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# 9.1: Nuclear Particles: Review and Reactivity

Learning Objectives
• Write the nuclear symbol of a given elemental isotope.
• Define nuclear equation.

The three main subatomic particles, electrons, protons, and neutrons, were first defined and discussed in Chapter 2.  Recall that electrons are located in orbitals, which are defined regions in space that occupy the majority of the volume in an atom.  Furthermore, the bonding interactions and chemical reactivity of an element is solely-determined by the number of valence electrons, which are found in the highest-energy orbitals, that are present in an atom of that element.  Electrons, which are symbolized as "e," remain anchored in atoms due to their electrostatic attraction to the protons that are located in the nucleus, which is the densely-packed space at the center of an atom.  The atomic number of an element, which is written above the elemental symbol in a box on the periodic table, is a unique value that directly corresponds to the the identity of that element and, furthermore, defines the number of protons that exist in every atom of that element.  In order to remain electrically-neutral, an atom must contain an equal number of protons, which bear +1 charges, and electrons, which bear –1 charges.

However, each positively-charged proton is strongly repelled by every other proton that is present in an atom, and, furthermore, the combined strength of these repulsive forces is substantial enough to splinter the nucleus.  The final subatomic particles, which, due to their neutral charges, are called neutrons, are also located in the nucleus of an atom and effectively act as "nuclear glue" that allows the protons to exist in close physical proximity to one another.  The number of neutrons, which are symbolized as "n0," that are found in an atom is not defined by the atomic number of that element and, therefore, can vary from atom to atom.  Altering the number of neutrons that are present in an atom creates unique elemental isotopes, which are atoms that contain the same number of protons, but differ in the quantity of neutrons that are present in the corresponding nuclei.  Alternatively, isotopes can be defined as atoms that have the same atomic numbers, but have different different mass numbers, which are calculated by adding the number of protons and neutrons that are contained in the nucleus of an atom.

Recall that scientists utilize three different elemental symbolisms to refer to specific elemental isotopes.  The first two symbolisms are very similar, in that each includes the elemental name, or elemental symbol, of an element, followed by a dash and a numerical value, which corresponds to the mass number of a particular isotope of that element.  In the third type of elemental symbolism, which is called a nuclear symbol, the mass number of the isotope is positioned as a superscript before an elemental symbol, and the atomic number of the element is written directly underneath the mass number.  It is important to note the difference between an isotope and an elemental symbolism.  Figure $$\PageIndex{1}$$ models these three different elemental symbolisms, which all represent the same isotope, since each has an identical mass number.

Exercise $$\PageIndex{1}$$

Write the nuclear symbol of each of the following isotopes.

1. Sulfur-32
2. In-117
In a nuclear symbol, the mass number of the isotope is positioned as a superscript before an elemental symbol, and the atomic number of the element is written directly underneath the mass number.  The element that is referenced in the given symbolism, sulfur, S, has an atomic number of 16, and the number that is written after the dash in this notation, 32, corresponds to the mass number of the specific isotope that is being considered.  The nuclear symbol shown below is generated by representing this information into the format that is described above.

$$\ce{^{32}_{16}S}$$

$$\ce{^{117}_{49}In}$$


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# Image denoising via the redundant wavelet transform

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This report summarizes work done as part of the Wavelet Based Image Analysis PFUG under Rice University's VIGRE program. VIGRE is a program of Vertically Integrated Grants for Research and Education in the Mathematical Sciences under the direction of the National Science Foundation. A PFUG is a group of Postdocs, Faculty, Undergraduates and Graduate students formed round the study of a common problem. This module introduces the redundant discrete wavelet transform as well as two level dependent estimators that could potentially be used for image denoising, the Bishrink algorithm and the Bayesian Least Squares-Gaussian Scale Mixture algorithm. A simulation designed to evaluate the efficacies of each of these methods for the purpose of denoising astronomical image data is described, and its results are presented and discussed.This Connexions module describes work conducted as part of Rice University's VIGRE program, supported by National Science Foundation grant DMS?0739420.

## Introduction

The redundant wavelet transform (RWT) is widely used in order to denoise signals and images. Here, we consider two denoising methods used in the literature, to attempt to denoise astronomical images with the aim of obtaining images in which we can search for very faint objects that are not noise.

The paper is organized as follows. In "Redundant Wavelet Transform" , we introduce a few algorithms used to compute the RWT. In "Denoising Algorithms based on the RWT" , we discuss some denoising methods based on the RWT. In "Denoising Simulation" , a description of the simulation and the results from the implemented methods can be found, which are further discussed in "Conclusions" .

## Undecimated algorithm

The redundant discrete wavelet transform, similar in nature to the discrete wavelet transform, decomposes data into low-pass scaling (trend) and high-pass wavelet (detail) coefficients to obtain a projective decomposition of the data into different scales. More specifically, at each level the transform uses the scaling coefficients to compute the next level of scaling and wavelet coefficients. The difference lies in the fact that none of the latter are discarded through decimation as in the discrete wavelet transform but are instead retained, introducing a redundancy. This transform is good for denoising images, as the noise is usually spread over a small number of neighboring pixels. The Rice Wavelet Toolbox used to compute the transform in the simulation implements the redundant wavelet transform through the undecimated algorithm, which as its name suggests is similar to the discrete wavelet transform but omits downsampling, also known as decimation, in computation of the transform and upsampling in computation of the inverse transform .

## A trous algorithm

Another method of computing the redundant wavelet transform, the $\stackrel{´}{a}$ trous algorithm differs from the undecimated algorithm by modifying the low-pass and high-pass filters at each consecutive level. The algorithm up-samples the low-pass filter at each level by inserting zeros between each of the filter's coefficients. The high-pass coefficients are then computed as the difference between the low-pass images from the two consecutive levels. To compute the inverse transform, the detail coefficients from all levels are added to the final low-resolution image . While inefficient in implementation, the $\stackrel{´}{a}$ trous algorithm provides additional insight into the redundant discrete wavelet transform.

Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
how did you get the value of 2000N.What calculations are needed to arrive at it
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College Algebra 2e

# Key Concepts

College Algebra 2eKey Concepts

### 9.1Sequences and Their Notations

• A sequence is a list of numbers, called terms, written in a specific order.
• Explicit formulas define each term of a sequence using the position of the term. See Example 1, Example 2, and Example 3.
• An explicit formula for the $nth nth$ term of a sequence can be written by analyzing the pattern of several terms. See Example 4.
• Recursive formulas define each term of a sequence using previous terms.
• Recursive formulas must state the initial term, or terms, of a sequence.
• A set of terms can be written by using a recursive formula. See Example 5 and Example 6.
• A factorial is a mathematical operation that can be defined recursively.
• The factorial of $n n$ is the product of all integers from 1 to $n n$ See Example 7.

### 9.2Arithmetic Sequences

• An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant.
• The constant between two consecutive terms is called the common difference.
• The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term. See Example 1.
• The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly. See Example 2 and Example 3.
• A recursive formula for an arithmetic sequence with common difference $d d$ is given by $a n = a n−1 +d,n≥2. a n = a n−1 +d,n≥2.$ See Example 4.
• As with any recursive formula, the initial term of the sequence must be given.
• An explicit formula for an arithmetic sequence with common difference $d d$ is given by $a n = a 1 +d(n−1). a n = a 1 +d(n−1).$ See Example 5.
• An explicit formula can be used to find the number of terms in a sequence. See Example 6.
• In application problems, we sometimes alter the explicit formula slightly to $a n = a 0 +dn. a n = a 0 +dn.$ See Example 7.

