text:
# 8.1 Tracking inflation  (Page 3/16)

 Page 3 / 16

## Index numbers

The numerical results of a calculation based on a basket of goods can get a little messy. The simplified example in [link] has only three goods and the prices are in even dollars, not numbers like 79 cents or $124.99. If the list of products was much longer, and more realistic prices were used, the total quantity spent over a year might be some messy-looking number like$17,147.51 or $27,654.92. To simplify the task of interpreting the price levels for more realistic and complex baskets of goods, the price level in each period is typically reported as an index number , rather than as the dollar amount for buying the basket of goods. Price indices are created to calculate an overall average change in relative prices over time. To convert the money spent on the basket to an index number, economists arbitrarily choose one year to be the base year , or starting point from which we measure changes in prices. The base year, by definition, has an index number equal to 100. This sounds complicated, but it is really a simple math trick. In the example above, say that time period 3 is chosen as the base year. Since the total amount of spending in that year is$107, we divide that amount by itself ($107) and multiply by 100. Mathematically, that is equivalent to dividing$107 by 100, or $1.07. Doing either will give us an index in the base year of 100. Again, this is because the index number in the base year always has to have a value of 100. Then, to figure out the values of the index number for the other years, we divide the dollar amounts for the other years by 1.07 as well. Note also that the dollar signs cancel out so that index numbers have no units. Calculations for the other values of the index number, based on the example presented in [link] are shown in [link] . Because the index numbers are calculated so that they are in exactly the same proportion as the total dollar cost of purchasing the basket of goods, the inflation rate can be calculated based on the index numbers, using the percentage change formula. So, the inflation rate from period 1 to period 2 would be This is the same answer that was derived when measuring inflation based on the dollar cost of the basket of goods for the same time period. Calculating index numbers when period 3 is the base year Total Spending Index Number Inflation Rate Since Previous Period Period 1$100
Period 2 $106.50 Period 3$107
Period 4 $117.50 If the inflation rate is the same whether it is based on dollar values or index numbers, then why bother with the index numbers? The advantage is that indexing allows easier eyeballing of the inflation numbers. If you glance at two index numbers like 107 and 110, you know automatically that the rate of inflation between the two years is about, but not quite exactly equal to, 3%. By contrast, imagine that the price levels were expressed in absolute dollars of a large basket of goods, so that when you looked at the data, the numbers were$19,493.62 and \$20,009.32. Most people find it difficult to eyeball those kinds of numbers and say that it is a change of about 3%. However, the two numbers expressed in absolute dollars are exactly in the same proportion of 107 to 110 as the previous example. If you’re wondering why simple subtraction of the index numbers wouldn’t work, read the following Clear It Up feature.

where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Researchers demonstrated that the hippocampus functions in memory processing by creating lesions in the hippocampi of rats, which resulted in ________.
The formulation of new memories is sometimes called ________, and the process of bringing up old memories is called ________.
Equilibrium price is a stable price and it must stay.discuss
Card 14 / 21: What are the similarities between a consumer’s budget constraint and society’s production possibilities frontier, not just graphically but analytically?


text:
# Why is $\sup_{t \in A} |x_m(t)-x(t)| \leq \sup_{t \in A} \lim_{k \rightarrow \infty} |x_m(t)-x_k(t)|$?

Why is $$\sup_{t \in A} |x_m(t)-x(t)| \leq \sup_{t \in A} \lim_{k \rightarrow \infty} |x_m(t)-x_k(t)|$$?

$$(x_n)$$ is Cauchy, it's in $$BoundedLinear(A,\mathbb{K})$$, $$t \in A$$.

I'm thinking that this assumes that the sequence's distances near each other and thus one can assume that $$x_k(t)$$ will remain "more distanced" from $$x_m(t)$$ than the limit point $$x_k \rightarrow x$$ itself, since $$|x_m(t)-x(t)|$$ would be where the distance is smallest. But this sounds a bit inexact.

• What is $x(t)$? – angryavian May 9 at 18:15

format: plain


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# MandysNotes

Tuesday, 18 March 2014 00:00

## Negation

By  Gideon
Rate this item
(1 Vote)

Suppose that the only information you have about a certain animal is that:

"This animal does not have a duck bill."

If A is the set of animals that have a duck bill, and C is the set of all animals, then the set of all animals that do not have a duck bill is the negation of A (with respect to C).

This is the set "not A," written as:

$\lnot A$

Note that if an animal is a member of the set $$\lnot A,$$ then it cannot be a duckling, and it cannot be a platypus.

Next. conside the set of all animals that are not mammals. This is the set:

"not B"

$= \lnot B.$

A member of the set $$\lnot B$$ cannot be a rabbit, and it cannot be a platypus.

The negation of the set "A or B" is the set

"not ( A or B)"

$= \lnot \left ( A \cup B \right).$

It is the set of animals that are not either duck-billed or mammals:

It is illustrated below:

A member of the set $$\lnot \left( A \cap B\right)$$ cannot be a duckling or a rabbit or a platypus.

Finally, the set animals that are not both duck-billed and mammals is the set:

"not ( A and B)"

$= \lnot \left( A \cap B \right).$

It is the set of all animals that are not platypuses.


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Activity 7: Use the MIN Command

Workshop Resources

Task 7: Find the Key to Get to the Confetti Cannon!

You found the location of the Legendary Confetti Cannon but it is locked away behind a door made of titanium and can only be unlocked with the smallest key on the planet!

To help you out, the Galactic Federation has sent you a reminder on how to use MIN() from a previous activity.


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# Difference between CBOW and SKIP Gram word vectors

I have gone through several links but was not able to understand how CBOW and Skip Gram is trained from scratch?

Any good link/blogs or books would be very helpful.

Word2Vec - CBOW and Skip-Grams

What's the difference between Skipgram word2vec and CBOW word2vec during training and when to use CBOW Skip-gram. ?

Example or Application where CBOW would be preferable choice but not Skip-gram and vice versa.

## 2 Answers

Here is a example for what @Christos Karatsalos is describing:

If you take a statement like "The cat jumped over the puddle",

In CBOW's case, the neural network's word embeddings are trained by passing a input set of words to the neural network and making the network to predict an output word as shown below:

$$\{ cat, jumped, over, the, puddle\} \rightarrow The$$

$$\{ The, jumped, over, the, puddle\} \rightarrow cat$$

$$\{ The, cat, over, the, puddle\} \rightarrow jumped$$

$$\{ cat, jumped, the, puddle\} \rightarrow over$$

$$\{ The, cat, jumped, over, puddle\} \rightarrow the$$

$$\{ The, cat, jumped, over, the\} \rightarrow puddle$$

In skip-gram's case, the neural network's word embeddings are trained by passing an input word to the neural network and teaching the network to predict a set of words as shown below:

$$The \rightarrow \{ cat, jumped, over, the, puddle\}$$

$$cat \rightarrow \{ The, jumped, over, the, puddle\}$$

$$jumped \rightarrow \{ The, cat, over, the, puddle\}$$

$$over \rightarrow \{ cat, jumped, the, puddle\}$$

$$the \rightarrow \{ The, cat, jumped, over, puddle\}$$

$$puddle \rightarrow \{ The, cat, jumped, over, the\}$$

For usage, if you are thinking of a fill-in-the-blanks kind of problem CBOW Word2Vec would be a suitable vector to use, on the other hand if you have a word and you are trying to come up with a new sentence with it then skip-gram Word2Vec will be useful.

Take a look this lecture notes for more information.

CBOW model is able to learn to predict the word by the context, which means that it tries to maximize the probability of the target word by looking at the context. On the other hand, the Skip-gram model is designed to predict the context given the word.

Skip-gram model works well with small amount of training data, moreover it represents well even rare words or phrases.

CBOW model is several times faster to train than the Skip-gram model, and achieves slightly better accuracy for the frequent words.

• Why Skip-gram model works well with small amount of training data? – MAC Oct 13 '20 at 6:42
• According to the following paper, Skip-gram works well with small amount of the training data. – Christos Karatsalos Oct 13 '20 at 21:06
• Tomas Mikolov, Ilya Sutskever, Kai Chen, Gregory S. Corrado, and Jeffrey Dean. 2013b. Distributed representations of words and phrases and their compositionality. In Advances in Neural Information Processing Systems 26: 27th Annual Conference on Neural Information Processing Systems 2013. Proceedings of a meeting held December 5-8, 2013, Lake Tahoe, Nevada, United States, pages 3111–3119. – Christos Karatsalos Oct 13 '20 at 21:06


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# zbMATH — the first resource for mathematics

The law of the iterated logarithm for locally univalent functions. (English) Zbl 1078.30009
Author’s abstract: In this paper we prove a sharp version of the Makarov law of the iterated logarithm. In particular, we show that the constant in the right side of this law depends on an asymptotic behaviour of the integral means of the derivative of an analytic function. Also, we establish that this constant is equal to the asymptotic variance for some domains with fractal type boundaries.

##### MSC:
 30C55 General theory of univalent and multivalent functions of one complex variable
Full Text:


text:
# How do I decide whether ring will expand in this reaction or not?

Attempt:

After forming the carbocation adjacent to the cyclopropyl ring I had two options,

1. $$\ce{Cl-}$$ then and there without any changes
2. Expand ring, form a new carbocation and add $$\ce{Cl-}$$ there.

I chose the first option because of exceptional stability of cyclopropyl methyl carbocation. However, answer given is option C which follows mechanism (2).

Can someone explain how to judge and decide how to proceed after the carbocation formation?

• Cyclopentyl ring does not let the cyclopropyl ring to provide stability. Think why? The reason is due the cyclopentyl ring the p-orbital of carbon (containing positive charge) becomes perpendicular to p-oribital on carbon of cyclpropyl ring. So the double bond character cannot be achieved there is a image in my answer see it. – Saurabh Singh Apr 20 at 6:32
• This is a challenging question because you can form a nonclassical carbocation (homoallyl) by protonating farther from the cyclopropyl ring... – Zhe Apr 20 at 17:31

Explanation

In case 1. You can see it is a 2° carbocation (or secondary carbonation) with angle strain of ~25° on cyclopropane

In case 2. You can see it is 3° carbocation (or tertiary carbocation) with angle strain of ~10° on cyclopropane

It is obvious that case 2 will be major intermediate yielding $$\ce{1-chlorobicyclic [3,2,0] heptane}$$ as main product that is C.

Edit1: After reading your question more thoroughly I found that I missed one thing about 'exceptional' stability of cyclopropyl methyl carbocation.

The stability that cyclopropyl provides to carbocation is due to interaction of p-orbitals which is not possible in your case because p-orbitals of carbon having '+' charge becomes out of plane with respect to p-orbitals of cyclopropyl.

Edit2: See image below for reference

Then in case of cyclopropyl methyl carbocation the p-orbitals of both carbon containing '+' charge and the carbon on ring are parallel

In your case the p-orbitals of carbon containing '+' charge is perpendicular to plane of cyclopentyl ring but the p-orbitals of carbon on cyclopropyl ring is parallel to plane cyclopentyl ring, so stability is not provided by cyclopropyl ring.

To sum up you can say that the cyclopentyl ring does not let cyclopropyl ring to provide stability to carbocation which makes ring expansion a preferred option.

