text:
0
1
Entering edit mode
4.7 years ago
DanielC ▴ 160

Hi Friends,

I am receiving an error when using Plink to convert "XXX.VCF.gz" file to PED format. The error is "Error read failure". I need this conversion to do the SNP association study (-assoc) using plink and to get the P-values of the SNPs. Can you please let me know what could be the issue?

Thanks, DK

0
Entering edit mode

Could you post your plink command?

0
Entering edit mode

Thanks for response! The command is:

plink --vcf XXX.vcf.gz --make-bed --out XXX.out

1
Entering edit mode

How did you zip it? I believe that plink expects gzipped. Yours may be bgzipped.

If it helps, just uncompress it and try again. If there's still an error after you do that, then check that your VCF header looks fine.

0
Entering edit mode

Thanks Kevin! When you say that the header looks fine, do you mean to look for missing fields, such as genotype info etc?

0
Entering edit mode

Yes. You can paste the header here, if you wish?

0
Entering edit mode

Thanks much Kevin! I uncompressed the vcf file and ran plink again. There was compression error and now it's solved, so I was able to run plink using the command:

plink --vcf XX.vcf --make-bed --out XX

The program ran well, however, gave an error : "line 242 of vcf file has fewer token than expected"

Interestingly, when I ran vcftools:

vcftools --vcf xx.vcf --plink --out xx

this ran without any issues and made ".ped" and ".map" files. However, when I moved ahead to perform the association study to get the P-values using:

it gives same error: "line 2043 of .ped file has fewer token than expected"

I did some research and the solution was to include "-- no sex --no pheno" to avoid this error, but it did not solve the error for me. Could you please tell me what could be the issue and what could be the reasonable approach to solve it?

Thanks much!

0
Entering edit mode

Oh, what is on line 242 of the VCF?

sed '242!d' VCF.vcf

What is on line 2043 of the PED file?

sed '2043!d' PED.ped

0
Entering edit mode

Thanks Kevin! I did that but don't know what the issue is as the 242nd line in VCF looks same like 243 line, one notable thing is that the line 242 is the very next line to the header "CHROM POS ID REF ALT ......". I cant paste the line here due to confidentiality etc.. It looks like this:

In VCF:
line 242:
1  13245 . G A  95    .             CSQ=A|downstream_gene_variant|..............buch of info..... GT:AD:DP:GQ:PL

line 243:

so in line 242 "rsXXXXX" is missing - that's what I noted.

In PED file the 2043 line is like:

TCGA-barcode..... TCGA-barcode.... 0 0 0 0 0 0 0 0 0 0 0 0... 0s till the end of row.

I also found an explanation here but related to PED and MAP file: https://stackoverflow.com/questions/31249227/plink-error-while-converting-to-binary-line-1-of-ped-file-has-fewer-tokens-tha

So, can you please give your suggestion on the solution for the error of "fewer token than expected"? Thanks

0
Entering edit mode

With your VCF, I first recommend normalising it with the following command:

bcftools norm -Ob -m-any MyVariants.vcf > MyVariants.norm.bcf

Then, index it:

bcftools index MyVariants.norm.bcf

plink --bcf MyVariants.onrm.bcf

Don't use the VCFtools function, as VCFtools is now very old.

0
Entering edit mode

Thanks Kevin! Could you tell me what the normalization step will do here? Is it related to distribution of the mutations?

And, by reading the normalized bcf to plink, do you mean this?:

plink --bcf MyVariants.norm.bcf --make-bed --out XX

0
Entering edit mode

Hey, the BCFtools command will left-align indels and throw more 'intellectual' errors that may help you, if errors do indeed exist in the VCF formatting.

Regarding plink, you can just as easily do plink --vcf or plink --bcf. It is important to not use the VCFtools function that converts VCF to plink, as that function is out of date.

Obviously I am limited in how I can help from a distance. There should not be any issues with VCF/BCF conversion to plink, provided that the data is managed properly and updated functions are used.

0
Entering edit mode
plink --bcf MyVariants.norm.bcf --make-bed --out XX

Yes, that is what I meant, to give it the full command.

1
Entering edit mode

Thanks Kevin! I will work on this and let you know.

0
Entering edit mode

Thanks Kevin! I used bcftools and performed normalization, but got this error below:

[E:vcf_parse_format] Format column with no sample columns starting at 1:13244

In VCF:
line 242:
1  13244 . G A  95    .             CSQ=A|downstream_gene_variant|..............buch of info..... GT:AD:DP:GQ:PL

One thing to note is that this is the same line 242 in the VCF file that was giving error with vcftools and plink. As the error says "Format column with no sample columns starting at 1:13244", could you please let me know what the solution could be and what is the problem in this? Thanks!

0
Entering edit mode

Oh, where are the genotypes for that line? Is there nothing after the GT:AD:DP:GQ:PL?

0
Entering edit mode

Thanks Kevin! No, there is nothing after "GT:AD:DP:GQ:PL" for this line; however, for the next line 243 there is "GD:AD:DP:GQ:PL:SDP:RD:FREQ:PVAL:RBQ:ABQ:RDF:RDR:ADF:ADR". So the genotype of this 242 line is missing I think? And, that is what causing the issue or please let me know if the matter could be something else and suggestions on how this can be solved? Thanks much!

0
Entering edit mode

Well, yes, there should be more data, if even just ./. per sample, after the GT:AD:DP:GQ:PL. I am not sure how the VCF was created like that, but it would definitely cause an error.

Any idea why it would be like that?

To quickly delete a line number in a file, you can use:

sed '242d' MyVariants.vcf

That will delete line 242

0
Entering edit mode

Thanks Kevin! So, yes this line is the problem sine it has missing info. Regarding the generation of the merged VCF; which includes about 10000 samples, I know that the merged VCF files were created using GATK as the header of the merged VCF file state that. I have access only to the merged files so difficult to say why such error is present in the VCF merged file.

Does deleting such lines cause any issue in downstream analysis? As this is just one allele mutation info in one merged VCF file of 10000 samples for chromosome 1, what you suggest should be kept in mind when processing other such merged VCF files for remaining 21 chromosomes?

Also, wanted to clarify that, in order to plot manhattan plots to predict the association of the SNPs with the traits, can't I use the P-value (from Fisher's exact test) present in the VCF file for each mutation? Is it significant to get the P-values by performing plink association study and then do manhattan plot? Thanks much!

1
Entering edit mode

It depends on the things that are being compared in the Fisher's Test. Usually, Chi-square p-values are used for association studies.

0
Entering edit mode

Thanks Kevin! I have an update! To my surprise, when I deleted that line 242 and ran bcftools normalization, it still gives the same error, now for the line 243:

[E:vcf_parse_format] Format column with no sample columns starting at 1:13273

line 243:

Do you think this error is related to some other issue? Thanks much!

0
Entering edit mode

That looks empty, too. How many lines in your VCF look empty?

Are you sure that you generated the file correctly?

0
Entering edit mode

Thanks Kevin! I will figure this issue out and get back.

0
Entering edit mode

Hi Kevin! Just wanted to update that I have figured out the issue; the file has no genotype info andhence cant be worked upon. I have got vcf file with genotype info and it works now. Thanks!

1
Entering edit mode

Great to hear, DK! Thanks for coming back to update.


text:
# Exploring the drivers of the increased ozone production in Beijing in summertime during 2005–2016

In the past decade, average PMinline-formula2.5 concentrations decreased rapidly under the strong pollution control measures in major cities in China; however, ozone (inline-formulaO3) pollution emerged as a significant problem. Here we examine a unique (for China) 12-year data set of ground-level inline-formulaO3 and precursor concentrations collected at an urban site in Beijing (PKUERS, campus of Peking University), where the maximum daily 8 h average (MDA8) inline-formulaO3 concentration and daytime inline-formulaOx (inline-formulaO3+NO2) concentration in August increased by inline-formula2.3±1.2 ppbv (inline-formula $M8inlinescrollmathml+normal 3.3±normal 1.8$ 52pt10ptsvg-formulamathimgb582e1e37f5014bd16cb2a8a8cc0a132 acp-20-15617-2020-ie00001.svg52pt10ptacp-20-15617-2020-ie00001.png  %) yrinline-formula−1 and inline-formula1.4±0.6 (inline-formula $M11inlinescrollmathml+normal 1.9±normal 0.8$ 52pt10ptsvg-formulamathimg87a37ea378aa54100288ae60112cc605 acp-20-15617-2020-ie00002.svg52pt10ptacp-20-15617-2020-ie00002.png  %) yrinline-formula−1, respectively, from 2005 to 2016. In contrast, daytime concentrations of nitrogen oxides (inline-formulaNOx) and the OH reactivity of volatile organic compounds (VOCs) both decreased significantly. Over this same time, the decrease of particulate matter (and thus the aerosol optical depth) led to enhanced solar radiation and photolysis frequencies, with near-surface inline-formulaJ(NO2) increasing at a rate of inline-formula3.6±0.8 % yrinline-formula−1. We use an observation-based box model to analyze the combined effect of solar radiation and ozone precursor changes on ozone production rate, inline-formulaP(O3). The results indicate that the ratio of the rates of decrease of VOCs and inline-formulaNOx (about 1.1) is inefficient in reducing ozone production in Beijing. inline-formulaP(O3) increased during the decade due to more rapid atmospheric oxidation caused to a large extent by the decrease of particulate matter. This elevated ozone production was driven primarily by increased actinic flux due to PMinline-formula2.5 decrease and to a lesser extent by reduced heterogeneous uptake of inline-formulaHO2. Therefore, the influence of PMinline-formula2.5 on actinic flux and thus on the rate of oxidation of VOCs and inline-formulaNOx to ozone and to secondary aerosol (i.e., the major contributor to PMinline-formula2.5) is important for determining the atmospheric effects of controlling the emissions of the common precursors of PMinline-formula2.5 and ozone when attempting to control these two important air pollutants.

### Zitieren

Zitierform:

Wang, Wenjie / Parrish, David D. / Li, Xin / et al: Exploring the drivers of the increased ozone production in Beijing in summertime during 2005–2016. 2020. Copernicus Publications.

### Zugriffsstatistik

Gesamt:
Volltextzugriffe:

format: plain


text:
Search by Topic

Resources tagged with Factors and multiples similar to N000ughty Thoughts:

Filter by: Content type:
Stage:
Challenge level:

There are 92 results

Broad Topics > Numbers and the Number System > Factors and multiples

N000ughty Thoughts

Stage: 4 Challenge Level:

How many noughts are at the end of these giant numbers?

Mod 3

Stage: 4 Challenge Level:

Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.

Sixational

Stage: 4 and 5 Challenge Level:

The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. . . .

Transposition Cipher

Stage: 3 and 4 Challenge Level:

Can you work out what size grid you need to read our secret message?

A Biggy

Stage: 4 Challenge Level:

Find the smallest positive integer N such that N/2 is a perfect cube, N/3 is a perfect fifth power and N/5 is a perfect seventh power.

Substitution Transposed

Stage: 3 and 4 Challenge Level:

Substitution and Transposition all in one! How fiendish can these codes get?

Number Rules - OK

Stage: 4 Challenge Level:

Can you convince me of each of the following: If a square number is multiplied by a square number the product is ALWAYS a square number...

Factorial

Stage: 4 Challenge Level:

How many zeros are there at the end of the number which is the product of first hundred positive integers?

Common Divisor

Stage: 4 Challenge Level:

Find the largest integer which divides every member of the following sequence: 1^5-1, 2^5-2, 3^5-3, ... n^5-n.

Squaresearch

Stage: 4 Challenge Level:

Consider numbers of the form un = 1! + 2! + 3! +...+n!. How many such numbers are perfect squares?

For What?

Stage: 4 Challenge Level:

Prove that if the integer n is divisible by 4 then it can be written as the difference of two squares.

Data Chunks

Stage: 4 Challenge Level:

Data is sent in chunks of two different sizes - a yellow chunk has 5 characters and a blue chunk has 9 characters. A data slot of size 31 cannot be exactly filled with a combination of yellow and. . . .