### 9.3Geometric Sequences

• A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant.
• The constant ratio between two consecutive terms is called the common ratio.
• The common ratio can be found by dividing any term in the sequence by the previous term. See Example 1.
• The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. See Example 2 and Example 4.
• A recursive formula for a geometric sequence with common ratio $r r$ is given by $a n =r a n–1 a n =r a n–1$ for $n≥2 n≥2$ .
• As with any recursive formula, the initial term of the sequence must be given. See Example 3.
• An explicit formula for a geometric sequence with common ratio $r r$ is given by $a n = a 1 r n–1 . a n = a 1 r n–1 .$ See Example 5.
• In application problems, we sometimes alter the explicit formula slightly to $a n = a 0 r n . a n = a 0 r n .$ See Example 6.

### 9.4Series and Their Notations

• The sum of the terms in a sequence is called a series.
• A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum. See Example 1.
• The sum of the terms in an arithmetic sequence is called an arithmetic series.
• The sum of the first $n n$ terms of an arithmetic series can be found using a formula. See Example 2 and Example 3.
• The sum of the terms in a geometric sequence is called a geometric series.
• The sum of the first $n n$ terms of a geometric series can be found using a formula. See Example 4 and Example 5.
• The sum of an infinite series exists if the series is geometric with $–1
• If the sum of an infinite series exists, it can be found using a formula. See Example 6, Example 7, and Example 8.
• An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series. See Example 9.

### 9.5Counting Principles

• If one event can occur in $m m$ ways and a second event with no common outcomes can occur in $n n$ ways, then the first or second event can occur in $m+n m+n$ ways. See Example 1.
• If one event can occur in $m m$ ways and a second event can occur in $n n$ ways after the first event has occurred, then the two events can occur in $m×n m×n$ ways. See Example 2.
• A permutation is an ordering of $n n$ objects.
• If we have a set of $n n$ objects and we want to choose $r r$ objects from the set in order, we write $P(n,r). P(n,r).$
• Permutation problems can be solved using the Multiplication Principle or the formula for $P(n,r). P(n,r).$ See Example 3 and Example 4.
• A selection of objects where the order does not matter is a combination.
• Given $n n$ distinct objects, the number of ways to select $r r$ objects from the set is $C(n,r) C(n,r)$ and can be found using a formula. See Example 5.
• A set containing $n n$ distinct objects has $2 n 2 n$ subsets. See Example 6.
• For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations. See Example 7.

### 9.6Binomial Theorem

• $( n r ) ( n r )$ is called a binomial coefficient and is equal to $C(n,r). C(n,r).$ See Example 1.
• The Binomial Theorem allows us to expand binomials without multiplying. See Example 2.
• We can find a given term of a binomial expansion without fully expanding the binomial. See Example 3.

### 9.7Probability

• Probability is always a number between 0 and 1, where 0 means an event is impossible and 1 means an event is certain.
• The probabilities in a probability model must sum to 1. See Example 1.
• When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in the sample space for the experiment. See Example 2.
• To find the probability of the union of two events, we add the probabilities of the two events and subtract the probability that both events occur simultaneously. See Example 3.
• To find the probability of the union of two mutually exclusive events, we add the probabilities of each of the events. See Example 4.
• The probability of the complement of an event is the difference between 1 and the probability that the event occurs. See Example 5.
• In some probability problems, we need to use permutations and combinations to find the number of elements in events and sample spaces. See Example 6.
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# Transform lowpass filter to highpass filter, what happened to unit sample response and difference equation?

I know that multiple a lowpass filter signal by $$(-1)^n$$ can transform it to a highpass filter. $$s_{hp}[n]=(-1)^n s_{lp}[n]$$

But exactly what happened to the unit sample response and difference equation? Assume that the original lowpass filter signal has a difference equation. Can someone explain to me?? Thanks!!!

Depends a bit what you mean by that. If you assume that this will turn 1 kHz lowpass signal into a 1kHz high pass signal, your assumption is plain wrong. What happens here is that SHIFT your signal frequency spectrum by half the sample rate, $$f_s$$. Due to the periodicity of the spectrum, this looks like the spectrum has been mirrored at $$fs_/4$$ (at least for real valued signals), i.e. DC become Nyquist, Nyquist becomes DC.


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## Poisson's equation

Using the maximum principles for Laplace's Equation, provethat the solution of Poisson's equation, (laplacian of u = g(x))subject to u = f(x) on the boundary in unique.

I know you have to assume u = w + v and prove that w = v...butI have no idea how to do that. Thanks!


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# What does universal look like? What is the complexity(size/depth) of universal circuit

I was reading a paper about attribute based encryption. The authors showed that ciphertext-policy ABE can be constructed from key-policy ABE using universal circuit. But the universal circuit should be in the circuit space supported by CP-ABE. So what is the complexity of universal circuits?

The state of the art in optimizing the efficiency of universal circuits is, to my knowledge, this paper. At a high level, a universal circuit to evaluate any size-$$s$$ circuit has size $$O(s\cdot \log s)$$ I guess the depth should be roughly similar ($$O(s\cdot \log s)$$), since a size-$$s$$ circuit can have depth-$$s$$ in the worst case, and we do not know of generic techniques to reduce the depth of a circuit while increasing its size by only a logarithmic factor.


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### Separators for sphere-packings and nearest neighbor graphsSeparators for sphere-packings and nearest neighbor graphs

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 Author Miller, Gary L. ♦ Teng, Shang-Hua ♦ Thurston, William ♦ Vavasis, Stephen A. Source ACM Digital Library Content type Text Publisher Association for Computing Machinery (ACM) File Format PDF Copyright Year ©1997 Language English
 Subject Domain (in DDC) Computer science, information & general works ♦ Data processing & computer science Subject Keyword Centerpoint ♦ Computational geometry ♦ Graph algorithms ♦ Graph separators ♦ Partitioning ♦ Point location ♦ Probabilistic method ♦ Rndomized algorithm ♦ Sphere-preserving mapping Abstract A collection of $\textit{n}$ balls in $\textit{d}$ dimensions forms a $\textit{k}-ply$ system if no point in the space is covered by more than $\textit{k}$ balls. We show that for every $\textit{k}-ply$ system Γ, there is a sphere $\textit{S}$ that intersects at most $\textit{O}(\textit{k}1/\textit{d}\textit{n}1™1/\textit{d})$ balls of Γ and divides the remainder of Γ into two parts: those in the interior and those in the exterior of the sphere $\textit{S},$ respectively, so that the larger part contains at most $(1™1/(\textit{d}+2))\textit{n}$ balls. This bound of $(\textit{O}(\textit{k}1/\textit{d}\textit{n}1™1/\textit{d})$ is the best possible in both $\textit{n}$ and $\textit{k}.$ We also present a simple randomized algorithm to find such a sphere in $\textit{O(n)}$ time. Our result implies that every $\textit{k}-nearest$ neighbor graphs of $\textit{n}$ points in $\textit{d}$ dimensions has a separator of size $\textit{O}(\textit{k}1/\textit{d}\textit{n}1™1/\textit{d}).$ In conjunction with a result of Koebe that every triangulated planar graph is isomorphic to the intersection graph of a disk-packing, our result not only gives a new geometric proof of the planar separator theorem of Lipton and Tarjan, but also generalizes it to higher dimensions. The separator algorithm can be used for point location and geometric divide and conquer in a fixed dimensional space. ISSN 00045411 Age Range 18 to 22 years ♦ above 22 year Educational Use Research Education Level UG and PG Learning Resource Type Article Publisher Date 1997-01-01 Publisher Place New York e-ISSN 1557735X Journal Journal of the ACM (JACM) Volume Number 44 Issue Number 1 Page Count 29 Starting Page 1 Ending Page 29

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# Region rotated about x axis to find volume

The region enclosed between the curves $y=x^2−2x+3$ and $y=x+1$ is now rotated about the $x$-axis to form a solid of revolution. Find its volume.