Edit3 (Bonus): I asked to my professor about this, he told me that there are two conformational isomers of cyclopropyl methyl carbocation. One is bisected conformation and other is perpendicular conformation. The bisected one allows the carbocation to be exceptionally stable but the perpendicular one isn't that much stable (like in your case) due to less interaction of p-orbital and the partial double bond character cannot be achieved.


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# [texhax] raisebox in longtable

Sat Dec 26 11:45:02 CET 2009

Hi,

To properly align images and text within a longtable, I use raisebox like the following:

---------
\latexhtml{\begin{longtable}{p{0,5\textwidth}p{0,45\textwidth}}}{\begin{longtable}{lp{20cm}}}
\latex{\multicolumn{2}{r}{\footnotesize{Fortsetzung folgt auf der nächsten Seite\dots}}\\}
\latex{\endfoot}
\latex{\endlastfoot}
\raisebox{-0.8\height}{\includegraphics[scale=0.5]{foo}} & some text goes here\\
\raisebox{-0.8\height}{\includegraphics[scale=0.5]{boom}} & some text goes here too\\
\end{longtable}
---------

Unfortunately, the images nearly stick together vertically. So there is no space between the first and the second row and therefore between the images.

How could I create a little space between the rows?

--
Christian

-The next generation of interesting software will be done on the Macintosh, not the IBM PC.-

Bill Gates, Nov 1984


text:
# JavaScript Logical Operator

Logical operator is one that compare two boolean operands and gives a boolean output. They are of high importance in computer science, especially in digital design and computer architectures.

They operands of logical operator could be constants, variables of all types and expressions involving comparison operators. First, we must understand the working of these logical operators.

### Types of Logical Operators

There are three types of logical operators and they are:

Now, out of these three, you must be familiar with logical not operator which is also a unary operator. To learn more about unary operators visit following link.

JavaScript Unary Operator

Let us try to understand each logical operator separately.

#### Logical AND operator

The logical AND operator takes two or more operands and convert them to equivalent boolean value and return true if all of them are true, otherwise, if at least one operand is false, the result is false.

You can understand the working of this operator with the help of a truth table. A truth table gives output for all combination of variable inputs.

The above is a two variable truth table for logical AND which shows that the output is true when all operands( X and Y) are true.

Example #1

//variable declaration
var x = prompt("Enter value for X:");
var y = prompt("Enter value for Y:");
var z = prompt("Enter value for Z:");
var real_x = Number(x);
var real_y = Number(y);
var real_z = Number(z);
//Check greatest of all
if (real_x > real_y && real_x > real_z)
{
}
else if ( real_y > real_x && real_y > real_z)
{
alert( "Y is the greatest number");
}
else if ( real_z > real_x && real_z > real_y)
{
}
else
{
alert("Invalid Input! Numbers cannot be same");
}

Output #1

#### Logical OR operator

The logical OR operator takes two or more operands and convert them into their boolean equivalent. If either of the operands is true, the logical outcome is also true, otherwise it is false.

Here is the truth table for logical OR.

In the above truth table, the outcome is false, when both operands (X and Y) are false. The rest of the entries are true.

Example #2

In this example, we will check if a number is within the range 1000 to 5000 including both numbers.

//variable declarations
var x = prompt("Enter your number:");
var real_x= Number(x);
//check range
if ( x >= 1000 || x <= 5000)
{
}
else
{
}


text:
# A team of ecologists is tracking mountain lions throughout California. They count 4543 mountain lions in the entire state. The area of California is 423,970 km. What is the population density of mountain lions per 250km2?

Jun 3, 2018

$2.68$ mountain lions per $250$ square metres.

#### Explanation:

This is a simple proportion problem :

In ${\text{423,970 m}}^{2}$, there are $4543$ mountain lions, so in ${\text{250 km}}^{2}$, there are $x$ mountain lions.

"423,970 m"^2/("4543 lions") = "250 km"^2/(x \ "lions")

So

$x = \text{250 km"^2/"423970 km"^2 xx "4543 mountain lions}$

$x = \text{2.68 mountain lions }$ (to 2 decimals)


text:
## Zentralblatt MATH

Publications of (and about) Paul Erdös

Zbl.No:  041.36807
Autor:  Erdös, Pál
Title:  Some asymptotic formulas in number theory. (In English)
Source:  J. Indian Math. Soc., n. Ser. 12, 75-78 (1948).
Review:  If n is a positive integer, let f(n) denote the number of solutions of the equation 2k+p = n, k being a non-negative integer and p a prime number. The author proves that:
(I) there is a positive constant c0 such that f(n) > c0 log log n for infinitely many positive integers n,
(II) if r is a fixed positive integer, there exists a positive number c(r) such that sumn = 1x fr(n) < c(r)x,
(III) if n\equiv 7629217 (mod 11184810); f(n) = 0.
(II) contains a theorem of N.P.Romanoff (Zbl 009.00801) to the effect that the positive integers expressible in the form 2k+p have positive density. The author generalizes Romanoff's theorem by proving, that if a1 < a2 < ··· is an infinite sequence of positive integers such that ak | ak+1 for each k, then the positive integers expressible in the form p+ak have positive density if and only if there exist positive numbers c1 and c2 such that {log ak \over k} < c1 and sumd/ak 1/d < c2 for every k.
Reviewer's remark: The proof of (I) uses a result of A.Page (Zbl 011.14905) on the number of prime numbers in an arithmetic progression with relatively large difference. In applying this result, the author forgets to take account of a possible exceptional real primitive residue character which occurs in Page's work. This difficulty can be easily overcome in much the same manner as an analogous difficulty was handled in a joint paper of the author, the reviewer, and S.Chowla (Zbl 036.30702, see p. 170 of the work).
Reviewer:  P.T.Bateman
Classif.:  * 11N56 Rate of growth of arithmetic functions
Index Words:  Number theory

© European Mathematical Society & FIZ Karlsruhe & Springer-Verlag


text:
# Find the linearization L(x) of the function at a. f(x)=\sqrt{x}, a=4

Find the linearization L(x) of the function at a.
$$\displaystyle{f{{\left({x}\right)}}}=\sqrt{{{x}}},{a}={4}$$

• Questions are typically answered in as fast as 30 minutes

### Plainmath recommends

• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.

brawnyN
$$\displaystyle{f{{\left({x}\right)}}}=\sqrt{{{x}}}$$
$$\displaystyle{f{{\left({4}\right)}}}=\sqrt{{{4}}}={2}$$
$$\displaystyle{f}'{\left({x}\right)}={\frac{{{1}}}{{{2}\sqrt{{{x}}}}}}$$
$$\displaystyle{f}'{\left({4}\right)}={\frac{{{1}}}{{{2}\sqrt{{{4}}}}}}={\frac{{{1}}}{{{4}}}}$$
The linearization L(x) of f at a=4:
L(x)=f(a)+f'(a)(x-a)
$$\displaystyle{L}{\left({x}\right)}={2}+{\frac{{{1}}}{{{4}}}}{\left({x}-{4}\right)}$$
$$\displaystyle{L}{\left({x}\right)}={\frac{{{1}}}{{{4}}}}{x}+{1}$$
Results:
$$\displaystyle{L}{\left({x}\right)}={\frac{{{1}}}{{{4}}}}{x}+{1}$$
###### Have a similar question?

• Questions are typically answered in as fast as 30 minutes


text:
This family of functions counts different types of relationships between two ivs. It works similar to base::match(), where needles[i] checks for a relationship in all of haystack.

• iv_count_overlaps() counts instances of a specific type of overlap between the two ivs.

• iv_count_precedes() counts instances when needles[i] precedes (i.e. comes before) any interval in haystack.

• iv_count_follows() counts instances when needles[i] follows (i.e. comes after) any interval in haystack.

These functions return an integer vector the same size as needles containing a count of the times a particular relationship between the i-th interval of needles and any interval of haystack occurred.

## Usage

iv_count_overlaps(
needles,
haystack,
...,
type = "any",
missing = "equals",
no_match = 0L
)

iv_count_precedes(
needles,
haystack,
...,
closest = FALSE,
missing = "equals",
no_match = 0L
)

iv_count_follows(
needles,
haystack,
...,
closest = FALSE,
missing = "equals",
no_match = 0L
)

## Arguments

needles

[iv]

Interval vectors used for relation matching.

• Each element of needles represents the interval to search for.

• haystack represents the intervals to search in.

Prior to comparison, needles and haystack are coerced to the same type.

haystack

[iv]

Interval vectors used for relation matching.

• Each element of needles represents the interval to search for.

• haystack represents the intervals to search in.

Prior to comparison, needles and haystack are coerced to the same type.

...

These dots are for future extensions and must be empty.

type

[character(1)]

The type of relationship to find. One of:

• "any": Finds any overlap whatsoever between an interval in needles and an interval in haystack.

• "within": Finds when an interval in needles is completely within (or equal to) an interval in haystack.

• "contains": Finds when an interval in needles completely contains (or equals) an interval in haystack.

• "equals": Finds when an interval in needles is exactly equal to an interval in haystack.

• "starts": Finds when the start of an interval in needles matches the start of an interval in haystack.

• "ends": Finds when the end of an interval in needles matches the end of an interval in haystack.

missing

[integer(1) / "equals" / "error"]

Handling of missing intervals in needles.

• "equals" considers missing intervals in needles as exactly equal to missing intervals in haystack when determining if there is a matching relationship between them.

• "error" throws an error if any intervals in needles are missing.

• If a single integer value is provided, this represents the count returned for a missing interval in needles. Use 0L to force missing intervals to never match.

no_match

[integer(1) / "error"]

Handling of needles without a match.

• "error" throws an error if any needles have zero matches.

• If a single integer is provided, this represents the count returned for a needle with zero matches. The default value gives unmatched needles a count of 0L.

closest

[TRUE / FALSE]

Should only the closest relationship be returned?

If TRUE, will only return the closest interval(s) in haystack that the current value of needles either precedes or follows. Note that multiple intervals can still be returned if there are ties, which can be resolved using multiple.

## Value

An integer vector the same size as needles.

## Examples

library(vctrs)
#>
#> Attaching package: ‘vctrs’
#> The following object is masked from ‘package:dplyr’:
#>
#>     data_frame

x <- iv_pairs(
as.Date(c("2019-01-05", "2019-01-10")),
as.Date(c("2019-01-07", "2019-01-15")),
as.Date(c("2019-01-20", "2019-01-31"))
)

y <- iv_pairs(
as.Date(c("2019-01-01", "2019-01-03")),
as.Date(c("2019-01-04", "2019-01-08")),
as.Date(c("2019-01-07", "2019-01-09")),
as.Date(c("2019-01-10", "2019-01-20")),
as.Date(c("2019-01-15", "2019-01-20"))
)

x
#> <iv<date>[3]>
#> [1] [2019-01-05, 2019-01-10) [2019-01-07, 2019-01-15) [2019-01-20, 2019-01-31)
y
#> <iv<date>[5]>
#> [1] [2019-01-01, 2019-01-03) [2019-01-04, 2019-01-08) [2019-01-07, 2019-01-09)
#> [4] [2019-01-10, 2019-01-20) [2019-01-15, 2019-01-20)

# Count the number of times x overlaps y at all
iv_count_overlaps(x, y)
#> [1] 2 3 0

# Count the number of times y is within an interval in x
iv_count_overlaps(y, x, type = "within")
#> [1] 0 0 2 0 0

# Count the number of times x precedes y
iv_count_precedes(x, y)
#> [1] 2 1 0

# ---------------------------------------------------------------------------

a <- iv(c(1, NA), c(2, NA))
b <- iv(c(NA, NA), c(NA, NA))

# Missing intervals are seen as exactly equal by default, so they are
# considered to overlap
iv_count_overlaps(a, b)
#> [1] 0 2

# If you'd like missing intervals to be treated as unmatched, set
# missing = 0L
iv_count_overlaps(a, b, missing = 0L)
#> [1] 0 0

# If you'd like to propagate missing intervals, set missing = NA
iv_count_overlaps(a, b, missing = NA)
#> [1]  0 NA


text:
# How do I splice two series together?

