text:
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# FileOpen 1077 Error with Webbatch

Keywords:   FileOpen 1077 Error with Webbatch

### Question:

I'm evaluating WebBatch and I'm getting this on a file open:
******************************
WebBatch Error

1077: FileOpen: Open failed

WebBatch 97C
WIL Version: 2.4cbp
******************************

I initialize "inifile" with this statement:
   inifile = StrCat(DirHome(), "sbstat.ini")

sbstat.ini is in the same directory with the webbatch script.

Here's my code fragment:

   inifile = StrCat(DirHome(), "sbstat.ini")
FileClose(hf)

(Note: I've examined 'inifile' and it contains the correct path+filename value)

This is running on an NT Workstation using Personal Web Server. the directory is on a fat drive... haven't tried this on a production nt server yet...

Any recommendations?

1. Sounds like a security issue. Does the Generic web user have access to the file? Check the NTFS file settings.

2. Also I worry that you are mixing iniwrite and FileOpen operations. Usually a dicey thing to di, as NT caches the ini file contents for long periods, resulting is stale files.

To force an ini file to disk do a:

   iniwritepvt("","","","filenam.ini")

### Question (continued):

ya, I started with inireadpvt() and when that failed I switched to fileread(). (I was concerned about the ini being held open and tried the iniwritepvt("","",""..) trick but no go.

I'm going to go ahead and try the webbatch script on a 'development' nt server (fresh environment) I've given the script directory (where this puppy & it's ini are located) 'Everyone Change', 'Administrators Full'. well see what happens -- keep you posted.

Later...

I'm not having any luck. besides the problem stated above (which still doesn't work) here's the real crux of the matter: I want to execute reskit utility pulist.exe from within my webbatch script, using redirection to generate an output file for examination of three tasks in particular. the problem seems to be with creating an output file.

The relevant code is:

   outlist = 'out put file path'
thisdata = 'set to server name'

command = "pulist.exe \\%thisdata% >%outlist%"
RunWait('cmd.exe', '/c %command%')

if !FileExist(outlist)
returncode = 1
return
endif

I'm getting a return code of 1. The outlist is in the script directory where this particular guy lives.

It's running under IIS 3.0. I've even tried putting the IUSR_servername account in the Administrator's group, no go.

1. Get it doen to a SMALL test case

2. Test it as a logged on user first.

3. I don't see return code getting itself reset to 0 anyplace.

4. Try a full path to cmd.exe

5. Try a full path to the utility you are trying to run

### Resolution:

Got it!

I had to move the temporary file that I was creating and the parameter I was trying to read OUT of the directory that contained the .web file and into a directory elsewhere on the disk drive... interesting. NTFS permissions basically the same, go figure.

## Another 1077 FileOpen Error (Pathing problems)

### Question:

I looked at the 1077 error under TechSupport/WebBatch/Error Messages and tried the solution listed there but still have the problem.

I am acquiring data from a form and need to write it out to an ascii file. Everything debugged good up to the point where I open the file for appending...

tablefile = FileOpen("table.dat", "APPEND")
filewrite(tablefile,"%field1%")

I double-checked the NTFS rights for generic visitors, (ok). I moved the file that I'm trying to write to another folder(?) as described in the solution mentioned above but got the same results.

What am I missing? Do I need to build a more specific path to "table.dat"?

I would try a more specific path.

It does not seem like you are into debugging the sharing problems that will come next.

### Question (cont'd):

Sharing problems? Tell me more. Save me the pain... ;)

OK. Well, you get your script working somehow. Tests *just fine*.

But then when 80 million people start using it yuu start getting occassional errors in the FileOpen. "File could not be opened" or somesuch. What could be happening is that two (or more) copies of your script could be running at the same time (for two different users) and both of the scripts try to append to the file at the same time.

This is called a "resource conflict" and the solution is a "resource lock". The easiest solution is to use the file itself as the lock.

while 1   ; wait for resource to become avaiable
ErrorMode(@off)
tablefile = FileOpen("d:\right\there\table.dat", "APPEND")
ErrorMode(@cancel)
if tablefile!=0 then break   ; resource lock obtained. exit loop
TimeDelay(0.5)               ; no?  delay.  try again in 1/2 second
endwhile

filewrite(tablefile,field1)
FileClose(tablefile)            ; Close file AND also free resource lock

Article ID:   W12466

Filename:   FileOpen 1077 Error with Webbatch.txt
File Created: 2017:07:28:12:58:11
Last Updated: 2014:06:17:12:57:59


text:
Math Help - Matrix operation

1. Matrix operation

Determine A^18

dont know what to do besides to just count. But i had a similar question before where i could se it was a rotation matrix and then it was easy. Is there an easy way for this one to?

Thanks!

2. Decompose

$A=2I+N$

and apply the Newton's binomial theorem. The matrix N is nilpotent of order 3.

format: plain


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Synopsis

# Magnetizing an Atomic Gas with Light

Physics 15, s51
Theorists predict that an atomic gas could be magnetized using only lasers, something that could provide a noninvasive way to quickly manipulate the magnetic properties of the gas.

Nanoscale manipulations of a material’s magnetic moment are tricky, as the charge currents used for this process are difficult to control at this small length scale. Now, Jonas Wätzel of the Martin-Luther-University Halle-Wittenberg in Germany and his colleagues report that they might have a solution to that problem [1]. The researchers predict that the magnetization of an atomic gas can be altered by high-power lasers with patterned wavefronts. They say that the method could find use in studying and manipulating the magnetic properties of active materials.

The team modeled the interaction of two lasers with a gas of helium atoms. One laser is an infrared (IR) “vortex” laser with orbital angular momentum and a lateral intensity profile that drops to zero at the beam’s center. The other laser is an extreme-ultraviolet (XUV) pulsed laser that co-propagates within the IR beam’s low-intensity region. When the lasers hit the gas, the XUV pulse excites the atoms’ valence electrons into a so-called Rydberg state. The IR vortex then transfers angular momentum to the electron charge density of this Rydberg state, causing it to rotate. This rotation creates a magnetic moment in the system that can last for several nanoseconds.

The researchers say that the direction, magnitude, and extent of this magnetization can be fully controlled by adjusting the properties of the IR vortex laser. While the team’s preliminary experiments show that magnetization of a gas can be photoinduced using their new method, they have yet to apply the method to magnetically active materials. They say that such experiments, while realizable with current setups, will be challenging, as the method requires the lasers to irradiate atoms simultaneously and their beams to be aligned with nanoscale precision.

–Katherine Wright

Katherine Wright is the Deputy Editor of Physics Magazine.

## References

1. J. Wätzel et al., “Light-induced magnetization at the nanoscale,” Phys. Rev. Lett. 128, 157205 (2022).

Magnetism

## Related Articles

Magnetism

### A Jiggling Ultracold Atomic Gas Simulates Spin Dynamics

Researchers produce analogues of hard-to-study quantum phenomena in a gas of strontium atoms near absolute zero. Read More »

Atomic and Molecular Physics

### Probing Molecular Magnetism Interferometrically

A matter-wave interferometer can probe the magnetism of a broad range of species, from single atoms to very large, weakly magnetic molecules. Read More »

Astrophysics

### Experiment Sees Elusive Magnetic-Fluid Instability

Magnetorotational instability—a process that might explain the dynamics of astrophysical accretion disks—has finally been observed in the laboratory. Read More »


text:
MathSciNet bibliographic data MR1340178 11F70 (11F46 11F72 11L05 22E50) Jacquet, Hervé; Ye, Yangbo Distinguished representations and quadratic base change for ${\rm GL}(3)$${\rm GL}(3)$. Trans. Amer. Math. Soc. 348 (1996), no. 3, 913–939. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
- Current Issue - Ahead of Print Articles - All Issues - Search - Open Access - Information for Authors - Downloads - Guideline - Regulations ㆍPaper Submission ㆍPaper Reviewing ㆍPublication and Distribution - Code of Ethics - For Authors ㆍOnline Submission ㆍMy Manuscript - For Reviewers - For Editors
 Self-homotopy equivalences of Moore spaces depending on cohomotopy groups J. Korean Math. Soc. 2019 Vol. 56, No. 5, 1371-1385 https://doi.org/10.4134/JKMS.j180680Published online September 1, 2019 Ho Won Choi, Kee Young Lee, Hyung Seok Oh Korea University; Korea University; Korea University Abstract : Given a topological space $X$ and a non-negative integer $k$, $\E_{k}^{\sharp}(X)$ is the set of all self-homotopy equivalences of $X$ that do not change maps from $X$ to an $t$-sphere $S^{t}$ homotopically by the composition for all $t \geq k$. This set is a subgroup of the self-homotopy equivalence group $\E(X)$. We find certain homotopic tools for computations of $\E_{k}^{\sharp}(X)$. Using these results, we determine $\E_{k}^{\sharp}(M(G,n))$ for $k\geq n$, where $M(G, n)$ is a Moore space type of $(G, n)$ for a finitely generated abelian group $G$. Keywords : self-homotopy equivalence, cohomotopy group, Moore space, co-Moore space MSC numbers : Primary 55P10, 55Q05, 55Q55 Downloads: Full-text PDF   Full-text HTML


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# degree of the differential equation -1

Calculus Level 3

What is the degree of the differential equation $y=1+\frac{dy}{dx} +\frac{1}{2!}.\left( \frac{dy}{dx} \right)^2+\frac{1}{3!}.\left( \frac{dy}{dx} \right)^3+\frac{1}{4!}.\left( \frac{dy}{dx} \right)^4+......$

Note : If degree is not defined, write $$1729$$ as your answer.

Check out more questions on differential equations

×


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# How do we know that fundamental constants don't slowly change in time? [duplicate]

Let me get one thing straight first. I am not saying that fundamental constants like the speed of light don't have the value that we know they have today. What I am asking is whether the value of a constant could change with time. For example, we measure today that the speed of light is almost $300000000\,m/s$, but this measurement was made over the last century. But what if a long time ago (billions of years ago) that constant had a different value?

So, it's always the limit of how fast something can move, but that limit slowly changes with time.
To conclude,i am asking if there is any kind of evidence that any of the fundamental constants like the speed of light remain constant in time or if we just took that for granted.