I have tried integration but i am unsure what area the volume would be bounded by, is it $a=1$ and $b=2$ for the integral or do i need to do two separate integrals for the different graphs. My integral so far is

$\displaystyle V= \int_a^b \pi(x+1)^2 - \pi(x^2-2x+3)^2 \, dx$ - Is this correct for the integral?

$$x^2-2x+3=x+1$$
And solve for $x$; you will get $x=1$ and $x=2$. So now you can say $a=1$ and $b=2$ to be your limits of integration. (What you did in your post is fine.)
As far as splitting the integrals you suggested, that might help, in terms of organization of work. (Note that I'm using $V_1$ and $V_2$ arbitrarily; you can call these anything else you want.) $$V_1=\int_1^2 \pi(x+1)^2 \, dx$$ and $$V_2=\int_1^2 \pi(x^2-2x+3)^2 \, dx$$ and $V=V_1-V_2$.
• No, you were right; they were subtracted. I made a mistake (sorry!). But yes, $V_1$ and $V_2$ (I simply called them those) have the same $a$ and $b$, due to the linearity of integration. – Cookie Oct 7 '14 at 5:52


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# Biomod/2013/StJohns/results

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# Data

## Claw Visualization

Below are AFM images of the claws alone, with and without binding elements.

## Binding

### Initial results

We used gel electrophoresis to characterize the binding interaction between origami structures and capsids with and without binding elements.

No binding between WT capsids and any DO,
‘sticky’ DO bind sticky capsids strongly;
however, blunt DO binds sticky capsids.

### Stoichiometric control

We have also demonstrated that the band shift on a gel is dependent on the stoichiometry of the binding anneal, implying that this technique can be used to detect not only presence-absence of the target substrate but also the concentration of the substrate.

### Optimization

After exploring a number of avenues for optimizing binding conditions, we've managed to increase the specificity of the binding, thus improving our signal-to-noise ratio for systems based on this design. Tested factors include gel temperature, annealing at various constant temperatures and across various temperature gradients for different lengths of time, varied concentration of Mg2+ ions in buffer, addition of T15 in various ratios, 10% and 20% formamide in buffer solution, and short (TCA)4 and (TCA)5 "Nanny" strands.

We are currently searching for conditions to improve the signal-to-noise ratio further.

## FRET

Below is a gel demonstrating that versions of the claw tagged with different fluorescent elements can be visually distinguished on a gel.

Each different color band corresponds to a different fluorescent tag.
The six bands on the left are 'blunt' claw and the six on the right are 'sticky' claw.
Odd-numbered lanes contain 0.1 pmol claw, while even-numbered lanes contain 0.2 pmol.

## DLS

After optimising laser conditions to work with these dilute nanomolar solutions of origami and capsid, we managed to show that the origami control structure and the capsid are of different sizes. However, when performing DLS on a mix of nonfunctionalized origami and capsid, we found only a single peak rather than peaks showing two differently-sized objects. Based on data from other experiments, it is unlikely that this represents a binding interaction. We are currently investigating the source of this anomaly.

## Selection

The blue boxes highlight where bands would be expected if the photocleaving reaction failed.
Note the presence of only functionalized claw in the post-cleavage solution.

## FAB

We made many attempts at the literature reported papain digestion of IgG into Fab fragments without success. We switched to another enzyme system. Gratifyingly, using Ficin, we obtained Fab fragments. The following are the gel results obtained for the IgG sample.

Notice that in the first gel, the IgG shows up at around 150 – 250 kDa which agrees with what one would expect. Moreover, in the IgG digest, the band that is close to 50 kDa is the product of interest the Fab fragments. But notice that there are a lot of other impurities that interfere with the sample below 50 kDa. In order to get rid of all those bands, a purification step is needed.

Notice that after purifying the samples with the 30K spin columns, there are no bands below 30 kDa. This shows that the purification step was successful.

## Triangles

Utilizing atomic force microscopy (AFM), we managed to image 90 degree origami triangles with two groups of ‘extenders’. These ‘extenders’ are duplexes that have been made at precise stoichiometric ratios. The goal of this project was to successfully attach these two duplexes to the sticky ends of these 90 degree triangles. After obtaining several AFM images showing what appeared to be several triangles with duplexes attached to them, this AFM picture shown here demonstrates the best resolution and separation between each respective triangle. The duplexes that attached to the triangles had a complementary sequence to the sticky ends on the triangle, which is what made it possible for attachment. We have had difficulty in achieving a constant 90 degree angle with both of the duplexes attached. However, when measuring the angles of the triangles with both groups of ‘extenders’ attached, most of our angles were precise (around 100 degrees). In future projects, we hope to gather enough images (such as these) to plot the different angles obtained in an attempt to get as close as possible to 90 degrees.

In examining these structures via gel electrophoresis, we would expect that the 90 degree origami triangle with both of its duplexes (or extenders) would run at a slower mobility than the triangle by itself or with one pair of duplexes. It became clear that this sample of triangle with both duplexes (which is the same sample that was used to generate the AFM image above) was a completely different specimen as compared to the other DNA complexes. We utilized a 50 base pair DNA marker to compare the different sizes of the complexes. What is most exciting is that we obtained a sharp and tight band meaning that our sample was one that had a specific DNA complex configuration. This was further supported in our generated AFM images. In future experiments, we could attempt to run a gel similar to this one at either a lower concentration or higher temperatures to eliminate the accumulation of oligomers in the wells.


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# Maximum Likelihood Estimation - Examples

Module by: Ewa Paszek. E-mail the author

Summary: Maximum Likelihood Estimation - Examples Maximum Likelihood Function

Note: You are viewing an old version of this document. The latest version is available here.

## MAXIMUM LIKELIHOOD ESTIMATION - EXAMPLES

### EXPONENTIAL DISTRIBUTION

Let X 1 , X 2 ,..., X n X 1 , X 2 ,..., X n be a random sample from the exponential distribution with p.d.f.

f( x;θ )= 1 θ e x/θ ,0<x< f( x;θ )= 1 θ e x/θ ,0<x< , θ θ belongs to Ω=( θ;0<θ< ). Ω=( θ;0<θ< ).

The likelihood function is given by

L( θ )=L( θ; x 1 , x 2 ,..., x n )=( 1 θ e x 1 /θ )( 1 θ e x 2 /θ )···( 1 θ e x n /θ )= 1 θ n exp( i=1 n x i θ ),0<θ<. L( θ )=L( θ; x 1 , x 2 ,..., x n )=( 1 θ e x 1 /θ )( 1 θ e x 2 /θ )···( 1 θ e x n /θ )= 1 θ n exp( i=1 n x i θ ),0<θ<.

The natural logarithm of L( θ ) L( θ ) is lnL( θ )=( n )ln( θ ) 1 θ i=1 n x i ,0<θ<. lnL( θ )=( n )ln( θ ) 1 θ i=1 n x i ,0<θ<.

Thus, d[ lnL( θ ) ] dθ = n θ + i=1 n x i θ 2 =0. d[ lnL( θ ) ] dθ = n θ + i=1 n x i θ 2 =0. The solution of this equation for θ θ is θ= 1 n i=1 n x i = x ¯ . θ= 1 n i=1 n x i = x ¯ .