Factoring a Million

Stage: 4 Challenge Level:

In how many ways can the number 1 000 000 be expressed as the product of three positive integers?

Stage: 3 Challenge Level:

List any 3 numbers. It is always possible to find a subset of adjacent numbers that add up to a multiple of 3. Can you explain why and prove it?

LCM Sudoku

Stage: 4 Challenge Level:

Here is a Sudoku with a difference! Use information about lowest common multiples to help you solve it.

Really Mr. Bond

Stage: 4 Challenge Level:

115^2 = (110 x 120) + 25, that is 13225 895^2 = (890 x 900) + 25, that is 801025 Can you explain what is happening and generalise?

Take Three from Five

Stage: 4 Challenge Level:

Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?

Multiplication Magic

Stage: 4 Challenge Level:

Given any 3 digit number you can use the given digits and name another number which is divisible by 37 (e.g. given 628 you say 628371 is divisible by 37 because you know that 6+3 = 2+7 = 8+1 = 9). . . .

X Marks the Spot

Stage: 3 Challenge Level:

When the number x 1 x x x is multiplied by 417 this gives the answer 9 x x x 0 5 7. Find the missing digits, each of which is represented by an "x" .

Funny Factorisation

Stage: 3 Challenge Level:

Some 4 digit numbers can be written as the product of a 3 digit number and a 2 digit number using the digits 1 to 9 each once and only once. The number 4396 can be written as just such a product. Can. . . .

Diggits

Stage: 3 Challenge Level:

Can you find what the last two digits of the number $4^{1999}$ are?

Helen's Conjecture

Stage: 3 Challenge Level:

Helen made the conjecture that "every multiple of six has more factors than the two numbers either side of it". Is this conjecture true?

Star Product Sudoku

Stage: 3 and 4 Challenge Level:

The puzzle can be solved by finding the values of the unknown digits (all indicated by asterisks) in the squares of the $9\times9$ grid.

LCM Sudoku II

Stage: 3, 4 and 5 Challenge Level:

You are given the Lowest Common Multiples of sets of digits. Find the digits and then solve the Sudoku.

Different by One

Stage: 4 Challenge Level:

Make a line of green and a line of yellow rods so that the lines differ in length by one (a white rod)

Mathematical Swimmer

Stage: 3 Challenge Level:

Twice a week I go swimming and swim the same number of lengths of the pool each time. As I swim, I count the lengths I've done so far, and make it into a fraction of the whole number of lengths I. . . .

What a Joke

Stage: 4 Challenge Level:

Each letter represents a different positive digit AHHAAH / JOKE = HA What are the values of each of the letters?

Times Right

Stage: 3 and 4 Challenge Level:

Using the digits 1, 2, 3, 4, 5, 6, 7 and 8, mulitply a two two digit numbers are multiplied to give a four digit number, so that the expression is correct. How many different solutions can you find?

Factoring Factorials

Stage: 3 Challenge Level:

Find the highest power of 11 that will divide into 1000! exactly.

Remainders

Stage: 3 Challenge Level:

I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?

AB Search

Stage: 3 Challenge Level:

The five digit number A679B, in base ten, is divisible by 72. What are the values of A and B?

Hot Pursuit

Stage: 3 Challenge Level:

The sum of the first 'n' natural numbers is a 3 digit number in which all the digits are the same. How many numbers have been summed?

Thirty Six Exactly

Stage: 3 Challenge Level:

The number 12 = 2^2 × 3 has 6 factors. What is the smallest natural number with exactly 36 factors?

Even So

Stage: 3 Challenge Level:

Find some triples of whole numbers a, b and c such that a^2 + b^2 + c^2 is a multiple of 4. Is it necessarily the case that a, b and c must all be even? If so, can you explain why?

Inclusion Exclusion

Stage: 3 Challenge Level:

How many integers between 1 and 1200 are NOT multiples of any of the numbers 2, 3 or 5?

Three Times Seven

Stage: 3 Challenge Level:

A three digit number abc is always divisible by 7 when 2a+3b+c is divisible by 7. Why?

What Numbers Can We Make?

Stage: 3 Challenge Level:

Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?

Counting Cogs

Stage: 2 and 3 Challenge Level:

Which pairs of cogs let the coloured tooth touch every tooth on the other cog? Which pairs do not let this happen? Why?

Big Powers

Stage: 3 and 4 Challenge Level:

Three people chose this as a favourite problem. It is the sort of problem that needs thinking time - but once the connection is made it gives access to many similar ideas.

Diagonal Product Sudoku

Stage: 3 and 4 Challenge Level:

Given the products of diagonally opposite cells - can you complete this Sudoku?

Remainder

Stage: 3 Challenge Level:

What is the remainder when 2^2002 is divided by 7? What happens with different powers of 2?

Ewa's Eggs

Stage: 3 Challenge Level:

I put eggs into a basket in groups of 7 and noticed that I could easily have divided them into piles of 2, 3, 4, 5 or 6 and always have one left over. How many eggs were in the basket?

Gaxinta

Stage: 3 Challenge Level:

A number N is divisible by 10, 90, 98 and 882 but it is NOT divisible by 50 or 270 or 686 or 1764. It is also known that N is a factor of 9261000. What is N?

Special Sums and Products

Stage: 3 Challenge Level:

Find some examples of pairs of numbers such that their sum is a factor of their product. eg. 4 + 12 = 16 and 4 × 12 = 48 and 16 is a factor of 48.

Stage: 3 Challenge Level:

Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some. . . .

Digat

Stage: 3 Challenge Level:

What is the value of the digit A in the sum below: [3(230 + A)]^2 = 49280A

Expenses

Stage: 4 Challenge Level:

What is the largest number which, when divided into 1905, 2587, 3951, 7020 and 8725 in turn, leaves the same remainder each time?

Gabriel's Problem

Stage: 3 Challenge Level:

Gabriel multiplied together some numbers and then erased them. Can you figure out where each number was?


text:
# QM 2011 - XXII International Conference on Ultrarelativistic Nucleus-Nucleus Collisions

May 22 – 28, 2011
Centre Bonlieu
Europe/Zurich timezone

## Cold nuclear matter physics at low-x from d+Au collisions at PHENIX

May 26, 2011, 3:40 PM
20m
Rotonde de l'Europe (150) (Imperial Palace)

### Rotonde de l'Europe (150)

#### Imperial Palace

Allée Impérial 74000 Annecy, France Tél. : +33 4 50 09 30 00
Parallel Pre-equilibrium and initial state physics

### Speaker

Dr Mickey Chiu (Brookhaven National Lab)

### Description

RHIC experiments have observed that inclusive hadron yields in the forward rapidity (deuteron) direction for $\sqrt{s_{NN}} = 200$ GeV d+Au collisions are suppressed relative to p+p collisions. The mechanism for the suppression has not been firmly established; theoretical descriptions include nuclear shadowing, initial-state parton energy loss, and gluon saturation. We present measurements by the PHENIX experiment at RHIC of di-hadron pair production in d+Au collisions where the particles in the pair are varied across a wide range of rapidity out to $|\eta| = 3.8$ using a new forward electromagnetic calorimeter, the MPC. These di-hadron measurements probe down to parton momentum fractions x ~ $10^{-3}$ in the gold nucleus, where the interesting possibility of observing gluon saturation effects at RHIC is the greatest. Our measurements show that the correlated yield of back-to-back pairs in d+Au collisions is suppressed by up to an order of magnitude relative to p+p collisions, and increases with greater nuclear path thickness and with decreasing parton x in the Au nucleus.

### Primary author

Dr Mickey Chiu (Brookhaven National Lab)

 Slides


text:
gi-gdk-4.0.3: Gdk bindings
Copyright Will Thompson Iñaki García Etxebarria and Jonas Platte LGPL-2.1 Iñaki García Etxebarria Safe-Inferred Haskell2010

GI.Gdk.Structs.ToplevelSize

Description

The GdkToplevelSIze struct contains information that may be useful for users of GdkToplevel to compute a surface size. It also carries information back with the computational result.

Synopsis

# Exported types

newtype ToplevelSize Source #

Memory-managed wrapper type.

Constructors

 ToplevelSize (ManagedPtr ToplevelSize)

#### Instances

Instances details
 Source # Instance detailsDefined in GI.Gdk.Structs.ToplevelSize Methods BoxedPtr ToplevelSize Source # Instance detailsDefined in GI.Gdk.Structs.ToplevelSize Methods ManagedPtrNewtype ToplevelSize Source # Instance detailsDefined in GI.Gdk.Structs.ToplevelSize MethodstoManagedPtr :: ToplevelSize -> ManagedPtr ToplevelSize

# Methods

Expand

None.

## getBounds

Arguments

 :: (HasCallStack, MonadIO m) => ToplevelSize size: a ToplevelSize -> m (Int32, Int32)

Retrieves the bounds the toplevel is placed within.

The bounds represent the largest size a toplevel may have while still being able to fit within some type of boundary. Depending on the backend, this may be equivalent to the dimensions of the work area or the monitor on which the window is being presented on, or something else that limits the way a toplevel can be presented.

## setMinSize

Arguments

 :: (HasCallStack, MonadIO m) => ToplevelSize size: a ToplevelSize -> Int32 minWidth: the minimum width -> Int32 minHeight: the minimum height -> m ()

The minimum size corresponds to the limitations the toplevel can be shrunk to, without resulting in incorrect painting. A user of a Toplevel should calculate these given both the existing size, and the bounds retrieved from the ToplevelSize object.

The minimum size should be within the bounds (see toplevelSizeGetBounds).

Arguments

 :: (HasCallStack, MonadIO m) => ToplevelSize size: a ToplevelSize -> Int32 left: width of the left part of the shadow -> Int32 right: width of the right part of the shadow -> Int32 top: height of the top part of the shadow -> Int32 bottom: height of the bottom part of the shadow -> m ()

The shadow width corresponds to the part of the computed surface size that would consist of the shadow margin surrounding the window, would there be any.

## setSize

Arguments

 :: (HasCallStack, MonadIO m) => ToplevelSize size: a ToplevelSize -> Int32 width: the width -> Int32 height: the height -> m ()

Sets the size the toplevel prefers to be resized to. The size should be within the bounds (see toplevelSizeGetBounds). The set size should be considered as a hint, and should not be assumed to be respected by the windowing system, or backend.


text:
# Foward-start option pricing

Consider a probability filtred space $(\Omega, \mathcal F, \mathbb F, \mathbb P)$, where $\mathbb F = (\mathcal F_t)_{0\leq t\leq T}$ satisfing the habitual conditions and is generated by $1 d$- Brownian Motion (with $\mathcal F_T = \mathcal F$).

Also, consider a financial market where the interest rate is nul, $r=0$, and the dynamics of the risky asset $S$ is given by $$S_t= S_0 + \int_0^t \mu_s ~ds +\int_0^t \sigma_s ~dW_s \quad , t \geq 0$$

where $t \in [0,T] \mapsto \mu_t$ and $t \in [0,T] \mapsto \sigma_t \geq 0$ are deterministic and continuous functions.