## marked as duplicate by ACuriousMind♦, dmckee♦Jul 14 '15 at 14:25

• Just to be clear, are you asking us what evidence we have that the speed of light hasn't changed? Or are you asking us what it would mean for physics if it had changed and we just haven't noticed? – Jim Jul 14 '15 at 13:05
• I know that there is experimental evidence for the speed of light and it being a fundamental constant for velocity.I have read some interesting points from some physicists that explain some of the early processes or the universe(or some of the paradoxes in some of our measurements from the universe) and they say that the speed of light might not have always have the value that we know that it has today.So,i am asking if there is evidence that the speed of light is constant through time – TheQuantumMan Jul 14 '15 at 13:09
• i edited the question – TheQuantumMan Jul 14 '15 at 13:10
• Possible duplicates: physics.stackexchange.com/q/21721/2451 and links therein. – Qmechanic Jul 14 '15 at 13:15
• The question and title are clear to me. We have evidence that the distance between objects increases because the nature of space is changing - space is expanding. This is too small to measure in a room, but shows up for distant galaxies. You are asking if similar effects are present for fundamental constants like the speed of light. What evidence is there? – mmesser314 Jul 14 '15 at 13:16


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Floating point numbers and roundoff¶

As is well, known, mathematical addition is an associative operation. That is, for all real numbers $x$, $y$, and $z$, we have: $$(x+y)+z = x+(y+z).$$ Floating point addition, as specified by the IEEE standard for floating point arithmetic, is sadly not associative:

In [1]:
((0.1+0.2)+0.3) - (0.1+(0.2+0.3))

Out[1]:
1.1102230246251565e-16

Ultimately, this is due to computational error. Here's another example with some error:

In [2]:
x = 1
h = 10**(-15)
((x+h)**1000 - x)/h

Out[2]:
1110.2230246251565

What do you suppose the answer should be??

These examples have two ingredients in common:

• The difference of two numbers that are very close together must be treated with caution and
• The fact that real numbers are only represented to a finite precision must cause round off error.

This second issue is complicated by the fact that the computer's representation of numbers is not in base 10 but in base 2. Thus, numbers like 0.1, which you expect to be exactly representable, are not.

In spite of all this, computer error is not something to be feared but something to be understood.

Floating point numbers in general¶

Floating point numbers are represented in terms of a base $\beta$, a precision $p$, and an exponent $e$. If $\beta=10$ and $p=2$, then the number $0.034$ can be represented as $3.4\times 10^{-2}$. With this representation, its exponent is $e=-2$. With this specification, floating point numbers are not unique, as this same number can be represented as $34\times 10^{-3}$ or $0.34\times 10^{-1}$. Thus, we typically take a floating point number to have the specific form $$\pm(d_0+d_1\beta^{-1}+\cdots+d_{p-1}\beta^{-(p-1)})\beta^e,$$ where each $d_i$ is an integer in $[0,\beta)$ and $d_0\neq0$. Such a representation is said to be a normalized floating point number.

Computers work more naturally in binary (or base 2) and most are hardwired this way. Thus, in the equation above, we have $\beta=2$ and each digit $d_i$ is zero or one. Taking this into account, here are some easily representable numbers with $p=5$ and $-5\leq e \leq 5$:

• $8 = 1.0000 \times 2^3 = 1000_{\dot 2}0$
• $\frac{1}{4} = 1.0000 \times 2^{-2} = 0_{\dot 2}01$
• $\frac{5}{16} = 1.0100 \times 2^{-2} = 0_{\dot 2}0101$
• $42 = 1.0101 \times 2^{5} = 101010_{\dot 2}$

64 bit IEEE numbers¶

The IEEE standard for double precision floating point numbers specifies that 64 bits be used to represent the number. Specifically, it requires

• $1$ bit for the sign $s$.
• $11$ bits for the exponent $e$ giving a range of $2^{11} = 2048$ choices for the exponent, which is assumed to be between $-1023$ and $1024$.
• $53$ bits for the significant or mantissa $c=1.d_1d_2\cdots d_{52}$.

Since the leading digit must be a $1$, we only need to store 52 digits. Given those choices for $s$, $e$, and $c$, our number is $$(-1)^s\times c \times \beta^e.$$ Since the exponent $e$ is expressed in binary with no sign, it might be more natural to think of it in the range $0\leq e < 2048$. In this case, we might write the number $$(-1)^s\times c \times \beta^{e-1023}.$$ Note that the standard leaves quite a bit freedom allowing differences and optimizations between implementations. Often, a few strings are reserved for special numbers or, more often, there are a few extra bits included.

As an example, suppose we'd like to compute the number with bit string $$1\,\,1010000000000000000000000000000000000000000000000101\,\,10100000101$$ To compute this, we might do the following:

In [3]:
e = 2**10 + 2**8 + 2**2 + 2**0
c = 1 + 2**-1 + 2**-3 + 2**-50 + 2**-52
s = 1
(-1)**s * c * 2.0**(e-1023)

Out[3]:
-1.2042377280680893e+79

Exploring numbers with Python¶

Python provides some useful tools for exploring floating point numbers. Two such tools are float.hex, which displays the hexadecimal expansion of a number and its inverse float.fromhex. For example, here's my computer's representation of 0.1:

In [4]:
float.hex(0.1)

Out[4]:
'0x1.999999999999ap-4'

This indicates that the internal representation of $0.1$ is $$2^{-4}\left(1+\sum_{k=1}^{12} \frac{9}{16^k} + \frac{10}{16^{13}}\right) \approx 0.10000000000000000555.$$

We can use this information to fully understand the discrepancy between $(0.1+0.2)+0.3$ and $0.1+(0.2+0.3)$. Not surprisingly, the hexadecimal representation of $0.2$ is 0x1.999999999999ap-3. The hexadecimal representation of $0.3$ is 0x1.3333333333333p-2, which to 17 digits is $0.29999999999999999$. The sum of these three numbers yields $0.6000000000000001$, which is slightly larger than the desired value of $0.6$. By contrast, the computer's computation of $0.2+0.3$ yields exactly $0.5$, which is not surprising since $0.5$ easily representable in binary. Then, $0.1+0.5$ yields exactly the computer representation of $0.6$ after rounding.

Rounding¶

Any set of floating point numbers is not algebraically closed. That is, you can do arithmetic with floating point numbers and potentially generate numbers not in your set. Given a real number x (perhaps generated via a computation with floating point numbers), the computer will generally store it as fl(x), which is the floating point number closest to x. This phenomenon is called rounding.

For example, when we compute 1/10, we generate the number slighly larger than 0.1 that's above.

Distribution¶

It's worth pointing out that normalized floating point numbers are not uniformly distributed. Let's plot, for example, the binary floating point numbers with precision 4 and maximum exponent size 3:

The numbers of the form $1.dddd$ (called lead_ones in the code) are shown in red. The other postive numbers have the form $2^nx$ for some $x$ with a leading 1. That leads to the scaling effect that we see.

Note, also, the hole near zero. That's because the smallest, postive, representable number (in this system) is $1.0000 \times 2^{-3} = 2^{-3}$. The next smallest number is $1.0001\times2^{-3} = (1+1/16)\times2^{-3}$ and the difference between these is quite a bit smaller than the smallest postive.

The difference between the smallest representable number that's larger than one and one is typically called machine epsilon.

Python's decimal library¶

Python has a fun little library called decimal that allows you to explore decimal floating point arithmetic. We can use it to help us solve problems like our Two digit partial pivoting problem on our forum. To do so, we first load it and set the precision:

In [5]:
from decimal import Decimal, getcontext
getcontext().prec = 2

Here's how to use it to peform simple subtraction:

In [6]:
a = Decimal('32')
b = Decimal('2.3')
a-b

Out[6]:
Decimal('30')

Now, let's use it to define a function that implements the first type of row operation on $2\times n$ matrices. While clearly very limited, this illustrates the basic ideas behind how linear algebra might be implemented in software.

In [7]:
def row_op1(A,s):
row1 = A[0]
row2 = A[1]
new_row2 = [s*row1[i] + row2[i] for i in range(len(row1))]
return [row1,new_row2]

OK, here's the matrix in the problem and it's row reduction.

In [8]:
A = [
[Decimal('0.1'), Decimal('2.7'), Decimal('10')],
[Decimal('1.0'), Decimal('0.5'), Decimal('-6')]
]
reduced = row_op1(A,-10)
reduced

Out[8]:
[[Decimal('0.1'), Decimal('2.7'), Decimal('10')],
[Decimal('0.0'), Decimal('-26'), Decimal('-1.1E+2')]]

Using back-substitution, it's pretty easy to now solve the system:

In [9]:
y = reduced[1][2]/reduced[1][1]
x = (reduced[0][2] - reduced[0][1]*y)/reduced[0][0]
[x,y]

Out[9]:
[Decimal('-1E+1'), Decimal('4.2')]

I'll leave it to you to find the solution after permutation.


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# Boltzmann Transport Equation nonlinear terms

Let's take into consideration the classical Boltzmann Equation for gas kinetics (i'm writing down the one obtained in Huang K. - Statistical mechanics, equation (3.36)):

$$\left(\frac{\partial}{\partial t} +\frac{\mathbf p_1}{m}\nabla_r+\mathbf F\nabla_{p_1} \right)f_1=\int d^3p_2 d^3p'_1 d^3p'_2 \delta^4(P_f-P_i)|T_{fi}|^2(f'_2f'_1-f_1f_2)$$

which is an integro-differential nonlinear (due to the $f'_2f'_1$ and $f_1f_2$ terms) equation in the distribution function $f_1$.

In a textbook I'm reading it is said that another feature of such equation is that

the nonlinear terms $f'_2f'_1$ and $f_1f_2$ are calculated for different values of the momentum.

What could be the meaning of this statement? Could "the momentum" be the $\mathbf p_1$ in $f_1(\mathbf x_1, \mathbf p_1, t$)?

• I think it means that $f_1$ is a shortcut notation for $f(\vec{x}, \vec{p}_1,t)$; then $f'_1 = f(\vec{x},\vec{p}'_1,t)$ etc... – Matteo Feb 4 '18 at 17:34

The meaning is that in the scattering term on the RHS of the Boltzmann transport equation you have the products of the distribution functions $f_1$ and $f_2$ for different momenta $p_1$ and $p_2$.
• Thank you. But in the case of $f'_2f'_1$ we take into consideration every value of $p'_1$ and $p'_2$, due to integration... – Lo Scrondo Feb 4 '18 at 21:54


text:
Which solution below has the highest concentration of hydroxide ions?

Which solution below has the highest concentration of hydroxide ions?
A) pH= 3.21
B) pH= 7.00
C) pH= 7.93
D) pH= 12.59
E) pH= 9.82
Please select the best answer from the above multiple choice and explain thanks

General guidance

Concept and reason

The concept used is to calculate the hydroxide ion concentration.

Fundamentals

is the measure of hydrogen ion concentration. Lower the , higher is the hydrogen ion concentration and lower is the hydroxide ion concentration.

Higher the , lower is the hydrogen ion concentration and higher is the hydroxide ion concentration.

