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Home » Math Theory » Fractions » Improper Fractions

# Improper Fractions

## Introduction

Numbers that represent a portion of a whole are called fractions. Let us use the square below as an example.

Since it has been cut into 4 equal parts, each piece represents one out of four. Mathematically, each piece is described as $\frac{1}{4}$. The number at the top is referred to as the numerator, and its bottom number is the denominator. $\frac{1}{4}$ is an example of a proper fraction because the denominator is greater than the numerator.

Otherwise, we will have an improper fraction when the numerator exceeds the denominator. Let us suppose that we instead have two squares and that each square is divided into four equal parts, as illustrated below.

There are 7 yellow parts out of 8 equal parts, which can be expressed as $\frac{7}{8}$. This number is an example of an improper fraction.

The topic of improper fractions will be explored in this article, along with methods for converting them to mixed fractions, simplifying them, and finding solutions to problems involving them.

## What is an Improper Fraction?

### Definition

When the numerator exceeds the denominator, the fraction is called improper. Suppose $\frac{a}{b}$ is an improper fraction; thus, a must be greater than b. Examples of improper fractions are $\frac{4}{3}$, $\frac{7}{4}$, $\frac{8}{5}$, and $\frac{11}{2}$.

A fraction has two main types, proper fractions and improper fractions. We have an improper fraction when the numerator is higher than the denominator; otherwise, it is a proper fraction.

## Simplifying Improper Fractions

Simplifying improper fractions is the same thing as reducing the fractions. We do this by dividing the denominator and the numerator by their greatest common factor.

Let us say, for example; we have the improper fraction $\frac{18}{12}$. The numerator in the fraction is 18, while the denominator is 12. These are the factors of 18 and 12.

Factors of 18: { 2, 3, 6, 9 }

Factors of 12: { 2, 3, 4, 6 }

From the list above, the common factors of 18 and 12 are 2, 3, and 6. To simplify the $\frac{18}{12}$, we will use 6 as the divisor since it is the greatest common factor. Hence, we have,

$\frac{18÷6}{12÷6}$ = $\frac{3}{2}$

Therefore, $\frac{18÷6}{12÷6}$ is $\frac{3}{2}$ when simplified.

## Improper Fraction vs Mixed Number

When the numerator exceeds the denominator, the fraction is called an improper fraction. A mixed number is a way of writing fractions that are greater than one. A whole number and a proper fraction make up a mixed number.

Examples of improper fractions are:

Examples of mixed numbers are:

## Improper Fraction Conversions

We can easily convert improper fractions to mixed numbers or decimals by following simple steps.

### Converting Improper Fractions to Mixed Numbers

We shall divide the numerator by the denominator to make an improper fraction into a mixed number. To complete the mixed number, the quotient is the whole number part; the numerator is the remainder, while the divisor is the denominator.

For example, let us say that we must convert the improper fraction $\frac{23}{7}$ to a mixed number. When we divide 23 by 7, we will get 3 with the remainder of 2. We can form 3 equal groups of 7 with the remainder of 2. The image below illustrates the division process.

To form the equivalent mixed number, we will use the quotient ( 3 ) as the whole number part, and the fractional part is $\frac{2}{7}$. Notice that we use the same denominator.

$\frac{23}{7}$ =3$\frac{2}{7}$

To check the conversion, we must multiply the denominator by the whole number part and then add the answer to the numerator. The denominator stays the same. Hence, we have,

( 7 × 3 ) + 2 = 21 + 2 = 23

Therefore, the improper fraction $\frac{23}{7}$ is equal to 3$\frac{2}{7}$ in mixed number.

For another example, let us convert $\frac{9}{2}$ to a mixed number. When we divide 9 by 2, we will have 4 with the remainder of 1. The figure below shows the division process.

To check the conversion, we must multiply the denominator by the whole number part and then add the answer to the numerator. The denominator stays the same. Hence, we have,

( 2 × 4 ) + 1 = 8 + 1 = 9

Therefore, the improper fraction $\frac{9}{2}$ is equal to 4$\frac{1}{2}$ in mixed number.

### Converting Improper Fractions to Decimals

To convert improper fractions to decimals, we must divide the numerator by the denominator. The process works the same way in converting improper fractions to mixed numbers.

For example, let us say that we must convert the improper fraction $\frac{11}{2}$ to decimals.

Therefore, $\frac{11}{2}$ is equal to 5.5 in decimals.

## Addition and Subtraction of Improper Fractions

When adding or subtracting improper fractions, it is important to identify if you are adding or subtracting like or unlike terms. Like fractions have the same denominators; otherwise, they are unlike fractions. For example, $\frac{4}{3}$ and $\frac{5}{3}$ are like improper fractions, while $\frac{5}{4}$ and $\frac{7}{6}$ are unlike improper fractions.

To add like improper fractions, we must add the numerators of the given fractions and then copy the same denominator. Simplify the answer when necessary.

Let us say, for example, let us add the like improper fractions $\frac{7}{4}$ and $\frac{9}{4}$. Hence we have,

$\frac{7}{4}$+$\frac{9}{4}$=$\frac{16}{4}$

Since $\frac{16}{4}$ can still be simplified, the sum of $\frac{7}{4}$ and $\frac{9}{4}$ is equal to 16.

For another example, let us say we have $\frac{11}{3}$+$\frac{14}{3}$. Since they have the same denominators, it is easy to get the sum of these fractions by adding the numerators and copying the same denominator.

$\frac{11}{3}$+$\frac{14}{3}$=$\frac{25}{3}$

Thus, the sum of $\frac{11}{3}$ and $\frac{14}{3}$ is equal to $\frac{25}{3}$.

To add unlike improper fractions, we must first make the given improper fractions into proper fractions, then proceed to the process of adding like fractions. Identifying the least common multiple ( LCM ) of the denominators in converting the improper fraction to proper fractions is important. Then, simplify the fraction when needed.

Let us say, for example, we have the $\frac{5}{2}$+$\frac{13}{6}$.

Let us first make the given fractions to like improper fractions by identifying the LCM of the denominators.

Multiples of 2 are: { 2, 4, 6, 10, 8, 12, 14, 16, 18, 20, … }

Multiples of 6 are: { 6, 12, 24, 18, 36 30, 40, … }

From the list, 6, 12, and 18 are the common multiples, but we have to get the least common multiple ( LCM ), which is 6.

Thus, to make the given improper fractions into proper fractions, we must identify the numbers to multiply to these fractions, making the denominators equal to 6.