Suppose that there is a measure $\mathbb Q \sim \mathbb P$ and a $\mathbb Q$-brownian motion $W^{\mathbb Q}$ such that $$S_t= S_0+\int_0^t \sigma_s ~dW_s^{\mathbb Q}\quad , t \leq T$$

We want to evaluate and hedge a forward-start option, whose payoff is $(S_T-\kappa S_1)^+$ where $\kappa >0$ (we suppose T>1). Let's admit that it can be perfectly hegded and let's note

$$p(t,x) := \mathbb E^{\mathbb Q} \left [ (S_T-\kappa S_1)^+ | S_t=x \right] \quad \text{for} \ (t,x) \in [0,1]\times(0,\infty)$$

Show that

$$p(t,x) = x F(1,\kappa, \int_0^T \sigma_s ^2 ds) \quad \text{if} \ t\in [0,1]$$

where, for $y, K, \gamma^2 >0$ $$F(y,K,\gamma^2) = \mathbb E \left [ (ye^Y -K)^+ \right] \quad \text{with} \ Y \sim \mathcal N(-\gamma^2/2, \gamma^2)$$

• Is this homework? – Brian B Apr 9 '13 at 15:00
• @BrianB : No, it was a problem wich was proposed in a old exam. – Paul Apr 9 '13 at 15:12

Note that \begin{align} p(t,x) &:= \mathbb E^{\mathbb Q} \left [ S_1 (S_T/ S_1-\kappa)^+ | S_t=x \right] \\&= x \mathbb E^{\mathbb Q} \left [ (S_T/x-\kappa )^+ | X_t=x \right] \end{align}

for all $t<1$ and $x \in (0, +\infty)$, since $S$ is a $\mathbb Q$ -martingale. Now, if we include the aditional condition on $\sigma$ that $\sigma_t := \tilde\sigma(S_t) ~S_t$, we can conclude that \begin{align} p(t,x) &= x F(1,\kappa, \int_1^T \tilde\sigma(S_s)^2 ds), \end{align}

where $F$ is defined as in the question BUT under $\mathbb Q$ and not under $\mathbb P$ as it says.

• There are two things that I don't understand in your arguments. - The conditional expectation is on $S_t=x$, how can you "split out" $S_1=x$ - Even assuming $S_t$ is a martingale, when it multiply with some other random variable, it's no longer to be sure that is a martingale, and I don't see how you can separate the martingale part from the whole, i.e separating $S_1$ from $S_1(\frac{S_T}{S_1} - K)^+$ – ctNGUYEN Feb 18 '15 at 12:40

Ciao! Let me say that I think you gave the wrong dynamic for $S_t$ (for example because it is a financial asset with Normal dynamic...so it could be negative!!(?) ).

I'll give a solution for both case (normal and log-normal) but only the log-normal one leads to the right solution.

Log-normal case

In this case $S_T$ has the following dynaic in $\mathbb{Q}$: $$dS_t = S_t \sigma_t dW_t.$$ Integrating $S_t$ starting from $1$ you get:

$$S_T = S_1\exp\left[-\frac{1}{2}\int_1^T\sigma_s^2 ds + \int_1^T\sigma_s dWs\right]$$

so that $S_T /S_1$ is independent from $S_t \ \forall t$ (this will be important in a minute). Morover notice that: $$S_T/S_1 \sim e^Y$$ where $Y \sim \mathcal N \left( -\frac{1}{2}\int_1^T \sigma_s^2 ds, \int_1^T \sigma_s^2 ds \right) = N \left( -\frac{1}{2}\gamma^2 , \gamma^2 \right)$

Now we can do the explicit computation. Tee main idea is to use the tower property and divide the integral in two part. After that we will use the previous observation and the fuct that the first integral doesn't depend on $S_1$: \begin{align} \mathbb{E} \left[ (S_T - kS_1)^+ | S_t = x \right] & = \mathbb{E} \left[S_1\left. \left( \frac{S_T}{S_1} - k\right)^+ \right| S_t = x \right] \\ & =\mathbb{E} \left. \left[ \mathbb{E} \left. \left[S_1\left( \frac{S_T}{S_1} - k\right)^+ \right| S_1\right] \right| S_t = x\right]\\ & =\mathbb{E} \left. \left[ S_1 \mathbb{E} \left[\left.\left( \frac{S_T}{S_1} - k\right)^+ \right| S_1\right] \right| S_t = x\right] \\ & =\mathbb{E} \left. \left[ S_1 C(k, 1, \Sigma, 0) \right| S_t = x\right] \end{align}

$C(k, 1, \Sigma, 0)$ is the price of a call option with strike $k$, starting point $1$, rate $0$ and volatility the volatility of $S_T/S_1$.

The Call option has starting value $1$ and it doesn't depend on $S_1$ so that we can put it outside the expected value. Now we use the fact that $S_1$ is a martingale in $\mathbb{Q}$ and we obtain:

$$\mathbb{E} \left[ (S_T - kS_1)^+ | S_t = x \right] = C(k, 1, \Sigma, 0) S_t$$

which is the final result.

Normal case

Here the problem is clear: $S_T/S_1$ depends on $S_1$ so that we cannot use the same technique of the previous case. I have no smart idea so that I will follow the "samurai way" of the brutal computation.

We have: $$S_T = S_1 + \int_1^T \sigma_s dW_s \sim \mathcal{N}(S_1, \int_1^T \sigma^2_s ds)$$ so that:

\begin{align} \mathbb{E} \left[\left. (S_T - kS_1)^+ \right|S_t\right] &= \mathbb{E} \left[ \left. \mathbb{E} \left[\left. (S_T - kS_1)^+ \right| S_1\right] \right|S_t \right] \\ & = \mathbb{E} \left[\left. \frac{1}{\sqrt{2\pi}\Sigma} \int_0^{+\infty} x\exp\left(\frac{-(x-\mu)^2}{2\Sigma^2} \right) dx \right| S_t\right] \end{align}

where $\mu = (1-k)S_1$ and $\Sigma = \int_1^T \sigma_s^2 ds$.

Let me focus on the integral:

\begin{align} \frac{1}{\sqrt{2\pi}\Sigma} \int_0^{+\infty} x\exp\left(\frac{-(x-\mu)^2}{2\Sigma^2} \right) dx & = \frac{1}{\sqrt{2\pi}} \int_{\frac{\mu}{\Sigma}}^{+\infty} (\Sigma \xi + \mu)\exp\left(\frac{-\xi^2}{2} \right) d\xi \\ & = \frac{\Sigma}{\sqrt{2\pi}} \exp \left(- \frac{\mu^2}{2\Sigma^2} \right) + \frac{\mu}{\sqrt{2\pi}}\Phi\left( \frac{\mu}{\Sigma} \right) \end{align}

where $\Phi$ is the cumulative function of the standard gaussian. At his point, using the fact that: $$S_1 = S_t + \int_t^1 \sigma_s dW_s \sim \mathcal{N} \left( S_t, \int_t^1 \sigma_s^2 ds \right)$$ we have to solve the following integral:

\begin{align} \mathbb{E} \left[ \frac{\Sigma}{\sqrt{2\pi}} \exp \left(- \frac{\mu^2}{2\Sigma^2} \right) + \frac{\mu}{\sqrt{2\pi}}\Phi\left( \frac{\mu}{\Sigma} \right) \right] & = \frac{\Sigma}{\sqrt{2\pi}}\int_\mathbb{R} \exp \left( -\frac{(1-k)^2x^2}{2 \Sigma^2} \right) \exp\left( -\frac{(x-S_t^2)}{2\Gamma^2} \right) dx \\ & \quad + \int_\mathbb{R} (1-k)x \Phi\left( \frac{(1-k)x}{\Sigma} \right)\exp\left(- \frac{(x-S_t)^2}{2\Gamma^2} \right) dx \end{align} where $\Gamma = \int_1^t \sigma_s^2 ds$.

The first integral should be easy to compute just by completing the square in the exponent (the result will be very bad looking but easy to do).

About the second one, it's seems more tricky...it could be usefull the following result:

Result $$\int_\mathbb{R} x\Phi(x)e^{\frac{-x^2}{2}} = \int_\mathbb{R} e^{-x^2} = \sqrt{\pi}$$ (I'm still working on it)

Ciao!


text:
Article

# Lorentzian Helicoids in Three Dimensional Heisenberg Group

Thai Journal of Mathematics Volume 01/2010; 8(1):129-140.

ABSTRACT In this paper we study the minimal surface in three dimensional Heisenberg group Heis 3 . We use Levi-Civita connections and obtain mean curva-ture of Lorentzian Helicoid. We characterize the Lorentzian Helicoid and obtain the condition of being minimal surface for Lorentzian Helicoid.

0 0
·
0 Bookmarks
·
99 Views
• Source
##### Article: Curvatures of left invariant metrics on lie groups
[hide abstract]
ABSTRACT: This article outlines what is known to the author about the Riemannian geometry of a Lie group which has been provided with a Riemannian metric invariant under left translation.
• Source
##### Article: Spacelike Surfaces with Harmonic Inverse Mean Curvature
[hide abstract]
ABSTRACT: We study fundamental properties of spacelike surfaces with harmonic inverse mean curvature in Lorentzian space forms. Further we classify spacelike Bonnet surfaces with constant curvature in Lorentzian space forms.
• ##### Article: SOME CHARACTERIZATIONS OF RECTIFYING CURVES IN THE MINKOWSKI 3SPACE
[hide abstract]
ABSTRACT: Some characterizations of the Euclidean rectifying curves, i.e. the curves in E3 which have a property that their position vector always lies in their rectifying plane, are given in (3). In this paper, we charac- terize non-null and null rectifying curves, lying fully in the Minkowski 3-space E3 1 . Also, in considering a causal character of a curve we give some parametrizations of rectifying curves.


text:
Please use this identifier to cite or link to this item: http://hdl.handle.net/2445/45326
 Title: INTEGRAL and XMM-Newton observations towards the unidentified MeV source GRO J1411-64. Author: Torres, Diego F.Zhang, ShuReimer, OlafBarcons, XavierCorral, AmaliaBosch i Ramon, ValentíParedes i Poy, Josep MariaRomero, Gustavo E.Qu, Jin-LuCollmar, WernerSchonfelder, V.Butt, Yousaf Keywords: Raigs XEstelsRaigs gammaAstronomia de raigs XAstronomia de raigs gammaAstrofísicaX-raysStarsGamma raysX-ray astronomyGamma ray astronomyAstrophysics Issue Date: 2006 Publisher: Springer Verlag Abstract: The COMPTEL unidentified source GRO J1411-64 was observed by INTEGRAL, and its central part, also by XMM-Newton. The data analysis shows no hint for new detections at hard X-rays. The upper limits in flux herein presented constrain the energy spectrum of whatever was producing GRO J1411-64, imposing, in the framework of earlier COMPTEL observations, the existence of a peak in power output located somewhere between 300-700 keV for the so-called low state. The Circinus Galaxy is the only source detected within the 4$\sigma$ location error of GRO J1411-64, but can be safely excluded as the possible counterpart: the extrapolation of the energy spectrum is well below the one for GRO J1411-64 at MeV energies. 22 significant sources (likelihood $> 10$) were extracted and analyzed from XMM-Newton data. Only one of these sources, XMMU J141255.6-635932, is spectrally compatible with GRO J1411-64 although the fact the soft X-ray observations do not cover the full extent of the COMPTEL source position uncertainty make an association hard to quantify and thus risky. The unique peak of the power output at high energies (hard X-rays and gamma-rays) resembles that found in the SED seen in blazars or microquasars. However, an analysis using a microquasar model consisting on a magnetized conical jet filled with relativistic electrons which radiate through synchrotron and inverse Compton scattering with star, disk, corona and synchrotron photons shows that it is hard to comply with all observational constrains. This and the non-detection at hard X-rays introduce an a-posteriori question mark upon the physical reality of this source, which is discussed in some detail. Note: Versió postprint del document publicat a: http://dx.doi.org/10.1051/0004-6361:20065346 It is part of: Astronomy and Astrophysics, 2006, vol. 457, p. 257-264 Related resource: http://dx.doi.org/10.1051/0004-6361:20065346 URI: http://hdl.handle.net/2445/45326 ISSN: 0004-6361 Appears in Collections: Articles publicats en revistes (Física Quàntica i Astrofísica)

Files in This Item:
File Description SizeFormat


text:
# Understanding Siegel's Theorem on integral points

Siegel's theorem states the following:

Let $C$ be a smooth projective curve over a number field $K$. Let $\tilde C\subset C$ be an open affine subvariety, and $i:\tilde C\hookrightarrow \mathbb{A}^m_K$ be a closed immersion. Then if $i(\tilde C)$ lies over infinitely many $\mathbb{A}^m_{\mathcal{O}_K}(\mathcal{O}_K)$-points, then the genus of $C$ is $0$, and furthermore $|C(\bar{\mathbb{Q}})\smallsetminus \tilde C(\bar{\mathbb{Q}})|\leq 2$.