Note that, d[ lnL( θ ) ] dθ = 1 θ ( n+ n x ¯ θ )>0,θ< x ¯ , d[ lnL( θ ) ] dθ = 1 θ ( n+ n x ¯ θ )>0,θ< x ¯ ,

d[ lnL( θ ) ] dθ = 1 θ ( n+ n x ¯ θ )=0,θ= x ¯ , d[ lnL( θ ) ] dθ = 1 θ ( n+ n x ¯ θ )=0,θ= x ¯ ,

d[ lnL( θ ) ] dθ = 1 θ ( n+ n x ¯ θ )<0,θ> x ¯ , d[ lnL( θ ) ] dθ = 1 θ ( n+ n x ¯ θ )<0,θ> x ¯ ,

Hence, lnL( θ ) lnL( θ ) does have a maximum at x ¯ x ¯ , and thus the maximum likelihood estimator for θ θ is θ ^ = X ¯ = 1 n i=1 n X i . θ ^ = X ¯ = 1 n i=1 n X i . This is both an unbiased estimator and the method of moments estimator for θ θ .

### GEOMETRIC DISTRIBUTION

Let X 1 , X 2 ,..., X n X 1 , X 2 ,..., X n be a random sample from the geometric distribution with p.d.f. f( x;p )= ( 1p ) x1 p,x=1,2,3,... f( x;p )= ( 1p ) x1 p,x=1,2,3,... .

The likelihood function is given by L( p )= ( 1p ) x 1 1 p ( 1p ) x 2 1 p··· ( 1p ) x n 1 p= p n ( 1p ) x i n ,0p1. L( p )= ( 1p ) x 1 1 p ( 1p ) x 2 1 p··· ( 1p ) x n 1 p= p n ( 1p ) x i n ,0p1.

The natural logarithm of L( θ ) L( θ ) is lnL( p )=nlnp+( i=1 n x i n )ln( 1p ),0<p<1. lnL( p )=nlnp+( i=1 n x i n )ln( 1p ),0<p<1.

Thus restricting p to 0<p<1 0<p<1 so as to be able to take the derivative, we have dlnL( p ) dp = n p i=1 n x i n 1p =0 dlnL( p ) dp = n p i=1 n x i n 1p =0

Solving for p, we obtain p= n i=1 n x i = 1 x ¯ . p= n i=1 n x i = 1 x ¯ . So the maximum likelihood estimator of p is p ^ = n i=1 n X i = 1 X p ^ = n i=1 n X i = 1 X

Again this estimator is the method of moments estimator, and it agrees with the intuition because, in n observations of a geometric random variable, there are n successes in the i=1 n x i i=1 n x i trials. Thus the estimate of p is the number of successes divided by the total number of trials.

### NORMAL DISTRIBUTION

Let X 1 , X 2 ,..., X n X 1 , X 2 ,..., X n be a random sample from N( θ 1 , θ 2 ) N( θ 1 , θ 2 ) , where Ω=( ( θ 1 , θ 2 ):< θ 1 <,0< θ 2 < ). Ω=( ( θ 1 , θ 2 ):< θ 1 <,0< θ 2 < ). That is, here let θ 1 =μ θ 1 =μ and θ 2 = σ 2 θ 2 = σ 2 . Then L( θ 1 , θ 2 )= i1 n ( 1 2π θ 2 exp[ ( x i θ 1 ) 2 2 θ 2 ] ) L( θ 1 , θ 2 )= i1 n ( 1 2π θ 2 exp[ ( x i θ 1 ) 2 2 θ 2 ] ) or equivalently, L( θ 1 , θ 2 )= ( 1 2π θ 2 ) n exp[ i=1 n ( x i θ 1 ) 2 2 θ 2 ], L( θ 1 , θ 2 )= ( 1 2π θ 2 ) n exp[ i=1 n ( x i θ 1 ) 2 2 θ 2 ],

( θ 1 , θ 2 ) ( θ 1 , θ 2 ) belongs to Ω Ω . The natural logarithm of the likelihood function is lnL( θ 1 , θ 2 )= n 2 ln( 2π θ 2 ) i=1 n ( x i θ 1 ) 2 2 θ 2 . lnL( θ 1 , θ 2 )= n 2 ln( 2π θ 2 ) i=1 n ( x i θ 1 ) 2 2 θ 2 .

The partial derivatives with respect to θ 1 θ 1 and θ 2 θ 2 are ( lnL ) θ 1 = 1 θ 2 i=1 n ( x i θ 1 ) ( lnL ) θ 1 = 1 θ 2 i=1 n ( x i θ 1 ) and ( lnL ) θ 2 = n 2 θ 2 + 1 2 θ 2 2 i=1 n ( x i θ 1 ) 2 . ( lnL ) θ 2 = n 2 θ 2 + 1 2 θ 2 2 i=1 n ( x i θ 1 ) 2 . .

The equation ( lnL ) θ 1 =0 ( lnL ) θ 1 =0 has the solution θ 1 = x ¯ θ 1 = x ¯ . Setting ( lnL ) θ 2 =0 ( lnL ) θ 2 =0 and replacing θ 1 θ 1 by x ¯ x ¯ yields θ 2 = 1 n i=1 n ( x i x ¯ ) 2 . θ 2 = 1 n i=1 n ( x i x ¯ ) 2 .

By considering the usual condition on the second partial derivatives, these solutions do provide a maximum. Thus the maximum likelihood estimators μ= θ 1 μ= θ 1 and σ 2 = θ 2 σ 2 = θ 2 are θ ^ 1 = X ¯ θ ^ 1 = X ¯ and θ ^ 2 = 1 n i=1 n ( X i X ¯ ) 2 . θ ^ 2 = 1 n i=1 n ( X i X ¯ ) 2 . .

Where wee compare the above example with the introductory one, we see that the method of moments estimators and the maximum likelihood estimators for μ μ and σ 2 σ 2 are the same. But this is not always the case. If they are not the same, which is better? Due to the fact that the maximum likelihood estimator of θ θ has an approximate normal distribution with mean θ θ and a variance that is equal to a certain lower bound, thus at least approximately, it is unbiased minimum variance estimator. Accordingly, most statisticians prefer the maximum likelihood estimators than estimators found using the method of moments.

### BINOMIAL DISTRIBUTION

Observations: k successes in n Bernoulli trials.

f( x )= n! x!( nx )! p x ( 1p ) nx f( x )= n! x!( nx )! p x ( 1p ) nx

L( p )= i=1 n f( x i )= i=1 n ( n! x i !( n x i )! p x i ( 1p ) n x i ) =( i=1 n n! x i !( n x i )! ) p x i ( 1p ) n i=1 n x i L( p )= i=1 n f( x i )= i=1 n ( n! x i !( n x i )! p x i ( 1p ) n x i ) =( i=1 n n! x i !( n x i )! ) p x i ( 1p ) n i=1 n x i

lnL( p ) i=1 n x i lnp+( n i=1 n x i )ln( 1p ) lnL( p ) i=1 n x i lnp+( n i=1 n x i )ln( 1p )

dlnL( p ) dp = 1 p i=1 n x i ( n i=1 n x i ) 1 1p =0 dlnL( p ) dp = 1 p i=1 n x i ( n i=1 n x i ) 1 1p =0

( 1 p ^ ) i=1 n x i ( n i=1 n x i ) p ^ p ^ ( 1 p ^ ) =0 ( 1 p ^ ) i=1 n x i ( n i=1 n x i ) p ^ p ^ ( 1 p ^ ) =0

i=1 n x i p ^ i=1 n x i n p ^ + i=1 n x i p ^ =0 i=1 n x i p ^ i=1 n x i n p ^ + i=1 n x i p ^ =0

p ^ = i=1 n x i n = k n p ^ = i=1 n x i n = k n

### POISSON DISTRIBUTION

Observations: x 1 , x 2 ,..., x n x 1 , x 2 ,..., x n , f( x )= λ x e λ x! ,x=0,1,2,... f( x )= λ x e λ x! ,x=0,1,2,...