The hydrogen ion concentration is calculated as follows:

The hydroxide ion concentration and are related as follows:

Step-by-step

Step 1 of 2

1.

Consider

Calculate the hydroxide ion concentration.

Calculate the hydroxide ion concentration.

Consider

Calculate the hydroxide ion concentration.

Calculate the hydroxide ion concentration.

Consider

Calculate the hydroxide ion concentration.

Calculate the hydroxide ion concentration.

Consider

Calculate the hydroxide ion concentration.

Calculate the hydroxide ion concentration.

The given solutions with these values are having lower concentrations of hydroxide ion.

[Hint for the next step]

Identify the highest hydroxide ion concentration.

Step 2 of 2

Consider

Calculate the hydroxide ion concentration.

Calculate the hydroxide ion concentration.

Part 1

Therefore, the solution with highest hydroxide ion concentration is .

The given solution with this value is having highest concentration of hydroxide ion.


text:
# Central and Inscribed Angles in Complex Numbers

This is probably one of the more popular "math facts" that the central angle in a circle is twice the inscribed angle subtended by the same arc. The original formulation comes from Euclid III.20:

In a circle the angle at the center is double the angle at the circumference when the angles have the same circumference as base.

Euclid's proof has been illustrated elsewhere based on the following diagram:

An important corollary of this fact is that, in a circle, all inscribed angles subtended by the same arc are equal. The importance of the statement and the corollary is underscored by the inclusion of a less elementary proof that employs complex numbers in a classic book on advanced geometry of plane curves. Of course, additionally, the proof serves to illustrate basic complex number techniques.

The proof is based on the following diagram.

Without loss of generality, the circle is assumed to have radius $1$ and be centered at the origin. The points are identified with complex numbers, so that, say, the equation of the circle is $z=e^{i\psi},$ where $\psi$ is a real number, the angle measured from the horizontal $x\mbox{-axis}.$ Let $B$ correspond to $\psi =0,$ $A$ to $\psi =\phi,$ and $P$ to $\psi =\nu:$ $B=1,$ $A=e^{i\phi},$ $P=e^{i\nu}.$

We find, for vectors, $AP= e^{i\nu}-e^{i\phi},$ $BP= e^{i\nu}-1.$

The argument of the ratio of this two complex numbers is $\angle APB=\alpha.$ Dividing the ratio by its conjugate eliminates the unimportant length of the ratio, but doubles the argument. Thus we have

\displaystyle \begin{align} e^{2i\alpha} &=\frac{e^{i\nu}-e^{i\phi}}{e^{i\nu}-1} : \frac{e^{-i\nu}-e^{-i\phi}}{e^{-i\nu}-1}\\ &=\frac{e^{i\nu}-e^{i\phi}}{e^{i\nu}-1} \cdot \frac{e^{-i\nu}-1}{e^{-i\nu}-e^{-i\phi}}\\ &=\frac{e^{i\nu}-e^{i\phi}}{e^{i\nu}-1} \cdot \frac{e^{i\phi}(1-e^{i\nu})}{e^{i\phi}-e^{i\nu}}\\ &=e^{i\phi}. \end{align}

Assuming all angles are between $0$ and $\pi,$ $2\alpha=\phi.$

### References

1. C. Zwikker, The Advanced Geometry of Plane Curves and Their Applications, Dover, 2005, p. 59


text:
# Bound A Sum Of Squares - Not As Easy As You Think

Algebra Level 5

Consider polynomials of the form $$x^{2014} + a_{2013} x^{2013} + \ldots + a_1x + 1$$, with real coefficients and having at least one real zero. Determine the smallest possible value of $$\sum_{i=1}^{2013} a_i^2$$.


text:
# Probability that a continuous random variable assumes a fixed point

I'm in an introductory statistics class in which the probability density function for continuous random variables has been defined as $P\left\{X\in B\right\}=\int_B f\left(x\right)dx$. I understand that the integral of $\int\limits_a^af(x)dx=0$ but I can't rectify this with my intuition of a continuous random variable. Say X is the random variable equal to the number of minutes from time t that the train arrives. How do I calculate the probability that the train arrives exactly 5 minutes from now? How can this probability be zero? Is it not possible? What if the train does arrive exactly 5 minutes from now, how could it occur if it had probability 0?

Thanks.

• Standing some of these questions on their head is helpful. E.g., if your intuition says every possible time must have some strictly positive probability, then--because there is an uncountable set of possible times in any interval--your intuition implies the total probability is infinite. Obviously that intuition is wrong. One thing that has to be given up is the idea that a probability of zero implies an impossibility: that is not true. Similarly, a probability of one does not imply a certainty. – whuber Nov 6 '13 at 4:17
• @whuber That's what I can't rectify. If the probability of an event taking place is 0, then it should never happen. For example, if I have a standard six-sided die, the probability I roll any number $\mathcal{Z}\setminus\left\{1,2,3,4,5,6\right\}$ is 0 and therefore will never happen. Moreover, how can an event with probability 1 not be a certainty in the subsequent experiment? Could you provide an example? – geofflittle Nov 7 '13 at 0:04
• Suppose you see a circle in which a chord is shown and it appears to be a diameter, prompting you to wonder "what was the chance that a randomly selected chord would not have been a diameter?" When the chord is obtained by choosing a pair of points uniformly and independently along the circumference, the answer is $1$, but this event did not occur. That provides (pretty strong!) evidence that the chord was not the result of the random process that you posited. One lesson afforded by such thought experiments is that intuitions based on finite probability spaces do not always generalize. – whuber Nov 7 '13 at 18:00

## 5 Answers

You may be falling into the trap of regarding 'five minutes from now' as lasting some finite period of time (which would have a nonzero probability).

"Five minutes from now" in the continuous variable sense is truly instantaneous.

Imagine that the arrival of the next train is uniformly distributed between 8:00 and 8:15. Further imagine we define the arrival of a train as occurring at the instant the front of the train passes a particular point on the station (perhaps the midpoint of the platform if there's no better landmark). Consider the following sequence of probabilities:

a) the probability a train arrives between 8:05 and 8:10

b) the probability a train arrives between 8:05 and 8:06

c) the probability a train arrives between 8:05:00 and 8:05:01

d) the probability a train arrives between 8:05:00 and 8:05:00.01 (i.e. in the space of one hundredth of a second

e) the probability a train arrives between 8:05 and one billionth of a second later

f) the probability a train arrives between 8:05 and one quadrillionth of a second later

... and so on

The probability that it arrives precisely at 8:05 is the limiting value of a sequence of probabilities like that. The probability is smaller than every $\epsilon>0$.

• I understand this, but, assuming the train arrives, it arrives at some point in time. Why can't this limit still converge to some probability? – geofflittle Nov 7 '13 at 0:06
• If you understand it, as you say, you can compute the probability in the indicated fashion. Let me make it easier: Imagine for convenience of calculation that the exact time that a train "arrives" (however we define it, as long as it's actually continuous) at a uniformly distributed time on the interval (0,1) (at whatever is a convenient time unit). What's the probability that the train arrives before time $x$, for some $x$ inside the interval? What's the probability that it arrives after time $x$? What's the probability that it arrives between $x$ and $x+dx$? ...(ctd) – Glen_b Nov 7 '13 at 0:59
• (ctd)... To say it 'arrives at time $x$' for a continuous variable, means "what is the limit of that last probability as $dx\to 0?$. So, what is that limit? Work it out! That is the probability it converges to. This feature is intimately related to what makes a continuous pdf continuous. – Glen_b Nov 7 '13 at 0:59
• Further note that if that last limit is anything but zero, your three probabilities (before $x$, after $x$ and "at" $x$) won't add to 1. – Glen_b Nov 7 '13 at 1:05

What if the train does arrive exactly 5 minutes from now, how could it occur if it had probability 0?

A probabilistic statement is not a statement about the possibility/feasibility of an event. It only reflects our attempt to quantify our uncertainty about it happening. So when a phenomenon is continuous (or is modeled as one), then our tools and current state of knowledge do not permit us to make a probabilistic statement about it taking a specific value. We can only make such a statement related to a range of values. Of course the usual trick here is to discretize the support, to consider"small" intervals of values rather than single values. Since continuous random variables bring great benefits and flexibility compared to discrete random variables, this has been found to be a rather small price to pay, perhaps as small as the intervals we are forced to consider.

• These statements are puzzling, perhaps because they could be interpreted in so many different ways. In some places you seem to deny the validity of the use of continuous distributions to model phenomena--and make a sharp distinction between the phenomenon and the model--and in other places you seem to drop that distinction altogether. My reading of it, which I suspect was not intended, is that you contend that the mathematical fact that $\Pr(X=a)=0$ for any continuous RV $X$ is in reality always false, but that makes it seem like you are denying the applicability of probability theory! – whuber Nov 6 '13 at 4:14
• Hi @whuber. Regarding the distinction between model and phenomenon, a map of the earth is not earth, but it may help you roam the earth. This is how I think about models, when I don't treat them as objects of pure intellectual pleasure (which they also are). As for the "zero probability" issue, it is an imperfection -after all, wouldn't it be great to have all the benefits of continuity and be able to make a probability statement about a single value? But being imperfect does not render something inapplicable of course, and as I write, this imperfection has proven of little importance. – Alecos Papadopoulos Nov 6 '13 at 10:09
• You implicitly suppose that probability is some objective thing "out there" in your mapping analogy, but it's not. Probability has a meaning only within a model. I see no "imperfections" in the axioms of probability and indeed, one can make accurate, consistent statements about probabilities of single values: often they are zero. – whuber Nov 6 '13 at 14:09
• @whuber No I don't suppose that, and I don't understand where did you see that in what I wrote. I said "the map is not the earth", which means "what is in the model does not exist in reality", so how can you infer from that the exact opposite? The "imperfection" does not refer to the axioms of probability, but to what tools these axioms lead us to, and to how effectively these tools can be put to use in order to model, study and understand the real world. And it is obvious that I believe that probability is an effective tool. – Alecos Papadopoulos Nov 6 '13 at 14:58

To give you some intuition for the above, try the following (thought) experiment:

Draw a real line around zero with a ruler. Now take a sharp dart and let it fall from above randomly on the line(let's assume you will always hit the line and only the lateral positioning matters for the sake of the argument).

However many times you let the dart fall randomly on the line, you will never hit the point zero. Why? Think what is the point zero, think what is its width. And after you recognise that its width is 0, do you still think you can hit it?

Will you be able to hit point 1, or -2? Or any other point you pick on the line for that matter?