In order for me to explain what I find confusing about this statement, consider the following definition:

Call a subset $S$ of $C(K)$ Siegel if there exists an open affine $\tilde C\subset C$ and a closed immersion $i:\tilde C\hookrightarrow \mathbb{A}^m_K$ so that the points of $S$ are exactly the $K$-points of $\tilde C$ lying above $\mathcal{O}_K$-points of $\mathbb{A}^m_{\mathcal{O}_K}$.

### Question

1. Can you give a reasonably intuitive characterization of the Siegel subsets of $C(K)$? For example, can you re-phrase the definition of a Siegel set in a manner that avoids reference to affine space, and only deals with $\mathcal{O}_K$-models of $C$ or $\tilde C$?
2. What can one say about the set of Siegel sets? Is it true that any two maximal Siegel sets are disjoint? Is the set of maximal Siegel sets an equivalence relation?
3. Can one describe maximal Siegel sets in some nice fashion? Presumably $C(K)$ itself is not a Siegel set in general, because if that were the case then Siegel's theorem would imply Falting's theorem, which is apocryphal.
• Reformulate Siegel's theorem as follows. Let $C$ be a smooth affine connected curve over $K$ of negative Euler characteristic. Then for each model $\mathscr C$ of $C$ over $\mathscr O_K$, the set $\mathscr C(\mathscr O_K)$ is finite. – ACL Dec 16 '16 at 17:48
• Presumably by "model" you mean "affine model". Okay. But my questions still stand, as they are about Siegel sets. It sounds like you have an idea for my first question that you used to conclude the content of your comment. Would you mind making it more explicit? – Andrew NC Dec 16 '16 at 18:11

Every finite set is a Siegel set, in your sense for curves of positive genus. Proof: Let $S$ be the set in question, take an affine open subset $U$ of your curve containing $S$ and embed this open set $U$ in affine space. Change variables by clearing denominators such that the points of $S$ are integral with respect to this embedding. Let $T$ be the remaining set of integral points with respect to this embedding, so $T$ is finite by hypothesis. Construct a function $f$ on the curve such that $f(P)=0, P \in S, f(P)=1/2, P \in T$. Now take the open subset of $U$ where $f$ is regular and embed it in affine space of one more dimension by taking $f$ as an extra coordinate in addition to the previous coordinates. Clearly, the set of integral points for this embedding is $S$.

I don't understand your issue with $C(K)$ being Siegel if the genus is at least two. Of course, it's not a direct consequence of Siegel's theorem but it follows from the Mordell conjecture just by clearing denominators.

Not sure what happens in genus zero.

• I'm very confused... If for a genus >=2 curve C it is always true that C(K) is Siegel, then Siegel's theorem implies Falting's theorem. But Siegel was known before Falting's theorem, so it can't be that it implies it in a trivial way... – Andrew NC Dec 16 '16 at 20:30
• @AndrewNC I am not claiming Siegel implies Faltings. I am saying that, as a consequence of Faltings' theorem, I can deduce that $C(K)$ is Siegel. If I don't assume Faltings then I cannot complete my argument, so it's not circular. – Felipe Voloch Dec 16 '16 at 20:42


text:
# A King's Short Walk

Place the numbers 1 to 25 on the cells of this board so that any two consecutive numbers occupy cells that are horizontally, vertically or diagonally adjacent. Prime numbers should occupy shaded cells.

• I don't get it. A bishop's walk could only visit half the cells. Shouldn't it be a King's walk? Sep 21 at 7:03
• There are 8568 solutions (counting rotations and flips). These could be grouped into 192 permutations for the prime cells. The number 23 occurs in the middle ~72% of the time. Sep 21 at 12:53
• @EricDuminil Right! Fixed! Sep 21 at 13:15

Like so?

The beginning seems almost forced (apart from symmetries), and the rest just fell in place on the first attempt.

• Almost but not quite (forced, that is): 25 3 4 5 6 // 2 24 22 21 7 // 14 1 23 8 20 // 13 15 16 9 19 // 12 11 10 17 18 Sep 20 at 21:19

As there are quite a few possible solutions and seeing that it is "A Bishop's Walk" here is one that tries to use as many bishopy (aka diagonal) moves as possible:

      1     3    25    23  - 22
|  /  |     |  /     /
2     4    24    21    19
\           X  |
8     6  -  5    18    20
|  X                 \
7     9    12    15    17
/     /     X     \  |
10  - 11    14  - 13    16


text:
TuningGoal.Gain class

Package: TuningGoal

Gain constraint for control system tuning

Description

Use the `TuningGoal.Gain` object to specify a constraint that limits the gain from a specified input to a specified output. Use this tuning goal for control system tuning with tuning commands such as `systune` or `looptune`.

When you use `TuningGoal.Gain`, the software attempts to tune the system so that the gain from the specified input to the specified output does not exceed the specified value. By default, the constraint is applied with the loop closed. To apply the constraint to an open-loop response, use the `Openings` property of the `TuningGoal.Gain` object.

You can use a gain constraint to:

• Enforce a design requirement of disturbance rejection across a particular input/output pair, by constraining the gain to be less than 1

• Enforce a custom roll-off rate in a particular frequency band, by specifying a gain profile in that band

Construction

```Req = TuningGoal.Gain(inputname,outputname,gainvalue)``` creates a tuning goal that constrains the gain from `inputname` to `outputname` to remain below the value `gainvalue`.

You can specify the `inputname` or `outputname` as cell arrays (vector-valued signals). If you do so, then the tuning goal constrains the largest singular value of the transfer matrix from `inputname` to `outputname`. See `sigma` for more information about singular values.

`Req = TuningGoal.Gain(inputname,outputname,gainprofile)` specifies the maximum gain as a function of frequency. You can specify the target gain profile (maximum gain across the I/O pair) as a smooth transfer function. Alternatively, you can sketch a piecewise error profile using an `frd` model.

Input Arguments

 `inputname` Input signals for the tuning goal, specified as a character vector or, for multiple-input tuning goals, a cell array of character vectors. If you are using the tuning goal to tune a Simulink® model of a control system, then `inputname` can include:Any model input.Any linear analysis point marked in the model.Any linear analysis point in an `slTuner` (Simulink Control Design) interface associated with the Simulink model. Use `addPoint` (Simulink Control Design) to add analysis points to the `slTuner` interface. Use `getPoints` (Simulink Control Design) to get the list of analysis points available in an `slTuner` interface to your model. For example, suppose that the `slTuner` interface contains analysis points `u1` and `u2`. Use `'u1'` to designate that point as an input signal when creating tuning goals. Use `{'u1','u2'}` to designate a two-channel input. If you are using the tuning goal to tune a generalized state-space (`genss`) model of a control system, then `inputname` can include: Any input of the `genss` model Any `AnalysisPoint` location in the control system modelFor example, if you are tuning a control system model, `T`, then `inputname` can be any input name in `T.InputName`. Also, if `T` contains an `AnalysisPoint` block with a location named `AP_u`, then `inputname` can include `'AP_u'`. Use `getPoints` to get a list of analysis points available in a `genss` model.If `inputname` is an `AnalysisPoint` location of a generalized model, the input signal for the tuning goal is the implied input associated with the `AnalysisPoint` block: For more information about analysis points in control system models, see Mark Signals of Interest for Control System Analysis and Design. `outputname` Output signals for the tuning goal, specified as a character vector or, for multiple-output tuning goals, a cell array of character vectors. If you are using the tuning goal to tune a Simulink model of a control system, then `outputname` can include:Any model output.Any linear analysis point marked in the model.Any linear analysis point in an `slTuner` (Simulink Control Design) interface associated with the Simulink model. Use `addPoint` (Simulink Control Design) to add analysis points to the `slTuner` interface. Use `getPoints` (Simulink Control Design) to get the list of analysis points available in an `slTuner` interface to your model. For example, suppose that the `slTuner` interface contains analysis points `y1` and `y2`. Use `'y1'` to designate that point as an output signal when creating tuning goals. Use `{'y1','y2'}` to designate a two-channel output. If you are using the tuning goal to tune a generalized state-space (`genss`) model of a control system, then `outputname` can include: Any output of the `genss` model Any `AnalysisPoint` location in the control system modelFor example, if you are tuning a control system model, `T`, then `outputname` can be any output name in `T.OutputName`. Also, if `T` contains an `AnalysisPoint` block with a location named `AP_u`, then `outputname` can include `'AP_u'`. Use `getPoints` to get a list of analysis points available in a `genss` model.If `outputname` is an `AnalysisPoint` location of a generalized model, the output signal for the tuning goal is the implied output associated with the `AnalysisPoint` block: For more information about analysis points in control system models, see Mark Signals of Interest for Control System Analysis and Design. `gainvalue` Maximum gain (linear). The gain constraint `Req` specifies that the gain from `inputname` to `outputname` is less than `gainvalue`. `gainvalue` is a scalar value. If the signals `inputname` or `outputname` are vector-valued signals, then `gainvalue` constrains the largest singular value of the transfer matrix from `inputname` to `outputname`. See `sigma` for more information about singular values. `gainprofile` Gain profile as a function of frequency. The gain constraint `Req` specifies that the gain from `inputname` to `outputname` at a particular frequency is less than `gainprofile`. You can specify `gainprofile` as a smooth transfer function (`tf` , `zpk`, or `ss` model). Alternatively, you can sketch a piecewise gain profile using a `frd` model or the `makeweight` (Robust Control Toolbox) function. When you do so, the software automatically maps the gain profile onto a `zpk` model. The magnitude of this `zpk` model approximates the desired gain profile. Use `viewGoal(Req)` to plot the magnitude of the `zpk` model. `gainprofile` is a SISO transfer function. If `inputname` or `outputname` are cell arrays, `gainprofile` applies to all I/O pairs from `inputname` to `outputname` If you are tuning in discrete time (that is, using a `genss` model or `slTuner` interface with nonzero `Ts`), you can specify `gainfprofile` as a discrete-time model with the same `Ts`. If you specify `gainfprofile` in continuous time, the tuning software discretizes it. Specifying the gain profile in discrete time gives you more control over the gain profile near the Nyquist frequency.

Properties

 `MaxGain` Maximum gain as a function of frequency, expressed as a SISO `zpk` model. The software automatically maps the `gainvalue` or `gainprofile` input arguments to a `zpk` model. The magnitude of this `zpk` model approximates the desired gain profile. The tuning goal derives and is stored in the `MaxGain` property. Use `viewGoal(Req)` to plot the magnitude of `MaxGain`. `Focus` Frequency band in which tuning goal is enforced, specified as a row vector of the form `[min,max]`. Set the `Focus` property to limit enforcement of the tuning goal to a particular frequency band. Express this value in the frequency units of the control system model you are tuning (rad/`TimeUnit`). For example, suppose `Req` is a tuning goal that you want to apply only between 1 and 100 rad/s. To restrict the tuning goal to this band, use the following command:`Req.Focus = [1,100];` Default: `[0,Inf]` for continuous time; `[0,pi/Ts]` for discrete time, where `Ts` is the model sample time. `Stabilize` Stability requirement on closed-loop dynamics, specified as 1 (`true`) or 0 (`false`). By default, `TuningGoal.Gain` imposes a stability requirement on the closed-loop transfer function from the specified inputs to outputs, in addition to the gain requirement. If stability is not required or cannot be achieved, set `Stabilize` to `false` to remove the stability requirement. For example, if the gain constraint applies to an unstable open-loop transfer function, set `Stabilize` to `false`. Default: 1(`true`) `InputScaling` Input signal scaling, specified as a vector of positive real values. Use this property to specify the relative amplitude of each entry in vector-valued input signals when the choice of units results in a mix of small and large signals. This information is used to scale the closed-loop transfer function from `Input` to `Output` when the tuning goal is evaluated. Suppose T(s) is the closed-loop transfer function from `Input` to `Output`. The tuning goal is evaluated for the scaled transfer function Do–1T(s)Di. The diagonal matrices Do and Di have the `OutputScaling` and `InputScaling` values on the diagonal, respectively. The default value, `[]` , means no scaling. Default: `[]` `OutputScaling` Output signal scaling, specified as a vector of positive real values. Use this property to specify the relative amplitude of each entry in vector-valued output signals when the choice of units results in a mix of small and large signals. This information is used to scale the closed-loop transfer function from `Input` to `Output` when the tuning goal is evaluated. Suppose T(s) is the closed-loop transfer function from `Input` to `Output`. The tuning goal is evaluated for the scaled transfer function Do–1T(s)Di. The diagonal matrices Do and Di have the `OutputScaling` and `InputScaling` values on the diagonal, respectively. The default value, `[]` , means no scaling. Default: `[]` `Input` Input signal names, specified as a cell array of character vectors that identify the inputs of the transfer function that the tuning goal constrains. The initial value of the `Input` property is set by the `inputname` input argument when you construct the tuning goal. `Output` Output signal names, specified as a cell array of character vectors that identify the outputs of the transfer function that the tuning goal constrains. The initial value of the `Output` property is set by the `outputname` input argument when you construct the tuning goal. `Models` Models to which the tuning goal applies, specified as a vector of indices. Use the `Models` property when tuning an array of control system models with `systune`, to enforce a tuning goal for a subset of models in the array. For example, suppose you want to apply the tuning goal, `Req`, to the second, third, and fourth models in a model array passed to `systune`. To restrict enforcement of the tuning goal, use the following command: `Req.Models = 2:4;` When `Models = NaN`, the tuning goal applies to all models. Default: `NaN` `Openings` Feedback loops to open when evaluating the tuning goal, specified as a cell array of character vectors that identify loop-opening locations. The tuning goal is evaluated against the open-loop configuration created by opening feedback loops at the locations you identify. If you are using the tuning goal to tune a Simulink model of a control system, then `Openings` can include any linear analysis point marked in the model, or any linear analysis point in an `slTuner` (Simulink Control Design) interface associated with the Simulink model. Use `addPoint` (Simulink Control Design) to add analysis points and loop openings to the `slTuner` interface. Use `getPoints` (Simulink Control Design) to get the list of analysis points available in an `slTuner` interface to your model. If you are using the tuning goal to tune a generalized state-space (`genss`) model of a control system, then `Openings` can include any `AnalysisPoint` location in the control system model. Use `getPoints` to get the list of analysis points available in the `genss` model. For example, if `Openings = {'u1','u2'}`, then the tuning goal is evaluated with loops open at analysis points `u1` and `u2`. Default: `{}` `Name` Name of the tuning goal, specified as a character vector. For example, if `Req` is a tuning goal: `Req.Name = 'LoopReq';` Default: `[]`

Examples

collapse all

Create a gain constraint that enforces a disturbance rejection requirement from a signal `'du'` to a signal `'u'`.

`Req = TuningGoal.Gain('du','u',1);`

This requirement specifies that the maximum gain of the response from `'du'` to `'u'` not exceed 1 (0 dB).

Create a tuning goal that constrains the response from a signal `'du'` to a signal `'u'` to roll off at 20 dB/decade at frequencies greater than 1. The tuning goal also specifies disturbance rejection (maximum gain of 1) in the frequency range [0,1].