Hence, we have,

$\frac{5}{2}$ × $\frac{3}{3}$ = $\frac{15}{6}$

$\frac{13}{6}$  × $\frac{1}{1}$ = $\frac{13}{6}$

We shall then add the fractions $\frac{15}{6}$ and $\frac{13}{6}$ and add like fractions.

$\frac{15}{6}$+$\frac{13}{6}$=$\frac{28}{6}$

Since $\frac{28}{6}$ can still be simplified since 28 and 6 are divisible by 2, we have $\frac{28÷2}{6÷2}$=$\frac{14}{3}$.

Therefore $\frac{5}{2}$+$\frac{13}{6}$=$\frac{14}{3}$.

### Subtracting Like Improper Fractions

To subtract like improper fractions, we must subtract the numerators of the given fractions and then copy the same denominator. Simplify the answer when necessary.

Let us say, for example, we subtract the like improper fractions $\frac{12}{5}$ and $\frac{9}{5}$. Hence we have,

$\frac{12}{5}$-$\frac{9}{5}$=$\frac{3}{5}$

Therefore, the difference between $\frac{12}{5}$ and $\frac{9}{5}$ is equal to $\frac{3}{5}$.

For another example, let us say we have $\frac{10}{7}-\frac{8}{7}$. Since they have the same denominators, it is easy to get the sum of these fractions by subtracting the numerators and copying the same denominator.

$\frac{10}{7}$-$\frac{8}{7}$=$\frac{2}{7}$

Thus, the difference between $\frac{10}{7}$ and $\frac{8}{7}$ is equal to $\frac{2}{7}$.

### Subtracting Unlike Improper Fractions

To subtract, unlike improper fractions, we must first make the given improper fractions into proper fractions, then proceed to the process of subtracting like fractions. Finding the denominators’ least common multiple (LCM) is crucial for converting an improper fraction to a proper fraction. Then, simplify the fraction when needed.

Let us say, for example, we have the $\frac{15}{4}-\frac{9}{5}$.

Let us first make the given fractions to like improper fractions by identifying the LCM of the denominators.

Multiples of 4 are: { 4, 8, 16, 12, 20, 24, 28, 32, 36, 40, … }

Multiples of 5 are: { 5, 15, 10, 20, 25, 30, 35, 40, … }

From the list, 20 and 40 are the common multiples of 4 and 5. but we have to get the least common multiple ( LCM ), which is 20.

Thus, to make the given improper fractions into proper fractions, we must identify the numbers to multiply to these fractions, which will make the denominators equal to 20.

Hence, we have,

$\frac{15}{4}$ × $\frac{5}{5}$ = $\frac{75}{20}$

$\frac{9}{5}$  × $\frac{4}{4}$ = $\frac{36}{20}$

We shall then subtract the fractions $\frac{45}{20}$ and $\frac{36}{20}$ and proceed to add like fractions.

$\frac{75}{20}$ – $\frac{36}{20}$ = $\frac{39}{20}$

Therefore $\frac{15}{4}$ – $\frac{9}{5}$ = $\frac{39}{20}$.

## Multiplying Improper Fractions

When multiplying improper fractions, we must obtain the product of the numerators of the provided fractions over the product of the denominators. And then simply the answer when necessary.

Let us say, for example; we have $\frac{7}{5}$ x $\frac{9}{8}$.

To get the product, we must multiply the numerators 7 and 9 and then multiply the denominators 5 and 8. Thus,

$\frac{7}{5}$ x $\frac{9}{8}$ =6340

Since 63 and 40 do not have common factors, $\frac{63}{40}$ is simplified already.

Thus, $\frac{7}{5}$ x $\frac{9}{8}$ is equal to $\frac{63}{40}$.

## More Examples

Example 1

Which of the following are improper fractions?

Solution

When the denominator exceeds the numerator, we have a proper fraction; otherwise, it is improper. Hence, the improper fractions above are as follows:

In each of the improper fractions, we have,

7 > 2 ; 11 > 7; 9 > 4
12 > 5 ; 5 > 3 ; 21 > 4

Example 2

Convert the following improper fractions to mixed numbers.

( a ) $\frac{7}{3}$ ( b ) $\frac{16}{5}$ ( c ) $\frac{27}{2}$ ( d ) $\frac{39}{4}$ ( e ) $\frac{50}{11}$

Solution

The improper fractions can be converted into mixed numbers by simply dividing the numerator by the denominator. To complete the mixed number, the quotient is the whole number part, while the numerator is the remainder. The denominator is the divisor or the same denominator of the given improper fraction.

( a ) $\frac{7}{3}$

When we divide 7 by 3, we will have 2 as the quotient and 1 as the remainder. Therefore, the improper fraction $\frac{7}{3}$ is equal to 2$\frac{1}{3}$ in mixed number.

( b ) $\frac{16}{5}$

When we divide 16 by 5, we will have 3 as the quotient and 1 as the remainder. Therefore, the improper fraction $\frac{16}{5}$ is equal to 3$\frac{1}{5}$ in mixed number.

( c ) $\frac{27}{2}$

When we divide 27 by 2, we will have 13 as the quotient and 1 as the remainder. Therefore, the improper fraction $\frac{27}{2}$ is equal to 13$\frac{1}{2}$ in mixed numbers.

( d ) $\frac{39}{4}$

When we divide 39 by 4, we will have 9 as the quotient and 3 as the remainder. Therefore, the improper fraction $\frac{39}{4}$ is equal to 9$\frac{3}{4}$ in mixed numbers.

( e ) $\frac{50}{11}$

When we divide 50 by 11, we will have 4 as the quotient and 6 as the remainder. Therefore, the improper fraction $\frac{50}{11}$ is equal to 4$\frac{6}{11}$ in mixed numbers.

Example 3

Write the following mixed numbers to improper fractions.

( a ) 2$\frac{3}{5}$ ( b ) 3$\frac{9}{11}$ ( c ) 7$\frac{4}{7}$ ( d ) 5$\frac{5}{9}$ ( e ) 12$\frac{1}{4}$

Solution

When we convert a mixed number into an improper fraction, we must multiply the denominator by the whole number part and then add the result to the numerator. The improper fraction and the mixed number have the same denominator.

( a ) 2$\frac{3}{5}$

We must multiply the denominator, 5, by the whole number part, 2. The product of 5 and 2 is10, then add the numerator, 3, which results in 13. Hence, we have ( 5 × 2 ) + 3 = 13. The answer gives the numerator of the improper fraction and then uses the same denominator.

2$\frac{3}{5}$= $\frac{(2×5)+3}{5}$=$\frac{13}{5}$

Therefore, the mixed number 2$\frac{3}{5}$ is equal to $\frac{13}{5}$ in improper fraction.

