF. At this resolution, we can begin to study star formation on a clump-by-clump basis, toward the goal of understanding feedback mechanisms and the buildup of exponential disks at high redshift.


text:
## Babyslapmafro 3 years ago Please help. How do you find the partial derivative of arctan(x/y) with respect to x?

1. hartnn

do you know the meaning of partial derivative ?

2. Babyslapmafro

yes

3. hartnn

then have you tried to find the partial derivative ? where r u stuck ?

4. hartnn

do you know derivative of inverse tan function ?

5. Babyslapmafro

$\frac{ 1 }{ x ^{2}+1 }$

6. hartnn

good! now since 'y' is constant, its like taking derivative of inverse tan (ax) {where 'a' =1/y} so, apply the chain rule (know it?) $$[\tan^{-1}{ax}]'= \dfrac{1}{1+(ax)^2}\dfrac{d}{dx}(ax)=....?$$

7. Babyslapmafro

$\frac{ 1 }{ 1+(\frac{ 1 }{ y ^{2} }+\frac{ 2 }{ y }x+x ^{2} )}$

8. hartnn

where does the expression 1/y^2 +2y/x +x^2 come from ???

9. Babyslapmafro

i replaced a with 1/y

10. hartnn

right, so, it was not (a+x)^2 , it was (ax)^2 so, the denominator would just be $$1+x^2/y^2$$ got this ?

11. hartnn

what about the numerator ?? d/dx (ax) =...?

12. Babyslapmafro

numerator = (1/y)

13. hartnn

no....in chain rule, we need to take derivative of the inner function (here ax or x/y) which is, as you correctly said, 1/y :)

14. hartnn

one last step of simplification : $$\dfrac{1/y}{1+x^2/y^2}=....?$$ multiply numerator and denominator by y^2 to simplify this :)

15. Babyslapmafro

y^2/(y^2+x^2)

16. hartnn

remember that the numerator was 1/y and y^2/y would just be 'y' so your final answer would look like this : $$\large \dfrac{y}{x^2+y^2}$$

17. Babyslapmafro

oh right ya, thanks

18. hartnn

welcome ^_^

Find more explanations on OpenStudy


text:
# Give a Balanced Equation for the Formation of the Following Oxide – Zinc Oxide from a Metal - Chemistry

Give a balanced equation for the formation of the following oxide –

Zinc oxide from a metal

#### Solution

$\ce{2Zn + O2->2ZnO}$

Concept: Classification of Oxides - Metal Oxides Are Basic
Is there an error in this question or solution?

#### APPEARS IN

Viraf J. Dalal Class 8 New Simplified Middle School Chemistry
Chapter 6 Chemical Reactions
Exercise | Q 10.2


text:
by xeon123
Tags: percentage
 P: 81 I'm trying to understand this problem. If a lollipop costed $6 and now it costs$8, how much percent the price grow? I think the answer is ((8-6)/8)*100=25% So I'm trying to apply this solution to the new problem. If a lollipop costed $1, and now it costs$2, how much percent the price grow? ((2-1)/2)*100=50%, but it should be 100%. Can anyone help me with this problems?
 HW Helper Thanks P: 4,676 I think you already know the answer: what must be in the denominator in order that your second example gives you 100%? RGV
 P: 81 In the denominator should be 1. But in the first equation it should be 8. This is the part that I don't get it. I'm putting in the denominator the current price.
HW Helper
Thanks
P: 4,676

 Mentor P: 21,067 The denominator should have the price we're calculating the percent increase/decrease for. For your two examples, the starting prices were $6 and$1, respectively, so those are the numbers you need to use in the denominator. On the other hand, if the price of something goes down from from $10 to$8, then the percent decrease is (10 - 8)/10 * 100 = 20 %. If the price happened to go back up by $2, then the percent increase would be (10 - 8)/8 * 100 = 25%. The reason we're getting a different number is we're using a different base or starting point.  P: 81 I wasn't understanding the problem because I got different percentage values. In the example of Mark44 the same price decreased 20% to pass from$10 to $8, and the increased 25% to pass from$8 to $10. Projecting my question with Mark 44 explanation, my confusion was in the different percentage values. I was thinking why a price decreases 20%, and it has to increase 25%, and not 20%, to get to the initial value? Now, I understand. P: 232 $\frac{new-old}{old}$(100)  Quote by xeon123 I wasn't understanding the problem because I got different percentage values. In the example of Mark44 the same price decreased 20% to pass from$10 to $8, and the increased 25% to pass from$8 to $10. Projecting my question with Mark 44 explanation, my confusion was in the different percentage values. I was thinking why a price decreases 20%, and it has to increase 25%, and not 20%, to get to the initial value? Because the amount you're multiplying the percent by is smaller. Math Emeritus Sci Advisor Thanks PF Gold P: 38,899  Quote by xeon123 In the denominator should be 1. But in the first equation it should be 8. This is the part that I don't get it. I'm putting in the denominator the current price. That's because the answer you gave to the first problem is wrong. If the lollypop increased from 6$ to 8$then it increased by$2 and that is 2/6= 1/3 of the original price. The price of the lollypop increased by 1/3 or 33 and 1/3 percent, NOT 25%.


text:
Alternate Representation for the Winding Number

Let $$c: I \rightarrow \mathbb{R}^2$$ be a closed curve which is parameterized by arc length. Further, let $$P$$ be the period of $$c$$ and $$Q \not\in \mathtt{trace}(c)$$.

The winding number of $$c$$ w.r.t. $$Q$$ is defined by $$W(c, Q) := \frac{1}{2\pi} (\vartheta(P) - \vartheta(0))$$

where $$\vartheta$$ is a lift of $$(c-Q)/\lvert c-Q \rvert$$.

Now, what I want to show is that $$W(c, Q) = -\frac{1}{2\pi} \int_0^P \frac{\langle c(t) - Q, N(t) \rangle}{\lvert c(t) - Q \rvert^2} dt.$$

Attempt: Let $$\kappa$$ be the curvature of $$(c-Q)/\lvert c-Q \rvert$$. Then we would have to show that $$\kappa$$ is the integrand multiplied by $$-1$$. This would prove the statement, since $$\kappa = \vartheta'$$.


text:
# Unit 15: CITIES

26 tháng 12 2020 lúc 22:10

Life in the countryside is quite different from that in the city. First, it is the atmosphere that differentiates the two living environments. In the countryside, the air and water are much fresher and safer, thus living there is totally comfortable and beneficial for our health. In contrast, people who live in big cities suffer from health problems and diseases related to poor air quality and sources of polluted water. Second, prices and costs of living are cheaper in the countryside than in big cities. Vegetables, fruits and meats are home-grown and home-produced. However, in cities, people have to go shopping for them. Third, the interpersional relationships are more harmonious and friendlier. Those who live in the same village, especially neighbours, for example, know each other very well. There is a close-knit cooperation between people in working and many other daily activities. On the contrary, negatively speaking, city people only live and know their own business. They hardly care about who their neighbours are and what they do. Sending a helping hand, as a result, is never seen among city people. In conclusion, although the city life holds many advantages that make life more convenient, I always prefer living in the countryside where the environment is nice, the costs of living is affordable, and the intimate relationships among people are really meaningful. Personally, If I had a choice between a villa in the city and a small house with garden surrounding, the latter would seem better to me.