You can use one of the “Join” functions to combine two series together.

Join(series1, series2, observation)
• This function will join “series1” and “series2” at a date that you specify. The “observation” parameter is optional. If not provided, the junction will be done at the end of series1.

Example:

Join(imf_q69466___zf, ngprod0001)

Join(imf_q69466___zf, ngprod0001, Date(2007, 1, 1))

JoinMoreHistory(series1, series2)
• This function will complement series1 with the values of series2 that are before the start of series1. You would typically use this function when you need to add history to a series.

Example:

JoinMoreHistory(ngprod0001, imf_q69466___zf)


text:
# Separation of Constituents of Inorganic Mixture Containing Two Cations

In this experiment, a paper chromatography procedure uses similar principles to separate several metal cations. Comparing the Rf values and colors of the stained spots of known solutions will identify an unknown solution containing two cations.

## Aim:

To separate the constituents present in an inorganic mixture containing two cations (Pb2+ and Cd2+) only and the constituents should have large difference in Rf values.

## Theory:

Chromatography is a modern and sensitive technique used to analyze and separate components from a mixture and purify compounds quickly and efficiently.

The characteristic rate of movement of each substance on the chromatography paper can be determined at a given temperature and for a given solvent as the moving phase moves. This is represented by the Rf value of the relative front or retardation factor.

Although the mobile phase (solvent) is the same, the Rf values of different compounds are different. In addition, a compound’s Rf value may differ in different solvents. Using the following expression, Rf values can be calculated.

Rf = $\frac{distance\, travelled\, by\, the\, substance\, from\, reference \, line\, (cm)}{distance\, travelled\, by\, the \,solvent\, front\, from\, reference\, line\, (cm)}$

In this experiment the two cations to be separated in this case are colorless. A developer is therefore needed. In this case, ammonium sulphide (NH4)2S may be used to locate these ions in chromatographic paper or plate position.

## Materials Required:

1. Whatman’s filter paper
2. Glass jar
3. 1-2% of lead nitrate solution
4. 1-2% of cadmium nitrate solution
5. Ammonium sulphide
6. Ethanol
7. 6M nitric acid
8. Rubber cork fixed with hook in the centre
9. Test tubes
10. Capillary tubes
11. Distilled water
12. Petroleum ether

## Procedure:

1. Mark a line at a distance of 3 cm from one end of whatman filter paper with the help of a pencil.
2. Use a fine capillary to place a spot of the mixture on the marked line.
3. In a jar containing a mixture of ethanol, 6.0 M HNO3 and distilled water, hang the filter paper in the ratio 8:1:1.
4. Keep the jar as such until the paper length rises to two-thirds of the mobile phase (solvent).
5. Now remove the filter paper and mark the solvent front.
6. Spray ammonium sulphide solution to obtain yellow and black spots on the chromatography paper.
7. Mark with a pencil the position of the spots and let the paper dry.
8. Measure the distance to the reference line moved by the solvent front and the cations ‘ different spots.
9. The distance between the reference line and the center of different spots is the shortest distance.
10. Record in tabular form the observations and calculate the Rf value for each cation.

## Observation and Inference:

 S.No Colour of the spot Distance travelled by different components from reference line. (line/cm) Distance travelled by the solvents from reference line. (line/cm) Rf value 1 2 3

## Results and Discussions:

1. The Rf values of Cd2+ ions is _________.
2. The Rf values of Pb2+ ions is _________.

## Precautions:

1. Always make use of a fine capillary tube.
2. Use the capillary finely drawn to place the spot on the paper.
3. Do not allow spots to spread while spotting the test solution on the paper.
4. Do not disturb the jar once the experiment is set as long as the chromatogram is developed.
5. Carefully handle organic solvents.
6. Before developing the spots, make the paper strip perfectly dry.

## Viva Questions:

1. What are the types of solvent generally used in chromatography?

Ans: There are two types of solvents used in chromatography. One is polar solvent which dissolves polar molecules and the other is non polar solvent used to dissolve non polar molecules.

2. List the advantages of chromatography.

Ans: Chromatography is a simple technique can be handled by a single person. The use of chromatography separates components from complex mixtures. There are different types of equipment for chromatography to separate different compounds.

3. What is the value of chromatography?

Ans: In chromatography, the Rf value is used to quantify a sample’s retardation amount in a stationary phase relative to a mobile phase.

4. Name the scientist who introduced chromatographic techniques.

Ans: Russian botanist M.Tswett introduced chromatography in 1906.

5. What is meant by the term “developer” in chromatography?

Ans:The mobile phase is called the developer and it is referred to as the movement of the liquid along the bed.


text:
# Using flashlight

#### 2020-06-20

library(ggplot2)
library(flashlight)      # model interpretation
library(MetricsWeighted) # metrics
library(dplyr)           # data prep
library(moderndive)      # data
library(ranger)          # random forest

## Introduction

In contrast to classical statistical modelling techniques like linear regression, modern machine learning approaches tend to provide black box results. In areas like visual computing or natural language processing, this is not an issue since their focus is usually on predicting things. Either the predictions are sufficiently useful in practice or the model won’t be used. However, in areas where the purpose of a statistical model is also to explain or validate underlying theories (e.g. in medicine, economics, and biology), black box models are of little use.

Thus, there is need to shed light into these black boxes resp. to explain machine learning models as good as possible. An excellent reference is the online book (Molnar 2019). Of special interest are model agnostic approaches that work for any kind of modelling technique, e.g. a linear regression, a neural net or a tree-based method. The only requirement is the availability of a prediction function, i.e. a function that takes a data set and returns predictions.

This is the purpose of the R package flashlight, which is inspired by the beautiful DALEX package, see (https://CRAN.R-project.org/package=DALEX).

The main props of flashlight:

1. It is simple, yet flexible.

2. It offers model agnostic tools like model performance, variable importance, global surrogate models, ICE profiles, partial dependence, further effects plots, scatter plots, interaction strength, and variable contribution breakdown for single observations.

3. It allows to assess multiple models in parallel without any redundancy in the code.

4. It supports “group by” operations.

5. All methods are able to utilize case weights.

Currently, models with numeric or binary response are supported.

We will now give a brief introduction to machine learning explanations and then illustrate them with the flashlight package.

## Background

Important model agnostic machine learning explanations include the following aspects, amongst many others.

### Model performance

How precise are models if applied to unseen data? This aspect is of key interest of basically any supervised machine learning model and helps to identify the best models or, if applied to subgroups, identify problematic segments with low performance.

### Variable importance

Which variables are particularly relevant for the model? This aspect is helpful in different ways. Firstly, it might help to simplify the full modelling process by eliminating difficult to assess input variables with low explanatory power. Secondly, its pure information. Thirdly, it might help to identify problems in data structure: if one variable is extremely relevant and all others are not, then there might be some sort of information leakage from the response. Finally, variable importance considerations are very relevant for quality assurance as well.

Different modelling techniques offer different ways of variable importance. In linear models, we consider for example F-test statistics or p values, in tree-based methods the number of splits or average split gains etc. A model agnostic way to assess this is called permutation importance: For each input variable $$X$$, its values are randomly shuffled and the drop in performance with respect to a scoring function is calculated. The more important a variable, the larger the drop. If a variable can be shuffled without any impact on model precision, it is completely irrelevant. The method is described in (Fisher, Rudin, and Dominici 2018). Multiple permutations lead to more stable estimates of variable importance and allow the calculation of standard errors.

### Effects of input variables

In linear regression, the fitted model consists of an affine linear function in the inputs. Its coefficients immediately tell us how the response is expected to change if the value of one single input variable $$X$$ is systematically being adapted. How to describe such effects of a variable $$X$$ for more complex models that include non-linearities and high-order interactions?

One approach is to study Individual Conditional Expectation (ICE) profiles of selected observations: They show how predictions of observation $$i$$ react when the input variable $$X$$ is systematically being changed, see (Goldstein et al. 2015). The more different the profiles are in shape or slope, the stronger are the interaction effects. For a linear regression without interactions, all such profiles would be parallel. Centered ICE profiles (“c-ICE”) help to detect interactions even better.

If many ICE profiles are averaged, we get partial dependence profiles which can be viewed as the average effect of variable $$X$$, pooled over all interactions. Partial dependence plots where introduced in Friedman’s seminal 2001 article on gradient boosting (Friedman 2001). Partial dependence profiles (as well as ICE profiles) are not limited to one single variable $$X$$. Also multiple variables can be systematically varied on a grid in order to study the multivariate impact on the typical response.

Studying ICE and partial dependence profiles means investigating the effect of $$X$$ while holding all other predictors fixed. The more unnatural this Ceteris Paribus assumption is, the less reasonable are the results of ICE and partial dependence. Accumulated local effects or ALE profiles (Apley and Zhu 2016) try to overcome this weakness of ICE and partial dependence considerations.

ALE profiles at positions $$x_i$$ are calculated as follows:

1. First, left derivatives $$\Delta_i$$ are estimated for all $$i$$: Calculate slope of partial dependence line from $$x_i-\varepsilon$$ to $$x$$ based on data points in $$[x_i-\varepsilon, x_i]$$.

2. The uncalibrated ALE value at $$x_i$$ is the (integrated/accumulated) partial sum $$\sum_{j \le i} \Delta_j$$.

3. In the calibration step, the effects are shifted to the level of the response variable or its predictions.

An alternative to partial dependence and ALE profiles is to look at combined effects of $$X$$ including the effects from all other predictors. Such effects are estimated by averaging the predictions within values of the predictor $$X$$ of interest. If such prediction profile differs considerably from the observed average response, this might be a sign of model underfit. This visualization is sometimes called “marginal plot” or “M plot”, see (Apley and Zhu 2016). Residual profiles immediately show such misfits as well. In classical statistical modelling, this sort of plots are called “fitted versus covariable” plots or “residual versus covariable” plots.

If SHAP values have been calculated, plotting their average contributions of $$X$$ against $$X$$ is an elegant way to visualize the effect of $$X$$, see Subsection “Variable contribution breakdown for single observations”.

Besides looking at average profiles, it is often also revealing to consider quartiles or to visualize partial dependence, response and prediction profiles in the same plot.

Instead of studying profiles, it might be revealing to look at scatter plots of responses, predictions, residuals, or SHAP values versus $$X$$.