( b ) 3$\frac{9}{11}$

We must multiply the denominator, 11, by the whole number part, 3. The product of 3 and 11 is 33, then add the numerator 9, which results in 42. Hence, we have ( 11 × 3 ) + 9 = 42. The answer forms the numerator of the improper fraction and then copies the same denominator.

3$\frac{9}{11}$= $\frac{(11×3) +9}{11}$=$\frac{42}{11}$

Therefore, the mixed number 3$\frac{9}{11}$ is equal to $\frac{42}{11}$ in improper fraction.

( c ) 7$\frac{4}{7}$

We must multiply the denominator, 7, by the whole number part, 7. The product of 7 and 7 is 49, then add the numerator 4, which results in 53. Hence, we have ( 7 × 7 ) + 4 = 53. The answer forms the numerator of the improper fraction and then copies the same denominator.

7$\frac{4}{7}$= $\frac{(7×7) +4}{7}$=$\frac{53}{7}$

Therefore, the mixed number 7$\frac{4}{7}$ is equal to $\frac{53}{7}$ in improper fraction.

( d ) 5$\frac{5}{9}$

We must multiply the denominator, 9, by the whole number part, 5. The product of 5 and 9 is 45, then add the numerator, 5, which results in 50. Hence, we have ( 9 × 5 ) + 5 = 50. The answer forms the numerator of the improper fraction and then copies the same denominator.

5$\frac{5}{9}$= $\frac{(9×5) +5}{9}$=$\frac{50}{9}$

Therefore, the mixed number 5$\frac{5}{9}$ is equal to $\frac{50}{9}$ in improper fraction.

( e ) 12$\frac{1}{4}$

We must multiply the denominator, 4, by the whole number part, 12. The product of 4 and 12 is 48, then add the numerator 1, which results in 49. Hence, we have ( 4 × 12 ) + 1 = 49. The answer forms the numerator of the improper fraction and then copies the same denominator.

12$\frac{1}{4}$= $\frac{(4×12) +1}{4}$=$\frac{49}{4}$

Therefore, the mixed number 12$\frac{1}{4}$ is equal to $\frac{49}{4}$ in improper fraction.

Example 4

Perform the indicated operations and simplify.

( a ) $\frac{5}{3}$+$\frac{8}{3}$

( b ) $\frac{16}{5}$-$\frac{9}{5}$

( c ) $\frac{7}{4}$ x $\frac{5}{3}$

( c ) $\frac{5}{2}$ x $\frac{8}{3}$

Solution

( a ) $\frac{4}{3}$+$\frac{8}{3}$=$\frac{12}{3}$=4

( b ) $\frac{16}{5}$ – $\frac{9}{5}$=$\frac{7}{5}$

( c ) $\frac{7}{4}$ x $\frac{5}{3}$ = $\frac{35}{12}$

( c ) $\frac{5}{2}$ x $\frac{8}{3}$ = $\frac{40}{6}$ = $\frac{20}{3}$

## Summary

Numbers that represent a portion of a whole are called fractions.

When the numerator exceeds the denominator, the fraction is called an improper fraction. Suppose $\frac{a}{b}$ is an improper fraction; thus, a must be greater than b.

Examples of improper fractions are:

A mixed number is a way of writing fractions that are greater than one. A whole number and a proper fraction make up a mixed number.

Examples of mixed numbers are:

To create a mixed number from an improper fraction, divide the numerator by the denominator. To form the mixed equivalent mixed number, the quotient is the whole number part; the numerator is the remainder, while the divisor is the denominator.

Simplifying Improper Fractions

Simplifying improper fractions is the same thing as reducing the fractions. We do this by dividing the denominator and the numerator by their greatest common factor.

## Frequently Asked Questions on Improper Fractions ( FAQs )

### How do proper and improper fractions differ from one another?

Numbers that represent a portion of a whole are called fractions. Fractions are written with a numerator, a fraction bar, and a denominator. When the denominator is greater than the numerator, we have a proper fraction; otherwise, it is improper.

Examples of proper fractions ( Numerator <  Denominator )

Examples of improper fractions ( Numerator > Denominator )

### How do you convert mixed numbers to improper fractions?

A mixed number has two parts: the whole number and the proper fraction. To make a mixed number into an improper fraction, we must multiply the denominator by the whole number part, then add the answer to the numerator. The denominator stays the same.

For example, let us say that we must convert the mixed number 5$\frac{2}{3}$ to an improper fraction. We will have,

( 3× 5 ) + 2  = 15 + 2 = 17 , this is the numerator of the improper fraction

Therefore, the mixed number 5$\frac{2}{3}$ is equal to $\frac{17}{3}$.

### How do like and unlike fractions

differ from one another?

If two fractions have the same denominator, they are said to be like; otherwise, they are unlike.

Examples of like fractions ( same denominators )

$\frac{5}{2}$ and $\frac{3}{2}$ are like fractions.

$\frac{9}{7}$ and $\frac{10}{7}$ are like fractions.

Examples of unlike fractions ( different denominators )

$\frac{11}{3}$ and $\frac{11}{2}$ are unlike fractions.

$\frac{12}{5}$ and $\frac{16}{9}$ are unlike fractions.

### How do we simplify improper fractions?

When simplifying improper fractions is the same thing as reducing the fractions. We do this by dividing the denominator and the numerator by their greatest common factor.

Let us say, for example; we have the improper fraction $\frac{50}{10}$. Since the greatest common factor of 10 and 50 is 10, we have,

$\frac{50÷10}{10÷10}$ = $\frac{5}{1}$ = 5

Therefore, the simplified form of  $\frac{50}{10}$ is 5.

### How do we add and subtract improper fractions?

When adding or subtracting improper fractions, it is important to identify if you are adding or subtracting like or unlike terms. Like fractions have the same denominators; otherwise, they are unlike fractions. For example, $\frac{4}{3}$ and $\frac{5}{3}$ are like improper fractions, while $\frac{5}{4}$ and $\frac{7}{6}$ are unlike improper fractions.

To add like improper fractions, we must add the numerators of the given fractions and then copy the same denominator. Simplify the answer when necessary.

When adding unlike improper fractions, it is important to make the fractions proper using the least common multiple ( LCM ) of the denominators. Then, proceed to the process of adding like fractions.

### How to multiply improper fractions?

When multiplying improper fractions, we must obtain the product of the numerators of the provided fractions over the product of the denominators. And then simply the answer when necessary.

Let us say, for example; we have $\frac{10}{2}$ x $\frac{5}{3}$.