Bình luận (1)
4 tháng 6 2020 lúc 22:56

1. The present which you gave me was very useful

2. The Vietnamese athletics participated in this event and played with great enthusiasm

Bình luận (0)
4 tháng 2 2020 lúc 9:07

À loại câu D do "was" là ở qk mà đg nói ở HT

Bình luận (0)
4 tháng 2 2020 lúc 9:14

T thây A.hợp lý nhất

Despite +N k có Despite of +N

Bình luận (0)
4 tháng 2 2020 lúc 9:20

Bình luận (0)
20 tháng 5 2017 lúc 20:10

Finish the second sentence so that it has the same meaning as the first.

1. Keeping the environment clean is very important.

-> It’s very important to keep the environment clean.

2. They will show the time machine to the public when they finish it.

-> The time machine will be shown to the public when it is finished.

3. He was delighted to receive his aunt’s letter.

-> He was delighted that his aunt sent him a letter.

4. Old car tires are recycled to make shoes and sandals.

-> People recycle old car - tires to make shoes and sandals.

5. I’m very sad that I wasn’t accepted in that group.

-> I’m very sad not to be accepted in that group.

6. Van keeps forgetting his homework.

-> Van is always forgetting his homework.

7. I would like you to help me to put the chairs away.

-> Do you mind helping me to put the chairs away?

8. Please don’t make any noise; I’m very tired.

-> I’d rather you didn’t make any noise; I’m very tired.

9. Having a vacation abroad is very interesting.

-> It is very interesting to have a vacation abroad.

10. The water was so cold that the children couldn’t swim in it.

-> The water wasn’t warm enough for the children to swim.

Bình luận (1)
20 tháng 5 2017 lúc 20:06

Give the correct form or tense of the verbs in the brackets.

1. My uncle (not/ be) hasn’t been out of his house since he (buy) bought a color TV.

2. I don’t think I (go) am going out tonight. I (be) am too tired.

3. John (wash) is washing his hands. He (just/ repair) has just repaired the TV set.

4. Tell him (bring) to bring his bike inside. If he (leave) leaves it there, someone (steal) will steal it.

5. Where (you/ spend) did you spend your summer holiday last year?

Bình luận (0)
20 tháng 5 2017 lúc 20:02

Give the correct form of the words in the brackets.

1. He fell off the bike, but his injuries were not serious. (injure)

2. She gave me a demonstration of the new house. (demonstrate)

3. The shop assistant showed me a lot of new models. (assist)

4. You must read the following safety precautions carefully. (safe)

5. You shouldn’t talk about her character only through her appearance (appear)

6. We are waiting for the arrival of our uncle’s train. (arrive)

7. The increase in population has led to overcrowding in many cities. (crowd)

8. We had a terrible trip because the weather was unpleasant (please)

10. It’s a good restaurant; the food there is very tasty (taste)

Bình luận (0)
16 tháng 3 2018 lúc 12:40

Stamps / a child/ minh / since/ collected / was / has/ he

=>Minh has collected stamps since he was a child.

Bình luận (0)
16 tháng 3 2018 lúc 12:42

Stamps / a child/ minh / since/ collected / was / has/ he

=> He has collected stamps since he was a child.

Bình luận (0)
16 tháng 3 2018 lúc 12:45

minh has collected stamps since he was a child

Bình luận (0)
20 tháng 5 2017 lúc 20:13

Use the given words to make complete sentences.

1. Linh/ parents/ proud/ him/ because/ he/ always/ get/ good marks.

$\Rightarrow$ Linh’s parents are very proud of him because he always gets good marks.

2. We/ very interested/ play/ soccer/ when/ live/ countryside.

$\Rightarrow$We were very interested in playing soccer when we lived in the countryside.

3. I/ not talk/ uncle/ since/ he/ buy/ new house/ city center.

$\Rightarrow$ I haven’t talked to my uncle since he bought a new house in the city center.

4. The Browns/ buy/ lot/ food/ because/ they/ go/ have/ party.

$\Rightarrow$ The Browns have bought a lot of food because they are going to have a party.

5. It/ only/ small car/ so/ there/ not/ enough room/ all/ us.

$\Rightarrow$ It is only a small car, so there isn’t enough room for all of us.

Bình luận (0)


text:
# Showing $H\unlhd G$ when $[G\colon H]$ is not the smallest prime dividing $|G|$

I recently read about the theorem that for a finite group $G$, if $p$ is the least prime dividing $|G|$, then any subgroup $H$ with $[G\colon H]=p$ is normal in $G$.

Going over some exercises, this theorem made it easy to see that a subgroup $H$, with $|H|=35$, is normal in $G$ when $|G|=70$.

A similar situation wasn't as obvious. If $|H|=35$, but $|G|=140$, then $H$ has index $4$. How then would one go about proving $H\unlhd G$? Thanks.

-
Really, the theorem you quote is a very special case. In "most" circumstances, you aren't nearly so lucky as to have a subgroup of prime index, let alone a subgroup whose prime index is the smallest prime dividing $|G|$. So proving that a subgroup is normal from purely numerical considerations is usually difficult, if not impossible. (E.g., it would be impossible to prove that a subgroup of order $12$ in a group of order $60$ is normal). –  Arturo Magidin Sep 1 '11 at 23:52
@Arturo, ah ok, that's too bad. I thought the approach may be similar. Is there another way to prove $H\unlhd G$ in this case using a different method? –  yunone Sep 1 '11 at 23:53
When one can do it, it is usually through the application of the Sylow theorems or Hall's generalization. –  Arturo Magidin Sep 1 '11 at 23:59

A $7$-Sylow subgroup of $G$ has order $7$; the number of $7$-Sylow subgroups must be congruent to $1$ modulo $7$, and must divide $140 = 7\times 5\times 4$. Thus, it must divide $20$ and be congruent to $1$ modulo $7$; the only possibility is that the number of $7$-Sylow subgroups of $G$ is one (hence it is normal; in fact, characteristic) Thus, the subgroup $H$ of order $35$ must contain the unique $7$-Sylow subgroup of $G$.