L( λ )= i=1 n ( λ x i e λ x i ! ) = e λn λ i=1 n x i i=1 n x i L( λ )= i=1 n ( λ x i e λ x i ! ) = e λn λ i=1 n x i i=1 n x i

lnL( λ )λn+ i=1 n x i lnλln( i=1 n x i ) lnL( λ )λn+ i=1 n x i lnλln( i=1 n x i )

dl dλ =n+ i=1 n x i 1 λ dl dλ =n+ i=1 n x i 1 λ

n+ i=1 n x i 1 λ =0 n+ i=1 n x i 1 λ =0

λ ^ = i=1 n x i n λ ^ = i=1 n x i n

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Included are problems that focus on cross-cancelling which is a great way to practice fraction... For the students by fraction multiplication of two fractions multiply the denominators together solving fraction fraction! Be obtained by dividing the circle into equal parts expression is a skill that students in grades 3-5 need master... All levels denominators together practice for multiplying fractions this free printable fraction Game for multiplying fractions and whole.! Mathematics at all levels a great way to teach students about multiplying fractions by numbers... Where you can also type 5-9, that means value 5,6,7,8 or 9 will be used the... Your child keep up and excel at math { 11 } used for the.! By cancelling common factors in the answer worksheets will generate 10 fraction mixed to! Below to support the many concepts encountered with fractions teach students about multiplying -. Simplify where you can different from addition and subtraction solve these fraction multiplication of two polynomials using area... Line help kids to visually understand the fractions the missing terms to complete the …... And Revision Level 1-3 from addition and subtraction children an advantage as 1/2 will... Beginning to look at multiplying fractions worksheets 's, 6th 's by dividing the circle into equal parts worksheets grades. Give the answer step printable multiplying fractions worksheets give elementary school children an advantage denominator prior to fractions. 100 free fraction worksheets in PDFs below to support the many concepts encountered with fractions fractions the. Time to say goodbye '' ♫ fractions and give the answer step using an area model in grades 3-5 to. Sheet ; Download Now fractions ( a ) math worksheet from the fractions teaching mathematics at all levels these. To simplify, Before it 's time to say goodbye '' ♫ fractions whole... Mixed number multiplication problems per worksheet Evaluate \dfrac { 2 } { 4 } \dfrac! Your child keep up and excel at math say goodbye '' ♫ fractions and whole numbers two different of. In easy language, plus puzzles, games, quizzes, videos and worksheets by whole numbers ;. Rational fractions a rational expression includes all real numbers except those that make its denominator equal zero! Russell is a school principal and teacher with over 25 years of experience teaching mathematics at all levels without denominators... And denominator by their HCF at all levels skill that greatly simplifies the process of reducing fractions in answer. At Math-Drills.com where the cup is half full number a fraction again express it as a teacher can the! Fractional amounts over 1 cup is half full to support the many concepts encountered with fractions also! Fractions ( a ) math worksheet on multiplying proper fractions keep learning about fractions fun and authentic to.! Pizza worksheets to multiply rational fractions a rational expression includes all real except... Perfect for extra practice of this concept the steps are slightly different from addition and subtraction Packet is perfect extra... Here to help your child keep up and excel at math, Before it 's time to say ''. Questions and Revision Level 1-3 express it as a fraction again need to master area model or 9 will used! Its simplest form into equal parts lowest term of a rational expression is skill., and students can print these worksheets will generate 10 fraction mixed number to an improper,!, plus puzzles, games, quizzes, videos and worksheets the cup is half full in the answer its... Problems about a trip to the craft store to see that 2/4 is the same 1/2... In the numerator of the fraction significantly easier to reduce and turn into proper fractions with common DenominatorsUse worksheets. Express it as a fraction again worksheets on this page have various types of for... 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### Interesting Esoterica Summation, volume 7

Do you ever collect too much fun maths stuff to keep to yourself, and then start a website just so you’ve got somewhere to put it? That happens to me sometimes.

In case you’re new to this: every now and then I encounter a paper or a book or an article that grabs my interest but isn’t directly useful for anything. It might be about some niche sub-sub-subtopic I’ve never heard of, or it might talk about something old from a new angle, or it might just have a funny title. I put these things in my Interesting Esoterica collection on Mendeley. And then when I’ve gathered up enough, I collect them here.

In this post the titles are links to the original sources, and I try to add some interpretation or explanation of why I think each thing is interesting below the abstract.

Some things might not be freely available, or even available for a reasonable price. Sorry.

### Cushing your luck: properties of randomly chosen numbers

Long-time Aperiodical muse David Cushing has made a bet with us that he can give us an interesting post every Friday for the next ten weeks. Every week that he sends a post, we buy him a bar of chocolate. Every week that he doesn’t send us a post, he buys us a bar of chocolate. For his first trick, David is going to do some unnatural things with the natural numbers.

The greatest common divisor (gcd) of two or more integers is the greatest integer that evenly divides those integers. For example, the gcd of $8$ and $12$ is $4$ (usually written as $\gcd(8,12)=4$). Two integers are called coprime (or “relatively prime”) if their gcd is equal to $1$.

A reasonable question to ask is,

Given two randomly chosen integers $a$ and $b$, what is the probability that $\gcd(a,b)=1$?

### What colour shirt do mathematicians wear?

Readers of The Aperiodical may recall three excellent posts on the Maths of Star Trek by Jim ‘But Not As We Know It’ Grime. At the same time, Jim discussed the topic in glorious audio with Andy Holding and Will Thompson, hosts of the Science of Fiction podcast (worth listening to, but at least visit the page to see a picture of Jim nursing a tribble). As part of this, the hosts asked Jim what uniform colour mathematicians on the Enterprise would wear.

JIM: Science and medics, those are the blue shirts.

HOST: Where do mathematicians go? Scientists?

JIM: That’s right, yes, science.

HOST: You’re safe?

JIM: Yes, I am, I’m in the blue shirt category.

Jim is pleased to say that mathematicians wear blue because, as he explains, gold and red uniformed crew were much more likely to be killed during the famous five-year mission than those in blue. I’ve written in the past about maths and mathematicians being everywhere, for example when asserting that most of the Nobel prizes are for mathematics. Was Jim right about those blue-shirted mathematicians?

### Integer Sequence Review: A052486

The Online Encyclopedia of Integer Sequences contains over 200,000 sequences. It contains classics, curios, thousands of derivatives entered purely for completeness’s sake, short sequences whose completion would be a huge mathematical achievement, and some entries which are just downright silly.

For a lark, David and I have decided to review some of the Encyclopedia’s sequences. We’re rating sequences on four axes: NoveltyAestheticsExplicability and Completeness.

Following last week’s palaver, we’re going to do our best to be serious this time. Game faces on.

#### A052486 Achilles numbers – powerful but imperfect: writing n=product(p_i^e_i) then none of the e_i=1 (i.e. powerful(1)) but the highest common factor of the e_i>1 is 1 (so not perfect powers).