To get the product, we must multiply the numerators 10 and 5, then multiply the denominators 2 and 3. Thus,

$\frac{10}{2}$ × $\frac{5}{3}$ = $\frac{50}{6}$

Since 2 is the greatest common factor of 50 and 6,  $\frac{50}{6}$ can be simplified as $\frac{25}{3}$.Thus, $\frac{10}{2}$ × $\frac{5}{3}$ is equal to $\frac{25}{3}$.


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## Boundary RG flows for fermions and the mod 2 anomaly

Philip Boyle Smith, David Tong

SciPost Phys. 10, 010 (2021) · published 19 January 2021

### Abstract

Boundary conditions for Majorana fermions in d=1+1 dimensions fall into one of two SPT phases, associated to a mod 2 anomaly. Here we consider boundary conditions for 2N Majorana fermions that preserve a $U(1)^N$ symmetry. In general, the left-moving and right-moving fermions carry different charges under this symmetry, and implementation of the boundary condition requires new degrees of freedom, which manifest themselves in a boundary central charge, $g$. We follow the boundary RG flow induced by turning on relevant boundary operators. We identify the infra-red boundary state. In many cases, the boundary state flips SPT class, resulting in an emergent Majorana mode needed to cancel the anomaly. We show that the ratio of UV and IR boundary central charges is given by $g^2_{IR} / g^2_{UV} = {\rm dim}\,({\cal O})$, the dimension of the perturbing boundary operator. Any relevant operator necessarily has ${\rm dim}({\cal O}) < 1$, ensuring that the central charge decreases in accord with the g-theorem.

### Authors / Affiliation: mappings to Contributors and Organizations

See all Organizations.
Funders for the research work leading to this publication

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# Studio One 4 and above with original LX/iX controllers

Nektar products are compatible with Studio One 4 and above. The latest installers for the Impact GX, LX+, and SE25/SE49 controllers can be found on your account page.

Please follow the instructions below if you are using an original Impact LX or iX controller.

## Original Impact Manual Installation Instructions – Windows

...
1. Copy the ‘Nektar’ folder from the .zip file above and paste it into the following directory, as appropriate for your version:
• C:\Program Files\PreSonus\Studio One 4\devices
or
• C:\Program Files\PreSonus\Studio One 5\devices

Once you have done that, your Impact controller will be detected automatically the next time you start Studio One.


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# What is the double bond equivalence of the following compound?

What is the double bond equivalence of the given compound? The given answer is 14 (8 pi electrons and 6 rings = 14 DBE). But why isn't the lone pair on nitrogen considered?? The lone pair on nitrogen involves in resonance with the other 2 bonds making the ring as aromatic... so shouldn't we consider that lone pair also???

• If some double bonds are in resonance, that does not make more or fewer of them. It is still the same number of double bonds. Resonance is about as irrelevant as the weather in the street outside. So is the lone pair of N. – Ivan Neretin Apr 2 at 12:00
• Thank you so much – Srini Apr 2 at 13:17

As Ivan Neretin pointed out in the comments, $$\pi$$ bonds will remain same whether or not in resonance.
However, if you have trouble counting the double bond equvalent, you can directly use the formula for it given as $$\ce{DU=\dfrac{2C+2-H+N-X}{2}}$$ so here number of carbons are C=23, H=21, N=1 which also gives 14 as the answer.


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 Analysis revealed that AIDS is most often constructed as integral to community and activism in the AIDS Culture. The buddy relationship is most often constructed as a friendship rather than a caregiver/client relationship. The participants also revealed that buddy dyads are both life-affirming and significant relationships. The examination of buddies' metaphors and descriptors further suggests that their lived experience with AIDS is uniquely different than the construction of AIDS most often made by the media and the rest of U.S. culture.

• #### Ways To Help And Ways To Hinder: Climate, Health, And Food Security In Alaska

This dissertation explores various ecological, socioeconomic, sociopolitical, and biophysical dimensions food security in Alaska. The context for this work is dramatic climatic change and ongoing demographic, socioeconomic and cultural transitions in Alaska's rural and urban communities. The unifying focus of the papers included here are human health. I provide multiple perspectives on how human health relates to community and ecosystem health, and of the roles of managers, policy makers, and researchers can play in supporting positive health outcomes. Topics include methylmercury (MeHg) contamination of wild fish, the impacts of changes to Alaskan landscapes and seascapes on subsistence and commercial activities, and on ways to design sustainable natural resource policies and co-management regimes such that they mimic natural systems. The operating premise of this work is that sustainability is ostensibly a matter of human health; the finding is that human health can provide a powerful point of integration for social and ecological sustainability research.


text:
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The four sentences (labelled $1, 2, 3, 4$) given below, when properly sequenced would yield a coherent paragraph. Decide on the proper sequence of the order of the sentences and key in the sequence of the four numbers as your answer.

1. Such a belief in the harmony of nature requires a purpose presumably imposed by the goodness and wisdom of a deity
2. These parts, all fit together into an integrated, well-ordered system that was created by design
3. Historically, the notion of a balance of nature is part observational, part metaphysical, and not scientific in any way
4. It is an example of an ancient belief system called teleology, the notion that what we call nature has a predetermined destiny associated with its component parts

edited | 55 views

The right order is CDBA.

Remember, the right order is always the most coherent paragraph.

Here, Option C is the most apt opening argument. Option D follows this by defining teleology. Option B follows this, and the entire paragraph is concluded by Option A.

by (874 points) 1 3 5


text:
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# reaction of kmno4 with h2o2 in acidic medium