### Interaction strength

Besides measuring overall variable importance, an interesting aspect is to measure the strength of non-additivity associated with each covariable (i.e. the overall interaction strength) and/or between pairs of covariables. A model-agnostic way to assess such measures is based on partial dependence curves, see (Friedman and Popescu 2008). A computationally less demanding alternative to their H-statistic of measuring total interaction strength per covariable can be extracted from centered ICE: How much of their variability is unexplained by the main effect?

### Variable contribution breakdown for single observations

Instead of studying global variable importance and effects, there are different techniques to entangle how a single prediction can be explained: LIME, LIVE, SHAP and breakdown, see (Molnar 2019) and (Gosiewska and Biecek 2019) for an overview. The flashlight package currently supports the breakdown method as well as approximate SHAP. These two methods are based on additive decomposition of a prediction.

The breakdown algorithm works as follows: First, the visit order $$(x_1, ..., x_m)$$ of variables is specified. Then, in the query data, the column $$x_1$$ is set to the constant value of $$x_1$$ of the observation to be explained. The change in the (weighted) average predicted value on the query data measures the contribution of $$x_1$$ on the prediction. This procedure is iterated over all $$x_i$$ until eventually, all rows in the query data are identical to the observation to be explained.

A complication with this approach is that the visit order is relevant, at least for non-additive models. Ideally, the algorithm could be repeated for all $$m!$$ possible visit orders and its results averaged per variable. This is what SHAP does, see e.g. (Gosiewska and Biecek 2019) for an explanation. Unfortunately, there is no efficient way to do this in a model agnostic way. Thus we offer two approximations. The first one is the short-cut described in (Gosiewska and Biecek 2019) and implemented in the ibreakdown package (https://CRAN.R-project.org/package=iBreakDown): There, the variables $$x_i$$ are sorted by the size of their contribution in the same way as the breakdown algorithm but without iteration, i.e. starting from the original query data for each variable $$x_i$$. We call this approach “breakdown”. The second, computationally more intensive way to approximate SHAP values is based on running the breakdown algorithm on a small number (e.g. 12) random permutations and then averaging the results. We call this approach (approximate) “SHAP”.

### Global surrogate models

As suggested in (Molnar 2019), one way to explain a complex model is to fit an easy to interprete model like a decision tree to the predictions of the complex model and then study this simple surrogate model. The quality of its approximation can be measured by the R-squared of the surrogate model.

### flashlight

The flashlight package offers these tools in a very simple way.

## Installation of flashlight

From CRAN:

install.packages("flashlight")

library(devtools)
install_github("mayer79/flashlight")

## Teaser

Let’s start with an iris example. For simplicity, we do not split the data into training and testing/validation sets.

# Fit model
fit <- lm(Sepal.Length ~ ., data = iris)

# Make flashlight
fl <- flashlight(model = fit, data = iris, y = "Sepal.Length", label = "ols",
metrics = list(rmse = rmse, R-squared = r_squared))

# Performance: rmse and R-squared
plot(light_performance(fl), fill = "darkred")

plot(light_performance(fl, by = "Species"), fill = "darkred")

# Variable importance by increase in rmse
imp <- light_importance(fl, m_repetitions = 4)
plot(imp, fill = "darkred")

plot(light_importance(fl, by = "Species"))

most_important(imp, 2)
#> [1] "Petal.Length" "Species"

# ICE profiles for Petal.Width
plot(light_ice(fl, v = "Petal.Width"))

plot(light_ice(fl, v = "Petal.Width", center = "first"))

plot(light_ice(fl, v = "Petal.Width", by = "Species"))

# Partial dependence profiles for Petal.Width
plot(light_profile(fl, v = "Petal.Width"))

plot(light_profile(fl, v = "Petal.Width", by = "Species"))

# Accumulated local effects (ALE) profiles for Petal.Width
plot(light_profile(fl, v = "Petal.Width", type = "ale"))

plot(light_profile(fl, v = "Petal.Width", by = "Species", type = "ale"))

# Prediction, response and residual profiles
plot(light_profile(fl, v = "Petal.Width", type = "response", stats = "quartiles"))

plot(light_profile(fl, v = "Petal.Width", type = "predicted"))

plot(light_profile(fl, v = "Petal.Width", type = "residual", stats = "quartiles"))

# Response profiles, prediction profiles, partial dependence, and ALE profiles in one
plot(light_effects(fl, v = "Petal.Width"), use = "all")

# Scatter plots
plot(light_scatter(fl, v = "Petal.Width", type = "predicted"))

# Variable contribution breakdown for single observation
plot(light_breakdown(fl, new_obs = iris[2, ]))

# Global surrogate
plot(light_global_surrogate(fl))

## flashlights and multiflashlights

The process of using the flashlight package is as follows:

1. Define a flashlight object for each model. This is basically a list with optional components relevant for model interpretation:

• model: The fitted model object as the one returned by lm.

• data: A data set used to evaluate model agnostic tools, e.g. the validation data.

• y: The name of the variable in data representing the model response.

• predict_function: A function taking model and data and returning numeric predictions.

• linkinv: Inverse link function used to retransform the values returned by predict_function. Defaults to the identity function function(z) z.

• w: The name of the variable in data representing the case weights.

• by: A character vector of names of grouping variables in data. These will be used to stratify all results.

• metrics: A named list of metrics. These functions need to be available in the workspace and require arguments actual, predicted, w (case weights) as well as a placeholder … for further arguments. All metrics available in R package MetricsWeighted are suitable.

• label: The name of the model as shown in plots. This is the only required input when building the flashlight.

2. Calculate relevant information by calling the key functions:

• light_performance: Calculates performance measures regarding different metrics, possibly within subgroups and weighted by case weights.

• light_importance: Calculates variable importance (worsening in performance by random shuffling) for each or a subset of variables. Possibly within subgroups and using case weights. The most important variable names can be extracted by the function most_important on the result of light_importance.

• light_ice: Calculates ICE profiles across a couple of observations, possibly within groups.

• light_scatter: Calculates values (e.g. predictions) to be plotted against a variable of interest.

• light_profile: Calculates partial dependence profiles across a covariable, possibly within groups. Generated by calling light_ice and aggregating the results. The function is flexible: it can also be used to generate ALE, response, residual, prediction, or SHAP profiles and calculate (weighted) quartiles instead of (weighted) means.

• light_effects: Combines partial dependence, response and prediction profiles. ALE profiles can be added as well.

• light_interaction: Calculates overall and pairwise interaction strength based on Friedman’s H statistic.

• light_breakdown: Calculates variable contribution breakdown (approximate SHAP) for a single observation.

• light_global_surrogate: Fits an easy to interprete decision tree to the predictions of the model.

3. Plot the result: Each of these functions offer a plot method with minimal visualization of the results through ggplot2. The resulting plot can be customized by adding theme and other ggplot elements. If customization is insufficient, you can extract the data slot in the object returned by above key functions and build your own plot.

In practice, multiple flashlights are being defined and evaluated in parallel. By the help of a multiflashlight object, The flashlight packages provides as much support as possible to avoid any redundancy. It can combine fully specified flashlights or, and that is the more interesting option, take minimally defined flashlights (e.g. only label, model and predict_function) and add common arguments like y, by, data and/or w (case weights) in calling multiflashlight. If necessary, the resulting completed flashlights contained in the multiflashlight can be extracted again by $. All key functions are defined for both flashlight and multiflashlight objects. ## SHAP SHAP (see (Lundberg and Lee 2017)) offer a very elegant way to summarize the additive contribution of each variable to a single prediction, see . If such decompositions are available for many predictions, they can also be used to study global properties of the model such as variable importance (by averaging the absolute SHAP values of each variable) or effects (by showing SHAP values per variable). The problem with SHAP values is that they are computationally too demanding to be computed in general. Only for certain model techniques (e.g. tree based methods, see (Lundberg et al. 2020)), they can be computed exactly. Nevertheless, approximations exist, see (Gosiewska and Biecek 2019) for details. Unfortunately, even these approximations take time to compute. The approach followed by flashlight is to compute SHAP values once and store them in the flashlight object. All available methods for SHAP values will then use these values without recalculating them. Thus, if e.g. the inverse link function is updated, this has no effect on calculated SHAP values. The workflow is as follows: # Fit models fit1 <- lm(Sepal.Length ~ ., data = iris) fit2 <- lm(Sepal.Length ~ . + Species:Petal.Length, data = iris) # Create multiflashlight fl1 <- flashlight(model = fit1, label = "additive") fl2 <- flashlight(model = fit2, label = "non-additive") fls <- multiflashlight(list(fl1, fl2), data = iris, y = "Sepal.Length") # Add SHAP values fls <- add_shap(fls) # Use them in different methods plot(light_importance(fls, type = "shap")) plot(light_scatter(fls, v = "Petal.Length", type = "shap"), alpha = 0.2) plot(light_scatter(fls, v = "Petal.Length", type = "shap", by = "Species"), alpha = 0.2) plot(light_profile(fls, v = "Petal.Length", type = "shap")) For larger data and/or slow prediction functions, even computing approximate SHAP values might take too much time. The following options in add_shap are available to reduce the running time: • Reduce the number of variables (v). • Reduce the number of SHAP values (n_shap). • Reduce the size of the reference data set (n_max). • Select visit_strategy = "importance" instead of “permutation”. • If visit_strategy = "permutation", reduce the number of permutations (n_perm). • Avoid unnecessary flashlights in a multiflashlight. Once calculated, we suggest do store the resulting values. For a single flashlight x, use saveRDS(x$shap), file = ... resp. x$shap <- readRDS(...) to do so. ## Example As illustration, we use the data set house_prices with information on 21613 houses sold in King County between May 2014 and May 2015. It is shipped along with R package moderndive. The first few observations look as follows: head(house_prices) #> # A tibble: 6 x 21 #> id date price bedrooms bathrooms sqft_living sqft_lot floors #> <chr> <date> <dbl> <int> <dbl> <int> <int> <dbl> #> 1 7129~ 2014-10-13 2.22e5 3 1 1180 5650 1 #> 2 6414~ 2014-12-09 5.38e5 3 2.25 2570 7242 2 #> 3 5631~ 2015-02-25 1.80e5 2 1 770 10000 1 #> 4 2487~ 2014-12-09 6.04e5 4 3 1960 5000 1 #> 5 1954~ 2015-02-18 5.10e5 3 2 1680 8080 1 #> 6 7237~ 2014-05-12 1.23e6 4 4.5 5420 101930 1 #> # ... with 13 more variables: waterfront <lgl>, view <int>, condition <fct>, #> # grade <fct>, sqft_above <int>, sqft_basement <int>, yr_built <int>, #> # yr_renovated <int>, zipcode <fct>, lat <dbl>, long <dbl>, #> # sqft_living15 <int>, sqft_lot15 <int> Thus we have access to many relevant features like size, condition as well as location of the objects. We want to use these variables to predict the (log) house prices by the help of the following regression techniques and shed some light on them: • linear regression, • random forests, and • boosted trees. We use 70% of the data to calculate the models, 20% for evaluating their performance and for explaining them. 10% we keep untouched. ### Data preparation Let’s do some data preparation common for all models under consideration. prep <- mutate(house_prices, log_price = log(price), grade = as.integer(as.character(grade)), year = as.numeric(format(date, '%Y')), age = year - yr_built, year = factor(year), zipcode = as.factor(as.character(zipcode)), waterfront = factor(waterfront, levels = c(FALSE, TRUE), labels = c("no", "yes"))) x <- c("grade", "year", "age", "sqft_living", "sqft_lot", "zipcode", "condition", "waterfront") ### Modelling The random forest can directly work with this data structure. However, for the linear model, we need a small function with additional feature engineering. We log transform some input variables and categorize the zipcode in large groups. Similarly, for XGBoost, such a wrapper function turns non-numeric input variables to numeric. We will make use of these functions for both data preparation and prediction. # Data wrapper for the linear model prep_lm <- function(data) { data %>% mutate(sqrt_living = log(sqft_living), sqrt_lot = log(sqft_lot)) } # Data wrapper for xgboost prep_xgb <- function(data, x) { data %>% select_at(x) %>% mutate_if(Negate(is.numeric), as.integer) %>% data.matrix() } Then, we split the data and train our models. # Train / valid / test split (70% / 20% / 10%) set.seed(56745) ind <- sample(10, nrow(prep), replace = TRUE) train <- prep[ind >= 4, ] valid <- prep[ind %in% 2:3, ] test <- prep[ind == 1, ] (form <- reformulate(x, "log_price")) #> log_price ~ grade + year + age + sqft_living + sqft_lot + zipcode + #> condition + waterfront fit_lm <- lm(update.formula(form, . ~ . + I(sqft_living^2)), data = prep_lm(train)) # Random forest fit_rf <- ranger(form, data = train, respect.unordered.factors = TRUE, num.trees = 100, seed = 8373) cat("R-squared OOB:", fit_rf$r.squared)
#> R-squared OOB: 0.8654302