To get back to maths, this is the difference between the physical world, and a mathematical concept such as real numbers (represented by the real line in my example). Probability theory has quite a bit more complicated definition of probability than you will see in your lecture. To quantify the probability of events and any combination of their outcomes, you need a probability measure. Both the Borel measure and Lebesgue measure are defined for an interval [a, b] on the real line as: $$\mu([a,b])=b-a$$ from this definition you can see what happens with the probability if you reduce the interval to a number (setting a = b).

The bottom line is that based on our current definition of probability theory (dating back to Kolmogorov) the fact that an event has 0 probability does not mean it cannot occur.

And as far as your example with the train goes, if you will have an infinitely precise watch, your train will never arrive exactly on time.

• You say "you will never hit the point zero", but what can you say of the point that I hit in my first dart throw? Let $x$ be the point that I hit. Before throwing my dart, you would have said "you will never hit the point $x$", but I've just hit it. Now what? – geofflittle Nov 7 '13 at 0:09
• I think that you have to differentiate between the question: What is the probability that I will hit some point? If we agree that you always throw a dart and it always hits somewhere along the line, that probability is 1. Also, I am not just saying you won't hit 0. I am saying that the probability that you hit ANY point you pick BEFORE throwing the dart is 0. In fact you can pick any finite set of points and the probability will still be 0. – means-to-meaning Nov 7 '13 at 1:22
• Regarding your question, I get your point, but to ask about probabilities of events after their occurrence is non-sensical. A statement such as P(X=x) refers to the future realisation of a random variable X. So AFTER you hit some point, I won't be saying anything about it. (large caps used just to point out the time-flow, not to shout…) – means-to-meaning Nov 7 '13 at 1:22

A probability distribution has to have an area of unity. If the measure is continuous then there is an infinite number of values that it can take (i.e. an infinite number of values along the x-axis of the distribution). The only way that the total area of the probability distribution can be finite is for the value at each of the infinite number of values to be zero. One divided by infinity.

In 'real life' there can be no measures that take an infinite number of values (by several different philosophical arguments that don't matter much here) so no value need take a probability of exactly zero. A useful practical argument is based on the finite precision of real-world measurements. If you use a stopwatch that measures to one tenth of a second, the train will have one tenth of a second in which to arrive in 'exactly' five minutes.

• The first paragraph here provides some vague intuition, though the deductive steps are incorrect. There are plenty of distributions that admit an infinite number of values yet each value has strictly positive probability. The second paragraph might profit from a rewording that emphasizes that to each measurement value is associated a (small) interval of possible values of the underlying quantity of interest. – cardinal Nov 6 '13 at 5:46
• What is the difference between a strictly positive value (of a finite value divided by infinity?) and zero in this context? – Michael Lew Nov 6 '13 at 5:56
• My point, probably poorly made, is that the argument in the first paragraph is based on the false premise that because the random variable can take on infinitely many values, each individual outcome must have probability zero. This is, of course, incorrect (Poisson, geometric, etc.); the concept of "infinity" is not a strong enough one here, we require uncountability. – cardinal Nov 6 '13 at 14:31

Other people have answered why the probability is zero (if you approximate time as being continuous, which it is effectively not, but anyway...) so I will just echo it briefly. To answer the last question that the OP asked---"how could it occur if it had probability 0?"---lots and lots of things can occur if they have probability zero. All a set of probability zero $$A$$ means is that, in the space of possible things that could happen, the set $$A$$ takes up no space. That is all. It is not more meaningful than this.

I am writing this to hopefully address something else that the OP said in the comments:

You say "you will never hit the point zero", but what can you say of the point that I hit in my first dart throw? Let 𝑥 be the point that I hit. Before throwing my dart, you would have said "you will never hit the point 𝑥", but I've just hit it. Now what?

This is a very good question and one that, when I began to learn about probability, I struggled with. Here is the answer: it isn't equivalent to the question that you originally asked! What you have done is bring time into the analysis, and that means that the underlying probability structure changes to become much more intricate. Here is what you need to know. A probability space $$(\Omega, A, \mu)$$ consists of three things: an underlying space $$\Omega$$, such as $$\mathbb{R}$$ or $$\mathbb{Z}$$; a set of all possible outcomes on this space, such as the set of all half-open intervals on $$\mathbb{R}$$, and a measure $$\mu$$ that satisfies $$\mu(\Omega) = 1$$. Your original problem lives in the space $$([a,b], \text{all half open intervals on [a,b]}, \nu)$$ where $$\nu$$ is Lebesgue measure (this means that $$\nu( [c,d) ) = \frac{1}{d - c}$$). In this space, the probability that you hit any single point $$x \in [a,b]$$ is zero for the reasons discussed above---I think we have this cleared up. But now, when you say things like the quoted passage above, you are defining something called a filtration, which we will write as $$\mathcal{F} = \{\mathcal{F}_t\}_{t \geq 0}$$. A filtration in general is a collection of subsets of $$A$$ that satisfy $$\mathcal{F}_t \subseteq \mathcal{F}_s$$ for all $$t < s$$. In your case, we can define the filtration $$\mathcal{F}_t = \{x \in [a,b]: \text{dart hit x at time t' < t} \}.$$ Now, in this new subset of your outcome space, guess what---you're right! You have hit it and, after your first throw, your probability of having hit that point when restricted to the filtration $$\mathcal{F}_1$$ is 1.

• Because you are using technical language, it would be best to employ standard meanings for the terms. In particular, what you call an "outcome" is usually termed a (basic) event: the outcomes are the elements of $\Omega.$ Your formula for (normalized) Lebesgue measure is incorrect: I suspect you intended $\nu([c,d])=(d-c)/(b-a).$ On a more fundamental level, it isn't clear why you need to invoke the machinery of stochastic processes in order to discuss a random variable modeling the time of a single event, nor is it evident this provides any insight. – whuber Apr 23 at 21:39


text:
# Latex Matrix

A matrix is defined as a set of numbers called elements arranged in the form of rows and columns so as to from the rectangular array. It has wide applications and are used for various purposes.

The environment is used to implement the matrix, which is given below:

The ampersand (&) symbol separates the columns in the matrix.

We can add as many rows and columns in the matrix according to the requirements.

The rows are separated by the \\ command, which is commonly used to specify the text in the new line.

The matrix environment is implemented using the \ and [.

It is written as:

Without the use of the above environment, the Texmaker will give an error.

Let's understand with a simple example. The example is a 2 x 2 matrix. It is a matrix that has two rows and two columns.

The code for such an example is given below:

Output:

There are different matrix environments used to create different styles of brackets for a matrix, which are listed below:

Environment Surrounding Bracket
pmatrix ( )
Bmatrix { }
bmatrix [ ]
Vmatrix || ||
vmatrix | |

The starred version of the above environments is used to align the columns of the matrix, which is listed below:

Environment Surrounding Bracket
pmatrix* ( )
Bmatrix* { }
bmatrix* [ ]
Vmatrix* || ||
vmatrix* | |

Let's understand the above aligning process by an example. The example is of 3 x 3 matrix having three rows and columns.

The code for such an example is given below:

Output:

You can also design the arbitrary sized matrix. The dots between the matrices can be specified by using the \ddots (⋱), \vdots (⋮), and \cdots (⋯) command.

The code to create such matrix is given below:

Output:

## Matrix with Text

In the running text, we can insert the matrix. The matrix of such type is termed as the small matrix.

The environment used to create small matrix is given below:

But here, we need to specify that the size of text should be greater than that of the matrix.

Let's understand with an example.

The code is given below:

Output:

## Border Matrix

The border matrix is used to create the borders around the matrix. The \bordermatrix command is used to implement such matrices.

The code for such an example is given below:

Output:

## Fraction Matrix

The fraction matrix consists of the fraction elements. We can also insert non-fraction elements according to the requirements.

Let's understand with a simple example. Here, we have used the Vmatrix environment.

The code for such an example is given below:

Output:

## Arrays

The array is similar to the matrix. But here, we can insert lines between rows and columns.

The environment used for such purpose is given below:

Let's understand with two examples.

The code for the first example is given below:

Output:

The code for the second example is given below:

Output:

### Cases

1) The curly brackets in mathematics are defined by using the \left{ and \right} command.

Let's understand by two examples.

The code for the first example is given below:

Output:

2) It is the crucial part of the cases. Here, the cases environment is used to implement such methods, which is written as:

The above environment is commonly used in mathematics to write two equations with certain conditions inside only single brackets.

The code for the second example is given below:

Output:

Next TopicLatex Integral


text:
Harrison Chapman

Math 317: Homework 10

Due: Friday, April 26, 2019
1. (33.2) Let $S$ be a nonempty bounded subset of $\mathbb R$. For fixed $c > 0$, let $cS = \{ cs : s \in S \}$. Show that $\sup(cS) = c\cdot\sup(S)$ and $\inf(cS) = c\cdot\inf(S)$.

Proof. Let $M = \sup S$. Then $cM$ is an upper bound for $cS$ as for all $cs \in cS$ we have that $M \ge s$ so $cM \ge cs$ as $c > 0$.

Now, if $\alpha \ge cs$ for all $cs \in cS$ (that is, $\alpha$ is an upper bound for $cS$), then $\alpha/c \ge s$ for all $s \in S$ so $\alpha$ is an upper bound for $S$. But then $M \le \alpha/c$ as $M$ is the least upper bound for $S$, so $cM \le \alpha$. So $cM = c\sup S = \sup(cS)$.

Now notice that $\inf(cS) = -\sup(-cS) = -c\sup(-S) = c\inf S$. $\Box$

2. (33.6) Let $f$ be integrable on $[a,b]$. Prove that, for any subset $S \subseteq [a, b]$ we have

Hint. For $x_0, y_0 \in S$, we have $\vert f(x_0)\vert - \vert f(y_0)\vert \le \vert f(x_0) - f(y_0)\vert \le M(f, S) - m(f, S)$.

Proof. The first part of the hint follows from the triangle inequality, as

The second part of the hint comes from how $M(f, S) \ge f(x_0)$ and $m(f, S) \le f(y_0)$ for all $x_0, y_0 \in S$.

On the other hand, we have that for any $\epsilon > 0$, properties of the supremum and infimum guarantee that there exists $x_0, y_0 \in S$ so that,

So we have that,

This holds for all $\epsilon > 0$, so we conclude that the desired inequality holds. $\Box$

3. (33.5) Show that $\left\vert \int_{-2\pi}^{2\pi} x^2 \sin^8(e^x)\; dx \right\vert \le \frac{16\pi^3}{3}$.

Proof. Using Theorems 33.4 and 33.5, together with the FTC I (for example), we have that (as $\left\vert \sin^8(e^x )\right\vert \le 1$)

4. (33.8) Let $f$ and $g$ be integrable functions on $[a, b]$.