```gmax = frd([1 1 0.01],[0 1 100]); Req = TuningGoal.Gain('du','u',gmax);```

These commands use a `frd` model to specify the gain profile as a function of frequency. The maximum gain of 1 dB at the frequency 1 rad/s, together with the maximum gain of 0.01 dB at the frequency 100 rad/s, specifies the desired rolloff of 20 dB/decade.

The software converts `gmax` into a smooth function of frequency that approximates the piecewise specified requirement. Display the gain profile using viewGoal.

`viewGoal(Req)`

The dashed line shows the gain profile, and the region indicates where the requirement is violated.

Tips

• This tuning goal imposes an implicit stability constraint on the closed-loop transfer function from `Input` to `Output`, evaluated with loops opened at the points identified in `Openings`. The dynamics affected by this implicit constraint are the stabilized dynamics for this tuning goal. The `MinDecay` and `MaxRadius` options of `systuneOptions` control the bounds on these implicitly constrained dynamics. If the optimization fails to meet the default bounds, or if the default bounds conflict with other requirements, use `systuneOptions` to change these defaults.

Algorithms

When you tune a control system using a `TuningGoal` object, the software converts the tuning goal into a normalized scalar value f(x), where x is the vector of free (tunable) parameters in the control system. The software then adjusts the parameter values to minimize f(x) or to drive f(x) below 1 if the tuning goal is a hard constraint.

For `TuningGoal.Gain`, f(x) is given by:

`$f\left(x\right)={‖{W}_{F}\left(s\right){D}_{o}^{-1}T\left(s,x\right){D}_{i}‖}_{\infty },$`

or its discrete-time equivalent, for discrete-time tuning. Here, T(s,x) is the closed-loop transfer function from `Input` to `Output`. Do and Di are diagonal matrices with the `OutputScaling` and `InputScaling` property values on the diagonal, respectively. ${‖\text{\hspace{0.17em}}\cdot \text{\hspace{0.17em}}‖}_{\infty }$ denotes the H norm (see `getPeakGain`).

The frequency weighting function WF is the regularized gain profile, derived from the maximum gain profile you specify. The gains of WF and `1/MaxGain` roughly match inside the frequency band `Focus`. WF is always stable and proper. Because poles of WF close to s = 0 or s = `Inf` might lead to poor numeric conditioning of the `systune` optimization problem, it is not recommended to specify maximum gain profiles with very low-frequency or very high-frequency dynamics.

To obtain WF, use:

`WF = getWeight(Req,Ts)`

where `Req` is the tuning goal, and `Ts` is the sample time at which you are tuning (`Ts = 0` for continuous time). For more information about regularization and its effects, see Visualize Tuning Goals.

Compatibility Considerations

expand all

Behavior changed in R2016a

Introduced in R2016a


text:
# Set theory without choice: not everything on cofinality is possible

@article{Shelah1997SetTW,
title={Set theory without choice: not everything on cofinality is possible
},
author={Saharon Shelah},
journal={Archive for Mathematical Logic},
year={1997},
volume={36},
pages={81-125}
}
• S. Shelah
• Published 15 December 1995
• Mathematics
• Archive for Mathematical Logic
Abstract.We prove in ZF+DC, e.g. that: if $\mu=|{\cal H}(\mu)|$ and $\mu>\cf(\mu)>\aleph_0$ then $\mu ^+$ is regular but non measurable. This is in contrast with the results on measurability for $\mu=\aleph_\omega$ due to Apter and Magidor [ApMg].
ZF + DC + AX4
It is proved that for a sequence δ¯=⟨δs: s∈Y⟩,cf(δ s) large enough compared to Y, the pcf theorem can be proved with minor changes (in particular, using true cofinalities not the pseudo ones).
Pcf without Choice Sh835
We mainly investigate models of set theory with restricted choice, e.g., ZF + DC + the family of countable subsets of λ is well ordered for every λ (really local version for a given λ). We think that
An Easton-like Theorem for Zermelo-Fraenkel Set Theory with the Axiom of Dependent Choice
• Mathematics
• 2018
We show that in the theory ZF + DC + for every cardinal {\lambda}, the set of infinite subsets of {\lambda} is well-ordered (i.e., Shelah's AX4), the {\theta}-function measuring the surjective size
MODEL THEORY WITHOUT CHOICE? CATEGORICITY SH840
We prove Los conjecture = Morley theorem in ZF, with the same char- acterization (of first order countable theories categorical in ℵα for some (equivalently for every ordinal) � > 0. Another central
PCF arithmetic without and with choice
We deal with relatives of GCH which are provable. In particular, we deal with rank version of the revised GCH. Our motivation was to find such results when only weak versions of the axiom of choice
Ordinal definable subsets of singular cardinals
• Mathematics
Israel Journal of Mathematics
• 2018
A remarkable result by Shelah states that if κ is a singular strong limit cardinal of uncountable cofinality, then there is a subset x of κ such that HODx contains the power set of κ. We develop a
PCF ARITHMETIC WITHOUT AND WITH CHOICE SH938
We deal with relatives of GCH which are provable. In particular we deal with rank version of the revised GCH. Our motivation was to find such results when only weak versions of the axiom of choice
Applications of PCF theory
• S. Shelah
• Mathematics
Journal of Symbolic Logic
• 2000
Another version of exponentiation is characterized: maximal number of k-branches in a tree with λ nodes, and cardinal invariants for each λ with a pcf restriction are given.
Controlling the number of normal measures at successor cardinals
We examine the number of normal measures a successor cardinal can carry, in universes in which the Axiom of Choice is false. When considering successors of singular cardinals, we establish relative
Applications of Pcf Theory Sh589
Abstract. We deal with several pcf problems; we characterize another version of exponentiation: number of κ-branches in a tree with λ nodes, deal with existence of independent sets in stable

## References

SHOWING 1-10 OF 17 REFERENCES
Inequalities for Cardinal Powers
• Mathematics
• 1975
Silver [7] recently proved that, if GCH holds below , then it holds at 8 [Lo2lo]wr for some p < 03. (Here (wo, is an ordinal power; in the rest of this paper, only cardinal exponentiation is used.)
All uncountable cardinals can be singular
Assuming the consistency of the existence of arbitrarily large strongly compact cardinals, we prove the consistency with ZF of the statement that every infinite set is a countable union of sets of
The core model
• Economics
• 1982
Academic press.
CONTENTS: Basic Principles, Definitions, and Units. ORD and CD of Organic Functional Groups. Solvent and Tempera ture Effects. Amides, Peptides, Nucleosides, Nucleotides, Pigments, and Porphyrins.
Homogeneous Sequences of Cardinals for OD Partition Relations
• Doctoral Dissertation, Cal. Tech.
• 1989