The skeleton reaction – KMnO 4 + H 2 SO 4 + Na 2 SO 3 = K 2 SO 4 + MnSO 4 + Na 2 SO 4 + H 2 O. Balance redox equations using the ion-electron method in an acidic solutions. Click hereto get an answer to your question  Hydrogen peroxide in its reaction with KIO4 and NH2OH respectively is acting as: Sulfuric acid is a mineral acid with molecular formula H₂SO₄. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. Home Reactions Blog. ... Picture of reaction: Сoding to search: 3 H2O2 + 2 KMnO4 = 3 O2 + 2 MnO2 + 2 KOH + 2 H2O. The half equations for this reaction is : Fe2+ ---> Fe3 + e-MnO4- + 8H + 5e- ---> Mn2 + 4H2O. As part of the introduction to a lab dealing with the kinetics of the reaction between KMnO4 and H2O2 in acidic medium, we look at the reaction if it is not deliberately made acidic. For decolourisation of 1 mole of kmno4, the moles of h2o2 required is 2 See answers mindfulmaisel mindfulmaisel The balance chemical equation for the reaction of with can be written as: Here 5 moles of react with 2 moles of . C. 5 to 2. What products are formed when the following compound is treated with $Br_2$ in the presence of $FeBr_3?$. Which of the following molecules has the maximum dipole moment? Solution containing MnO 4 – ions are purple in colour and the solution containing Mn 2+ ions are colourless and hence permanganate solution is decolourised when added to a solution of a reducing agent. AIPMT 2014: The reaction of aqueous KMnO4 with H2O2 in acidic conditions gives (A) Mn4+ and O2 (B) Mn2+ and O2 (C) Mn2+ and O3 (D) Mn4+ and M Hence, aqueous solution of KMnO4 with H2O2 yields Mn2+ and O2 in acidic … Among the transition metals of 3d series, the one that has the highest negative $M^{2+} /M$ standard electrode potential is, If $Cu^{2+}$ ions is treated with excess of $KCN$, the complex formed is. What is the magnitude of this change? Identify compound X in the following sequence of reactions: Identify a molecule which does not exist. For the best answers, search on this site https://shorturl.im/axu6J. Classification of Elements and Periodicity in Properties, The reaction of aqueous KMnO4 with H2O2 in acidic conditions gives, Artificial sweetener which is stable under cold conditions only is. All reactants and products must be known. wt. 3 However, the product is predominantly octanoic acid, with only a small amount of aldehyde, resulting from overoxidation. * 2 KMnO4 + 3 H2SO4====> K2SO4+ 2MnSO4 + 3 H2O + 5(O) 1. In the reaction of KMnO4 with an oxalate in acidic medium, MnO4 is reduced to Mn2+ and C2O42- is oxidised to ... M H2C2O4 (d) 25 mL of 0.1 M H2C2O4 #NEETCHEMISTRY #IITJEECHEMISTRY The reaction when one mole of FeC2O4 is oxidised by how many no of mole KMnO4 in an acidic medium? Classification of Elements and Periodicity in Properties, The reaction of aqueous KMnO4 with H2O2 in acidic conditions gives, Artificial sweetener which is stable under cold conditions only is. Become a Patron! These requests choose an electrical potential lower than the permanganate's one. This method is suitable for measuring aqueous solutions of H2O2 ranging from 0.25 to 70 wt.%. which generically would be: 2 KMnO4 + … When acid (H2SO4) is added the reaction proceeds to Mn2+ with the production of O2 gas. After doing this experiment in my school I have a few questions. Using the Gibbs energy change, $\Delta G^\circ = +63.3kJ$ The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. 1.630 g of iron ore is dissolved in an acidic solution. #NEETCHEMISTRY #IITJEECHEMISTRY The reaction when one mole of FeC2O4 is oxidised by how many no of mole KMnO4 in an acidic medium? Balance the following redox equation using the half-reaction method: MnO41 H2O2 Mn2+ +O2 (acidic medium) Reduction half Mno Mn Oxidation hatf H20, O2 2f 8H MnO H20, 02° 2 t 4H2O No + 2Ht 2/8H 5e 5 (He O 2.1 T + MnOy Mn 4 H2O) + O2 4 2H 2e ) IoHt 10e + 2 Mna 50, t 2Mn2 8 H,O 5 H2O + + 8 HOt50, + 1OH 2Mn2 +8 H20 t50, + IOH lo Ht 2 MnOy- 6. What products are formed when the following compound is treated with $Br_2$ in the presence of $FeBr_3?$. Thermodynamic properties of substances The solubility of the substances Periodic table of elements. Find another reaction. asked Feb 14, 2020 in Chemistry by SurajKumar ( 66.2k points) 1. Check Answer and Solution for above question from Chemistry in The d-and f-Block Elements - Tardigrade 6 to 2. Alcohols Primary alcohols such as octan-1-ol can be oxidized efficiently by KMnO 4 , in the presence of basic copper salts. The atomic radiusis: Find out the solubility of $Ni(OH)_2$ in 0.1 M NaOH. In this reaction, it acts as an acidic medium only. Question 4. In acidic solution, MnO 4 – + 8H + + 5e – → Mn 2+ + 4H 2 O. 1 decade ago . Question 3. Calculate the energy in joule corresponding to light of wavelength 45 nm (Planck's constant, $h=6.63\times10^{-34}$ Js; speed of light, $c=3\times 10^{8}$. Half equations for H2O2 for its reducing and oxidising nature in acidic and alkaline medium. Which one of the following species has plane triangular shape? Add / Edited: 27.09.2014 / Evaluation of information: 5.0 out of 5 / number of votes: 1. Potassium dichromate and hydrogen peroxide reaction in acidic medium. Hence, Eq. Any substance which reduces KMnO4 under acidic conditions will produce a positive interference. Click hereto get an answer to your question  (a) 4→3 280 ml of H202 reacts completely with 15.8 g of KMno4 in acidic medium, then the volume strength of H2O2 is : … 3. Hydrogen peroxide react with potassium permanganate to produce oxygen, manganese(IV) oxide potassium hydroxide and water. Reaction between acetone and methyl magnesium chloride followed by hydrolysis will give : Identify the correct statements from the following: Which of the following set of molecules will have zero dipole moment ? if you do not add an acid , your answer is: 2KMnO4 + H202 => 2KOH + MnO2 +O2 ===== but if it was done in an acidic medium: such as with sulfuric acid: 2 KMnO4 + 3 H2SO4 + 5 H2O2 … Log in: Chemical reactions Сhemical tables. The reaction of aqueous KMnO4 with H2O2 in acidic medium is 3H2SO4 + KMnO4 +5 H2O2 --> 5O2 +2MnSO4 +8H2O +K2SO4In the reaction, KMnO4 oxidise H2O2 to O2 and itself [MnO4-] gets reduced to Mn2+ ions as MnSO4. Share 5. $Ag_2CO_3 \rightleftharpoons 2Ag^+_{(aq)}+CO^{2^−}_{3(aq)}$ Which of the following ions has the same number of unpaired electrons as those present in $V^{3+}$? Paper by Super 30 Aakash Institute, powered by embibe analysis.Improve your score by 22% minimum while there is still time. H2O2 + 2H2O + 2e- → 2H2O + 2OH-Step 5. KMnO 4 can oxidize many inorganic compounds. Chemistry. MEDIUM. H2O2 + 2H2O + 2e- → 2H2O + 2OH-Step 5. Among the following sets of reactants which one produces anisole? K 2 Cr 2 O 7 + 3H 2 O 2 + 4H 2 SO 4 = Cr 2 (SO 4) 3 + 3O 2 + K 2 SO 4 + 7H 2 O. Chemical reaction. That is a reaction between potassium permanganate and hydrogen peroxide. So, the reaction between Mohr’s salt and KMnO 4 is a redox reaction in which oxidation and reduction takes place simultaneously. of KMnO4 in neutral or weakly alkaline medium Mol. The atomic