1. It is a fact (see Exercise 33.7) that if $h$ is integrable on $[a,b]$, then so is $h^2$. Prove that $fg$ is integrable on $[a, b]$. Hint. Use that $4fg = (f+g)^2 - (f-g)^2$.

2. Show that $\max(f,g)$ and $\min(f,g)$ are integrable on $[a, b]$. You may use the results of Excercise 17.8 without proof.

Use that for integrable functions $f, g$ we have that $f+g$, $f-g$, $(f+g)^2$, $(f-g)^2$, $\vert f-g \vert$ are all integrable. Further applications of linearity of the integral prove (a), and together with $\min(f,g) = \frac 12 (f+g) - \frac 12 \vert f-g \vert$ prove the first part of (b). The second part then follows from $\max(f,g) = -\min(-f,-g)$ and integrability of $-f$, etc.

5. (33.10) Let $f$ be the function,

Prove that $f$ is integrable on $[-1, 1]$. Hint. See Excercise 33.11(c) and its solution in the textbook. (Why can we not apply the Dominated Convergence Theorem to prove that $f$ is integrable?)

Proof. Let $\epsilon > 0$. Notice that $f$ is continuous on each of $[-1, -\epsilon/8]$ and $[\epsilon/8, 1]$, and is hence integrable. This means that there exist partitions $Q_1$ of $[-1, -\epsilon/8]$ and $Q_2$ of $[\epsilon/8, 1]$ so that

Notice that $P = Q_1 \cup Q_2$ is a partition of $[-1, 1]$ and furthermore that

and,

(Why? The only terms in the two sums are those contributed by the interval $[-\epsilon/8, \epsilon/8]$ but we know that the supremum and infimum of $f$ on any such interval are $1$ and $-1$ respectively and that the width of this interval is $\epsilon/4$) So,

We can find any such $P$ for any $\epsilon > 0$, so we conclude that $f$ is integrable on $[-1, 1]$.


text:
Lemma 10.42.4. Let $k \subset K$ be a finitely generated field extension. There exists a diagram

$\xymatrix{ K \ar[r] & K' \\ k \ar[u] \ar[r] & k' \ar[u] }$

where $k \subset k'$, $K \subset K'$ are finite purely inseparable field extensions such that $k' \subset K'$ is a separably generated field extension.

Proof. This lemma is only interesting when the characteristic of $k$ is $p > 0$. Choose $x_1, \ldots , x_ r$ a transcendence basis of $K$ over $k$. As $K$ is finitely generated over $k$ the extension $k(x_1, \ldots , x_ r) \subset K$ is finite. Let $k(x_1, \ldots , x_ r) \subset K_{sep} \subset K$ be the subextension found in Fields, Lemma 9.14.6. If $K = K_{sep}$ then we are done. We will use induction on $d = [K : K_{sep}]$.

Assume that $d > 1$. Choose a $\beta \in K$ with $\alpha = \beta ^ p \in K_{sep}$ and $\beta \not\in K_{sep}$. Let $P = T^ n + a_1T^{n - 1} + \ldots + a_ n$ be the minimal polynomial of $\alpha$ over $k(x_1, \ldots , x_ r)$. Let $k \subset k'$ be a finite purely inseparable extension obtained by adjoining $p$th roots such that each $a_ i$ is a $p$th power in $k'(x_1^{1/p}, \ldots , x_ r^{1/p})$. Such an extension exists; details omitted. Let $L$ be a field fitting into the diagram

$\xymatrix{ K \ar[r] & L \\ k(x_1, \ldots , x_ r) \ar[u] \ar[r] & k'(x_1^{1/p}, \ldots , x_ r^{1/p}) \ar[u] }$

We may and do assume $L$ is the compositum of $K$ and $k'(x_1^{1/p}, \ldots , x_ r^{1/p})$. Let $k'(x_1^{1/p}, \ldots , x_ r^{1/p}) \subset L_{sep} \subset L$ be the subextension found in Fields, Lemma 9.14.6. Then $L_{sep}$ is the compositum of $K_{sep}$ and $k'(x_1^{1/p}, \ldots , x_ r^{1/p})$. The element $\alpha \in L_{sep}$ is a zero of the polynomial $P$ all of whose coefficients are $p$th powers in $k'(x_1^{1/p}, \ldots , x_ r^{1/p})$ and whose roots are pairwise distinct. By Fields, Lemma 9.28.2 we see that $\alpha = (\alpha ')^ p$ for some $\alpha ' \in L_{sep}$. Clearly, this means that $\beta$ maps to $\alpha ' \in L_{sep}$. In other words, we get the tower of fields

$\xymatrix{ K \ar[r] & L \\ K_{sep}(\beta ) \ar[r] \ar[u] & L_{sep} \ar[u] \\ K_{sep} \ar[r] \ar[u] & L_{sep} \ar@{=}[u] \\ k(x_1, \ldots , x_ r) \ar[u] \ar[r] & k'(x_1^{1/p}, \ldots , x_ r^{1/p}) \ar[u] \\ k \ar[r] \ar[u] & k' \ar[u] }$

Thus this construction leads to a new situation with $[L : L_{sep}] < [K : K_{sep}]$. By induction we can find $k' \subset k''$ and $L \subset L'$ as in the lemma for the extension $k' \subset L$. Then the extensions $k \subset k''$ and $K \subset L'$ work for the extension $k \subset K$. This proves the lemma. $\square$

Comment #738 by Keenan Kidwell on

The $L^{\prime\prime}$ in the last line should be $L^\prime$.

There are also:

• 4 comment(s) on Section 10.42: Separable extensions

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text:
# -XXaltjvm, dtruss and ASAN

Been a while since I’ve posted anything technical, but in any case it is good to be back! Aside from work, I hadn’t been doing any hacking on my own for a while, due to the lack of time and energy. So it’s a nice change for a break this year that I’ve finally gotten the motivation back to be developing for leisure again :)

The current little hack is simply a start from scratch project to build an alternative JVM runtime. The basis of this design will be based on what I’ve learnt from the DRLVM’s code base, and avoiding some of their design pitfalls that I’ve encountered with their code.

The additional challenge is that my new development environment is MacOSX - while UNIX-y enough, just simply does not have a number of the GNU tools which I miss (objcopy, ldd) and some new formats that I’m not familiar with (eg. Mach-O vs ELF) and some new tools that I have to Google to find the analogue equivalent of (eg. strace vs dtruss), but nothing insurmountable.

It’s been months since I’ve last touched VM’s code, and my rusty-ness shows, so this entry is more of an re-primer on how to load an alternative JVM runtime. This is one of those situations in which Googling works less well than code-diving, and the flag that I was after is -XXaltjvm=. This does not follow the -XX: semantics of the current day JVM, which is an indication of its age. os_bsd.cpp explains its use:

// Support for the java launcher's '-XXaltjvm=<path>' option. Typical
// value for buf is "<JAVA_HOME>/jre/lib/<arch>/<vmtype>/libjvm.so"
// or "<JAVA_HOME>/jre/lib/<vmtype>/libjvm.dylib". If "/jre/lib/"
// appears at the right place in the string, then assume we are
// installed in a JDK and we're done. Otherwise, check for a
// JAVA_HOME environment variable and construct a path to the JVM
// being overridden.

With my lack of a creativity, I’ve decided to name my project runtime; hence the command line to use this would be -XXaltjvm=runtime. The JVM didn’t like a symlink, which I checked with dtruss:

% sudo dtruss ...

stat64("/Users/vincent.liu/repos/graal-exp/labsjdk1.8.0_111-jvmci-0.23/jre/lib/runtime/libjvm.|", 0x7FFF5AAAB408, 0x1000)       = -1 Err#2
write_nocancel(0x2, "Error: missing runtime' JVM at /Users/vincent.liu/repos/graal-exp/labsjdk1.8.0_111-jvmci-0.23/jre/lib/runtime/libjvm.dylib'.\nPlease install or use the JRE or JDK that contains these missing components|", 0xCB)               = 203 0

% ls -lah /Users/vincent.liu/repos/graal-exp/labsjdk1.8.0_111-jvmci-0.23/jre/lib/runtime/libjvm.dylib
lrwxr-xr-x  1 vincent.liu  44B Dec 24 07:46 /Users/vincent.liu/repos/graal-exp/labsjdk1.8.0_111-jvmci-0.23/jre/lib/runtime/libjvm.dylib -> runtime/build/runtime/src/.libs/libjvm.dylib

So the actual binary must be copied to the \$JAVA_HOME path, which once after I’ve done that, I’m hitting it not finding the linked ASAN checks that I’ve compiled in:

Error: dl failure on line 674

No biggie, but I’ve got to stop to read more on how to deal with automake and libtool, all new things I’ve been learning. While autotools looks awfully complex, it’s still so much better than the old ‘Apache Ant’ build system!


text:
# How to obtain all minima? [duplicate]

If you Solve an equation, you obtain all solutions.

But If you NMinimize a function, you only get one minimum.

The question is: is there a way to obtain all minima of a given function?

Example:

Solve[((x - 2) (x + 2))^2 == 0, x]
Print["____________"]
NMinimize[((x - 2) (x + 2))^2, x]
NMinimize[ {((x - 2) (x + 2))^2, x >= 0}, x]

yields

{{x -> -2}, {x -> -2}, {x -> 2}, {x -> 2}}
____________
{0., {x -> -2.}}
{0., {x -> 2.}}

You can see that for Solve you obtain all solutions, even with their multiplicity, but in the minimization, you only obtain one solution.

• NMinimize does not perform a global optimization, and will return only one (quite possibly local) minimum. You can try Minimze which will return a global minimum for polynomial functions, but not other (local) ones. – Yves Klett Mar 25 '14 at 10:32
• related: How to find all the local minima/maxima in a range. About infinite domain, Solve for f' == 0 && f'' >0. – Kuba Mar 25 '14 at 10:35
• @Kuba: Nice answer in that link, thanks for posting, had not seen it before. – ciao Mar 25 '14 at 10:41


text:
# How to Find the Factorial of a Number using Python?

PythonServer Side ProgrammingProgramming

Factorial of a number is the product of all integers between 1 and itself. To find factorial of a given number, let us form a for loop over a range from 1 to itself. Remember that range() function excludes the stop value. Hence stop value should be one more than the input number.

Each number in range is cumulatively multiplied in a variable f which is initialised to 1

## Example

num=int(input('enter a number'))
f=1
for i in range(1,num+1):
f=f*i
print ('factorial of', num, '=',f)

## Output

Sample run of above code −

enter a number5
factorial of 5 = 120
Published on 24-Feb-2018 15:14:38


text:
# Task 4: Statistical testing

This is the second part of tick 3.