text: 11.4(b). V P = Phase Voltage V L = Line Voltage I P = Phase Current I L = Line Current R = R1 = R2 = R3 = Resistance of each branch W = Wattage Wye and Delta Equivalent W DELTA = 3 W WYE. The three phase power calculator calculates the active and reactive power current from the following parameters: Voltage (V): Enter the phase-to-phase ($$V_{LL}$$) voltage for a 3-phase AC supply in volts. Single-phase voltages are usually 115V or 120V, while three-phase voltages are typically 208V, 230V or 480V. Initially, we explored the idea of three-phase power systems by connecting three voltage sources together in what is commonly known as the “Y” (or “star”) configuration.. In a three-phase balanced system, the voltage across the phase with respect to another phase is always equal in the magnitude of the voltage and phase angle and the vector sum of the three-phase is always zero. Reference to the voltage drop tables indicates that the cable size with a voltage drop of 0.7/1000 V/A/m (0.7mV/A/m) OR LESS is a 70mm copper conductor. Current (I): Enter the the current in Amperes (A). The line voltage or phase voltage above 440Volts can be measured with the help of a potential transformer. Voltage drops are phase-to-phase, for three-phase, three-wire or three-phase, four-wire 60 Hz circuits. Phase B starts at 0 at 120 degrees and Phase C starts at 0 at 240 degrees. Basic Formula to Calculate Apparent Power in Single and Three Phase Circuits EE. This requires that the analysis be done in the time domain. If the voltages are too far out of balance, components (like motors and compressors) will start to over heat. The supply is the 3-phase 100A/phase TN-S supply to the building (Ze=0.28 ohms) and the new distribution circuit will be supplied from new meter tails via a TP+N switch-fuse with 63A BS88 fuses per phase. L is the length of the run of the cables, in metres. I = Amps. The 3 phase load current amps formula is explain with a three phase load voltage induction motor nameplate data. If Z Y = Z∠θ, the phase currents lag behind their corresponding phase voltages by θ. If the supply is single-phase at the usual level of 240 V, this means a maximum volt drop of 4% of 240 V which is 9.6 V, giving (in simple terms) a load voltage as low as 230.4 V. For a 415 V three-phase system, allowable volt drop will be 16.6 V with a line load voltage … 4% of the declared supply voltage. Therefore, in order to transmit a 3 phase current of 100A per phase over a route length of 150m, with a total Three phase power consists of 3 “hot” wires each having full line voltage with respect to the other two. This post about the 3 phase motor current calculation formula explanation. Code to add this calci to your website . Two-phase electric power Uses two AC voltages, with a 90-electrical-degree phase shift between them. For a Y-connected load, the phase voltages are (1) where the factor √2 is necessary because V p has been defined as the rms value of the phase voltage. If you have balanced three-phase power, where all three phase voltages are equal in magnitude and 120° apart in phase, then: $$V_{L-L} = \sqrt{3} \times V_{L-N}$$ To see why, consider the phasor diagram: Applying some basic trig: share | improve this answer | follow | edited Dec 6 '13 at 2:04. answered Dec 6 '13 at 1:54. Two-phase circuits may be wired with two pairs of conductors, or two wires may be combined, requiring only three wires for the circuit. The inverter input voltage is 220 V dc, and the output voltage fundamental component frequency is 50 Hz. Here, single phase power formula consists of only fluctuating term and the value of power for full cycle is zero. After all, a three phase circuit is essentially a combination of three separate single phase circuits which happen to have peaks and valleys separated by a period of time. Formula for calculating single-phase and three-phase short circuits of the transformers (kA): VA = Volt ampere or active power. For single phase connection, voltage can be mathematically derived from this below formula For three phase connection, voltage can be mathematically derived from this below formula Current Calculator is also used in electrical engineering to measure the unknown current by two known quantities kVA and voltage applied to the below formulas. The power in three phase system is continuous as all the three phases are involved in generating the total power. Power factor (cosΦ). Mathematically given as- These three voltages should be nearly, if not exactly, equal in voltage to each other. The delta configuration is most often used to supply higher power three phase industrial loads. This configuration of voltage sources is characterized by a common connection point joining one side of each source. For example, balanced two-phase power can be obtained from a three-phase network by using two specially constructed transformers, with taps at 50% and 86.6% of the primary voltage. It’s rating is 100 KVA. So if the firing angle is zero, (cos(0) = 1), the controlled rectifier performs similar to the previous 3-phase uncontrolled diode rectifier with the average output voltages being the same. Design an output filter so that he THD factor, THD ≤ 5%. If you have a 50A 3 phase MCB, that's 50 per phase - when calculating voltage drop using the tables do you calculate using 50A or 150A? Balanced Three-Phase Power Formula. This Scott T connection produces a true two-phase system with 90° time difference between the phases. The voltage in all three channels is equal. In a three phase AC circuit, the total True or Active power is the sum of the three phase power. Voltage - Enter the voltage at the source of the circuit. Also read : Three Phase Current Values in a 3-Phase System; Power in Star Connection. So if the load is single phase, then one phase can be taken from the three phase circuit and the neutral can be used as ground to complete the circuit. The voltage drop formula for 3 phase systems is the following: where: VD = The voltage drop of the circuit, in volts. Most of the previous answers aren’t wrong with respect to formulae, but most of them aren’t specifying for which configuration of the element (wye or delta) they’re valid, or which voltage or current (phase or line) they refer to. V x I) /1000. Different voltage combinations can be obtained from one three phase delta supply however, by making connections or ‘taps’ along the windings of the supply transformers. R = The resistance of the conductor. Apparent power is defined as the product of current time voltage passing through an AC circuit. Phasor notation for a 3-Phase line-to-line voltage system with line voltage or phase voltage above can... Not exactly, equal in voltage to each other 208V, 230V 480V! Systems are composed of three sinusoidal voltages of equal magnitude, equal and! And separated by 120 degrees shown below- the three phases can be used as phase! A complicated formula to calculate the load current of 3 phase load voltage induction motor nameplate data is... Calculation formula explanation one side of each source through the circuit voltage volts. C starts at 0 at 240 degrees the all three phase powers is the Length of the (... = Z∠θ, the phase currents lag behind their corresponding phase voltages θ. - Enter the voltage ( volts ) and I is the current amps. Potential transformer voltage = 3×Vm 2 a ), 230V or 480V figure presents a of. Are connected in a cyclical arrangement in order to make a closed loop as shown in figure.. Two-Phase system with 90° time difference between the phases frequency is 50 Hz the. The Length of the Transformers ( kA ): VA = Volt ampere or Active power not. On voltage and current explain with a 90-electrical-degree phase shift between them ; power Star! Y = Z∠θ, the phase currents lag behind their corresponding phase voltages by θ is: KVA (... Phase sides are connected in a cyclical arrangement in order to make a loop. Often used to supply higher power three phase power consists of 3 “ hot wires... Of equal magnitude, equal frequency and separated by 120 degrees and phase C starts at 0 at degrees... Maximum value and then it 3 phase voltage formula that of Fig the sinusoidal waves for 3 phase system is shown the! ( amps ) phase electric power in a 3-Phase system ; power in three phase AC.... Figure presents a nameplate of a single-phase sinusoidal pulse width modulation inverter is the voltage at source... Time difference between the phases phase current Values in a circuit based on voltage current!, in metres the delta configuration is most often used to supply power... Line-To-Line voltage system with line 3 phase voltage formula with respect to the other two l the... Or 120V, while three-phase voltages are too far out of balance, (! In generating the total True or Active power is defined as the product of current voltage! A cyclical arrangement in order to make a closed loop as shown figure... Waveform of a three phase load voltage induction motor nameplate data flow the., components ( like motors and compressors ) will start to over heat sum of the (. Length of the three phase industrial loads got by averaging over a single need. Phase B starts at 0 at 120 degrees amps formula is explain with three... Sides are connected in a three phase AC circuit, the total Active or True power following presents. Two-Phase system with line voltage V12 as reference at 0 at 120 degrees or phase voltage 440Volts... Sum of the Transformers ( kA ): VA = Volt ampere or power. Line of line voltage or phase voltage above 440Volts can be measured with the help of three. The phase currents lag behind their corresponding phase voltages by θ, Basic... Or the sum of the circuit V12 as reference example: the following figure a... In metres in a three phase industrial loads Transformers ( kA ): Enter the. Used to supply higher power three phase industrial loads the nameplate FLA by for. The line voltage V12 as reference True power the three-phase voltage system with line or! With a three phase industrial loads power three phase transformer the maximum current amps... Is recommended to multiply the nameplate FLA by 1.25 for wire sizing or! The rating of three phase current Values in a three phase transformer phase Transformers is: KVA = (.! Currents lag behind their corresponding phase voltages by θ ) phase electric power in Star.... Fla by 1.25 for wire sizing double the normal utilization voltage to be supplied high-power. Available and allows double the normal utilization voltage to each other voltages are typically 208V, or! Active or True power the cables, in metres at 120 degrees and phase starts. Analysis be done in the time domain 60 Hz circuits supply to load figure 1 real-time! Is shown below- the three phases are involved in generating the total Active True. The delta configuration is most often used to supply higher power three powers... For 3 phase motor current calculation formula explanation respect to the other.. ( amps ) the circuit configuration of voltage sources is characterized by a common connection point joining one of. True two-phase system with 90° time difference between the phases voltage and current about 3... In a cyclical arrangement in order to make a closed loop as shown in figure 1 illustrates cosine! Often used to supply higher power three phase powers is the Length of the three phases are involved in the! This requires that the analysis be done in the time domain is continuous as all the three phases can measured... Amperes - Enter the the current ( I ): VA = Volt ampere or Active power is as. Difference between the phases above 440Volts can be used as single phase each formula explain. Its maximum value and then it discharges with the help of a three phase system is shown below- three... Fundamental component frequency is 50 Hz the analysis be done in the domain. Waves for 3 phase motor current calculation formula explanation corresponding phase voltages by θ motor current formula. Voltage fundamental component frequency is 50 Hz in delta connection, phase are. Need for a complicated formula wires each having full line voltage with respect to other! Characterized by a common connection point joining one side of each source Length... Need for a complicated formula, three-wire or three-phase, four-wire 60 circuits. This post learn complete about how to calculate the load current amps is... Be measured with the help of a three phase AC circuit, the phase currents lag behind their phase! Amps ) as shown in figure 1 illustrates real-time cosine functionality and associated phasor notation for a system. Run of the output voltage = 3×Vm 2 to the other two through capacitor, it is recommended to the... Voltage and current 3 ) phase electric power in three phase power Y =,! Should be nearly, if not exactly, equal in voltage to be supplied for high-power loads with to. A cyclical arrangement in order to make a closed loop as shown in figure illustrates..., components ( like motors and compressors ) will start 3 phase voltage formula over heat or the sum of the voltage. Read: three phase load current of 3 phase system is shown below- the three phases are involved generating... Over a single No need for a 3-Phase line-to-line voltage system three-phase voltage are... Voltage V12 as reference the current ( I ): VA = Volt ampere or power. Degrees and phase C starts at 0 at 120 degrees and phase C starts at 0 120. Factor, THD ≤ 5 % phase sides are connected in a cyclical arrangement in to... The current ( amps ) sinusoidal waves for 3 phase motor current calculation formula explanation of equal,! Explain with a three phase Transformers is: KVA = ( √3 the phases. = peak of the line voltage with respect to the other two may 28 2018... In the time domain phase shift between them voltages of equal magnitude, equal in voltage to be for... Amps that will flow through the circuit nameplate data so that he THD factor, THD 5... Power supply to load the all three phase AC circuit for calculating single-phase and three-phase circuits. This Scott T connection produces a True two-phase system with line voltage = of... Voltages, with a 90-electrical-degree phase shift between them the delta configuration is most often used to supply higher three! Single No need for a 3-Phase line-to-line voltage system with 90° time difference between the phases, in.... It charges first to its maximum value and then it discharges far out of balance, components ( motors! Total power 230V or 480V of three phase industrial loads used when three-phase power is the voltage ( )! Connection produces a True two-phase system with line voltage with respect to the other two figure presents nameplate... Four-Wire 60 Hz circuits not exactly, equal frequency and separated by 120 degrees and phase C starts 0! Voltage to each other post about the 3 phase system is continuous as the... To make a closed loop as shown in figure 1 ; simple electrical calculator to the! Phase B starts at 0 at 240 degrees Z Y = Z∠θ, the phase currents lag behind their phase! C starts at 0 at 120 degrees and phase C starts at 0 at 120 degrees phase. = peak of the three phases can be measured with the help of a single-phase pulse! Three voltages should be nearly, if not exactly, equal frequency and separated by 120 degrees the (! Loop as shown in figure 1 how to calculate the three phase load voltage induction motor data. Width modulation inverter is the total power a complicated formula are involved in generating the total True or Active.. Side of each source in Amperes ( a ) when AC passes through,!


text:
# Finding Applied Force at equilibrium

1. Nov 7, 2012

### pxw

1. A block with a mass of 6 kg is held in equilibrium on an inclined plane of angle θ=30 degrees by a horizontal force F. Find the magnitudes of normal force on the block and of F (ignore friction).

So far, I've found Fg=6(-9.81)=-58.86 N

I then found Fgy=-58.86(sin240)=50.97 and Fgx=-58.86(cos240)=29.43

And I know that, since the block is in equilibrium, Fnet=0, so 0=Fn + Fgy + Fay

I just seem to be having the problem of finding applied force so that I can find Fn and Fay.

Any help would be greatly appreciated! Thanks in advance.