dtrain <- xgb.DMatrix(prep_xgb(train, x), label = train[["log_price"]])
dvalid <- xgb.DMatrix(prep_xgb(valid, x), label = valid[["log_price"]])

params <- list(learning_rate = 0.2,
max_depth = 5,
alpha = 1,
lambda = 1,
colsample_bytree = 0.8)

fit_xgb <- xgb.train(params,
data = dtrain,
watchlist = list(train = dtrain, valid = dvalid),
nrounds = 250,
print_every_n = 10,
objective = "reg:linear")
#> [1]  train-rmse:10.048225    valid-rmse:10.056726
#> [11] train-rmse:1.120189 valid-rmse:1.127692
#> [21] train-rmse:0.276046 valid-rmse:0.284578
#> [31] train-rmse:0.220493 valid-rmse:0.230205
#> [41] train-rmse:0.201741 valid-rmse:0.213547
#> [51] train-rmse:0.191478 valid-rmse:0.205166
#> [61] train-rmse:0.185060 valid-rmse:0.200389
#> [71] train-rmse:0.179371 valid-rmse:0.195981
#> [81] train-rmse:0.175560 valid-rmse:0.193149
#> [91] train-rmse:0.171573 valid-rmse:0.190594
#> [101]    train-rmse:0.169047 valid-rmse:0.189242
#> [111]    train-rmse:0.166089 valid-rmse:0.187361
#> [121]    train-rmse:0.164131 valid-rmse:0.186762
#> [131]    train-rmse:0.161985 valid-rmse:0.185919
#> [141]    train-rmse:0.159950 valid-rmse:0.185113
#> [151]    train-rmse:0.158342 valid-rmse:0.184533
#> [161]    train-rmse:0.156986 valid-rmse:0.184286
#> [171]    train-rmse:0.155399 valid-rmse:0.184044
#> [181]    train-rmse:0.154224 valid-rmse:0.183937
#> [191]    train-rmse:0.153022 valid-rmse:0.183618
#> [201]    train-rmse:0.151803 valid-rmse:0.183548
#> [211]    train-rmse:0.150595 valid-rmse:0.183108
#> [221]    train-rmse:0.149530 valid-rmse:0.182821
#> [231]    train-rmse:0.148626 valid-rmse:0.182916
#> [241]    train-rmse:0.147686 valid-rmse:0.183019
#> [250]    train-rmse:0.146648 valid-rmse:0.182994

### Creating the flashlights

Let’s initialize one flashlight per model. Thanks to individual prediction functions, any model can be used in flashlight, even h2o and keras models.

fl_mean <- flashlight(model = mean(train$log_price), label = "mean", predict_function = function(mod, X) rep(mod, nrow(X))) fl_lm <- flashlight(model = fit_lm, label = "lm", predict_function = function(mod, X) predict(mod, prep_lm(X))) fl_rf <- flashlight(model = fit_rf, label = "rf", predict_function = function(mod, X) predict(mod, X)$predictions)
fl_xgb <- flashlight(model = fit_xgb, label = "xgb",
predict_function = function(mod, X) predict(mod, prep_xgb(X, x)))
print(fl_xgb)
#>
#> Flashlight xgb
#>
#> Model:            Yes
#> y:            No
#> w:            No
#> by:           No
#> data dim:         No
#> predict_fct default:  FALSE
#> metrics:      rmse
#> SHAP:             No

What about all other relevant elements of a flashlight like the underlying data, the response name, metrics, retransformation functions etc? We could pass them to each of our flashlights. Or better, we can combine the flashlights to a multiflashlight and pass additional common arguments there.

fls <- multiflashlight(list(fl_mean, fl_lm, fl_rf, fl_xgb), y = "log_price", linkinv = exp,
data = valid, metrics = list(rmse = rmse, R-squared = r_squared))

We could even extract these completed flashlights from the multiflashlight as if the latter is a list (actually it is a list with additional class multiflashlight).

fl_lm <- fls$lm ### Assess performance Let’s compare the models regarding their validation performance. perf <- light_performance(fls) perf #> #> I am an object with class(es) light_performance_multi, light_performance, light, list #> #> data.frames (maximum 6 rows shown): #> #> data #> # A tibble: 6 x 3 #> label metric value #> <fct> <fct> <dbl> #> 1 mean rmse 0.527 #> 2 mean R-squared -0.000267 #> 3 lm rmse 0.188 #> 4 lm R-squared 0.872 #> 5 rf rmse 0.192 #> 6 rf R-squared 0.868 plot(perf) plot(perf, fill = "darkred") + xlab(element_blank()) The plot “politics” of flashlight is to provide simple graphics with minimal ggplot-tuning, so you are able to add your own modifications. If you are completely unhappy about the proposed plot (e.g. rather favour a scatterplot over a barplot), extract the data slot of perf and create the figure from scratch: head(perf$data)
#> # A tibble: 6 x 3
#>   label metric        value
#>   <fct> <fct>         <dbl>
#> 1 mean  rmse       0.527
#> 2 mean  R-squared -0.000267
#> 3 lm    rmse       0.188
#> 4 lm    R-squared  0.872
#> 5 rf    rmse       0.192
#> 6 rf    R-squared  0.868

perf$data %>% ggplot(aes(x = label, y = value, group = metric, color = metric)) + geom_point() The same logic holds for all other main functions in the flashlight package. For performance considerations, the minimum required info in the (multi-)flashlight are: “y”, “predict_function”, “model”, “data” and “metrics”. The latter two can also be passed on the fly. ### Variable importance Now let’s study variable importance of the explainers. (imp <- light_importance(fls, v = x)) #> #> I am an object with class(es) light_importance_multi, light_importance, light, list #> #> data.frames (maximum 6 rows shown): #> #> data #> # A tibble: 6 x 5 #> label metric variable value error #> <fct> <fct> <chr> <dbl> <lgl> #> 1 mean rmse grade 0 NA #> 2 mean rmse year 0 NA #> 3 mean rmse age 0 NA #> 4 mean rmse sqft_living 0 NA #> 5 mean rmse sqft_lot 0 NA #> 6 mean rmse zipcode 0 NA plot(imp, fill = "darkred") Note for permutation importance: If the calculation takes too long (e.g. large query data), set n_max to some reasonable value. light_importance will then randomly pick n_max rows and use only these for assessment of importance. On the other hand, if the calculations is very fast, consider setting m_repetitions to a value > 1 in order to repeat the permutations multiple times. The resulting variable importance corresponds to the average drop in performance and standard errors are added. If SHAP values have been precomputed by add_shap, use type = "shap" to show mean absolute shap values per covariable. ### Individual conditional expectation How do predictions change when sqft_living changes alone? We can investigate this question by looking at “Individual Conditional Expectation” (ICE) profiles of a couple of observations. cp <- light_ice(fls, v = "sqft_living", n_max = 30, seed = 35) plot(cp, alpha = 0.2) Note: Setting seed to a fixed value will ensure that the flashlights will consider the same rows. An alternative would be to pass a small subset of the data to light_ice and calculate all profiles or by passing row indices through indices for fixed selection. Centered ICE profiles (“c-ICE”) can help to increase visibility of interactions. cp <- light_ice(fls, v = "sqft_living", n_max = 30, seed = 35, center = "first") plot(cp, alpha = 0.2) ### Partial dependence profiles If many ICE profiles (in our case 1000) are averaged, we get an impression on the average effect of the considered variable. Such curves are called partial dependence profiles (PD) resp. partial dependence plots. pd <- light_profile(fls, v = "sqft_living") pd #> #> I am an object with class(es) light_profile_multi, light_profile, light, list #> #> data.frames (maximum 6 rows shown): #> #> data #> # A tibble: 6 x 5 #> sqft_living counts value label type #> <dbl> <int> <dbl> <fct> <fct> #> 1 500 1000 462845. mean partial dependence #> 2 1500 1000 462845. mean partial dependence #> 3 2500 1000 462845. mean partial dependence #> 4 3500 1000 462845. mean partial dependence #> 5 4500 1000 462845. mean partial dependence #> 6 5500 1000 462845. mean partial dependence plot(pd) The light_profile function offers different ways to specify the evaluation points of the profiles. One option is to explicitly pass such points. pd <- light_profile(fls, v = "sqft_living", pd_evaluate_at = seq(1000, 4000, by = 100)) plot(pd) For discrete variables: pd <- light_profile(fls, v = "condition") plot(pd) ### Accumulated local effects (ALE) An approximation of main effects without Ceteris Paribus clause are accumulated local effects profiles (Fisher, Rudin, and Dominici 2018). They are based on accumulating local partial dependence slopes. ale <- light_profile(fls, v = "sqft_living", type = "ale") ale #> #> I am an object with class(es) light_profile_multi, light_profile, light, list #> #> data.frames (maximum 6 rows shown): #> #> data #> # A tibble: 6 x 5 #> sqft_living counts value label type #> <dbl> <int> <dbl> <fct> <fct> #> 1 1500 1000 462845. mean ale #> 2 2500 1000 462845. mean ale #> 3 3500 859 462845. mean ale #> 4 4500 231 462845. mean ale #> 5 5500 52 462845. mean ale #> 6 6500 16 462845. mean ale plot(ale) Note: While equally sized x-breaks are easy to read, quantile binning usually leads to more stable results. plot(light_profile(fls, v = "sqft_living", type = "ale", cut_type = "quantile")) In order to calculate ICEs, PDs and ALEs, the following elements need to be available in the (multi-)flashlight: “predict_function”, “model”, “linkinv” and “data”. “data” can also be passed on the fly. ### Profiles of predicted values, residuals, and response We can use the function light_profile not only to create partial dependence profiles but also to get profiles of predicted values (“M plots”), responses or residuals. Additionally, we can either use averages or quartiles as summary statistics. Average predicted values versus the living area are as follows: format_y <- function(x) format(x, big.mark = "'", scientific = FALSE) pvp <- light_profile(fls, v = "sqft_living", type = "predicted", format = "fg", big.mark = "'") plot(pvp) + scale_y_continuous(labels = format_y) Note the formatting of y values as well as the formatC option format = "fg" and big.mark passed to the constructor of the x labels in order to improve basic appearance. We will recycle some of these settings for the next plots. Similarly, the average response profiles (identical for all flashlights, to we only show one of them): rvp <- light_profile(fl_lm, v = "sqft_living", type = "response", format = "fg") plot(rvp) + scale_y_continuous(labels = format_y) Same, but quartiles: rvp <- light_profile(fl_lm, v = "sqft_living", type = "response", stats = "quartiles", format = "fg") plot(rvp) + scale_y_continuous(labels = format_y) What about residuals? First, we remove the “mean” flashlight by setting it NULL. fls$mean <- NULL
rvp <- light_profile(fls, v = "sqft_living", type = "residual",
stats = "quartiles", format = "fg")
plot(rvp) +
scale_y_continuous(labels = format_y)

While the tree-based models have smaller residuals and medians close to 0, the linear model shows residual curvature that could be captured by representing sqft_living by more parameters.