## Step 1: Magnitude classifier

The lexicon classifier you created in Task 1 does not use all the information available in the lexicon. In this step you should modify the simple classifier to account for the strength/magnitude of the sentiment expressed by each word. The tester assumes the use of weight 2 for strong sentiment.

## Step 2: Sign test

To investigate whether the two systems are significantly different, we will use the sign test. In this case, the sign test is based on the binomial distribution.

Count all cases when system 1 is better than system 2, when system 2 is better than system 1, and when they are the same. Call these numbers Plus, Minus and Null. The sign test returns the probability that the null hypothesis is true.

This probability is called the p-value and it can be calculate for the two-sided sign test using the following formula:

$2\sum_{i=0}^{k}{n \choose i}q^i (1-q)^{n-i}$

where $$n=2 \lceil \frac{Null}{2} \rceil + Plus + Minus$$ is the total number of cases and $$k=\lceil \frac{Null}{2} \rceil + \min{\{Plus,Minus\}}$$ is the number of cases with the less common sign. In this experiment, $$q=0.5$$.

Here, we are dividing the null cases evenly between the two signs, rounding up if necessary (why is this better than rounding down? Hint: we are looking for statistical significance. We then use the formula as above. The formula is multiplied by two because this is a two-sided sign test and tests for the significance of differences in either direction, so a difference in one of the directions is half as significant as it would be otherwise.

The numbers you will calculate are high enough to cause overflow if you use standard double or int types. Use BigInteger class.

## Step 3: Test your classifiers

Is the difference in the scores obtained by your simple classifier and the magnitude classifier statistically significant? What about the magnitude and Naive Bayes classifiers? Make sure you compare the results on the same examples and that you don't test on the training set. As usual, please see Exercise4Tester.java as a guide for what is expected.


text:
# Documentation

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# ltePUCCH2DRS

PUCCH format 2 demodulation reference signal

## Syntax

• ``seq = ltePUCCH2DRS(ue,chs,ack)``
example
• ``````[seq,info] = ltePUCCH2DRS(ue,chs,ack)``````
example

## Description

example

````seq = ltePUCCH2DRS(ue,chs,ack)` returns a matrix containing demodulation reference signal (DRS) associated with PUCCH format 2 transmission, given a structure of UE-specific settings, a structure with channel transmission configuration settings, and hybrid ARQ (HARQ) indicator values, `ack`. ```

example

``````[seq,info] = ltePUCCH2DRS(ue,chs,ack)``` also returns a PUCCH information structure array, `info`.```

## Examples

collapse all

Generate PUCCH Format 2 DM-RS symbols for UE specific settings.

Initialize input configuration structures (`ue` and `chs`). Here no HARQ bits will be sent by inputting an empty `ack` vector. Generate the PUCCH Format 2 DM-RS symbols.

```ue.NCellID = 1; ue.NSubframe = 0; ue.CyclicPrefixUL = 'Normal'; ue.Hopping = 'Off'; chs.ResourceIdx = 0; chs.ResourceSize = 0; chs.CyclicShifts = 0; sym = ltePUCCH2DRS(ue,chs,[]); ```

Demonstrate Uplink Release 11 cooordinated multipoint (CoMP) operation. Intercell interference can be avoided by using a virtual cell identity for a potentially interfering UE in a neighboring cell.

Configuration for UE of interest, UE 1 in cell 1.

```ue1.NCellID = 1; ue1.NSubframe = 0; ue1.CyclicPrefixUL = 'Normal'; ue1.Hopping = 'Off'; chs1.ResourceIdx = 0; chs1.ResourceSize = 0; chs1.CyclicShifts = 0; ack1 = 0; ```

Configuration for interferer, UE 2 in cell 2.

```ue2.NCellID = 2; ue2.NSubframe = 0; ue2.CyclicPrefixUL = 'Normal'; ue2.Hopping = 'Off'; chs2.ResourceIdx = 1; chs2.ResourceSize = 0; chs2.CyclicShifts = 0; ack2 = 0; ```

Measure the interference between the DM-RS signals.

```interferenceNoCoMP = abs(sum(ltePUCCH2DRS(ue1,chs1,ack1).*conj(ltePUCCH2DRS(ue2,chs2,ack2)))) ```
```interferenceNoCoMP = 5.4903 ```

Reconfigure interferer for CoMP operation: use virtual cell identity equal to the cell identity for the UE of interest.

```ue2.NPUCCHID = ue1.NCellID; ```

Measure the interference between the DM-RS signals when using CoMP.

```interferenceUsingCoMP = abs(sum(ltePUCCH2DRS(ue1,chs1,ack1).*conj(ltePUCCH2DRS(ue2,chs2,ack2)))) ```
```interferenceUsingCoMP = 3.9627e-15 ```

Comparing the correlations between the DM-RS signals for two UEs with and without CoMP, `interferenceUsingCoMP` and `interferenceNoCoMP` respectively. Using CoMP, the interference is reduced to effectively zero.

Generate the PUCCH format 2 DM-RS sequences for two transmit antenna paths.

Initialize UE-specific and channel configuration structures. Provide an empty vector for the `ack`, indicating there are no HARQ bits for this PUCCH transmission. Generate PUCCH 2 DM-RS and information outputs.

```ue.NCellID = 1; ue.NSubframe = 0; ue.CyclicPrefixUL = 'Normal'; ue.Hopping = 'Off'; chs.ResourceIdx = [0 3]; chs.ResourceSize = 0; chs.CyclicShifts = 0; ack = []; [drsSeq,info] = ltePUCCH2DRS(ue,chs,ack); ```

Because there are two antennas, the DM-RS sequences are output as a two- column vector, and the `info` output structure contains two elements.

```drsSeq(1:10,:) size(info) ```
```ans = 0.5000 + 0.5000i 0.5000 + 0.5000i -0.1830 + 0.6830i -0.6830 - 0.1830i -0.1830 - 0.6830i 0.1830 + 0.6830i 0.5000 - 0.5000i -0.5000 - 0.5000i 0.6830 + 0.1830i 0.6830 + 0.1830i -0.1830 - 0.6830i 0.6830 - 0.1830i 0.5000 + 0.5000i -0.5000 - 0.5000i 0.1830 - 0.6830i -0.6830 - 0.1830i 0.6830 - 0.1830i 0.6830 - 0.1830i -0.5000 - 0.5000i 0.5000 - 0.5000i ans = 1 2 ```

View the contents of the two `info` structure elements.