Likewise, a $5$-Sylow subgroup of $G$ has order $5$, and the number of $5$-Sylow subgroups of $G$ must be congruent to $1$ modulo $5$ and divide $140$, hence must divide $7\times 4$. The only possibility is that there is a unique $5$-Sylow subgroup of $G$ (which is therefore characteristic, and thus normal), which must also be contained in $H$.

Thus, $H$ must be product of the unique $7$-Sylow subgroup and the unique $5$-Sylow subgroup of $G$. Since $H$ is the product of characteristic subgroups, it is in fact characteristic and therefore normal in $G$.

(Note that any group of order 35 is necessarily cyclic, by the by).

-
Thanks Arturo. I just had two questions. Why does $H$ necessarily contain both the Sylow subgroups? And why does a $p$-Sylow subgroup being the unique one of its order imply it is characteristic? –  yunone Sep 2 '11 at 1:18
@yunone: Since $H$ is of order $35$, it contains a subgroup of (of itself) of order $5$. This subgroup is also a subgroup of $G$, and is thus a $5$-Sylow subgroup of $G$ (it has the correct number of elements). But since $G$ has only one $5$-Sylow subgroup, then this subgroup of $H$ is the $5$-Sylow subgroup of $G$. Similar argument works for the $7$-Sylow subgroup. For your second question: if $K$ is a subgroup of $G$, and $K$ is the only subgroup of $G$ of order $|K|$, then $K$ is characteristic: for any $\phi\in\mathrm{Aut}(G)$, $\phi(K)$ is a subgroup of $G$ of order $|K|$. –  Arturo Magidin Sep 2 '11 at 2:24
Oh whoops, I should have known that. Thanks for your help. –  yunone Sep 2 '11 at 2:45

Since $[G:H]=4$, there is a homomorphism $\phi \colon G\rightarrow S_4$ with $ker(\phi)\subseteq H$. (See "Generalized Cayley's theorem- Introduction to theory of groups : Rotman )

If $ker(\phi)\neq H$, then it will be a subgroup (of $H$) of order $1,5$ or $7$.

• If $|Ker(\phi)|=7$ then $|Im(\phi)|=|G/ker(\phi)|=140/7=20$; but $S_4$ has no subgroup of order 20.

• If $|Ker(\phi)|\leq 5$ then $|Im(\phi)|=|G/ker(\phi)|\geq 140/5>24$, contradiction.

Therefore $ker(\phi)=H$ which is normal in $G$.

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Thanks group, I had been meaning to browse through Rotman's book soon. –  yunone Sep 3 '11 at 3:07

I think I may have thought up an answer which doesn't use a whole lot of group theory. If anyone reads this, I'd appreciate feedback on whether it's right or wrong.

Let $H$ act on $G/H$ by $h\cdot gH=hgH$. Since $[G:H]=4$, the orbit of any $gH\in G/H$ is at most $4$. But the order of any orbit must divide $|H|$, so must be $1$, $5$, $7$, or $35$. So each orbit is of order 1, and thus trivial since $egH=gH$ in particular. So $hgH=gH$ for any $g\in G$, $h\in H$, hence $g^{-1}hgH=g^{-1}gH=H$. Then $g^{-1}hg\in H$, so $H\unlhd G$?

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You cannot go from $g^{-1}hgH$ to $g^{-1}gH$ in general, though you can go, here, to $H$ (since $hgH = gH$ is already known). $g^{-1}hgH = H$ suffices to show $g^{-1}hg\in H$. –  Arturo Magidin Sep 8 '11 at 13:20
Thanks for the feedback. –  yunone Sep 8 '11 at 14:07
@yunone: in other words, your last string of equalities might confuse the grader. Maybe better, "so dividing on the left by $g$, we get $g^{-1}hgH = g^{-1}gH$, but the right hand side is simply $H$. Hence $g^{-1}hg \in H$ and $H \unlhd G$." –  Jack Schmidt Sep 8 '11 at 17:04
@Jack, ah ok, I was mostly just concerned on whether it was right or wrong, since I'm just learning. Thanks for pointing out what may be confusing to someone else, as the writer, I might not notice. –  yunone Sep 8 '11 at 22:02


text:
# Calculating with quantum numbers and shape of nodal shells

I'm attending in a chemistry course now, and in the beginning we scratch the surface of the quantum physics, which have led me to some problems. Mostly I understand all what we get until now, but there some blind spots, where I don't understand some things. Now I have two major problems, one with some calculations with quantum numbers, and now with the nodal shells and I hope somebody could explain these to me (who have never learned this part of physics before).

So the first problem was with some exercising. I've understood the questions and had the approximate way to solve them, but the solution by the teacher was a little bit surprising, and it left me in doubt and confused. Here is the tasks and the solution by the teacher:

I've thought that every orbital could take one electron, so in the (a) I've "calculated" 3 of them, but there is 6 in her solution. Before that she doesn't told us about how many electron could be on one orbit, but she said that in the (b), so okay... there are 2 electrons on each orbit.. in this case at least (hydrogen atom).. For the (b) I've put again my bad 3 electrons because of the lack of information from earlier. But if there was a surprise in (a), I was (and still) totally confused by the solution of (c).. where comes the 5??

Okay, next one. This time there is no calculation, but some basic knowledge (lack of mine). It's about the orbit's naming and it's "transformation". So firstly we checked the s orbital (1s, 2s, etc) which has a sphere shaped "appearance". Second the p orbital, where are 3 orientations (-1,0,1) and on the illustration, the signed areas were laid on the axes. But in the next page on the d orbital, she said one of the fig.s (which one I highlighted with green) is not correct, because the orbital areas should have been in between the axes, not on them.

And it caused some problem to me again. Why the orbital areas are between the axes when one orbital earlier we(/she) put them on the axes without problem? Furthermore there is an other orbital version which is the $d_{x^2-y^2}$ and she also doesn't tell why this is looks like that (areas placed on the axes).

I know these are the very basics and I would like to understand every aspects of this topic (partly because it is very interesting, but mainly because I don't want to fall behind because of "some" misunderstanding), so I hope and I would be very grateful if somebody could make me some explanation of these, or just give me a reliable site or book name where these topics are explained well (for someone like me, who just started this first time).

For the first part: Each orbital can hold two electrons with opposite spin states (this is the exclusion principle). That is why the answers to (a) and (b) are doubled. However, in case (c) the $m_s=1/2$ condition demands that the electron(s) be only spin-up (i.e. no spin-down states) so you don't double them.
Regarding the figures: the wavefunction includes a factor that's shown in the subscript. So for $d_{xy}$ the wavefunction is proportional to $xy$. Therefore, along, say, the $x$ axis, $y=0$ and therefore the wavefunction is 0. Likewise, for $d_{x^2-y^2}$, the wavefunction is zero when $|x|=|y|$ (i.e. the two diagonals).


text:
In this chapter you will learn to find numbers that make certain statements true. A statement about an unknown number is called an equation. When we work to find out which number will make the equation true, we say we solve the equation. The number that makes the equation true is called the solution of the equation.