If your are unhappy about the “group by” strategy, set swap_dim to TRUE.

plot(rvp, swap_dim = TRUE) +
scale_y_continuous(labels = format_y)

For less bars, set n_bins in light_profile:

rvp <- light_profile(fls, v = "sqft_living", type = "residual",
stats = "quartiles", format = "fg", n_bins = 5)
plot(rvp, swap_dim = TRUE) +
scale_y_continuous(labels = format_y)

Just as diverging ICE profiles give you a clou about the presence of interactions, we can use the option stats = "quartiles" (with pd_center) to show divergence of the mean-centered ICE profiles as boxes (predictions at log-scale to suppress interaction-like effects of the retransformation function):

rvp <- light_profile(fls, v = "sqft_living", use_linkinv = FALSE,
stats = "quartiles", pd_center = "mean")
plot(rvp)

For prediction profiles, the same elements as for ICE/PDs are required, while for response profiles we need “y”, “linkinv” and “data”. “data” can also be passed on the fly.

### Visualizing different types of profiles as “effects” plot

In assessing the model quality, it is often useful to visualize

• response profile (quartiles or means),

• average predictions, and

• model effects (partial dependence profiles, or accumulated local effects)

in the same plot and for each input variable. The flashlight package offers the function light_effects to combine such profile plots:

eff <- light_effects(fl_lm, v = "condition")
p <- plot(eff) +
scale_y_continuous(labels = format_y)
p

Let’s add counts to see if the gaps between response and predicted profiles are problematic or just due to small samples.

plot_counts(p, eff, alpha = 0.2)

The biggest gaps occur at very rare conditions, so the model looks quite fine.

Note: Due to retransformation from log scale, the response profile is slightly higher than the profile of predicted values. If we would evaluate on the modelled log scale, that gap would vanish.

eff <- light_effects(fl_lm, v = "condition", linkinv = I)
p <- plot(eff, use = "all") +
scale_y_continuous(labels = format_y) +
ggtitle("Effects plot on modelled log scale")
p

Besides adding counts to the figure, representing observed responses as boxplots (no whiskers and outliers in order to avoid too large y scale) might help to judge if there is a problematic misfit.

eff <- light_effects(fl_lm, v = "condition", stats = "quartiles")
p <- plot(eff, rotate_x = FALSE) +
scale_y_continuous(labels = format_y)
plot_counts(p, eff, fill = "blue", alpha = 0.2, width = 0.3)

The plot method of light_effects allows to hide certain plot elements if it looks too dense. If we would e.g. want to compare partial dependence and accumulated local effects plots, we can use the following approach.

eff <- light_effects(fls, v = "sqft_living", v_labels = FALSE,
cut_type = "quantile", n_bins = 25)
plot(eff, use = c("pd", "ale"), show_points = FALSE) +
scale_y_continuous(labels = format_y) +
coord_cartesian(xlim = c(1000, 4000), ylim = c(3e5, 9e5))

As an alternative to profile plots, we could also consider scatter plots.

pr <- light_scatter(fls, v = "sqft_living", n_max = 300)
plot(pr, alpha = 0.2)

### Interaction strength

Based on Friedman’s H statistic (Friedman and Popescu 2008), we can investigate which variable has strongest interaction effect relative to its full effect.

st <- light_interaction(fls, v = x, grid_size = 30, n_max = 50, seed = 42)
plot(st)

Since we have not used interaction terms in the linear model, all values are zero there.

Pairwise interactions are expensive to compute as each variable pair has to be assessed. We limit the search for interactions to the four variables with highest relative interaction strength.

st_pair <- light_interaction(fls, v = most_important(st, 4),
pairwise = TRUE, n_max = 50, seed = 42)
plot(st_pair)

### Variable contribution breakdown for single observation

Besides investigating global effects, we can use light_breakdown to calculate variable impact on one single (log) prediction.

bd <- light_breakdown(fl_lm, new_obs = valid[1, ], v = x, n_max = 1000, seed = 74)
plot(bd, size = 3)

We have set the size of the reference data set to n_max = 1000 in order to save time. The only variable with positive impact on the prediction is “age”. The other variables have a negative impact compared to the 1000 reference observations.

### Global surrogate trees

A very different, but intuitive approach is to fit the original predictions of the models by simple decision trees and then visualize these trees.

## Working with case weights

In many applications, case weights need to be taken into account. All main functions in flashlight are able to deal with them. The only thing you need to do is to pass the column name of the case weights to w when initializing the (multi-)flashlight.

Let’s go through the initial iris example again with (artificial) case weights:

# Add weight info to the flashlight
fl_weighted <- flashlight(fl, w = "Petal.Length", label = "ols weighted")
fls <- multiflashlight(list(fl, fl_weighted))

# Performance: rmse and R-squared
plot(light_performance(fls))

plot(light_performance(fls, by = "Species"))

# Variable importance by drop in rmse
plot(light_importance(fls, by = "Species"))

# ICE profiles for Petal.Width
# (not affected by weights because nothing is being aggregated)
indices <- seq(10, 150, by = 10)
plot(light_ice(fls, v = "Petal.Width", indices = indices), alpha = 0.2)

plot(light_ice(fls, v = "Petal.Width", by = "Species", indices = indices))

# c-ICE -> lines overlap, no interactions at all
plot(light_ice(fls, v = "Petal.Width", indices = indices, center = "first"), alpha = 0.2)

plot(light_ice(fls, v = "Petal.Width", by = "Species", indices = indices, center = "first"))

# Partial dependence profiles for Petal.Width
plot(light_profile(fls, v = "Petal.Width"))

plot(light_profile(fls, v = "Petal.Width", by = "Species"))

# ALE profiles for Petal.Width
plot(light_profile(fls, v = "Petal.Width", type = "ale"))

plot(light_profile(fls, v = "Petal.Width", by = "Species", type = "ale"))

# Observed, predicted, and partial dependence profiles
plot(light_effects(fls, v = "Petal.Width"))

eff <- light_effects(fls, v = "Petal.Width", stats = "quartiles")
plot(eff) %>%
plot_counts(eff, alpha = 0.2, fill = "blue")

# Variable contribution breakdown for single observation (on log-scale)
plot(light_breakdown(fls, new_obs = iris[2, ]), size = 2.5)

# Global surrogate
plot(light_global_surrogate(fls))


text:
## 2019N0718

### ̑̑@yɖŎ󂯂IQeXg

NbN11X肢܂B

https://iqexams.net/

posted by Line Segment at 18:00 | Comment(0) | L | |


text:
# Getting Started¶

Improving the performance of your quantum programs is only a few lines of code away.

This getting started shows examples using cirq cirq and qiskit. We’ll first test mitiq by running against the noisy simulator built into cirq. The qiskit example work similarly as you will see in Qiskit Mitigation.

## Multi-platform Framework¶

In mitiq, a “back-end” is a function that executes quantum programs. A “front-end” is a library/language that constructs quantum programs. mitiq lets you mix and match these. For example, you could write a quantum program in qiskit and then execute it using a cirq backend, or vice versa.

Back-ends are abstracted to functions called executors that always accept a quantum program, sometimes accept other arguments, and always return an expectation value as a float. You can see some examples of different executors for common packages here and in this getting started. If your quantum programming interface of choice can be used to make a Python function with this type, then it can be used with mitiq.

## Error Mitigation with Zero-Noise Extrapolation¶

We define some functions that make it simpler to simulate noise in cirq. These don’t have to do with mitiq directly.

import numpy as np
from cirq import Circuit, depolarize
from cirq import LineQubit, X, DensityMatrixSimulator

SIMULATOR = DensityMatrixSimulator()
# 0.1% depolarizing noise
NOISE = 0.001

def noisy_simulation(circ: Circuit) -> float:
""" Simulates a circuit with depolarizing noise at level NOISE.
Args:
circ: The quantum program as a cirq object.

Returns:
The expectation value of the |0> state.
"""
circuit = circ.with_noise(depolarize(p=NOISE))
rho = SIMULATOR.simulate(circuit).final_density_matrix
# define the computational basis observable
obs = np.diag([1, 0])
expectation = np.real(np.trace(rho @ obs))
return expectation

Now we can look at our example. We’ll test single qubit circuits with even numbers of X gates. As there are an even number of X gates, they should all evaluate to an expectation of 1 in the computational basis if there was no noise.

from cirq import Circuit, LineQubit, X

qbit = LineQubit(0)
circ = Circuit(X(qbit) for _ in range(80))
unmitigated = noisy_simulation(circ)
exact = 1
print(f"Error in simulation is {exact - unmitigated:.{3}}")

Error in simulation is 0.0506

This shows the impact the noise has had. Let’s use mitiq to improve this performance.

from mitiq import execute_with_zne

mitigated = execute_with_zne(circ, noisy_simulation)
print(f"Error in simulation is {exact - mitigated:.{3}}")

Error in simulation is 0.000519

print(f"Mitigation provides a {(exact - unmitigated) / (exact - mitigated):.{3}} factor of improvement.")

Mitigation provides a 97.6 factor of improvement.