```info(1) info(2) ```
```ans = struct with fields: Alpha: [1.0472 3.1416 1.5708 2.0944] SeqGroup: [1 1] SeqIdx: [0 0] NResourceIdx: [1 10] NCellCyclicShift: [193 89 101 234] Symbols: [1×4 double] OrthSeq: [2×2 double] ans = struct with fields: Alpha: [2.6180 4.7124 0 0.5236] SeqGroup: [1 1] SeqIdx: [0 0] NResourceIdx: [4 7] NCellCyclicShift: [193 89 101 234] Symbols: [1×4 double] OrthSeq: [2×2 double] ```

## Input Arguments

collapse all

UE-specific configuration settings, specified as a structure that can contain the following fields.

Parameter FieldRequired or OptionalValuesDescription
`NCellID`Required

Integer from 0 to 503

Physical layer cell identity

`NSubframe`Required

0 (default), nonnegative scalar integer

Subframe number

`CyclicPrefixUL`Optional

`'Normal'` (default), `'Extended'`

Cyclic prefix length

`Hopping`Optional

`'Off'` (default), `'Group'`

Frequency hopping method.

`NPUCCHID`Optional`NCellID` (default),

Integer from 0 to 503

PUCCH virtual cell identity. If this field is not present, `NCellID` is used as the identity.

Data Types: `struct`

PUCCH channel settings, specified as a structure that can contain the following fields.

Parameter FieldRequired or OptionalValuesDescription
`ResourceIdx`Optional

0 (default), integer from 0 to 1185 or vector of integers.

PUCCH resource indices which determine the physical resource blocks, cyclic shift, and orthogonal cover used for transmission. (${n}_{PUCCH}^{\left(2\right)}$). Define one index for each transmission antenna.

`ResourceSize`Optional

0 (default), integer from 0 to 98.

Size of resource allocated to PUCCH Format 2 (${N}_{RB}^{\left(2\right)}$)

`CyclicShifts`Optional

0 (default), integer from 0 to 7

Number of cyclic shifts used for format 1 in resource blocks (RBs) with a mixture of format 1 and format 2 PUCCH, specified as an integer from 0 to 7. (${N}_{cs}^{\left(1\right)}$)

Hybrid ARQ indicator values, specified as nonnegative integer vector. This vector is expected to be the block of bits b(0),...,b(Mbit–1) specified in TS 36.211 [1], Section 5.4.2. An Mbit value of 20, 21, or 22 corresponds to PUCCH format 2, 2a, or 2b, respectively, as described in TS 36.211 [1], Table 5.4-1.

Example: [ ] indicates that no HARQ are transmitted in the subframe.

## Output Arguments

collapse all

PUCCH format 2 DRS values, returned as a numeric matrix. The symbols for each antenna are in the columns of `seq`, with the number of columns determined by the number of PUCCH resource indices specified in `chs``.``ResourceIdx`.

 Note:   The standard does not support format 2a or 2b transmission with extended cyclic prefix. If the `ack` setting corresponds to format 2a or 2b transmission and extended cyclic prefix is set for `ue``.``CyclicPrefixUL`, the function returns an empty matrix for `seq`.

Data Types: `double`
Complex Number Support: Yes

PUCCH format 2 information, returned as a structure array with elements corresponding to each transmit antenna and containing these fields. When configured for format 2a or 2b transmission with extended cyclic prefix, the `info` structure contains all fields, but each field is empty.

Reference signal cyclic shift for each OFDM symbol, returned as a two-column vector. (α)

PUCCH base sequence group number for each slot, returned as two-column vector. (u)

PUCCH base sequence number for each slot, returned as two-column vector. (v)

PUCCH resource indices for each slot, returned as two-column vector. (n')

Cell-specific cyclic shift for each OFDM symbol, returned as vector. (${n}_{cs}^{cell}$)

Modulated data symbols, returned as a vector. There is one element for each OFDM symbol. (z)

Example: [0.7071 + 0.7071i,...]

Orthogonal sequence for each slot, returned as a 4-by-2 numeric matrix. ($\overline{w}$)

Example: [1.000 + 1.000i,...]

Data Types: `struct`

## References

[1] 3GPP TS 36.211. "Physical Channels and Modulation." 3rd Generation Partnership Project; Technical Specification Group Radio Access Network; Evolved Universal Terrestrial Radio Access (E-UTRA). URL: http://www.3gpp.org.


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off shell cross section

Asked by Wang Daohan on 2020-04-06

I have a problem with on shell and off shell cross sections. maybe it's naive, but I am confused for a long time. Take a SM example.
I generate p p > t t~ > b b~ w- j j, let's call it process 1. this process contain t and w+ off shell and on shell contributions. But the cross section is smaller than this process p p > t t~, t > w+ > b j j, t~ > b~ w- which only contain on shell contribution, let's call it process2.
But when I generate p p > t t~ > b b~ w+ w- in madgraph and write the decay in madspin. Let's call it process 3. The corresponding cross section then is larger than this process p p > t t~, t > w+ > b j j, t~ > b~ w-.
I want to know why the cross section of process 1 is smaller than process 2 and why the cross section of process 3 is larger than process 2.
I hope you can help me with it.
Thank you very much!
Dorhand

Question information

Language:
English Edit question
Status:
Solved
For:
Assignee:
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Solved by:
Olivier Mattelaer
Solved:
2020-04-06
Last query:
2020-04-06
2020-04-06
 Olivier Mattelaer (olivier-mattelaer) said on 2020-04-06: #1

Hi,

Did you look at the FAQ that I just linked?
I believe that what you observe is just related to different cut (and in particular to the variable cut_decays set to False in your case)

Cheers,

Olivier

FAQ #2442: “why production and decay cross-section didn't agree.”.

 Wang Daohan (dorhand) said on 2020-04-06: #3

Thanks very much for your reply. The variable cut_decays defaulted setting is False. But my question is still there. Why the cross section of this process p p > t t~ > b b~ w- j j is larger than p p > t t~, t > w+ > b j j, t~ > b~ w- ? The only difference is the latter t tbar is forced to be on shell. I really hope you can give me some hints.

 Olivier Mattelaer (olivier-mattelaer) said on 2020-04-06: #4

Hi,

When i run:
> p p > t t~ > b b~ w- j j

INFO: #***************************************************************************
#
# original cross-section: 241.955420632
# scale variation: +30.5% -21.9%
# central scheme variation: +5.02% -26.1%
# PDF variation: +4.38% -4.38%
#
# dynamical scheme # 1 : 202.875 +28.6% -21% # \sum ET
# dynamical scheme # 2 : 197.925 +28.4% -20.8% # \sum\sqrt{m^2+pt^2}
# dynamical scheme # 3 : 254.094 +30.9% -22.1% # 0.5 \sum\sqrt{m^2+pt^2}
# dynamical scheme # 4 : 178.787 +27.3% -20.3% # \sqrt{\hat s}
#***************************************************************************

When I run
> p p > t t~, t > w+ > b j j, t~ > b~ w-

(with cut_decays=True)
INFO: #***************************************************************************
#
# original cross-section: 237.297817773
# scale variation: +30.3% -21.7%
# central scheme variation: +4.91% -26%
# PDF variation: +4.29% -4.29%
#
# dynamical scheme # 1 : 199.166 +28.4% -20.8% # \sum ET
# dynamical scheme # 2 : 194.22 +28.2% -20.7% # \sum\sqrt{m^2+pt^2}
# dynamical scheme # 3 : 248.939 +30.7% -22% # 0.5 \sum\sqrt{m^2+pt^2}
# dynamical scheme # 4 : 175.718 +27.1% -20.1% # \sqrt{\hat s}
#***************************************************************************

So the difference is quite small (4pb) and you can see that the theoretical uncertainty is much bigger than that (30% of the corss-section). The fact is that the default dynamical scale depends of the exact syntax that you use and therefore can lead to difference in term of prediction which are within theoretical uncertainty.

Now if I run for HT/2 as the central dynamical scale (which does not depend of the syntax used)
> p p > t t~ > b b~ w- j j

INFO: #***************************************************************************
#
# original cross-section: 261.413583767
# scale variation: +30.9% -22%
# central scheme variation: + 0% -29.6%
# PDF variation: +4.31% -4.31%
#
# dynamical scheme # 1 : 208.918 +28.5% -20.9% # \sum ET
# dynamical scheme # 2 : 203.786 +28.3% -20.8% # \sum\sqrt{m^2+pt^2}
# dynamical scheme # 3 : 261.414 +30.9% -22% # 0.5 \sum\sqrt{m^2+pt^2}
# dynamical scheme # 4 : 184.028 +27.2% -20.2% # \sqrt{\hat s}
#***************************************************************************

When I run
> p p > t t~, t > w+ > b j j, t~ > b~ w-

(with cut_decays=True)
INFO: #***************************************************************************
#
# original cross-section: 250.52584976
# scale variation: +30.7% -22%
# central scheme variation: +9.78e-09% -29.3%
# PDF variation: +4.28% -4.28%
#
# dynamical scheme # 1 : 200.41 +28.4% -20.8% # \sum ET
# dynamical scheme # 2 : 195.454 +28.2% -20.7% # \sum\sqrt{m^2+pt^2}
# dynamical scheme # 3 : 250.526 +30.7% -22% # 0.5 \sum\sqrt{m^2+pt^2}
# dynamical scheme # 4 : 177.001 +27.1% -20.2% # \sqrt{\hat s}
#***************************************************************************

Finally If I try to remove the onshell requirement with this syntax (in order to match the above result) by setting bwcutoff to 1e9, I get:
INFO: #***************************************************************************
#
# original cross-section: 261.589530988
# scale variation: +30.8% -22%
# central scheme variation: +2.96e-09% -29.4%
# PDF variation: +4.26% -4.26%
#
# dynamical scheme # 1 : 209.256 +28.5% -20.9% # \sum ET
# dynamical scheme # 2 : 204.063 +28.2% -20.7% # \sum\sqrt{m^2+pt^2}
# dynamical scheme # 3 : 261.59 +30.8% -22% # 0.5 \sum\sqrt{m^2+pt^2}
# dynamical scheme # 4 : 184.723 +27.1% -20.2% # \sqrt{\hat s}
#***************************************************************************

This being said, using
> p p > t t~ > b b~ w- j j
or
> p p > t t~, t > w+ > b j j, t~ > b~ w-
with very large bwcutoff are both impacted by another source of theoretical error due to the fact that you neglect interference with other type of diagrams and by the handling of the width (which is using here a fix value, i.e no running).
Note that you can use complex mass scheme in madgraph (but this requires that you also decay the W for that scheme to make sense).

Cheers,

Olivier

> On 6 Apr 2020, at 09:22, Wang Daohan <email address hidden> wrote:
>
> Question #689732 on MadGraph5_aMC@NLO changed:
>
> Wang Daohan gave more information on the question:
> Thanks very much for your reply. The variable cut_decays defaulted
> setting is False. But my question is still there. Why the cross section
> of this process p p > t t~ > b b~ w- j j is larger than p p > t t~, t >
> w+ > b j j, t~ > b~ w- ? The only difference is the latter t tbar is
> forced to be on shell. I really hope you can give me some hints.
>
> --
> You received this question notification because you are an answer
> contact for MadGraph5_aMC@NLO.

 Wang Daohan (dorhand) said on 2020-04-06: #5

Thank you very much !

 Wang Daohan (dorhand) said on 2020-04-06: #6

Thanks Olivier Mattelaer, that solved my question.

To post a message you must log in.


text:
A finite classical group is best thought of as a group of linear operators on some vector space defined over a finite field. Which means, of course, that I can take a basis for this vector space and then represent the elements of my group as matrices.

However, I have almost never seen anyone do this in the literature. “Taking a basis” is thought of as a rather crude thing to do when doing linear algebra – one generally tries to write general arguments that do not refer to any particular basis. However, speaking for myself, when I’m trying to work out what the hell is going on inside a finite matrix group, I often end up trying to write down individual elements as matrices… And then hiding all this when it comes to writing up the paper!

This means that I have never publicly written down any of these calculations, despite using many of them over and over again. So this page is designed to be a little repository for me to note down interesting observations about such calculations… And perhaps they’ll be useful for someone else some time…

## Working with $\Omega_4^+(q)$

This family of groups has some weird behaviour, especially when $q$ is even. For instance, let us write $\mathcal{U}$ for the set of maximal totally singular subspaces in a formed space of type $O+$ and dimension $2m$. If $q$ is even, then, provided $(m,q)\neq (2,2)$, we can define $\Omega_{2m}^+(q)$ to be the group inducing odd permutations on $\mathcal{U}$…. If $(m,q)=(2,2)$, however, the group so defined is not $\Omega_4^+(2)$, and one needs to define it in a completely different way (see Kleidman and Liebeck, p.31).

# Finding some elements in $\Omega_4^+(q)$.

Let $(e_1,f_1)$ and $(e_2, f_2)$ be hyperbolic pairs, and consider the basis $\mathcal{B}={e_1, e_2, f_2, f_1}$ maintaining order. Then, one can calculate directly that $O_4^+(q)$ contains elements which are written with respect to $\mathcal{B}$ in the form

these elements form a subgroup $S$ of size $q^2$; indeed, for $q$ odd, they form a Sylow $p$-subgroup of $O_4^+(q)$.