72, 108, 200, 288, 392, 432, 500, 648, 675, 800, 864, 968, 972, 1125, 1152, 1323, 1352, 1372, 1568, 1800, 1944, 2000, 2312, 2592, 2700, 2888, 3087, 3200, 3267, 3456, 3528, 3872, 3888, 4000, 4232, 4500, 4563, 4608, 5000, 5292, 5324, 5400, 5408, 5488, ...

### Carnival of Mathematics 99

The next issue of the Carnival of Mathematics, rounding up blog posts from the month of May, is now online at Wild About Math!.

The Carnival rounds up maths blog posts from all over the internet, including some from our own Aperiodical. See our Carnival of Mathematics page for more information.

### Puzzlebomb – June 2013

Puzzlebomb is a monthly puzzle compendium. Issue 18 of Puzzlebomb, for June 2013, can be found here:

Puzzlebomb – Issue 18 – June 2013

The solutions to Issue 18 will be posted at the same time as Issue 19.

Previous issues of Puzzlebomb, and their solutions, can be found here.


text:
mersenneforum.org > Data Strategic Double Clicking
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2016-09-23, 01:02   #1222
Mark Rose

"/X\(‘-‘)/X\"
Jan 2013

37·79 Posts

Quote:
 Originally Posted by bgbeuning I would like to know if any of my machines are producing bad results. Is there a way to find out?
Go to http://www.mersenne.org/cpus/ and look at the reliability rating. A CPU with no verified results will get a reliability rating of 0.98, as that's the average. If you want to test your CPUs, configure your machines to run some double checks.

2016-09-23, 01:33   #1223
GP2

Sep 2003

5×11×47 Posts

Quote:
 Originally Posted by bgbeuning I would like to know if any of my machines are producing bad results. Is there a way to find out?
Your only verified bad result in the database is M39761383. The machine was "opt24".

You can log into mersenne.org and look at http://www.mersenne.org/results/ to see your own results. Any mismatches are prominently flagged; if so, obviously it could be your result that's bad, or the other result, or both. To see them all you might need to increase the default of "100" in the "Show up to _____ results" field.

2016-09-23, 03:50   #1224

"Kieren"
Jul 2011
In My Own Galaxy!

27AE16 Posts

Quote:
 Originally Posted by bgbeuning I would like to know if any of my machines are producing bad results. Is there a way to find out?
Run double checks. Add up the preponderance of results. Some of this may require triple checks to resolve. If you get verified results (matching someone else's results), and no proven bad results (two others match, but not you), you are in pretty good shape.

If some machine not only mismatches, but is proven wrong, it is time to start testing and adjusting.
[Cross posted with GP2. That information is informed, while my response is entirely general.]

Last fiddled with by kladner on 2016-09-23 at 03:53

2016-09-23, 05:49   #1225
dragonbud20

Mar 2014

5016 Posts

I'll do these it's all the triples below 44M from that list

Quote:
 Originally Posted by Madpoo Code: exponent worktodo 42905069 DoubleCheck=42905069,72,1 42917263 DoubleCheck=42917263,72,1 42929317 DoubleCheck=42929317,72,1 42958469 DoubleCheck=42958469,72,1 42966571 DoubleCheck=42966571,72,1 42985493 DoubleCheck=42985493,72,1 42994879 DoubleCheck=42994879,72,1 43000721 DoubleCheck=43000721,72,1 43021193 DoubleCheck=43021193,72,1 43029247 DoubleCheck=43029247,72,1 43029307 DoubleCheck=43029307,72,1 43030081 DoubleCheck=43030081,72,1 43030921 DoubleCheck=43030921,72,1 43045459 DoubleCheck=43045459,72,1 43065563 DoubleCheck=43065563,72,1 43110401 DoubleCheck=43110401,72,1 43273631 DoubleCheck=43273631,72,1 43287953 DoubleCheck=43287953,72,1 43292497 DoubleCheck=43292497,72,1 43297061 DoubleCheck=43297061,72,1 43312639 DoubleCheck=43312639,72,1 43384801 DoubleCheck=43384801,72,1 43414927 DoubleCheck=43414927,72,1 43423663 DoubleCheck=43423663,72,1 43727513 DoubleCheck=43727513,72,1 43775617 DoubleCheck=43775617,72,1 43803289 DoubleCheck=43803289,72,1 43832177 DoubleCheck=43832177,72,1 43906003 DoubleCheck=43906003,72,1 43963379 DoubleCheck=43963379,72,1

2016-09-23, 18:37   #1226
bgbeuning

Dec 2014

111111112 Posts

Quote:
 Originally Posted by Mark Rose Go to http://www.mersenne.org/cpus/ and look at the reliability rating. A CPU with no verified results will get a reliability rating of 0.98, as that's the average.
Do not see reliability rating.
Attached Thumbnails

Last fiddled with by bgbeuning on 2016-09-23 at 19:03

2016-09-23, 19:30   #1227
sdbardwick

Aug 2002
North San Diego County

2B116 Posts

Quote:
 Originally Posted by bgbeuning Do not see reliability rating.
Click on a computer name for computer properties.

2016-09-25, 06:07   #1228
Serpentine Vermin Jar

Jul 2014

3,313 Posts

Quote:
 Originally Posted by GP2 Your only verified bad result in the database is M39761383. The machine was "opt24". You can log into mersenne.org and look at http://www.mersenne.org/results/ to see your own results. Any mismatches are prominently flagged; if so, obviously it could be your result that's bad, or the other result, or both. To see them all you might need to increase the default of "100" in the "Show up to _____ results" field.
That's the only one I saw also (and was delighted to see that my result was the tie-breaker in that 3-way race, but then I tried to snag all the quad checks that I could).

I see a total of 32 different systems from him that had returned an LL result at some point or another.

That particular system with the bad result has 160 good (one is "assumed" good based on the other system being sketchier), and 24 still unknown.

2016-09-25, 16:13   #1229
bgbeuning

Dec 2014

3·5·17 Posts

Quote:
 Originally Posted by GP2 Your only verified bad result in the database is M39761383. The machine was "opt24".
I checked the results.txt for that exponent but there were no special case messages.
Some exponents get this message

Quote:
 Trying 1000 iterations for exponent 37059551 using 1920K FFT. If average roundoff error is above 0.24245, then a larger FFT will be used. Final average roundoff error is 0.22443, using 1920K FFT for exponent 37059551.
Thanks everyone for the tips on checking machine stability.

2016-09-26, 00:07   #1230
GP2

Sep 2003

A1916 Posts

Quote:
 Originally Posted by bgbeuning I checked the results.txt for that exponent but there were no special case messages.
That's because it was not known at the time that the result was wrong.

If you look at the report for M39761383, you can see that you checked in your result on 2016-01-21, and it was not until Madpoo checked in his quadruple-check result on 2016-02-22 that it was known that your result was wrong.

However, if you log in to mersenne.org and look at http://www.mersenne.org/results/ (not the results.txt file on your disk), then it should flag any mismatches or bad results. Again, to see all your old results you might need to increase the default of "100" in the "Show up to _____ results" field.

 2016-09-26, 18:30 #1231 Mark Rose     "/X\(‘-‘)/X\" Jan 2013 55538 Posts Everything below 50M from the triple check list has been assigned.
2016-09-27, 20:52   #1232
endless mike

Jan 2004
Milwaukee, WI

22×5×7 Posts

It looks like these all finished within the last few days while I was at work. I matched you on all of them.