You can also use mitiq to wrap your backend execution function into an error-mitigated version.

from mitiq import mitigate_executor

run_mitigated = mitigate_executor(noisy_simulation)
mitigated = run_mitigated(circ)
print(round(mitigated,5))

0.99948

Note

As shown here, mitiq wraps executor functions that have a specific type: they take quantum programs as input and return expectation values. However, one often has an execution function with other arguments such as the number of shots, the observable to measure, or the noise level of a noisy simulation. It is still easy to use these with mitiq by using partial function application. Here’s a pseudo-code example:

from functools import partial

def shot_executor(qprogram, n_shots) -> float:
...
# we partially apply the n_shots argument to get a function that just
# takes a quantum program
mitigated = execute_with_zne(circ, partial(shot_executor, n_shots=100))

The default implementation uses Richardson extrapolation to extrapolate the expectation value to the zero noise limit [1]. Mitiq comes equipped with other extrapolation methods as well. Different methods of extrapolation are packaged into Factory objects. It is easy to try different ones.

from mitiq import execute_with_zne
from mitiq.zne.inference import LinearFactory

fac = LinearFactory(scale_factors=[1.0, 2.0, 2.5])
linear = execute_with_zne(circ, noisy_simulation, factory=fac)
print(f"Mitigated error with the linear method is {exact - linear:.{3}}")

Mitigated error with the linear method is 0.00638

You can read more about the Factory objects that are built into mitiq and how to create your own here.

Another key step in zero-noise extrapolation is to choose how your circuit is transformed to scale the noise. You can read more about the noise scaling methods built into mitiq and how to create your own here.

## Qiskit Mitigation¶

Mitiq is designed to be agnostic to the stack that you are using. Thus for qiskit things work in the same manner as before. Since we are now using qiskit, we want to run the error mitigated programs on a qiskit backend. Let’s define the new backend that accepts qiskit circuits. In this case it is a simulator, but you could also use a QPU.

import qiskit
from qiskit import QuantumCircuit

# Noise simulation packages
from qiskit.providers.aer.noise import NoiseModel
from qiskit.providers.aer.noise.errors.standard_errors import depolarizing_error

# 0.1% depolarizing noise
NOISE = 0.001

QISKIT_SIMULATOR = qiskit.Aer.get_backend("qasm_simulator")

def qs_noisy_simulation(circuit: QuantumCircuit, shots: int = 4096) -> float:
"""Runs the quantum circuit with a depolarizing channel noise model at
level NOISE.

Args:
circuit (qiskit.QuantumCircuit): Ideal quantum circuit.
shots (int): Number of shots to run the circuit
on the back-end.

Returns:
expval: expected values.
"""
# initialize a qiskit noise model
noise_model = NoiseModel()

# we assume a depolarizing error for each
# gate of the standard IBM basis

# execution of the experiment
job = qiskit.execute(
circuit,
backend=QISKIT_SIMULATOR,
basis_gates=["u1", "u2", "u3"],
# we want all gates to be actually applied,
# so we skip any circuit optimization
optimization_level=0,
noise_model=noise_model,
shots=shots
)
results = job.result()
counts = results.get_counts()
expval = counts["0"] / shots
return expval

We can then use this backend for our mitigation.

from qiskit import QuantumCircuit
from mitiq import execute_with_zne

circ = QuantumCircuit(1, 1)
for __ in range(120):
_ = circ.x(0)
_ = circ.measure(0, 0)

unmitigated = qs_noisy_simulation(circ)
mitigated = execute_with_zne(circ, qs_noisy_simulation)
exact = 1
# The mitigation should improve the result.
print(abs(exact - mitigated) < abs(exact - unmitigated))

True

Note that we don’t need to even redefine factories for different stacks. Once you have a Factory it can be used with different front and backends.


text:
## On interacting bumps of semi-classical states of nonlinear Schrödinger equations.(English)Zbl 1217.35065

Summary: We study concentrated positive bound states of the following nonlinear Schrödinger equation $h^2 \Delta u \Gamma V (x)u + u^p = 0;~ u> 0;~ x\in\mathbb{R}^N$ where $$p$$ is subcritical. We prove that, at a local maximum point $$x_0$$ of the potential function $$V (x)$$ and for arbitrary positive integer $$K$$ ($$K>1$$), there always exist solutions with $$K$$ interacting bumps concentrating near $$x_0$$ . We also prove that at a nondegenerate local minimum point of $$V (x)$$ such solutions do not exist.

### MSC:

 35J60 Nonlinear elliptic equations 35B50 Maximum principles in context of PDEs 35Q55 NLS equations (nonlinear Schrödinger equations)


text:
Math Help - Max/min on a continuous interval.

1. Max/min on a continuous interval.

Suppose that $f(x)$ is continuous and $> 0$ on $I = \mathbb{R},$ and $\displaystyle\lim_{x\to \pm\infty}f(x) = 0.$

$(a)$ Prove $f(x)$ has no minimum on $I.$
$(b)$ Prove $f(x)$ has a maximum on $I$ (note that $I$ is not compact).
$(c)$ Prove $(b)$ under weaker hypothesis than positivity on all of $I$.

2. Originally Posted by cgiulz
Suppose that $f(x)$ is continuous and $> 0$ on $I = \mathbb{R},$ and $\displaystyle\lim_{x\to \pm\infty}f(x) = 0.$

$(a)$ Prove $f(x)$ has no minimum on $I.$
$(b)$ Prove $f(x)$ has a maximum on $I$ (note that $I$ is not compact).
$(c)$ Prove $(b)$ under weaker hypothesis than positivity on all of $I$.
For (a) I'm assuming you mean a global minimum (otherwise this is not true) and it's obvious since that minimum would have to be $0$ contradicting the fact that $f>0$.

For (c) we're going to assume $f \geq 0$ for some $x \in \mathbb{R}$. Take an interval of the form $[-K,K]$ and take the maximum of f there say $f(x_0)=y_0>0$ (if it's zero we'll deal with it later). Now since $f(x) \rightarrow 0$ when $x \rightarrow \pm \infty$ we have two constants $K_1,K_2>0$ such that $\vert f(x) \vert < y_0$ whenever $x>K_1$ or $x<-K_2$ take $m= \max \{K_1,K_2 \}$ and so we have that $f$ attains a maximum $y$ in $[-m,m]$ with $y \geq y_0$ and since out of this interval the function never exceeds $y_0$, $y$ is a global maximum.

If $y_0$ were $0$ and $f \leq 0$ in I we're finished. If it's $f>0$ on an unbounded interval apply a similar argument to the above, if $f>0$ on a bounded interval it's clear it has a maximum. And this covers all cases.

3. Can anyone show me where to start in order to prove the existence of a max by positivity?

Thanks.

[Forget it, I realized I can just pull it out of part (c)]


text:
Select language: En
0
2013
Impact Factor

# Statistical characteristics of attainable set of controllable system, non-wandering, and minimal attraction center

2009, Vol. 5, No. 2, pp.  265-288

Author(s): Rodina L. I., Tonkov E. L.

In the terms of Lyapunov functions we obtain the conditions that allow to estimate the relative frequency of occurrence of the attainable set of a controllable system in a given set $\mathfrak{M}$. The set $\mathfrak{M}$ is called statistically invariant if the relative frequency of occurrence in $\mathfrak{M}$ is equal to one. We also derive the conditions of the statistically weak invariance of $\mathfrak{M}$ with respect to controllable system, that is, for every initial point from $\mathfrak{M}$, at least one solution of the controllable system is statistically invariant. We obtain the conditions for the attainable set to be non-wandering as well as the conditions of existence of the minimal attraction center.

Keywords: controllable systems, dynamical systems, differential inclusions, attainability, invariance, non-wandering, recurrence
Citation: Rodina L. I., Tonkov E. L., Statistical characteristics of attainable set of controllable system, non-wandering, and minimal attraction center, Rus. J. Nonlin. Dyn., 2009, Vol. 5, No. 2, pp.  265-288
DOI:10.20537/nd0902008


text:
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# systems of linear equations word problems calculator

## systems of linear equations word problems calculator

When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. Systems of linear equations word problems — Basic example. Write a system of equations describing the following word problem: The Lopez family had a rectangular garden with a 20 foot perimeter. Lacy will have traveled about 1050 feet when the police car catches up to her. We can use either Substitution or Elimination, depending on what’s easier. You have learned many different strategies for solving systems of equations! What were the dimensions of the original garden? J.9 – Solve linear equations: mixed. One step equation word problems. solving systems of linear equations: word problems? shehkar pulls 31 coins out of his pocket. Or click the example. We need to find the intersection of the two functions, since that is when the distances are the same. To solve word problems using linear equations, we have follow the steps given below. Well, that or spending a semester studying abroad in Mathrovia. The problem has given us two pieces of information: if we add the number of cats the lady owns and the number of birds the lady owns, we have 21, and if we add the number of cat legs and the number of bird legs, we have 76. Section 2-3 : Applications of Linear Equations. Once you do that, these linear systems are solvable just like other linear systems.The same rules apply. Solver : Linear System solver (using determinant) by ichudov(507) Solver : SOLVE linear system by SUBSTITUTION by ichudov(507) Want to teach? third order linear equations calculator ; java "convert decimal to fraction" ... solving problems systems of equations worksheet log on ti 89 ... modeling word problems linear equations samples online algebra calculator html code The solutions are $$\left( {-.62,.538} \right)$$, $$\left( {.945,2.57} \right)$$ and $$\left( {4.281,72.303} \right)$$. Solve age word problems with a system of equations. Systems of Equations Calculator is a calculator that solves systems of equations step-by-step. Substituting the $$y$$ from the first equation into the second and solving yields: \begin{array}{l}\left. Solving word problems (application problems) with 3x3 systems of equations. We'd be dealing with some large numbers, though. We could also solve the non-linear systems using a Graphing Calculator, as shown below. Plug each into easiest equation to get $$y$$’s: First solve for $$y$$ in terms of $$x$$ in the second equation, and. They enlarged their garden to be twice as long and three feet wider than it was originally. On to Introduction to Vectors  – you are ready! 6-1. Instead of saying "if we add the number of cats the lady owns and the number of birds the lady owns, we get 21, " we can say: What about the second piece of information: "if we add the number of cat legs and the number of bird legs, we get 76"? This calculators will solve three types of 'work' word problems.Also, it will provide a detailed explanation. To describe a word problem using a system of equations, we need to figure out what the two unknown quantities are and give them names, usually x and y. Linear systems of equations word problems 4 examples study guide piecewise functions in the graphing calculator advanced matrix and solving with matrices she loves math mixture solutions questions s complete a table graph using mode gcse maths casio fx 83gt fx85gt plus absolute value khan academy on ti core lesson Linear Systems Of Equations Word Problems 4 Examples… Read More » if he has a total of 5.95, how many dimes does he have? "Solve Linear Systems Word Problems Relay Activity"DIGITAL AND PRINT: Six rounds provide practice or review solving systems of linear equations word problems in context. Read the given problem carefully; Convert the given question into equation. {\underline {\, Pythagorean theorem word problems. (b)  We can plug the $$x$$ value ($$t$$) into either equation to get the $$y$$ value ($$d(t)$$); it’s easiest to use the second equation: $$d\left( t \right)=4{{\left( {16.2} \right)}^{2}}\approx 1050$$. Time and work word problems. The distance that Lacy has traveled in feet after $$t$$ seconds can be modeled by the equation $$d\left( t\right)=150+75t-1.2{{t}^{2}}$$. Solve Equations Calculus. You really, really want to take home 6items of clothing because you “need” that many new things. \end{array}. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Our second piece of information is that if we make the garden twice as long and add 3 feet to the width, the perimeter will be 40 feet.


text:
## BdMO National 2012: Higher Secondary, Secondary 07

Moon
Posts: 751
Joined: Tue Nov 02, 2010 7:52 pm
Contact:

### BdMO National 2012: Higher Secondary, Secondary 07

Problem:
In an acute angled triangle $ABC$, $\angle A= 60^0$. Prove that the bisector of one of the angles formed by the altitudes drawn from $B$ and $C$ passes through the center of the circumcircle of the triangle $ABC$.
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photon
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### Re: BdMO National 2012: Higher Secondary, Secondary 07

Moon wrote:Problem:
In an acute angled triangle $ABC$, $\angle A= 60^0$. Prove that the bisector of one of the angles formed by the altitudes drawn from $B$ and $C$ passes through the center of the circumcircle of the triangle $ABC$.
what is that "angles formed by the altitudes drawn from $B$ and $C$" ?
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*Mahi*
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### Re: BdMO National 2012: Higher Secondary, Secondary 07

Let the feet of altitudes from $B,C$ be $E,F$. If $BE$ and $CF$ intersects at $H$ then the angles are $\angle BHC,\angle CHE,\angle EHF,\angle BHF$.