When $q$ is odd, one can look at orders to see directly that $S$ is actually a subgroup of $\Omega_4^+(q)$. Indeed, the same is true when $q$ is even, but to see this it is easiest to observe that $S$ is normalized by the element

This element clearly takes one element $U=\langle e_1, e_2\rangle \in \mathcal{U}_2$ to another element $W=\langle e_1, f_2\rangle\in \mathcal{U}_2$. What is more $U\cap W$ has codimension $1$ in both $U$ and $W$. This allows us to conclude that $g\in SO_4^+(q)\setminus \Omega_4^+(q)$ (see Kleidman and Liebeck, p. 30). Now order considerations imply that $S$ must be a subgroup of $\Omega_4^+(q)$.

# Finding all elements in $\Omega_4^+(q)$.

Let $R$ be the set of elements whose transpose is in $S$. One sees immediately that both $R$ and $S$ lie in $\Omega_4^+(q)$ and hence so does $\langle R,S\rangle$.

If one sets $a$ to equal $0$, while $b$ ranges across $\mathbb{F}_q$, in both $R$ and $S$, then one immediately observes a copy of $SL_2(q)$ in $\Omega_4^+(q)$. The same is true setting $b$ to equal $0$. Since these two copies effectively “avoid interaction” one immediately obtains a copy of $SL_2(q)\circ SL_2(q)$ inside $\Omega_4^+(q)$. (The central product is due to the fact that both copies of $SL_2(q)$ share $-I$ as an element.)

Now one can observe that $\Omega_4^+(q)$ also contains the element

Now order considerations allow us to conclude that $\Omega_4^+(q)\cong (SL_2(q)\circ SL_2(q)):2$, and we have written down all elements of the group.

## $Sp_{n-2}(2^a) < \Omega_n^+(2^a)$

When $q$ is even, $\Omega_n^+(q)$ has a maximal subgroup – the stabilizer of a non-degenerate $1$-space – isomorphic to $Sp_{n-2}(q)$. I will write down the elements of this subgroup for $n=4$; the general case follows similarly.

First, we adjust our basis from before to be $\mathcal{C}={z_1, x_1, x_2, y_2}$ where $z_1=x_1+y_1$.

With respect to this basis, our quadratic form becomes

and we see that $z_1$ is non-singular. Now simply observe that the following elements stabilize $z_1$:

Doing the “transpose trick” like before, one immediately obtains a copy of $SL_2(q)$ in this stabilizer and (using the perfectness of $SL_2(q)$ for $q>3$ if necessary), one obtains that the stabilizer of $z_1$ in $\Omega_4^+(q)$ is isomorphic to $SL_2(q)\cong Sp_{2}(q)$.

One can do this more generally for larger $n$ – one simply has to exhibit the root groups of $Sp_{n-2}(q)$ inside the stabilizer of $z_1$ in $\Omega_4^+(q)$. There are two sorts of root group here: the first already ocur in $\Omega_{n-2}^+(q)$ and so are easy to write down in $\Omega_n^+(q)$; the second all take the form given above.


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Infoscience

Journal article

# Search for Majorana Neutrinos in B- -> pi(+)mu(-)mu(-) Decays

A search for heavy Majorana neutrinos produced in the B- -> pi(+)mu(-)mu(-) decay mode is performed using 3 fb(-1) of integrated luminosity collected with the LHCb detector in pp collisions at center-of-mass energies of 7 and 8 TeV at the LHC. Neutrinos with masses in the range 250 to 5000 MeV and lifetimes from zero to 1000 ps are probed. In the absence of a signal, upper limits are set on the branching fraction B(B- -> pi(+)mu(-)mu(-)) as functions of neutrino mass and lifetime. These limits are on the order of 10(-9) for short neutrino lifetimes of 1 ps or less. Limits are also set on the coupling between the muon and a possible fourth-generation neutrino.


text:
# For the most recent year available, the mean annual cost to attend a private university in the United States was $50,900. Assume the distribution 3 answers ###### Question: For the most recent year available, the mean annual cost to attend a private university in the United States was$50,900. Assume the distribution of annual costs follows the normal probability distribution and the standard deviation is $4,500. Ninety-five percent of all students at private universities pay less than what amount ## Answers #### Similar Solved Works 3 answers ### Abullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a Abullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a mass of 0.095kg. the block rests on a frictionless surface, and is thin enough that the bullet passes completely through it. immediately after the bullet exits the block, the speed of the block is 23m/s. (a) what is... 4 answers ### Add and subtract the rational expressions. 7x+5/x-1 -(8x1/x-1+2x-4/x-1) Which statements are true about Add and subtract the rational expressions. 7x+5/x-1 -(8x1/x-1+2x-4/x-1) Which statements are true about the process for adding and subtracting the rational expressions? Select three options. 1. The expression is not defined when x = –1. The first step in the simplification is . 2. The denominator ... 5 answers ### In response to an increase in the number of immigrants filling jobs in northern factories in the early 19th century In response to an increase in the number of immigrants filling jobs in northern factories in the early 19th century... 10 answers ### At the restaurant, mike and three friends decided to divide the bill evenly. if each person paid$13

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text:
### Home > CC4 > Chapter 2 > Lesson 2.2.3 > Problem2-74

2-74.

Each part below represents a different situation. For each situation, determine the pattern of growth and the number of tiles in Figure 0 or the $y$-intercept. Homework Help ✎

1. How many more squares does Figure 3 have than Figure 2?

1. Make a slope triangle between two points on the graph to find the slope, or the growth rate.

Growth rate: $−2$

1. $y=3x−14$

$G = 3, y-int = –14$

2.  $\boldsymbol x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $\boldsymbol y$ $18$ $13$ $8$ $3$ $-2$ $-7$ $-12$

What is the difference between each $y$-value?
What is the value of $y$ when $x$ is $0$?

3. Which situation above is growing (increasing) at the fastest rate?


text:
# Given a matrix, determine whether linear transformation and find bases for kernel and range

Define T : $P_3(\mathbb{R}) → M_2(\mathbb{R})$ by $T(f)$ = $\begin{bmatrix}f'(0) & 2f(1)\\0 & f''(2)\end{bmatrix}$

a) Show that T is a linear transformation.

b) Find bases for Ker(T) and range(T)

I'm so confused with the matrix. What should be my approach?

• Your approach would be the same as always, don't "mind" the matrix. Verify, for part (a), whether $T(\alpha f + \beta g) = \alpha \, T( f )+ \beta \, T( g)$. Can you start? Jun 6 '17 at 21:03

If T is a linear transformation then for every pair polynomials $p,q \in\mathbb P_3 ; T(p + q) = T(p) + T(q)$

b) $Ker(T):$ what is the set of $p$ such that $T(p) = \mathbb 0$

$p(x) = ax^3 +bx^2 + cx + d$

Find $a,b,c,d$ such that

$p(1) = 0\\ p'(0) = 0\\ p''(2) = 0$

The range, are there any matrices that cannot be formed? you must have a $0$ in the bottom left hand corner. Any other restrictions?

• so that would be $T(f+g) = \begin{bmatrix}f'(0) & 2f(1)\\0 & f''(2)\end{bmatrix} + \begin{bmatrix}g'(0) & 2g(1)\\0 & g''(2)\end{bmatrix} = \begin{bmatrix}f'(0)+g'(0) & 2f(1) + 2g(1)\\0 & f''(2)+g''(2)\end{bmatrix} = T(f)+T(g)$ and $T(rf) = \begin{bmatrix}rf'(0) & 2rf(1)\\0 & rf''(2)\end{bmatrix} = r(T(f))$ but how about the range and kernel?
– Antt
Jun 6 '17 at 21:09


text:
Computer Science

Imagine you have a file that contains the following text:

This is an example file.
It has several lines of text.
Nothing terribly interesting.

You can read and print every line in this file using the following code:

file = open('example.txt')

for line in file:
print(line)

file.close()

When you call open('example.txt'), this will open a file in the same directory as the script you are writing. The open() function returns a file object, which we store in the file variable.

After you open the file, you can iterate over all of its lines using:

for line in file

You then have to remember to close the file, with file.close().

## A better way to open files

A better way to open files uses the with keyword:

with open('example.txt') as file:
for line in file:
print(line)

This code does the same thing! It opens the file in example.txt, sets the file variable to be a file object that can read this file, and then uses for line in file to iterate over the lines in the file.

An important feature of this syntax is that the file is automatically closed when you finish the with block. We will use this syntax in class because having the file be automatically closed is helpful.

## Iteration

We previously saw iteration with lists. Remember:

numbers = [2, 23, 17, 75]
for number in numbers:
print(number*2)

Iteration over the lines of a file uses the same for ... in statement:

with open('example.txt') as file:
for line in file:
print(line)

## Getting all the lines in a file

This is a useful function that we will reuse throughout the course:

def get_lines(filename):
with open(filename) as file:
lines = []
for line in file:
lines.append(line)
return lines

This function collects all the lines in the file into a list and then returns that list. If you call it:

print(get_lines('example.txt'))

You will see this output:

['This is an example file.\n', 'It has several lines of text.\n', 'Nothing terribly interesting.\n']

Notice that each line has a newline character at the end, \n. A file is just a long string of characters! The only thing that indicates a new line is the newline character.

We can write a shorter version of get_lines() as follows:

def get_lines(filename):
with open(filename) as file:
return list(file)

Because a file is iterable (we can use for ... in to loop over the lines), the list() function will, in one step, collect all of the lines into a list.

## Split

We can take any string and split it into a list. For example:

long_string = "this is a string containing several words"
result = long_string.split()
print(result)

This will print:

['this', 'is', 'a', 'string', 'containing', 'several', 'words']

You can do this when reading a file, because the lines in a file are just a string. For example, imagine we have a file called input.txt that contains the following:

5 6 3
2 7 10
5 2 1

We can split and then print these lines like this:

def split_lines(input_file):
with open(input_file) as infile:
for line in infile:
words = line.split()
print(words)

split_lines('input.txt')

This will print:

['5', '6', '3']
['2', '7', '10']
['5', '2', '1']

## Converting to integers

When we read numbers from a file, they are read as strings. See the above example, showing that each number, such as 5, is a string.

If we want to calculate something with these numbers, we need to convert them to integers first. We can do this with the int() function. For example:

a = '5'
num = int(a)
print(num + 5)

This will print:

10

We cannot do this without first converting a to an integer using int(). If we tried to add 5 to a, this would result in an error because we can’t add an integer to a string.

We can do this with an entire file of numbers as well. Using our same input.txt from above, we can add 2 to all of the numbers in the file:

def get_lines(filename):
with open(filename) as file:
return list(file)

words = line.split()
for word in words:
print(int(word) + 2)

def increment_by_2(input_file):
lines = get_lines(input_file)
for line in lines:

increment_by_2('input.txt')

This will print:

7
8
5
4
9
12
7
4
3

## Converting to floats

We can likewise convert a string to a float:

a = '5.3'
num = float(a)
print(num + 5)

This will print:

10.3

## Example

Here is an example that covers many of these concepts. We want to compute the mean of every line in a file. We have a data file called data.txt that contains:

1 2 3 4 5
4 5 6 7 8
9 10 11 12 13

We need to read each line, compute the mean, and then print it out. Here is one solution:

def get_lines(filename):
"""Returns a list of the lines in the file."""
with open(filename) as file:
return list(file)

def get_numbers(line):
"""Parses out the integers in line into a list.
"""
tokens = line.split()
numbers = []
for token in tokens:
numbers.append(int(token))
return numbers

def average(numbers):
"""Computes the average of a list of numbers
"""
total = 0
for number in numbers:
total = total + number

def print_means(filename):
"""Prints the average value for each line in the file."""
for line in get_lines(filename):
numbers = get_numbers(line)
ave = average(numbers)
print(ave)

filename = 'data.txt'
print_means(filename)

If you look at just the print_means() function, you can see the entire logic of the program:

• loop through all of the lines in a file
• for each line
• convert the line into a list of numbers
• compute the average of that list of numbers
• print the average

As you decompose problems, start with the big picture, like this. You should already have a get_lines() function. Then write and test the first function, get_numbers() Once that is working, write and test the average() function.