## Setting up equations

An equation is a mathematical sentence that is true for some numbers but false for other numbers. The following are examples of equations:

$x + 3 = 11 \text{ and } 2^x = 8$

$$x + 3 = 11$$ is true if $$x =8$$, but false if $$x =3$$.

When we look for a number or numbers that make an equation true we say that we are solving the equation. For example , $$x = 4$$ is the solution of $$2x + 8$$ because it makes $$2x +8$$true. (Check: $$2 \times 4 = 8$$

### Looking for numbers to make statements true

1. $$x - 3 = 0$$, if $$x = -3$$
2. $$x^3 = 8$$, if $$x = -2$$
3. $$3x = -6$$, if $$x = -3$$
4. $$3x = 1$$, if $$x = 1$$
5. $$6x + 5 = 47$$, if $$x = 7$$
2. Find the original number. Show your reasoning.
1. A number multiplied by 10 is 80.
3. Divide a number by 5 and the answer is 4.
4. Multiply a number by 4 and the answer is 20.
5. Twice a number is 100.
6. A number raised to the power 5 is 32.
7. A number raised to the power 4 is -81.
8. Fifteen times a number is 90.
9. 93 added to a number is -3.
10. Half a number is 15.
3. Write the equations below in words using "a number" in place of the letter symbol $$x$$. Then write what you think "the number" is in each case.

Example: $$4 +x = 23$$. Four plus a number equals twenty-three. The number is 19.

1. $$8x = 72$$
2. $$\frac{2x}{5} = 2$$

3. $$2x+ 5 = 21$$

4. $$12 + 9x = 30$$

5. $$30 - 2x = 40$$

6. $$5x + 4 = 3x+ 10$$

## Solving equations by inspection

### The answer is in plain sight

1. Seven equations are given below the table. Use the table to find out for which of the given values of $$x$$ it will be true that the left-hand side of the equation is equal to the right-hand side.

You can read the solutions of an equation from a table.

 $$x$$ -3 -2 -1 0 1 2 3 4 $$2x + 3$$ -3 -1 1 3 5 7 9 11 $$x + 4$$ 1 2 3 4 5 6 7 8 $$9 -x$$ 12 11 10 9 8 7 6 5 $$3x -2$$ -11 -8 -5 -2 1 4 7 10 $$10x -7$$ -37 -27 -17 -7 3 13 23 33 $$5x +3$$ -12 -7 -2 3 8 13 18 23 $$10 - 3x$$ 19 16 13 10 7 4 1 -2
1. $$2x + 3 = 5x + 3$$
2. $$5x + 3 = 9 - x$$
3. $$2x + 3 = x+4$$
4. $$10x -7 = 5x + 3$$
5. $$3x -2 = x+4$$
6. $$9 -x = 2x + 3$$
7. $$10 -3x = 3x - 2$$

Two or more equations can have the same solution. For example, $$5x = 10$$ and $$x + 2 = 4$$ have the same solution; $$x = 2$$ is the solution for both equations.

Two equations are called equivalent if they have the same solution.

1. Which of the equations in question 1 have the same solutions? Explain.
2. Complete the table below. Then answer the questions that follow.

You can also do a search by narrowing down the possible solution to an equation.

 $$x$$ 0 5 10 15 20 25 30 35 40 $$2x + 3$$ $$3x -10$$
1. Can you find a solution for $$2x + 3 = 3x - 10$$ in the table?
2. What happens to the values of $$2x + 3$$ and $$3x - 10$$ as $$x$$ increases? Do they become bigger or smaller?
3. Is there a point where the value of $$3x - 10$$ becomes bigger or smaller than the value of $$2x + 3$$ as the value of $$x$$ increases? If so, between which $$x$$-values does this happen?

This point where the two expressions are equal is called the break-even point.

4. Now that you narrowed down where the possible solution can be, try other possible values for $$x$$ until you find out for what value of $$x$$ the statement $$2x + 3 = 3x - 10$$ is true.

"Searching" for the solution of an equation by using tables or by narrowing down to the possible solution is called solution by inspection.

## More examples

### Looking for and checking solutions

1. What is the solution for the equations below?
1. $$x - 3 = 4$$
2. $$x + 2 = 9$$
3. $$3x = 21$$
4. $$3x + 1 = 22$$

When a certain number is the solution of an equation we say that the number satisfies the equation. For example, $$x=4$$ satisfies the equation $$3x=12$$ because $$3 \times 4 = 12$$.

1. Choose the number in brackets that satisfies the equation. Explain your choice.
1. $$12x = 84$$ {5; 7; 10; 12}
2. $$\frac{84}{x} = 12$$ {-7; 0; 7; 10}

3. $$48 = 8k + 8$$ {-5; 0; 5; 10}

4. $$19 - 8m = 3$$ {-2; -1; 0; 1; 2}

5. $$20 = 6y - 4$$ {3; 4; 5; 6}

6. $$x^3 = -64$$ {-8; -4; 4; 8}

7. $$5^x =125$$ {-3; -1; 1; 3}

8. $$2^x = 8$$ {1; 2; 3; 4}

9. $$x^2 = 9$$ {1; 2; 3; 4}

1. $$m + 8 = 100$$
2. $$80 = x + 60$$
3. $$26 - k = 0$$
4. $$105 \times y = 0$$
5. $$k \times 10 = 10$$
6. $$5x = 100$$
7. $$\frac{15}{t} = 5$$
8. $$3 = \frac{t}{5}$$
1. $$12x + 14 = 50$$
2. $$100 = 15m + 25$$
3. $$\frac{100}{x} =20$$
4. $$7m + 5 = 40$$
5. $$2x + 8 = 10$$
6. $$3x + 10 = 31$$
7. $$-1 + 2x = -11$$
8. $$2 + \frac{x}{7} = 5$$
9. $$100 = 64 + 9x$$
10. $$\frac{2x}{6}= 4$$


text:
# Misunderstanding of Sylow theory

I think I have a misunderstanding a part of Sylow theory for groups or I have made a big mistake in my reasoning below.

We have the following lemma in Sylow theory:

Let $$G$$ be a finite group and let $$P$$ be a Sylow-$$p$$ subgroup of $$G$$. If $$g \in G$$ such that the order of $$g$$ is $$p^k$$ for some $$k$$, then $$g \in P$$.