radiusis: Find out the solubility of $Ni(OH)_2$ in 0.1 M NaOH. These requests choose an electrical potential lower than the permanganate's one. Image Transcriptionclose. B. of moles (n) × nf = normality × volume (In litres) Kmn04 is an oxiding agent and will oxide O- of H2O2 to O2 (oxidation state=0). In such a case the solutions of KMnO4 and H2O2 are slightly basic and the rapid formation of solid MnO2 occurs. Balance the following redox equation using the half-reaction method: MnO41 H2O2 Mn2+ +O2 (acidic medium) Reduction half Mno Mn Oxidation hatf H20, O2 2f 8H MnO H20, 02° 2 t 4H2O No + 2Ht 2/8H 5e 5 (He O 2.1 T + MnOy Mn 4 H2O) + O2 4 2H 2e ) IoHt 10e + 2 Mna 50, t 2Mn2 8 H,O 5 H2O + + 8 HOt50, + 1OH 2Mn2 +8 H20 t50, + IOH lo Ht 2 MnOy- 6. Write down the unbalanced equation ('skeleton equation') of the chemical reaction. Reaction of kmno4 with oxalic acid in acidic medium Ask for details ; Follow Report by Khannaankit6132 12.06.2019 Log in to add a comment 2KMnO 4 + 5H 2 C 2 O 4 + 3H 2 SO 4 → 2MnSO … 2. In basic medium hydrogen peroxide reacts with KMnO4 giving manganese dioxide,2 KMnO4 + 3 H2O2 →2 MnO2 + 3 O2 + 2 KOH + 4 H2OIn acidic medium hydrogen peroxide reacts with K2Cr2O7 giving chromic sulphateK2Cr2O7 +4 H2SO4 + 3 H2O2 → K2SO4 + Cr2(SO4)3 + 7 H2O + 3 O2 AIPMT 2014: The reaction of aqueous KMnO4 with H2O2 in acidic conditions gives (A) Mn4+ and O2 (B) Mn2+ and O2 (C) Mn2+ and O3 (D) Mn4+ and MnO2 . The medium of solution plays an important role in determining the products of reaction. H 2 O 2 + KMnO 4 → MnO 2 + KOH + O 2 + H 2 O Step 2. How many mL of KMnO 4 (used in question 2) should be used for 2.00 g H 2 O 2 titration?. 2 FeSO4 + H2SO4=> Fe2(SO4)3 + H2O So here KMnO4 will get reduced to MnSO4. Calculate the energy in joule corresponding to light of wavelength 45 nm (Planck's constant, $h=6.63\times10^{-34}$ Js; speed of light, $c=3\times 10^{8}$. Most of the reactions with Potassium permanganate are redox reactions (a chemical reaction in which one substance is oxidized and another is reduced). Picture of reaction: Сoding to search: 3 H2O2 + 2 KMnO4 = 3 O2 + 2 MnO2 + 2 KOH + 2 H2O. On electrolysis of dil.sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be: An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. Given that the ionic product of $Ni(OH)_2$ is $2 \times 10^{-15}$. Convert the following redox reactions to the ionic form. Name the elements which can be used in the production of alloy steel. Hydrogen peroxide react with potassium permanganate and sulfuric acid H2O2 + H2SO4 + KMnO4 = O2 + H2 + MnSO4 + K2SO4 or, After balancing the Redox reaction 5 H 2 O 2 + 3 H 2 SO 4 + 2 KMnO 4 = 5 O 2 + 8 H 2 O + 2 MnSO 4 + K 2 SO 4 In acidic medium it changes from purple to faint permanent pink colour due Mn(2+), (5-electron change). 30g of KMnO4 sample containing inert impurity completely reacted with 60 ml of 56 volume of H2O2 in acidic medium, then what is the percentage purity of KMNO4? In the above reaction, K M n O 4 o x i d i s e s H 2 O 2 t o O 2 and itself [ M n O 4 −] gets reduced to M n 2 + ion as M n S O 4. The first evidence is derived from competition experiments. 2. In such a case the solutions of KMnO4 and H2O2 are slightly basic and the rapid formation of solid MnO2 occurs. 5. Sodium sulfite reacts with potassium permanganate and sulfuric acid. The Sodium sulfite reacts with potassium permanganate and sulfuric acid to form the potassium sulfate, Manganese(II) sulfate, water as well as sodium sulfate. is 28) in BM unit is, When $CuSO_4$ solution is treated with an excess of $KCN$, a colourless complex salt in which copper has a $d^{10}$ configuration is obtained. Steve O. Lv 7. Among the following sets of reactants which one produces anisole? $Be^{2+}$ is isoelectronic with which of the following ions? Corrosion is nessesarily a chemical reaction where what all changes occur and its cell reaction is... Give two examples of interstitial compounds. 158 52.67 3 3 Some oxidizing properties of KMnO4 in the neutral medium: These are given below: (i) It oxidizes hot manganous sulphate to manganese dioxide. Which of the following transition metals of 3 series has the lowest melting point? HCHO will be oxidized to formate / formic acid. In acidic or alkaline media you may note the oxygen evolvement, in respect of electrical potential comparison. It is a colorless, odorless, and syrupy liquid that is soluble in water. Which of the following will not be soluble in sodium hydrogen carbonate? $\ce{H2SO4}$ is stable to oxidization. Answer. reaction of kmno4 and h2o2 in acidic medium gives Share with your friends. if you do not add an acid , your answer is: 2KMnO4 + H202 => 2KOH + MnO2 +O2 ===== but if it was done in an acidic medium: such as with sulfuric acid: 2 KMnO4 + 3 H2SO4 + 5 H2O2 ==> K2SO4 + 2MnSO4 + 8H2O +5O2. 1. 7 to 2. a 1 M 1 V 1 = a 2 M 2 V 2. Check Answer and Solution for above question from Chemistry in The d-and f-Block Elements - Tardigrade Hydrogen peroxide react with potassium permanganate and sulfuric acid 5H 2 O 2 + 2KMnO 4 + 3H 2 SO 4 → 5O 2 + 2MnSO 4 + K 2 SO 4 + 8H 2 O [ Check the balance ] Sulfuric acid acts as a catalyst in the reaction $\ce{MnO4- + 8H+ + 5Fe2+ -> 5Fe3+ + Mn2+ + 4H2O}$ It initiates a 5e transfer in the ionic equation It used rather than $\ce{HCl}$ because $\ce{HCl}$ is volatile acid which will oxidize to $\ce{Cl}$ in presence of $\ce{KMnO4}$ (strong oxidizing agent). → K 2 MnO 4 – + 8H + + 5e – → Mn 2+ coefficients of species..., 8 months ago write the reaction being Mno4- +8H+5e -- -- >!, acidified KMnO4 oxidises FeSO4 to Fe2 ( SO4 ) 3 + H2O So here will... Acidic conditions will produce a positive interference formed when the following compound treated... 3 and it itself gets reduced to Mn 2+ May 3 '19 At 22:38 OH _2... Will get reduced to MnSO4 change ) decomposes to give Mn2+ ions which impart pink colour the. We use the ion-electron technique under the condition of stress which stimulates glycogenolysis the... Are formed when the following compound is treated with $Br_2$ in the reduction half-reaction present $... ( H2SO4 ) is added reaction of kmno4 with h2o2 in acidic medium reaction is as follows: 2KMnO 4 2MnSO. Minimum while there is still time H2O2 reduced 20 mL of KMnO →! Not be soluble in water the ionic form determining the products of reaction of solid MnO2 occurs gas. Salt while hydrogen peroxide in sulfuric acid is because in acidic medium is stable to oxidization: 5.0 out 5. Medium kmn04 decomposes to give Mn2+ ions which impart pink colour to the product... Oxidises FeSO4 to Fe2 ( SO4 ) 3 + H2O So here KMnO4 will reduced... + KMnO 4, in the production of alloy steel reaction will take 10 - 180 minutes depending reaction! What is the chemical reaction hydrogen carbonate H2SO4==== > K2SO4+ 2MnSO4 + 3 +! What is the chemical reaction acid, with only a small amount of aldehyde, resulting from.... Powered by embibe analysis.Improve your score by 22 % minimum while there is still time decomposition: when potassium! Of$ Ni ( OH ) _2 $is stable to oxidization balance redox equations using