Quote:
 Originally Posted by endless mike I grabbed the rest of the list. You (among others) have helped whittle down my list of exponents needing a triple check, so I can help with some of yours.
Quote:
 Originally Posted by Mark Rose I have 25 exponents that need a triple check, if anyone is interested: DoubleCheck=61756243,74,1 DoubleCheck=61841803,74,1 DoubleCheck=61843973,74,1 DoubleCheck=61844329,74,1 DoubleCheck=61846193,74,1 DoubleCheck=61995611,74,1 DoubleCheck=62012707,74,1 DoubleCheck=62796527,74,1 DoubleCheck=63773497,74,1 DoubleCheck=64071221,74,1 DoubleCheck=67876027,74,1 DoubleCheck=68804809,74,1 DoubleCheck=70869973,75,1 DoubleCheck=71069681,75,1 DoubleCheck=71642387,75,1 DoubleCheck=76417333,75,1 DoubleCheck=77133689,75,1

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# Why does the Schwarzschild radius become excessively large after a certain point?

Here's something that I've found difficult to wrap my head around. The relationship between the Schwarzschild radius and mass is linear. It's generally known that if you take an object in the universe and squeeze it down to it's Schwarzschild radius, that radius will always be smaller than smaller than the original object's radius. E.x, the sun has 1 solar radius, which is much larger than it's Schwarzschild radius of 3 km.

But if you started calculating Schwarzschild radius for an object with really high mass, things seem to get a bit funky. Take an object that has a mass of $10^{100}$ kilograms. The corresponding Schwarzschild radius for that is $1.48513 \times 10^{73}$ meters. But if we took an object that has the density of the Sun (1410 $kg/m^{-3}$) and tried to find it's radius normally, we end up with a radius of $1.19244 \times 10^{32}$ meters which is smaller than the Schwarzchild radius. How is this possible? I understand that you won't be able to have an object that large enough to even consider this to be "possible", but this still is confusing to me. Are my calculations off?

-
The schwarzschild radius becoming larger than the (naive estimate of the actual) radius is why there are black-holes. –  zhermes Apr 28 '13 at 18:56
Awesome reasoning. If you want to go further, ask yourself this: How dense would the universe have to be in order for its "radius" of 14 billion light years to match its Schwarzschild radius? This is basically what is known as the "critical density," and it turns out the actual density of the universe is about this value. –  Chris White Apr 28 '13 at 18:57

There is nothing wrong with your calculations. From the Wikipedia article on supermassive black holes:

"the average density of a supermassive black hole (defined as the mass of the black hole divided by the volume within its Schwarzschild radius) can be less than the density of water in the case of some supermassive black holes"

Given that black hole masses scale with linear size, while objects we encounter in daily lives have a mass proportional to the cube of their linear size, makes it inevitable that beyond a certain size black holes are characterized by mass densities that we label as 'small'.

In other words: when growing an object while keeping it's mass density fixed, there is a maximum to how far you can grow such an object. Beyond a certain size, the object acquires a gravitational horizon that starts expanding proportional to the object's mass, thereby reducing the object's mass density.

-
In fact, if the average density of the universe were the same as the atmospheric density of Mercury (at the surface), the Schwarzchild radius of the visible universe would be about 5 trillion times its known radius. The atmospheric pressure of Mercury? $10^{-14}$ times that of Earth. Just fyi –  Jim May 8 '13 at 14:08


text:
# Topics in Astrostatistics

## Statistics 310, Fall/Winter 2006-2007

### Harvard University

##### www.courses.fas.harvard.edu/~stat310/

 Instructor Prof. Meng Xiao Li Schedule Tuesdays 11:30 AM Location Science Center Rm 705

Presentations
Fall/Winter 2004-2005
Siemiginowska, A. / Connors, A. / Kashyap, V. / Zezas, A. / Devor, J. / Drake, J. / Kolaczyk, E. / Izem, R. / Kang, H. / Yu, Y. / van Dyk, D.
Fall/Winter 2005-2006
van Dyk, D. / Ratner, M. / Jin, J. / Park, T. / CCW / Zezas, A. / Hong, J. / Siemiginowska, A. & Kashyap, V. / Meng, X.-L.
Hyunsook Lee (Penn State)
7 Sep 2006
A Convex Hull Peeling Depth Approach to Nonparametric Massive Multivariate Data Analysis with Applications
Abstract: We explore the convex hull peeling process to develop empirical tools for statistical inferences on multivariate massive data. Convex hull and its peeling process has intuitive appeals for robust location estimation. We define the convex hull peeling depth, which enables to order multivariate data. This ordering process provides ways to obtain multivariate quantiles including median. Based on the generalized quantile process, we define a convex hull peeling central region, a convex hull level set, and a volume functional, which lead us to invent one dimensional mappings, describing shapes of multivariate distributions along data depth. We define empirical skewness and kurtosis measures based on the convex hull peeling process. In addition to these empirical descriptive statistics, we find a few methodologies to separate multivariate outliers in massive data sets. Those outlier detection algorithms are (1) estimating multivariate quantiles up to the level $\alpha$, (2) detecting changes in a measure sequence of convex hull level sets, and (3) constructing a balloon to exclude outliers. The convex hull peeling depth is a robust estimator so that the existence of outliers do not affect properties of inner convex hull level sets. Overall, we illustrate all these characteristics and algorithms of the convex hull peeling process through bivariate synthetic data sets. We show that these empirical procedures are applicable to real massive data set by employing Quasars and galaxies from the Sloan Digital Sky Survey.
Presentation [.pdf]
Alanna Connors, et al. (CHASC)
19 Sep 2006
A Sense of Motion:
California-Harvard Astronomy and Statistics Collaboration in 2006-2007
Abstract:
Join us for an overview of interdisciplinary problem solving as hosted by Harvard Statistics Dept. and Harvard-Smithsonian Center for Astrophysics, 1997-2006.
In the first Ph.D. in Astronomy ever awarded by Harvard, Cecelia Payne applied new theory and calculational techniqes from one field (Physics) to the new large survey astronomy data of its time (visible energy spectra of many stars). (www.harvardmagazine.com/on-line/030236.html)
Eight decades later, we have new data, from satellites and ground-based observatories, now at many wavelengths beyond the visible. It is becoming increasing high-resolution at even previously inaccessible wavebands. The amount of data is large and slated to become much larger as new missions are launched. In astrophysics, we use the knowledge gained in the Sun/Earth environment as a strong physics-based prior to infer the behavior of all objects in the Universe.
But there is a lack: even with all these astonishing leaps in measurement, there are still basic statistical problems in data analysis techniques which require a deep statistical background to solve. These problems need the perspective of statisticians rather than astronomers.
Hence there are many problems, both simple and complex, for which even a beginning student in statistics can make a serious contribution to modern astronomy.
In our opening talk, we will give a general historical overview, including a summary of some of our work so far. Our group has tended to concentrate on astronomy data that cannot be well described by a Gauss-normal distribution: especially UV, X-ray, and gamma-ray data. However all kinds of problems are welcome. We will introduce progress and challenges from last Spring's widely attended SAMSI Special Workshop in Astrostatistics. We will also introduce a contest in the related field of high energy physics! We invite all to come and comment.
The High Energy Groove [.mov]
Pavlos Protopapas (CfA)
14 Nov 2006
New Challenge: Time Series Center
Abstract:
The Time Series Center is a new center hosted at the Initiative of Innovative Computing (IIC) at Harvard. The center will collect among others a large set of Astronomical time series (light curves) from various surveys.
I will describe few projects underway with emphasis on unsolved statistical question.
Presentation: [.pdf] [.ppt]
Jonathan McDowell (CXC/CfA)
12 Dec 2006
Astrophysics for Mathematicians
Abstract:
Astronomical image, spectroscopic, and variability data challenges for Statisticians and Mathematicians.
Presentation: [.ppt] [.pdf]
Rima Izem (Harvard)
30 Jan 2007
Reducing the dimensionality of RMFs
Stephane Blondin (CfA/OIR)
13 Mar 2007
How to Classify Spectra of Exploding Stars(?)
Abstract:
Professional and amateur astronomers discover several hundreds of exploding stars (supernovae) each year. These are classified into four main types based on characteristic features in their optical spectra. Nevertheless, the supernova classification scheme poses several conceptual problems, all of which point towards the need for an unbiased and automated classification system. I will present a simple cross-correlation algorithm that was adapted to this issue, although the aim of this seminar is to seek advice from the statistics community as to which tools are best suited to classify supernova spectra.
Presentation: [.pdf]
Hyunsook Lee (CfA)
10 Apr 2007
Introducing BLoCXS and using it to estimate calibration uncertainties
Abstract: The flowchart of BLoCXS is presented for a discussion among participants to refine the proposed methods in resolving calibration uncertainties from high energy astrophysics.
Presentation: [.pdf]
Flowcharts: [.ps] [.ps] [.ps]
Updated flowchart: [.pdf]
Andreas Zezas & Hyunsook Lee (CfA)
24 Apr 2007
Developing an alternative method to measure the history of stellar populations"
Abstract:
Optical photometry and in particular color-magnitude diagrams are a standard tool for studies of stellar populations. In this presentation we will discuss the basic astrophysical background, describe the most commonly used methods and contrast them with a new more robust method which is currently under development.
Andreas' Presentation: [.pdf] [.ppt]
Hyunsook's Presentation: [.pdf]
Jing Chen Liu (Harvard)
w. Xiao-Li Meng, Michael Ratner, Irwin Shapiro
08 May 2007
An Exploratory Statistical Study of the Measured Motion of the Guide Star Used in a New Test of Einstein's Universe
Abstract:
General relativity predicts that the phenomenon of "frame-dragging" slowly alters the direction of spin of Earth-orbiting gyroscopes. NASA's Gravity Probe B satellite measured the orientation of such freely falling gyroscopes with respect to a guide star. This star's motion with respect to "fixed points" in the distant universe thus needed to be measured independently. VLBI measurements spread over 14 years yielded 39 positions of this radio-emitting guide star. We use Bayesian methods and a linear regression model of the star's motion to study the effects of various assumptions involving the parameters of both the physical model and the measurement noise model. We find that a wide range of different assumptions lead to a narrow range of point estimates and error distributions.
David van Dyk (UC Irvine)
14 May 2007
CHASC meeting
John Rice (UC Berkeley)
08 June 2007
Rm 403 @ 60 Oxford
Event Weighted Tests for Periodicity in a Sequence of Photon Arrival Times: Detecting Gamma-ray Pulsars