Now consider $$S_5$$, it has order $$120 = 2^3 \times 3 \times 5$$. Take a Sylow-$$2$$ subgroup $$H$$ of $$S_5$$. Sylow theory tells us that $$|H| = 2^3 = 8$$. Now any $$4$$-cycle in $$S_5$$ has order $$4 = 2^2$$. So any $$4$$-cycle in $$S_5$$ must be contained in $$H$$. But the number of $$4$$-cycles in $$S_5$$ is $$\frac{5!}{(5-4)!4} = \frac{120}{4} = 30$$, so $$S_5$$ has at least $$30$$ elements of order $$4$$ all of which must be contained in $$H$$ which has order $$8$$, an obvious contradiction.

What have I done wrong here?

• The lemma is just not right. One thing that would be true is that there exists $h\in G$ such that $g^h\in P$. Are you sure you have all the hypotheses of the Lemma down? – Arturo Magidin Apr 19 at 19:21
• The lemma is just false, you cannot expect any $p$-Sylow subgroup to contain all $p$-primary elements. – Captain Lama Apr 19 at 19:21
• Why do you think your Sylow p-subgroup $P$ is unique? – Anton Zagrivin Apr 19 at 19:25

The lemma you stated is false. It should be like this: if $$g\in G$$ has order $$p^k$$ for some $$k\in\mathbb{N}$$ then there exists a $$p$$-Sylow subgroup $$P\leq G$$ such that $$g\in P$$. So it doesn't say that such an element $$g$$ must be in all $$p$$-Sylow subgroups of $$G$$. The version that you wrote is true if there is only one $$p$$-Sylow subgroup.


text:
## Introduction

Most of the biomedical data are rectangular in shape, including those of ‘omics, large cohorts, population studies, and surveys. Few missing data were present in these datasets. An increasing number of human genomic and survey data have been produced in recent years [1]. Rigorous analyses on these large datasets demand considerably more efficient and more accurate algorithms, which are poorly understood.

Machine learning (ML) algorithms are aimed to produce a model that can be used to perform classification, prediction, estimation, or any other similar task [2, 3]. The unknown dependencies/associations are estimated based on a given dataset and later can be used to predict the output of a new system or dataset [2,3,4]. Therefore, ML algorithms have been used to analyze large biomedical datasets [5,6,7]. Studies have compared the accuracy of several ML algorithms for classifying microarray or genomic data [8,9,10], and show a superior performance of random forests (RF). However, only a few studies accessed the accuracy of ML algorithms in classifying multi-category outcomes, which are important for the more-in-depth understanding of biological and clinical processes. Only when we can accurately predict or identify the factors linked to multicategory outcomes, we can provide patients with more targeted prevention and treatment for the most likely outcome. Hence, we aimed to understand the performance and efficiency of RF, decision tree (DT), artificial neural networks (ANN), and support vector machine (SVM) algorithms in classifying multi-category outcomes of rectangular-shaped biomedical datasets.

As an example, we used a large population-based breast cancer dataset with long-term follow-up data to create a large rectangular dataset. The reasons were: (1) cancer is one of the most common diseases [11]. Knowledge gained by studying cancer can be easily generalized to other biomedical fields; (2) breast cancer is the second most common cancer in U.S. women [11, 12] and would provide a sufficiently large sample size (n = 45,085); (3) breast cancers diagnosed in 2004 had 12 years of follow-up on average and very few missing outcomes/follow-ups were anticipated; (4) breast cancers diagnosed in 2004 would have a moderately good prognosis and their outcomes will be rather diversified (i.e., many patients might not die of breast cancer). Therefore, this study is designed to systematically compare the performance and efficiency of DT, RF, SVM, and ANN algorithms in classifying multicategory causes of death (COD) in a large biomedical dataset (breast cancers).

## Methods

### Data analysis

We obtained individual-level data from the surveillance, epidemiology, and end results-18 (SEER-18) (www.seer.cancer.gov) SEER*Stat database with treatment data using SEER*Stat software (Surveillance Research Program, National Cancer Institute SEER*Stat software (seer.cancer.gov/seerstat) version <8.3.6>) as we did before [13,14,15]. SEER-18 is the largest SEER database including cases from 18 states and covering near 30% of the U.S. population. The SEER data were de-identified and publicly available. Therefore, this was an exempt study (category 4) and did not require an institutional review board (IRB) review. All incidental invasive breast cancers of SEER-18 diagnosed in 2004 were included and had the follow-up to December 2016. The individual deaths were verified via certified death certificates (2019 data-release). We chose the diagnosis year of 2004 with the consideration of the implementation of the sixth edition of the tumor, node, and metastasis staging manual (TNM6) of the American joint commission on cancer (AJCC) in 2004. Moreover, we only included the primary-cancer cases that had a survival time >1 month, age of 20+ years, and known COD.

The features (i.e., variables) were dichotomized for more efficiency and slightly better performance [7], while the actual values were also tuned for and analyzed using RF models. A total of 54 features were included (Supplementary Table 1). We conducted correlation analyses and produced a correlation matrix (Supplementary Fig. 1) to identify the closely correlated factors. The outcomes of the classification models were the patient’s five-category COD. The COD was originally classified using SEER’s recodes of the causes of death, which were collected through death certificates of deceased patients (https://seer.cancer.gov/). We simplified the SEER COD into five categories based on the prevalence of COD [16,17,18], including alive, non-breast cancer, breast cancer, cardiovascular disease (CVS), and other cause.

The most common task of ML techniques in the learning process is classification [19]. The tenfold cross-validation approach was used to tune all models, which is also termed as model optimization (see Supplementary methods) as described before [20, 21]. This approach is an approximation of leave-p-out cross-validation has the advantage of repeat the process to reduce variance and being less sensitive to the partitioning of the data than the holdout method. Briefly, the samples were randomly divided into ten same-size subsamples, among which one subsample remained as the validation set and the other nine as the test set. The (cross-)validation would be performed ten times as the test subsample was shuffled among the ten subsamples. The mean of the cross-validation’s performances would be calculated and used as the performance of the model. Multinomial logistic regression (MLR) and the ML analyses were carried out using MATLAB (version 2018a, MathWorks, Natick, MA).

### Model tuning

The detailed model tuning process is described in Supplementary material. Several DT methods, such as CHAID, CART, and exclusive CHAID are available with MATLAB [22]. We used classification and regression tree(CART) to predict the categories, using the Gini index as the split criterion and 100 iterations for each run (Supplementary Fig. 2). There are no default RF packages/toolboxes in MATLAB’s own toolbox. We thus used the Randomforest-Matlab open-source toolbox developed by Jaiantilal et al. [23, 24].