ion-electron. Potential comparison balance redox equations using the ion-electron technique molecules has the lowest point! Reaction being Mno4- +8H+5e -- -- - > Mn2+ + 4H2O redox reaction: and... Of can be shown as: hence, are required for one of. Reaction, it acts as an acidic medium the oxidising ability of 4. Of human beings with$ Br_2 $in the presence of basic copper salts and syrupy that..., acidified KMnO4 oxidises FeSO4 to Fe2 ( SO4 ) 3 + H2O So here KMnO4 will get to! Is represented by the following transition metal ions shows magnetic moment of BM... In this reaction, we use the ion-electron technique ranging from 0.25 to 70 wt. % role determining... O2 gas, 8 months ago reaction between potassium permanganate ( KMnO4 ) by hydrogen reaction... Number of unpaired electrons as those present in$ V^ { 3+ } $) _2$ in M... So here KMnO4 will get reduced to MnSO4 and sulfuric acid to produce manganese ( II ),... Minimum while there is still time Mn 2+ O2 gas because in acidic solution, MnO 4 – + +. Reaction with acid: when permanganate reacts with concentrated hydrochloric acid it produces chlorine should. For a better result write the reaction will take 10 - 180 minutes depending on temperature! Method is suitable for measuring aqueous solutions of H2O2 reduced 20 mL of KMnO4. Thermodynamic properties of substances the solubility of $Ni ( OH ) _2$ in the reduction.! For its reducing and oxidising nature in acidic medium kmn04 decomposes to give Mn2+ which! Acidified KMnO4 oxidises FeSO4 to Fe2 ( SO4 ) 3 and it itself gets reduced to.. A molecule which does not exist 0.1M KMnO4 solution in acidic medium it changes from purple to permanent. While hydrogen peroxide ( H2O2 )? following compound is treated with $Br_2$ in 0.1 M.! Must be equal the electrons gained in the oxidation half-reaction must be equal the electrons lost in the reduction.! Of $Ni ( OH ) _2$ in the oxidation half-reaction must be the! Ask question Asked 6 years, 8 months ago liver of human beings will produce a positive.! Of basic copper salts liver of human beings write down the unbalanced equation ( 'skeleton equation )... $Br_2$ in the following salts is coloured salt before heating (... Manganese ( II ) sulfate, potassium sulfate and water oxidation reactions KMnO4. Resulting from overoxidation '19 At 22:38 product is predominantly octanoic acid, with only a small amount aldehyde. In terms of molarity the following will not be soluble in water dichromate and hydrogen peroxide is... > Mn2+ + 4H2O redox reaction, it acts as an acidic medium the balance chemical equation 1. ) should be used in question 2 ) should be used for 2.00 H. To 70 wt. % Mno4- +8H+5e -- -- - > Mn2+ + 4H2O redox reaction it... Of 3 series has the same number of votes: 1 octan-1-ol can be shown as: hence are! School I have a few questions when one mole of, we use ion-electron! Oxidising nature in acidic medium only are slightly basic and the rapid formation solid... 8 H 2 O 2 + KMnO 4 ( used in question 2 ) should reaction of kmno4 with h2o2 in acidic medium used for g! As an acidic solutions: when solid potassium permanganate and sulfuric acid a... Doing this experiment in my school I have a few questions calculate the of! Reaction with acid: when solid potassium permanganate and sulfuric acid → 2MnSO … Image Transcriptionclose,... 2 + KMnO 4 → MnO 2 + 2 M n S O 4 + 2! Of KMnO 4, in the liver of human beings this method the. Main PRE-AIPMT 2016 table of elements oxidation number of votes: 1 formed when the sets... Rapid formation of solid MnO2 occurs added the reaction between potassium permanganate react with potassium permanganate KMnO4... S ) + O 2 titration? 0.1 M NaOH potassium dichromate solution is to... To 70 wt. % following sequence of reactions: identify a molecule which does not exist of! Months ago + 4H 2 O 2 + H 2 O determining the products of reaction not soluble. W/V of H2O2 ( hydrogen peroxide is oxidized to green colour chromium salt while hydrogen peroxide ( H2O2?... ) sulfate, carbon dioxide, potassium sulfate and water reaction will take 10 - 180 minutes depending on temperature. Many mL of KMnO 4 in terms of molarity the following redox reactions to the solution /! K 2 MnO 4 – + 8H + + 5e – → Mn 2+ the balance chemical for... Ability of KMnO 4, in the acid medium, the oxidation number of votes:.... To permanganate action that the ionic form, pH, and syrupy that... Hydrogen peroxide react with potassium permanganate and hydrogen peroxide react with oxalic acid acidic. In 0.1 M NaOH equations for H2O2 for its reducing and oxidising in. Hydrochloric acid it produces chlorine by Super 30 Aakash Institute, powered by embibe analysis.Improve your score by %... Produces chlorine 2OH-Step 5 because in acidic solution volume strength of 3 series has same. Ions shows magnetic moment of 5.92 BM made with KMnO 4, in the acid medium, product! K2So4+ 2MnSO4 + 3 H2SO4==== > K2SO4+ 2MnSO4 + 3 H2O + 5 ( O 1..., acidified KMnO4 oxidises FeSO4 to Fe2 ( SO4 ) 3 + H2O So here KMnO4 will reduced... On reduction of potassium permanganate and sulfuric acid ( 5-electron change ) pink colour the! Criterion states some requests about the half-reaction coupled to permanganate action integers producing the lowest melting?. The reduction half-reaction balance redox equations using the ion-electron method in an medium... Primary alcohols such as octan-1-ol can be oxidized efficiently by KMnO 4 terms. M 2 V 2 made with KMnO 4 is represented by the following ions the presence of $Ni OH... Formation of solid MnO2 occurs volume strength of 3 series has the melting! Reaction: KMnO4 and Fe2+ 2 S O 4 + 5H 2 C 2 O 4 + 8 2. Acidic solutions reddish brown precipitate of MnO2 ( 3- electron change ) H2O2 )? NEETCHEMISTRY IITJEECHEMISTRY! 2Mnso … Image Transcriptionclose hydrochloric acid it produces chlorine Evaluation of information: 5.0 out of /... The ionic product of$ FeBr_3? $acidic medium titration? '19 At.! What products are formed when the following sets of reactants which one anisole. Down the unbalanced equation ( 'skeleton equation ' ) of the following molecules has the maximum dipole moment ) hydrogen... To make the two equal, multiply the coefficients of all species by integers producing the common... ) _2$ is $2 \times 10^ { -15 }$ ion ( At → MnO (..., acidified KMnO4 oxidises FeSO4 to Fe2 ( SO4 ) 3 and it gets... -- -- - > Mn2+ + 4H2O redox reaction, it acts as an medium! Add / Edited: 27.09.2014 / Evaluation of information: 5.0 out of 5 / number of unpaired electrons those... + K 2 MnO 4 – + 8H + + 5e – → Mn 2+ + 4H 2 O in. Produce manganese ( II ) sulfate, potassium sulfate and water which impart pink due. Lowest common multiple between the half-reactions depending on reaction temperature, pH, and syrupy that! Pink colour to the solution ) 3 + H2O So here KMnO4 will reduced... Is represented by the following ions molecule which does not exist following sets of reactants which one of the Periodic... Soluble in water acidified orange colour potassium dichromate and hydrogen peroxide in sulfuric acid species has plane triangular?!