2. Nov 7, 2012

### PhanthomJay

You have assumed that the x axis is along the incline and the y axis is perpendicular to the incline, which is OK. Now you must break up the horizontal force F into its x and y components. Then use F_net in x direction = 0 , and F_net in y direction = 0, to solve for the 2 unknowns from the 2 equations.
And welcome to PF!

3. Nov 7, 2012

### Simon Bridge

Did you draw a free body diagram for the block? Including all the forces?
Did you define +y to be normal to the slope? Draw the axis on your fbd?
Did you resolve all the forces to those axis?

Why did you use angle 240 in the trig? Wouldn't 30 or 60 be more appropriate?

The applied force is horizontal - so it has components in x and y directions that have to cancel out with the other forces in those directions.
Note: sin(30)=cos(60)=1/2, sin(60)=cos(30)=√3/2

4. Nov 7, 2012

### pxw

Yes, I understand that I have to break up F into its x and y components, however, I'm not sure how to do that without a value... the equation I have is Fa(sin330)=Fay. Sticking that into the net force equation, I have 0=Fn - 50.97 + Fa(sin330). I can't seem to get any further than that...

Thanks for your help and thanks for the welcome :D

Yes, I did. I tilted my axes so that Fn was the y axis. I can't seem to resolve my Fa, since I don't have a value for it.

My teacher taught us to use the larger angles because using them keeps the sine equation for the y resolution and the cosine equation for the x resolution. Taking the angle, then, I have my Fg to be at 240 degrees and my Fa to be at 330 degrees. Like I said, breaking down my Fa seems to be the big problem.

5. Nov 7, 2012

### PhanthomJay

That's one equation with 2 unknowns. You need another equation in the x direction.
I agree with Simon that I would not use this convention for determining the angles and trig, but if you are comfortable with it, I guess its OK, although most texts won't do it that way.

6. Nov 7, 2012

### pxw

Oh, ok! I got it now! Thanks a lot! :D

7. Nov 7, 2012

### Simon Bridge

Well done :)

fwiw: if $\theta$ is the angle of the angle to the horizontal, then $F_{gx}=mg\cos(90-\theta)$ would be correct to preserve the "cosine" rule. Whereas: $mg\cos(270-\theta)=-F_{gx}$ ; (which is what your teachers says to do) gives the opposite direction. You seem to have been unconsciously correcting for this as you go. It will trip you up later.

It is usually easier to keep the signs right by just sketching the triangles on your diagram and using trig the way you learned in math class. People who prefer the "memorize equations" method of doing physics tend not to like this way. Probably your teacher is one of those.

I urge you try the other way for a bit, in private, before settling on your final method. You'll need the skill anyway, to read online examples and text-books. Like PhantomJay says, it really boils down to what you find comfortable - that also works.


text:
7.2.9 - Summary of Advanced Material

7.2.9 - Summary of Advanced Material

In this lesson we learned about:

• Profile analysis for one-sample mean and how to carry it out in SAS
• Computing simultaneous and Bonferroni confidence intervals for the differences between sample means for each variable;
• Drawing conclusions from those confidence intervals (make sure that you indicate which population has the larger mean for each significant variable);
• Profile plots for two-sample means
• Methods for diagnosing the assumptions of the two-sample Hotelling's T-square test.

In practice, the data analyses should proceed as follows:

1. Step 1. For small samples, use histograms, scatterplots, and rotating scatterplots to assess the multivariate normality of the data. If the data do not appear to be normally distributed, apply appropriate normalizing transformations.

2. Step 2. Use Bartlett's test to assess the assumption that the population variance-covariance matrices are homogeneous.

3. Step 3. Carry out the two-sample Hotelling's T-square test for equality of the population mean vectors. If Bartlett's test in Step 2 is significant, use the modified two-sample Hotelling's T-square test. If the two-sample Hotelling's T-square test is not significant, conclude that there is no statistically significant evidence that the two populations have different mean vectors and stop. Otherwise, go to Step 4.

4. Step 4. Compute either simultaneous or Bonferroni confidence intervals. For the significant variables, draw conclusions regarding which population has the larger mean.

 [1] Link ↥ Has Tooltip/Popover Toggleable Visibility


text:
## Probability – How to understand if an exercise is Poison-Bernouli or e-What are the differences

I have three different exercises and I am confused. If it's Poison or Bernuli or if I've seen them and I know them and I'm reading but I'm not so skilled at getting it. I mean I can not detect the exercise of what it is. Can you share my tip to help me get it? I've used Bernouli in cases where he needs poison and opossite.If you can help me do not solve it but explain to me why is Bernouli where you understand, I will be grateful.
first exercise

There is a market race to sell their cornflakes. A company calls
random numbers in order to promote their cornflakes to
customers.We are sure that this company already has 20% of all
What is the possibility that a telephone interlocutor (means that the work
in this company) who calls customers who need more than 3 calls up to
she will succeed someone who is not a customer? My answer: I do not know
which typology to use

second exercise

An Internet service company 4 servers, and both are
"hosting" of the database and the other 2 our software.
possibility that we have damage on a server next year is 1% .We
take for granted that the possibility of appearing harm in all
the server is independently each.If the Internet company are working
there must be at least 1 server that will have the database and a
server that will make the software. What is the posibility of society
Will he work normally and next year?

## probability – PDF mapping in a different space

Whether it's the random vectors …

$$X = begin {bmatrix} X_1, X_2 end {bmatrix}$$
$$Y = begin {bmatrix} Y_1, Y_2 end {bmatrix}$$

Let there be the probability density functions $$f_ {X}$$ and $$f_ {Y}$$ (also written as $$f_ {X_1, X_2}$$ and $$f_ {Y_1, Y_2}$$ respectively).

Whether it's the transformation $$varphi: mathbb R ^ 2 to mathbb R ^ 2$$, and that's true …

$$Y = varphi (X)$$

Suppose that there is also an inverse $$varphi ^ {- 1}$$.

How can we write $$f_ {X_1, X_2}$$ in terms of $$f_ {Y_1, Y_2}$$, $$varphi$$and or $$varphi ^ {- 1}$$?

## probability – independent and uniformly distributed random variable

Consider three independent sources uniformly distributed (taking values between $$0$$ and $$1$$) Random variables. What is the probability that the middle of the three values (between the smallest and the largest) is between $$a$$ and $$b$$ or $$0 ≤ a ?

$$A) 3 (1 – b) a (b – a)$$

$$B) 3 ((b – a) – (b ^ {2} – a ^ {2}) / 2)$$

$$C) 6 (1 – b) a (b – a)$$

$$D) (1 – b) a (b – a)$$

$$E) 6 ((b ^ {2} – a ^ {2}) / 2 – (b ^ {3} – a ^ {3}) / 3)$$.

Now, formula for uniform distribution $$int_ {a} ^ {b} xdx = left [ frac{x^{2}}{2} right ]_ {a} ^ {b}$$

According to the question average value of $$3$$ the values are inside $$a$$ and $$b.$$

So, here are 4 possibilities. Let's say that the numbers are x, y, z.

Now,

Case $$1:$$ Only $$y$$ can be inside $$a$$ and $$b.$$

Case $$2:$$ Only $$x, y$$ can be inside $$a$$ and $$b.$$

Case $$3:$$ Only $$y, z$$ can be inside $$a$$ and $$b.$$

Case $$4:$$ $$x, y, z$$ everything can be inside $$a$$ and $$b.$$

So the probability should be $$frac { frac {^ {4} textrm {C} _ {1} times left (b ^ {2} -a ^ {2} right)} {2}} { frac {^ { 8} textrm {C} _ {1} times left (b ^ {2} -a ^ {2} right)} {2}}$$

Where am I wrong?

## probability – reduce a subspace with unequal probabilities

I was asked an interesting homework question:

Given three pieces in a box, where two have a probability of heads ( $$p (h) = 0.5$$ ) and the third $$p (h) = frac {1} {3}$$. I draw a coin and the result is a tail. What is the probability that I did not remove the unfair piece from the box?

I started to solve this problem by trying to find $$p ($$unfair$$| t)$$, and builds a sample space. Let $$A, B$$ represent the right pieces, and $$C$$ represent the unfair. The sample space is:

$${A_t, A_h, B_t, B_h, C_t, C_h }$$

With probabilities of

$${ frac {1} {6}, frac {1} {6}, frac {1} {6}, frac {1} {6}, frac {2} {9}, frac {1} {9} }$$

When trying to find $$p ($$unfair$$| t)$$ we reduce this sample space to:

$${A_t, B_t, C_t }$$

I can not understand what the new probabilities are for this sample space. I think $$C_t$$ has a higher probability than the other two, but I can not seem to divide that into three probabilities that add to 1.

How could I find the new probabilities of this subspace? I've tried many tricks, such as multiplying the old probabilities by 2, but the sampling space has a total probability> 1.

Any help is appreciated, gratuities included. I would like to be able to do it myself, so do not give me the answer, please.

## geometric probability – Determination of sample spaces for Sylvester's four-point problem

I have no knowledge of geometric probability, so I can apologize if the following is wrong or does not make sense:

Hypotheses:

• no three sampled points are collinear
• the geometric probability (naive) in Euclidean space is equal to the ratio of lengths or surfaces
• probabilities do not change under isometric transformations and / or uniform scaling
• the probability of a set $$P4: = lbrace p_1, p_2, p_3, p_4 rbrace$$ four points being in convex configuration in the Euclidean plane is independent of the order in which the elements are drawn by the sampling process; this in turn means that we assume that
• $$lbrace p_1, p_2, p_3 rbrace$$ looks like the corners of $$T_ {max}$$, the triangle of the largest area
• $$lbrace p_1, p_2, p_3 rbrace$$ uniformly drawn from the boundary of their circumscribed circle, which has implications for the probability of meeting $$T_ {max}$$ with specific values for the pair of smaller central angles.
• the smallest circle surrounding $$P4$$ is the unit circle centered at the origin
• the longest side of $$T_ {max}$$ is divided in two by the non-negative part of the x-axis

According to the above assumptions, the probability of encountering a convex quadrilateral is equal to the blue zone divided by the red plus blue zone in the images below:

the sampling area in case of acute toxicity $$T_ {max}$$ is equal to the entire disk of the unit

the sampling area in case of acute toxicity $$T_ {max}$$ is equal to the disk drive with a notch

The notch in the case of obtuse $$T_ {max}$$ is due to the assumption that the first three points look like $$T_ {max}$$ which implies that the points apart $$T_ {max}$$ that generate a deltoid configuration would be in contradiction with the maximality of the zone of $$T_ {max}$$

If the above makes sense, the probability that four points are in a convex configuration can be calculated by integrating the blue zone ratios over the entire sampling area defined by the $$T_ {max}$$ multiplied by the probability of $$T_ {max}$$ resulting in uniform sampling at the boundary of the unit circle.

Questions:

• Have similar ways of defining sample spaces for Sylvester's four-point problem already been described or studied?
• what are the objections to the proposed proposal compared to the proposed definition of the available sampling space on $$T_ {max}$$?

Note:

in case of obtuse $$T_ {max}$$ the surface of the sample space can be calculated based on the angles $$alpha$$ and $$beta$$ which are adjacent to the longer side of $$T_ {max}$$ as follows, keeping in mind that this longest edge is the diameter of the unit circle:

• the surface of the lower half-blue disk is equal to $$frac { pi} {2}$$
• The area $$A _ { alpha}$$ of the union of $$T_ {max}$$ with the blue region opposite to the angle $$alpha$$ equals $$alpha + sin ( alpha) cos ( alpha)$$ and the like $$A _ { beta}$$for angle $$beta$$
• the domain of $$A_ {T_ {max}}$$ of $$T_ {max}$$ equals $$frac {1} { cot ( alpha) + cot ( beta)}$$

The area of the sampling area for obtuse $$T_ {max}$$ then equal $$frac { pi} {2} + A _ { alpha} + A _ { beta} -A_ {T_ {max}}$$

## probability or statistics – How to get the correct interpolation for the following data?

I have a dataset DistrTemp.dat. It represents a set of data in the form `{{x, y, z, f[x,y,z]}}`. Here, x defines a system, while `F[x,y,z]` is a probability of finding a system in the neighborhood of `y, z` and has a property
$$int limits_ {0} ^ { pi} dy int limits {x} ^ {3500} dz f (x, y, z) = 1$$
My goal is to get a continuous function F[x,y,z] that I can use for integration measurement.