In tuning RF models, the parameter nTrees, which was to set how many trees in a random forest, and may have an impact on the classification results. We set the value of nTrees from 1 to 600 separately, the results show that if this parameter is not too few (greater than ten), the accuracy of recognition can reach 69–70% (Supplementary Fig. 3). Therefore, we set this parameter to 136 in RF-based analyses. The value of Mtry node for the best model performance was identified by setting the parameter value from 1 to 20 with 1 as the interval. We found 5 was the value, which was indeed consistent with the default value generated using the MATLAB model’s default (i.e., Mtry = floor(sqrt(size(P_train,2)))).

Among different training algorithms of ANN, we used the Trainscg algorithm because it is the only conjugate gradient method that did not require a linear search. The number of input layer nodes is 54 and the number of output layer nodes is 6. To tune the ANN model, we conducted experiments of either single hidden or double hidden layers, with the node numbers ranged from 5 to 100. According to the tuning results of accuracy and mean squared error (MSE), we would set the model to double hidden layers, and the number of layers for the highest accuracy and lowest MSE.

We used the multi-class error-correcting output codes (ECOC) model the SVM modeling which allows classification in more than two classes; and the MATLAB fitcecoc function that creates and adjusts the template for SVM [25]. The Kernel functions considered in the SVM were: linear, radial basis function, Gaussian, and polynomial.

### Performance analysis

We analyzed the performance metrics of each proposed model, including accuracy, recall, precision, F1 score, and specificity [26, 27]. They were defined as follows:

$$\mathrm{Accuracy} = \frac{{\mathrm{TP + TN}}}{{\mathrm{All}}}$$
(1)
$$\mathrm{Precision} = \frac{{\mathrm{TP}}}{{\mathrm{TP + FP}}}$$
(2)
$$\mathrm{Recall} = \frac{{\mathrm{TP}}}{{\mathrm{TP + FN}}}$$
(3)
$$F1 = \frac{{2 \ast \mathrm{Precision} \ast \mathrm{Recall}}}{{\mathrm{Precision + Recall}}}$$
(4)

True positive (TP) and true negative (TN) were defined as the number of samples that are classified correctly. False positive (FP) and false negative (FN) were defined as the number of samples that are misclassified into the other mutational classes [26, 27]. The specificity or true negative rate (TNR) is defined as the percentage of mutations that are correctly identified:

$$\mathrm{Specificity} = \mathrm{TNR} = \frac{{\mathrm{TN}}}{{\mathrm{TN + FP}}}$$
(5)

The receiver operating curve (ROC) is a graph where recall is plotted as a function of 1-specificity. It can more objectively measure the performance of the model itself [25]. The model performance was also evaluated using the area under the ROC, which is denoted the area under the curve (AUC). An AUC value close to 1 highlight a high-performance model, while an AUC value close to 0 demonstrate a low-performance model [28, 29]. AUC is independent of the class prior distribution, class misclassification cost, and classification threshold, which can more stably reflect the model’s ability to sort samples and characterize the overall performance of the classification algorithm. The formula used to determine the AUC can be written as follows [29, 30]:

$$\mathrm{AUC} = \frac{{{\sum} {\mathrm{TP} + {\sum} {\mathrm{TN}} } }}{{P + N}}$$
(6)

where P is the total number of positive class and N is the total number of negative class.

### Dimension reduction based on the information entropy and information gain

Information entropy is an indicator to measure the purity of the sample set. The formula is as follows:

$$H(S) = - \mathop {\sum}\limits_{i = 1}^n {p_i\log } p_i$$
(7)

The measure of information gain is to see how much information a feature can bring to the classification system. The more information it brings, the more important the feature. Information gain IG(A) is the compute of the difference in entropy from start to end the set S is split on an attribute A, the information gain is defined as follows [31]:

$${\mathrm{IG}}(A,S) = H(S) - \mathop {\sum}\limits_{t \in T} {p(t)H(t)}$$
(8)

where H(S) is the entropy of set S, T is the subsets created from splitting set S by attribute A such that $$S = U_{t \in T}t$$, p(t) is the proportion of the number of elements in t to the number of elements inset S, and H(t) is the entropy of subset T.

The information gain of a feature can indicate how much information it brings to the classification system and can be used as a feature weight. When the model uses more features the classification time will be longer. Arbitrarily reducing the characteristics will likely reduce classification accuracy. Therefore, we screened the features based on the calculated information gains to achieve a balance between run time and classification accuracy. We then step-wise deleted the features of the least information of gain, which were likely the least important. Because DT and RF were both ensemble-based algorithms and had similar performances, we conducted dimension reduction with RF, ANN, and SVM models and expect similar results with DT models.

## Results

### Dataset characteristics and the model tuning

Of the 52,818 samples, we step-wise excluded the 5294 (~10%) cases missing tumor-grade data (_grade), 1770 (3.4%) missing survival-time and 352 (0.6%) missing laterality data (lat_bi), and 317 (0.6%) missing data of TNM6 N category. Overall, we included 45,085 cases of breast cancer diagnosed in 2004 that were included in the SEER-18 and had no missing data in all features (Table 1). The variables were dichotomized into 54 features. Likely due to the unique nature of the pathological and socioeconomic factors, the correlation analyses showed only a few features had a correlation coefficient >0.95 or <−0.95 (Supplementary Fig. 1).

For DT models, the minimum number of leaf nodes with the best performance of the DT range 20–300 and peak at 93 (Supplementary Fig. 2). Therefore, the minimum number of samples contained in the leaf node was 93 for optimization. The overall classification accuracy could reach 67.3%, which was about 4% higher than the original DT, and the cross-validation error has also decreased. However, due to the uneven distribution of the data samples, some categories such as non-breast cancer and other cause were pruned, causing the loss of some data information.

For RF models, the parameter nTrees set how many trees in a RF and may have an impact on the classification results. The Mtry is the number of features randomly sampled as candidates at each split and was optimized (Supplementary methods). After tuning the models, we set the nTrees parameter to 136 in RF-based analyses with the best Mtry node value of 6 (Supplementary Fig. 3).

According to the tuning results of accuracy and MSE in ANN models, when the number of layers was greater than 20, the models’ performance appeared stabilized (Table 2). Therefore, we set the model to double hidden layers, and the number of layers is 40.

For the SVM models of linear, radial basis function, Gaussian or polynomial function, we found the linear kernel function had the highest accuracy (69.06%) and shortest run-time (175.50 s, Table 3), with a one-vs-one approach.