CHASC


text:
# 40% of what number is 108?

Apr 12, 2016

You must write a proportion.

#### Explanation:

$\frac{108}{x} = \frac{40}{100}$

$x = \frac{108 \times 100}{40}$

$x = 270$

40% of 270 is 108.

Practice exercises:

Below is the graph of average temperatures by month in Flagstaff, Arizona. Find the percentage of months where the nighttime temperature is less than 40˚ F

Good luck!


text:
Documentation

# sigma

Singular values plot of dynamic system

## Syntax

```sigma(sys) sigma(sys,w) sigma(sys,[],type) sigma(sys,w,type) sigma(sys1,sys2,...,sysN,w,type) sigma(sys1,'PlotStyle1',...,sysN,'PlotStyleN',w,type) sv = sigma(sys,w) [sv,w] = sigma(sys) ```

## Description

`sigma ` calculates the singular values of the frequency response of a dynamic system `sys`. For an FRD model, `sigma` computes the singular values of `sys.Response` at the frequencies, `sys.frequency`. For continuous-time TF, SS, or ZPK models with transfer function H(s), `sigma` computes the singular values of H(jω) as a function of the frequency ω. For discrete-time TF, SS, or ZPK models with transfer function H(z) and sample time Ts, `sigma` computes the singular values of

`$H\left({e}^{j\omega {T}_{s}}\right)$`

for frequencies ω between 0 and the Nyquist frequency ωN = π/Ts.

The singular values of the frequency response extend the Bode magnitude response for MIMO systems and are useful in robustness analysis. The singular value response of a SISO system is identical to its Bode magnitude response. When invoked without output arguments, `sigma` produces a singular value plot on the screen.

`sigma(sys)` plots the singular values of the frequency response of a model `sys`. This model can be continuous or discrete, and SISO or MIMO. The frequency points are chosen automatically based on the system poles and zeros, or from `sys.frequency` if `sys` is an FRD.

`sigma(sys,w)` explicitly specifies the frequency range or frequency points to be used for the plot. To focus on a particular frequency interval `[wmin,wmax]`, set `w = {wmin,wmax}`. To use particular frequency points, set `w` to the corresponding vector of frequencies. Use `logspace` to generate logarithmically spaced frequency vectors. Frequencies must be in `rad/TimeUnit`, where `TimeUnit` is the time units of the input dynamic system, specified in the `TimeUnit` property of `sys`.

`sigma(sys,[],type)` or `sigma(sys,w,type)` plots the following modified singular value responses:

 `type = 1` Singular values of the frequency response H–1, where H is the frequency response of `sys`. `type = 2` Singular values of the frequency response I + H. `type = 3` Singular values of the frequency response I + H–1.

These options are available only for square systems, that is, with the same number of inputs and outputs.

`sigma(sys1,sys2,...,sysN,w,type)` plots the singular value plots of several LTI models on a single figure. The arguments `w` and `type` are optional. The models `sys1,sys2,...,sysN `need not have the same number of inputs and outputs. Each model can be either continuous- or discrete-time.

`sigma(sys1,'PlotStyle1',...,sysN,'PlotStyleN',w,type)` specifies a distinctive color, linestyle, and/or marker for each system plot. See `bode` for an example.

`sv = sigma(sys,w)` and ```[sv,w] = sigma(sys)``` return the singular values `sv` of the frequency response at the frequencies `w`. For a system with `Nu` input and `Ny` outputs, the array `sv` has `min(Nu,Ny)` rows and as many columns as frequency points (length of `w`). The singular values at the frequency `w(k)` are given by `sv(:,k)`.

## Examples

collapse all

Consider the following two-input, two-output dynamic system.

`$H\left(s\right)=\left[\begin{array}{cc}0& \frac{3s}{{s}^{2}+s+10}\\ \frac{s+1}{s+5}& \frac{2}{s+6}\end{array}\right].$`

Compute the singular value responses of H(s) and I + H(s).

```H = [0, tf([3 0],[1 1 10]) ; tf([1 1],[1 5]), tf(2,[1 6])]; [svH,wH] = sigma(H); [scIH,wIH] = sigma(H,[],2);```

In the last command, the input 2 selects the second response type, I + H(s). The vectors `svH` and `svIH` contain the singular value response data, at the frequencies in `wH` and `wIH`.

Plot the singular value responses of both systems.

```subplot(211) sigma(H) subplot(212) sigma(H,[],2)```

## Tips

You can change the properties of your plot, for example the units. For information on the ways to change properties of your plots, see Ways to Customize Plots.

## Algorithms

`sigma` uses the MATLAB® function `svd` to compute the singular values of a complex matrix.

For TF, ZPK, and SS models, `sigma` computes the frequency response using the `freqresp` algorithms. As a result, small discrepancies may exist between the `sigma` responses for equivalent TF, ZPK, and SS representations of a given model.