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photon
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### Re: BdMO National 2012: Higher Secondary, Secondary 07

$BC=\frac{2R}{sinA}=\sqrt{3}R$
$\angle BOC=2x60^O=120^O$ AND $\angle BHC=120^O$. so, $B,H,O,C$ are cyclic.
in tri $BOC$, $sin \angle OBC=\frac{sinA.R}{BC}=\frac{sin60^O.R}{\sqrt{3}R}=\frac{1}{2}$
so, $\angle OBC=30^O$, $\angle OHC=30^O$[$B,H,O,C$ are cyclic.]
so, $HO$ bisects $\angle CHE=60^O$.Done.
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Try not to become a man of success but rather to become a man of value.-Albert Einstein

SANZEED
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### Re: BdMO National 2012: Higher Secondary, Secondary 07

When the altitudes intersect, 2 acute angles and 2 obtuse angles are formed. If my check is right, then $OH$ always divides the acute ones.
Do we need to prove this?
And also, there is an easy euclidean proof taking this fact into account.
$\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$


text:
Sum of Telescoping Series

Another fascinating series is telescoping series. This series is quite different from a regular progression. If you came across to a problem that ask for the sum and you can’t find it by using arithmetic or geometric series. It could be the solve using telescoping technique. This series requires our knowledge to express fractional expression to its partial fraction form. I won’t be surprise if this is your first time to hear such series because the application of this topic is usually for Olympiad Mathematicians.

Ingenious way to express telescopic sum

$\displaystyle\frac{1}{n(n+1)}=\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}$

$\displaystyle\frac{k}{n(n+k)}=\displaystyle\frac{1}{k}[\frac{1}{n}-\frac{1}{n+k}]$

$\displaystyle\frac{k}{n(n+1)(n+k)}= \displaystyle\frac{1}{k}[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+k)}]$

Sample Problem 1:

Find the sum of the series $\frac{1}{1(2)}+\frac{1}{2(3)}+\frac{1}{2(3)}+\ldots +\frac{1}{2013(2014)}$

Solution:

We can express the nth term as $\displaystyle\frac{1}{n(n+1)}$

As mentioned above $\displaystyle\frac{1}{n(n+1)}=\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}$

Let n=1

$\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}=\displaystyle\frac{1}{1}-\displaystyle\frac{1}{2}$

Let n=2

$\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}=\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3}$

Let n=3

$\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}=\displaystyle\frac{1}{3}-\displaystyle\frac{1}{4}$

.

.

.

Let n=2013

$\displaystyle\frac{1}{n}-\displaystyle\frac{1}{n+1}=\displaystyle\frac{1}{2013}-\displaystyle\frac{1}{2014}$

Expressing $\frac{1}{1(2)}+\frac{1}{2(3)}+\frac{1}{2(3)}+\ldots +\frac{1}{2013(2014)}$ to its equivalent partial fractions we have

$(\displaystyle\frac{1}{1}-\displaystyle\frac{1}{2})+(\displaystyle\frac{1}{2}-\displaystyle\frac{1}{3})+(\displaystyle\frac{1}{3}-\displaystyle\frac{1}{4})+\ldots +(\displaystyle\frac{1}{2013}-\displaystyle\frac{1}{2014})$

Cancel out terms leaving

$1-\displaystyle\frac{1}{2014}=\displaystyle\frac{2013}{2014}$

Therefore we can conclude that $\sum_{n=1}^k \displaystyle\frac{1}{n(n+1)}=\displaystyle\frac{k}{k+1}$

Sample Problem 2:

Find the sum of

$\sum_{k=0}^{99} \log(\displaystyle\frac{k}{k+1})$

Solution:

Using the law of logarithm,

$\log(\displaystyle\frac{k}{k+1})=\log(k)-\log(k+1)$

Let k=1

$\log 1-\log 2$

Let k=2

$\log 2-\log 3$

Let k=3

$\log 3-\log 4$

.

.

.

Let k=99

$\log 99-\log 100$

$\sum_{k=0}^{99} \log(\displaystyle\frac{k}{k+1})=\sum_{k=0}^{99} \log(k)-log(k+1)$ $(\log 1-\log 2)+(\log 2-\log 3)+ (\log 3-\log 4)+\ldots +(\log 99-\log 100)$

Cancel out terms from log2 to log99 we have

$\log 1-log 100$ but log1=0

$-\log 100=-2$

Therefore we can conclude that the formula for Find the sum of

$\sum_{k=0}^{n} \log(\displaystyle\frac{k}{k+1})=-\log n$

Sample Problem 3:

Find the sum of

$\displaystyle\frac{1}{1\sqrt{2}+2\sqrt{1}}+\displaystyle\frac{1}{2\sqrt{3}+3\sqrt{2}}+\displaystyle\frac{1}{3\sqrt{4}+4\sqrt{3}}+\ldots +\displaystyle\frac{1}{840\sqrt{841}+841\sqrt{840}}$

Solution:

The nth term of this series is $\displaystyle\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}$

By rationalizing the denominator,

$\displaystyle\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}\cdot \displaystyle\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{(n+1)\sqrt{n}-n\sqrt{n+1}}$

$=\displaystyle\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{((n+1)\sqrt{n})^2-(n\sqrt{n+1})^2}$

$=\displaystyle\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)^2-n^2(n+1)}$

Factor the common terms

$=\displaystyle\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)[(n+1)-n]}$

$=\displaystyle\frac{n\sqrt{n+1}-(n+1)\sqrt{n}}{n(n+1)(1)}$

$=\displaystyle\frac{(n+1)\sqrt{n}-n\sqrt{n+1}}{n(n+1)}$

$=\displaystyle\frac{(n+1)\sqrt{n}}{n(n+1)}-\displaystyle\frac{n\sqrt{n+1}}{n(n+1)}$

$=\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}$

This means that

$\displaystyle\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}$

Let n=1

$\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}=1-\displaystyle\frac{\sqrt{2}}{2}$

Let n=2

$\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}=\displaystyle\frac{\sqrt{2}}{2}-\displaystyle\frac{\sqrt{3}}{3}$

Let n=3

$\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}=\displaystyle\frac{\sqrt{3}}{3}-\displaystyle\frac{\sqrt{4}}{4}$

.

.

.

Let n=840

$\displaystyle\frac{\sqrt{n}}{n}-\displaystyle\frac{\sqrt{n+1}}{n+1}=\displaystyle\frac{\sqrt{840}}{840}-\displaystyle\frac{\sqrt{841}}{841}$

We can express

$\displaystyle\frac{1}{1\sqrt{2}+2\sqrt{1}}+\displaystyle\frac{1}{2\sqrt{3}+3\sqrt{2}}+\displaystyle\frac{1}{3\sqrt{4}+4\sqrt{3}}+\ldots +\displaystyle\frac{1}{840\sqrt{841}+841\sqrt{840}}$

As

$(1-\displaystyle\frac{\sqrt{2}}{2})+(\displaystyle\frac{\sqrt{2}}{2}-\displaystyle\frac{\sqrt{3}}{3})+(\displaystyle\frac{\sqrt{3}}{3}-\displaystyle\frac{\sqrt{4}}{4})+\ldots +(\displaystyle\frac{\sqrt{840}}{840}-\displaystyle\frac{\sqrt{841}}{841})$

Cancel out terms leaving

$1-\displaystyle\frac{\sqrt{841}}{841}=\displaystyle\frac{28}{29}$

Dan

Blogger and a Math enthusiast. Has no interest in Mathematics until MMC came. Aside from doing math, he also loves to travel and watch movies.

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1. August 1, 2014

[…] already learned about arithmetic, geometric as well as telescoping series. Today let’s learn how to evaluate series. If you’re thinking to create your own formula, this […]


text:
# How do you write a polynomial with zeros: 2, 4 + sqrt5, 4-sqrt5?

Feb 9, 2016

${x}^{4} - 14 {x}^{3} + 67 {x}^{2} - 130 x + 88$

#### Explanation:

Note that: if $a$ is a zero of a polynomial, then $\left(x - a\right)$ is a root of the polynomial. Applying this to the zeros at hand, we see that that polynomial is equal to

$\left(x - 2\right) \left(x - 4\right) \left(x - \left(4 + \sqrt{5}\right)\right) \left(x - \left(4 - \sqrt{5}\right)\right)$

We can simplify the first two through distributing. We will get to the second two, which is the more challenging part of this problem, but first distribute the minus sign.

$= \left({x}^{2} - 6 x + 8\right) \left(x - 4 - \sqrt{5}\right) \left(x - 4 + \sqrt{5}\right)$

$= \left({x}^{2} - 6 x + 8\right) \left(\left(x - 4\right) - \sqrt{5}\right) \left(\left(x - 4\right) + \sqrt{5}\right)$

Notice that the last two terms are in the form $\left(a - b\right) \left(a + b\right)$, which equals ${a}^{2} - {b}^{2}$. Here, $a = x - 4$ and $b = \sqrt{5}$. This gives us

$= \left({x}^{2} - 6 x + 8\right) \left({\left(x - 4\right)}^{2} - 5\right)$

$= \left({x}^{2} - 6 x + 8\right) \left({x}^{2} - 8 x + 16 - 5\right)$

$= \left({x}^{2} - 6 x + 8\right) \left({x}^{2} - 8 x + 11\right)$

Now, we can distribute these terms. There will be a lot, so hang on tight.

$= {\overbrace{{x}^{4} - 6 {x}^{3} + 8 {x}^{2}}}^{\left({x}^{2} - 6 x + 8\right) \cdot {x}^{2}} + {\overbrace{- 8 {x}^{3} + 48 {x}^{2} - 64 x}}^{\left({x}^{2} - 6 x + 8\right) \cdot - 8 x} + {\overbrace{11 {x}^{2} - 66 x + 88}}^{\left({x}^{2} - 6 x + 8\right) \cdot 11}$

Combining all the like terms ${x}^{3}$ with ${x}^{3}$, and so on, our result is

$= {x}^{4} - 14 {x}^{3} + 67 {x}^{2} - 130 x + 88$

Note that this polynomial can be multiplied by any constant, and the roots will be the same. However, this is the simplest form of the polynomial.