### Performance analysis results

Based on the confusion matrices (Fig. 1), the 4 ML models appeared to have similar performance (Table 4) and all were more accurate than the MLR. The best classification accuracy of DT, RF, ANN, and SVM models in this study was 69.21%, 70.23%, 70.16%, and 69.06%, respectively, and higher than 68.12% of a conventional statistical algorithm (MLR). However, to evaluate the pros and cons of a model, it is not enough to just look at the accuracy. The values of recall, TNR, and F1 can specifically reflect the classification of each category. Further comparing the metrics among the four models show that the precision-recall and F1 in classifying breast cancer by RF model were better than the DT, ANN, and SVM models. This is mainly because the RF model is an integrated learning algorithm; through the voting mechanism, one can balance a certain error. Compared with ML algorithms, MLR had lower overall accuracy and a much lower recall (1.13% vs ~50%) and a lower specificity (Table 4) for predicting the death of breast cancer.

### ROC analysis results

After a comparative analysis of the above performance metrics, we found that the RF model was superior to the DT, ANN, and SVM models. However, in the classification of other causes, the ANN model has a higher recognition rate than the RF model, and the F1 values of non-breast cancer cannot be calculated. Therefore, we used the ROC curve to further analyze non-breast cancer and other cause in RF and ANN models.

The AUC of the 5 COD in the RF model was overall lower than those in the ANN model (Fig. 2). Because the AUC index can measure the performance of the model more objectively, the results show that the overall performance of the ANN model is similar or better than that of the RF model.

### Dimension reduction based on the information entropy and information gain

This dataset had 54 categorical/binary features. Based on the training datasets, the information entropy and information gain of 54 features were calculated (Supplementary Table 3). Using information entropy and information gain, we obtained the following important features: age >65; surgery; TNM6 metastasis subgroup1; AJCC stage 4; TNM6 metastasis subgroup2; surgery–other; TNM6 tumor subgroup1; TNM6 tumor subgroup2; AJCC stage 1; surgery-lumpectomy; TNM6 lymph node subgroup5; and TNM6 lymph node subgroup1. Then we characterized the key features in a step-wise fashion (Supplementary Fig. 4).

We successfully reduced the data dimension based on information gain and shortened the run times in RF, ANN, and SVM models, while maintaining the overall classification accuracy (Table 5 and Supplementary Tables 35). Removal of features with low information gain (0.0000–0.0005) in RF models led to a slight increase in the alive class and the overall accuracy rates, while no accuracy changes in CVS and breast cancer classes. The classification of CVS was always 100%; accuracy rate of alive class and the overall had a slight improvement, respectively about 0.5% and 0.3%; breast cancer and CVS had no significant changes; non-breast cancer accuracy was slightly reduced, and the classification effect is unstable. Therefore, the features with low information gain (0.0000–0.0005) may be considered as redundant features, and deleted in the models, while the running times were scientifically reduced. We also found similar changes in ANN and SVM models.

### Feature importance in RF models using the data of convention encoding or one-hot encoding

Our previous works have shown that one-hot encoding (dichotomization of features) led to a slight increase in prediction accuracy of the RF model on prostate cancer using Stata [7]. Consistent with that finding, our tuned RF model on actual values had a prediction accuracy of 69.70%, which was slightly lower than the accuracy of 70.23% on dichotomized features (Supplementary Table 6). However, the top-five important features were different in the two models, except age 65+ years (vs <65 years, Supplementary Figs. 5 and 6).

## Discussion

We here compared the performance and efficacy of DT, RF, ANN, and SVM in classifying five-category outcomes of a large rectangular database (54 features and ~45,000 samples). The accuracy in classifying five-category COD with DT, RF, ANN, and SVM was 69.21%, 70.23%, 70.16%, and 69.06%, respectively. It is noteworthy that the accuracy in classifying five-category outcomes is much more difficult than classifying binary-category outcomes since it depends on four sequential classification processes in one-over-all or one-over-rest algorithms (i.e., final accuracy is the accuracy multiplication for four times). Moreover, based on the information entropy and information gain of feature values, we could reduce the feature number in a model to 32 and maintained a similar accuracy, while the running time decreased from 55.57 s for 54 features to 25.99 s for 32 features in RF, from 12.92 s to 10.48 s in ANN and from 175.50 s to 67.81 s in SVM. The DT algorithm was not tested after dimension reduction for its lower performance than RF and its theoretical framework (ensemble-based) like RF.

Few studies to our knowledge investigated the dimension reduction of multicategory classification. A two-stage approach has been shown to effectively and efficiently select features for balanced datasets [32], but no specific reduction in run time was reported. We here show that dimension reduction and efficiency improvement can be achieved by removing features of low to medium information gain (<0.0005) in RF, ANN, and SVM models, which apparently have little effect on the overall classification performance. Such a strategy may be applied to other ML models in classifying unbalanced large rectangular datasets, while caution should be used when classifying outcomes in a balanced dataset.

Consistent with a previous report [7], one-hot encoding (i.e., dichotomization of the features) produced slightly better prediction accuracy in our data than conventional encoding and also has the advantage of no need for normalization. The top-five important features in the models based on the dichotomized and actual-value features differed considerably except age. This finding further confirms the important role of age in modeling five-category COD. However, five of the top ten important features shared in these two models, including advanced age, pathologic staging, surgery status, N category, and histology type. These features thus should be carefully examined, recorded, and considered for predicting or preventing five-category COD in breast cancer patients.

The four included ML algorithms each have their own theoretical frameworks. DT is a logical-based ML approach [33]. The structure of the DT is similar to a flowchart. Using top-down recursion, the classification tree produces the category output. Starting from the root node of the tree, test and compare property values on its internal node, then determine the corresponding branch, and finally reach a conclusion in the leaf node of the DT. This process is repeated at each node of the tree by selecting the optimal splitting features until the cut-off value is reached [34]. A leafy tree tends to overtrain, and its test accuracy is often far less than its training accuracy. By contrast, a shallow tree can be more robust and be easy to interpret [35].

DT works by learning simple decision rules extracted from the data features. But RF is a combination of tree predictors such that each tree depends on the values of a random vector sampled independently, and with the same distribution for all trees in the forest [36]. When RF used for a classification algorithm, the deeper the tree is, the more complex the decision rules and the fitter the model [36, 37]. The generalization error of a forest of tree classifiers depends on the strength of the individual trees in the forest and the correlation between them. Random decision forests overcome the problem of overfitting of the DTs and are more robust with respect to noise.

ANN technique is one of the artificial intelligence tools. It is a mathematical model that imitates the behavior characteristics of animal neural networks [38]. This kind of network carries on distributed parallel information processing by adjusting the connection between a large number of internal nodes, so as to achieve the purpose of processing information. After repeated learning and training, network parameters corresponding to the minimum error are determined, and the ANN model classifies the output automatically from the dataset.

SVM is another popular ML tool based on statistical learning theory, which was first proposed by Vapnik and his colleagues [39, 40]. Unlike traditional learning methods, SVMs are approximate implementations of structural risk minimization methods. The input vector is mapped to a high-dimensional feature space through some kind of non-linear mapping which was selected in advance. An optimal classification hyperplane is constructed in this feature space, to maximize the separation boundary between the positive and negative examples [39, 40]. Support vectors are the data points closest to the decision plane, and they determine the location of the optimal classification hyperplane.