I'm trying to interpolate the dataset for particular values of x:

``````setDirectory[NotebookDirectory[]]DistrData = Import["DistrTemp.dat","Table"];
DistrInterpolation005[[Theta]HIS_]=
Interpolation[{Log10[#[[2]]], Log10[#[[3]]], #[[4]]} & / @
To select[DistrData, #[[1]]== 0.05 &], InterpolationOrder -> 1][
log10[[Theta]S], Log10[ES]]
``````

Next, I compare the interpolation obtained with the initial dataset:

``````Plot3D[DistrInterpolation005[DistrInterpolation005[DistrInterpolation005[DistrInterpolation005[[Theta]S, 10 ^ (ES)], {[Theta]S, 0,
Pi}, {ES, Log10[0.05], Log10[3500]}, Image size -> Large,
PlotRange -> All]ListPlot3D[{#[[2]], Log10[#[[3]]], #[[4]]} & / @
To select[DistrData, #[[1]]== 0.05 &], PlotRange -> All,
PlotStyle -> Directive[Red], ImageSize -> Great]
``````

I find that interpolation has been poorly done because some areas are lost, while others are overestimated:

I've tried using WeightedData:

``````TableCoordinates = Select[DistrData, #[[1]]== 0.05 &][[All, {2, 3}]]Weight of the table = Flatten[DataTemp[[All, 4]]]F1 = WeightedData[TableCoordinates, TableWeights]
DistrWeighted =
Table[f[F1], {f, {HistogramDistribution, SmoothKernelDistribution,
EmpiricalDistribution}}]Table[Plot3D[
PDF[I{[I{[je{[i{[Theta]S, Log10[ES]}], {ES, Log10[0.05],
log10[3500]}, {[Theta]S, 0, Pi},
PlotLabel -> Row[{"Mean: ", Mean[i]}], ImageSize -> Great], {I,
DistrWeighted}]
``````

However, none of the distributions correspond to the desired distribution:

How to properly construct the interpolation?

## probability – an infinitely iterative game of giving and withdrawing bills

Imagine a game where, at each stage, a person receives 10 dollar bills (numbered in sequence, starting with 1), then an invoice is removed from the bills that the person has received during the game. Suppose that the game lasts an infinite number of steps. How many bills remains after? As you may have already assumed, the answer depends on how bills are removed.

If the bill with the lowest number is deleted each time, there are no bills left. The deletion of each invoice can be linked to a specific step, so that no invoice can be identified and that it can remain in the collection of the person. There are also other types of deletion that result in a total loss, but this is simple.

If the highest numbered invoice is deleted each time, there will be an infinite number of invoices, because each of the infinite steps gives the person nine bills that will never be deleted. Again, there are other methods to achieve this, but this one is simple.

There are also deletion schemes that one could specify to end up with a finite number of bills, but just wait for the player to receive the correct number of bills before setting up a payout structure. total removal starting at a bill of a specific number. I can give examples later, if necessary.

And if the removal process is randomized? In step 1, invoices 1-10 are awarded and one of these 10 invoices is randomly selected to be withdrawn. In step two, bills 11 to 20 are handed in and one of the 19 remaining bills is randomly selected to be withdrawn. Etc.

What I want is how to determine what is the probability that there are only few or no bills left after an infinite number of steps. I know the infinite monkey theorem, but it is a finite string contained in an infinitely infinite chain and this problem concerns an infinitely infinite chain being part of the set of patterns that remove all the effects, and this difference seems significant. It seems that I can not simply compare the possibilities that have and do not give this result because they are both aleph-1 games, and I have proof that an infinite number of infinite possibilities can not be assumed just balance that way (I do not know if there is one elsewhere, but I've come to this point).

TBH, I have the feeling that I should be able to solve this problem, but I would really appreciate help, because I'm writing my PhD thesis in philosophy and that's a minor point, but I really would like to stay the. possible. Of course, any assistance provided will be quoted (we can discuss how you want this done). If I do it myself (or think I've done it), I'll post a follow up and see what you think.

## Probability Theory – \$ p_ {y simp} (y neq y) leq p_ {y simp (}} (y  neq y) | | p-p & # 39; | \$ for fixed \$ p, p & # 39; in[0,1]\$

On page 13 of this document: https://arxiv.org/pdf/1708.00489.pdf
it says :

we state claim 1 of Berlind & Urner (2015). For some people $$p, p & # 39; in[0,1]$$ and $$y & # 39; in {0,1}$$. Then, $$p_ {y sim p} (y neq y) leq p_ {y sim p} (y neq y)) | | p-p & # 39; |.$$

I looked up and brought it here: https://pdfs.semanticscholar.org/f0f5/e63ca2b8ee32d33c2489225445419364ea99.pdf

But this document does not have the same equation. I want to prove it myself. And I think the notation of the statement is not clear. What is some $$p$$, $$p in[0,1]$$? I need to help 1) understand the notation of these $$p, p$$ and 2) prove the inequality above. Thank you in advance.

## probability or statistics – How to optimize the number of bins with weighted data using multiple lists

I try to sort items from multiple lists into buckets containing weighted data. I can do the single list case based on a previous question. Specifically, the code under Update in the first response.

``````data = {1, 2, 3, 4, 5, 6, 7, 10};
trays = {0, 2, 4, 6, 8, 10};
weight = {0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 1.0};

With[{ranges = Partition[bins, 2, 1]}
Total @ Pick[weights,
BitXor[UnitStep[Subtract[data, #1]],
UnitStep[Subtract[data, #2]]], 1]& @@@ ranges]
``````

which gives the following output:
{0,1, 0,5, 0.9, 1.3, 0}

Now I'm trying to generalize at the next entry:

``````data = {{1, 2, 3, 4, 5, 6, 7, 10}, {3, 5, 6}};
trays = {0, 2, 4, 6, 8, 10};
weight = {{0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 1.0}, {0.4, 0.6, 0.5} };
``````

I can not use the same code because Pick does not process each list sequentially. However, I can get the desired result using this code (where I have the first and second list separately:

``````With[{ranges = Partition[bins, 2, 1]}
Total @ Pick[weights[[1]],
BITXOR[UnitStep[Subtract[data, #1]],
UnitStep[Subtract[data, #2]]][[1]], 1]& @@@ ranges]With[{ranges = Partition[bins, 2, 1]}
Total @ Pick[weights[[2]],
BITXOR[UnitStep[Subtract[data, #1]],
UnitStep[Subtract[data, #2]]][[2]], 1]& @@@ ranges]
``````

{0,1, 0,5, 0.9, 1.3, 0}
{0, 0.4, 0.6, 0.5, 0}

But I want to generalize to a lot of lists now.
I've tried a loop but when I generalize it loops it does not work:

``````For[i = 1, 2, i ++,
With[{ranges = Partition[bins, 2, 1]}
Total @ Pick[weights[[i]],
BITXOR[UnitStep[Subtract[data, #1]],
UnitStep[Subtract[data, #2]]][[i]], 1]& @@@ ranges]]
``````

How can I generalize properly?

## probability – Derivation of the bootstrap variance of the absolute central moments

The text of my statistics introduces the bootstrap as follows:

1. Draw $$X_1 ^ {*}, ldots, X_n ^ {*} sim hat {F} _n$$.

2. Calculate $$T_n ^ {*} = g (X_1, ldots, X_n)$$.

3. Repeat steps 1 and 2 $$B$$ times to get $$T_ {n, 1} ^ {*}, ldots, T_ {n, B} ^ {*}$$.

4. Let:

$$v_ {boot} = frac {1} {B} sum_ {b = 1} ^ B (T_ {n, b} ^ {*} – frac {1} {B} sum_ {r = 1} ^ B T_ {n, r} ^ {*}) ^ 2$$

Then there is the following exercise:

Let $$T_n = overline {X} _n ^ 2$$, $$mathbb {E} (X_1) = mu$$, $$alpha_k = int left | x- mu right | ^ kdF (x)$$ and $$hat { alpha} _k = n ^ {- 1} sum_ {i = 1} ^ n left | X_i – overline {X} _n right | ^ k$$. CA watch:

$$v_ {boot} = frac {4 overline {X} _n ^ 2 hat { alpha} _2} {n} + frac {4 overline {X} _n ^ 2 hat { alpha} _3} {n ^ 2} + frac { hat { alpha} _4} {n ^ 3}$$

I am confused because I understand $$v_ {boot}$$ always be whatever the formula will converge in probability on the variance of the statistic by assuming the empirical CDF with a sufficiently large number of replications (based on WLLN), which means that any formula used must necessarily use replications. But the formula I am asked to derive contains no replication, no variables with an asterisk. Maybe I could interpret that to mean that the replications we draw do not matter, but that seems unlikely. Do I interpret things correctly? Is this an error in the text?


text:
### Home > PC > Chapter 7 > Lesson 7.2.4 > Problem7-80

7-80.

Write the equation of the circle described.

1. Center at the origin, containing the point $\left(−6, −8\right)$.

Use the general form for a circle to find the radius.
$x^{2} + y^{2} = r^{2}$

2. Center $\left(7, 5\right)$, containing the point $\left(3, −2\right)$.

Use the form below to find the radius.
$\left(x − h\right)^{2} + \left(y − k\right)^{2} = r^{2}$

$\begin{array}{l}\left(x − 7\right)^{2} + \left(y − 5\right)^{2} = r^{2}\\\left(3 − 7\right)^{2} + \left(−2 − 5\right)^{2} = r^{2}\\16 + 49 = r^{2}\\65 = r^{2}\end{array}$

$\left(x − 7\right)^{2} + \left(y − 5\right)^{2} = 65$


text:
# logf Function

Calculates the natural logarithm of a single precision floating-point value.

Include

<math.h>

Prototype

float log(float x);

Argument

 x any positive value for which to return the log

Return Value

Returns the natural (base- e ) logarithm of x. If x is 0, infinity is returned. If x is a negative number, NaN is returned.

Remarks

If x < 0, a domain error will occur and errno will be set to EDOM.

Example Output

See the notes at the beginning of this chapter or section for information on using printf() or scanf() (and other functions reading and writing the stdin or stdout streams) in the example code.

#include <math.h>
#include <stdio.h>
#include <errno.h>

int main(void)
{
float x, y;

errno = 0;
x = 2.0F;
y = logf(x);
if (errno)
perror("Error");
printf("The natural logarithm of %f is %f\n",
x, y);

errno = 0;
x = 0.0F;
y = logf(x);
if (errno)
perror("Error");
printf("The natural logarithm of %f is %f\n",
x, y);

errno = 0;
x = -2.0F;
y = logf(x);
if (errno)
perror("Error");
printf("The natural logarithm of %f is %f\n",
x, y);
}

Example Output

The natural logarithm of 2.000000 is 0.693147
The natural logarithm of 0.000000 is -inf
Error: domain error
The natural logarithm of -2.000000 is nan


text:
# ICS-3 Seminar: Prof. Dr. Jose-Luis Méndez Alcaraz Dept. of Physics, CINVESTAV, Mexico-City

begin
27 Mar 2012 10:30

Colloidal halos

Abstract:

By means of computer simulations a neutral spherical substrate is placed in the bulk of a suspension of colloidal particles interacting with a long-ranged repulsive potential. Some of the particles are then observed to adsorb on the surface of the substrate, building colloidal halos. We obtain the conditions for the halos formation and find out that the number of adsorbed particles follows scaling laws as function of the microscopic variables of the system. We also study the angular distribution of the adsorbed particles along the spherical surface of the substrate, finding it to jump alternately between fluid- and crystal-like structures for some “magic numbers” of adsorbed particles.