The 4 ML algorithms had different strengths and weaknesses, while all outperformed conventional statistical algorithm (MLR). The RF algorithm in our study seems to have the best overall performance for its lack of being unable to classify some CODs and the best overall accuracy. Despite the similar classification accuracy (~70%), the DT algorithm could not accurately classify the non-breast cancer group. Given the similar theoretical framework, we did not access its performance after dimension reduction. The ANN algorithm in our study is most efficient before and after dimension reduction. Surprisingly, we also notice a small increase in accuracy after dimension reduction which warrants further investigation. The SVM algorithm in our study appears very sensitive to the subgroup size (i.e., number of samples) and was not able to classify two of the five COD, although it also acceptable classification accuracy.

The study’s limitations should be noted when applying our findings to other databases. First, this type of rectangular database is typical in survey and population-study, but not so in computational biology. The major difference is the large p in ‘omics datasets vs the large n in epidemiological datasets, which referred to feature number and sample number, respectively. Second, some of the outcomes were not accurately classified. It is likely owing to the unbalanced outcome distribution. On the other hand, such an undesired situation reflexes real-world evidence/experience. Further studies are needed to improve the classification accuracy in the classes of fewer samples. Third, our works were exclusively based on the MATLAB platform and may not be applicable to other platforms such as R or python. Fourth, a molecular subtype of the cancer was not available due to the lack of human epidermal growth factor receptor 2 (Her2) data. Fifth, the prediction accuracy of the ML algorithms was moderately acceptable (~70%), although ML had much better recall and sensitivity than MLR. The challenge in predicting multi-category outcomes remains outstanding, despite the use of tuned ML algorithms. Thus, future works are needed to improve the prediction performance. Finally, ideally, we should use a large database to validate our models, but it is very difficult to curate and apply the tuned models to another large database that is similar to the SEER database.

In summary, we here show that RF, DT, ANN, and SVM algorithms had similar accuracy, but outperformed MLR, for classifying multi-category outcomes in this large rectangular dataset. Dimension reduction based on information gain will significantly increase the model’s efficiency while maintaining classification accuracy.


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# How is the pka related to the ph of the solution for an amine?

So was reading a magazine for competitive exams and this question is bugging me a lot.

How is pka values of a base associated with the pH of the solution. The solution for it is that the Nitrogens would remain this way and only the H from OH would be dissociated. That's obvious because the medium is basic and the acidic H would be dissociated, but what about the nitrogen atoms? How is the pka of a base represented? Is that the equilibrium constant for it's conjugate base?

• No the pKb value is for a base to accept a proton. For aqueous solutions: $$\mathrm{pKa + pKb = 14}$$ – MaxW Mar 20 '16 at 21:35


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# An equation for special relativity?

While surfing the internet, I came across a statement that this is the equation for special relativity:

Line 1: x = a + b
Line 2: x = a + b (c2/c2) with c = speed of light
Line 3: x = a + (y/c2) if y = b(c2)

Is this really the one? If not, is it relevant at all?

Related Special and General Relativity News on Phys.org
Matterwave
Gold Member
I have no idea what those equations are, I have never seen them before. Also, they seem to all be the same since the c's will cancel from top to bottom and you'll just get x=a+b for all 3 "lines".

pervect
Staff Emeritus
While surfing the internet, I came across a statement that this is the equation for special relativity:

Line 1: x = a + b
Line 2: x = a + b (c2/c2) with c = speed of light
Line 3: x = a + (y/c2) if y = b(c2)

Is this really the one? If not, is it relevant at all?
This seems to be some garbled version of the Lorentz transform, but it's not coherent enough for me to be to be sure, as there is no explanation of what the equations mean. For more details on the Lorentz transform, see for instance the wiki article at http://en.wikipedia.org/w/index.php?title=Lorentz_transformation&oldid=628048814

The Lorentz transform provides a transformation between the coordinates in two different inertial frames, moving relative to each other with velocity v. Because every event in space-time has one and only one set of coordinates, there is a 1:1 mapping between events and their coordinates. This implies there is also a 1:1 mapping between the coordinates between any two inertial frames, including the particular case we are interested in where the two inertial frames are in relative motion. The Lorentz transform provides this 1:1 mapping explicitly. Letting the coordinates in the first inertial frame (presumed stationary) be (t, x,y,z), and the coordinates in the second inertial frame (assumed to be moving with velocity v relative to the first inertial frame) be (t', x' y', z'), we can write the Lorentz transform as:

##t' = \gamma \left(t - vx/c^2\right) \quad x' = \gamma \left(x - vt \right) \quad y' = y \quad z'=z##

Here v is the velocity between frames, and ##\gamma = 1 / \sqrt{1 - v^2/c^2}##

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ghwellsjr


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# The Quasar SDSS J1536+0441: An Unusual Double-peaked Emitter

### Abstract

The quasar SDSS J153636.22+044127.0, exhibiting peculiar broad emission- line profiles with multiple components, was proposed as a candidate sub-parsec binary supermassive black hole system. More recently, imaging revealed two spatially distinct sources, leading some to suggest that the system was a quasar pair separated by åisebox-0.5ex 5 kpc. We present Palomar and Keck optical spectra of this system from which we identify a third velocity component to the emission lines. We argue that the system is more likely an unusual member of the class of active galactic nuclei known as double-peaked emitters’’ than a sub-parsec black hole binary or quasar pair. We find no significant velocity evolution of the two main peaks over the course of 0.95 yr, with a 3σ upper limit on any secular change of 70 km s$^-1$ yr$^-1$. We also find that the three velocity components of the emission lines are spatially coincident to within 0farcs015 along the slit, apparently ruling out the double-quasar hypothesis.

Publication
Astrophysical Journal Letters


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# Calculate amplifier gain?

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#### ayamingesh

##### Newbie level 4
Need help with calculating the gain.

1: Calculate the "raw gain" of the amplifier. Ar=Vut/Vin
2: Calculate the gain of the amplifier. Av=Vut/Vs

What is the difference between "raw gain" and gain?
Sorry for the bad translation, I don't even know if "raw gain" is the correct term in english, haha.

http://img827.imageshack.us/img827/9972/namnls2m.png

#### Junus2012

Hello
you mean to say the voltage gain Av or Avs

the difference between Av and Avs is expressed inthe equations below
Av=V0/Vin
where Vin = (Rin / Rin+Rs). Vs

Avs = Vo/Vs

in the first equation (Av) we considering the voltage drop on the source resistance Rs while in the later one (Avs) the gain is calculated prior the source resistance drop

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