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# k¶

• Available in: K-Means, PCA, GLRM

• Hyperparameter: yes

## Description¶

### K-Means¶

In K-Means, a “cluster” refers to groups of data in a dataset that are similar to one another. The K-Means algorithm finds clusters and cluster centers in a set of unlabled data.

This option specifies the maximum number of clusters that K-Means will form. If estimate_k is disabled, the model will find $$k$$ centroids, otherwise it will find up to $$k$$ centroids. This parameter is required and has no default value.

### PCA¶

In PCA, k specifies the rank of matrix approximation. This can be a value from 1 to 9 and defaults to 1.

## Example¶

library(h2o)
h2o.init()

# import the seeds dataset:
# this dataset looks at three different types of wheat varieties
# the original dataset can be found at http://archive.ics.uci.edu/ml/datasets/seeds
seeds <- h2o.importFile("https://s3.amazonaws.com/h2o-public-test-data/smalldata/flow_examples/seeds_dataset.txt")

# set the predictor names
# ignore the 8th column which has the prior known clusters for this dataset
predictors <- colnames(seeds)[-length(seeds)]

# split into train and validation
seeds_splits <- h2o.splitFrame(data = seeds, ratios = 0.8, seed = 1234)
train <- seeds_splits[[1]]
valid <- seeds_splits[[2]]

# try using the k parameter:
# build the model with three clusters
seeds_kmeans <- h2o.kmeans(x = predictors, k = 3, training_frame = train, validation_frame = valid, seed = 1234)

# print the total within cluster sum-of-square error for the validation dataset
print(paste0("Total sum-of-square error for valid dataset: ", h2o.tot_withinss(object = seeds_kmeans, valid = T)))

# select the values for k to grid over:
# Note: this dataset is too small to see significant differences between these options
# the purpose of the example is to show how to use grid search with k if desired
hyper_params <- list( k = c(2, 3, 7) )

# this example uses cartesian grid search because the search space is small
# and we want to see the performance of all models. For a larger search space use
# random grid search instead: list(strategy = "RandomDiscrete")
grid <- h2o.grid(x = predictors, training_frame = train, validation_frame = valid,
algorithm = "kmeans", grid_id = "seeds_grid", hyper_params = hyper_params,
search_criteria = list(strategy = "Cartesian"), seed = 1234)

## Sort the grid models by TotSS
sorted_grid <- h2o.getGrid("seeds_grid", sort_by  = "tot_withinss", decreasing = F)
sorted_grid

import h2o
from h2o.estimators.kmeans import H2OKMeansEstimator
h2o.init()

# import the seeds dataset:
# this dataset looks at three different types of wheat varieties
# the original dataset can be found at http://archive.ics.uci.edu/ml/datasets/seeds
seeds = h2o.import_file("https://s3.amazonaws.com/h2o-public-test-data/smalldata/flow_examples/seeds_dataset.txt")

# set the predictor names
# ignore the 8th column which has the prior known clusters for this dataset
predictors = seeds.columns[0:7]

# split into train and validation sets
train, valid = seeds.split_frame(ratios = [.8], seed = 1234)

# try using the k parameter:
# build the model with three clusters
# initialize the estimator then train the model
seeds_kmeans = H2OKMeansEstimator(k = 3, seed = 1234)
seeds_kmeans.train(x = predictors, training_frame = train, validation_frame=valid)

# print the total within cluster sum-of-square error for the validation dataset
print("Total sum-of-square error for valid dataset:",seeds_kmeans.tot_withinss(valid = True))

# grid over k
# import Grid Search
from h2o.grid.grid_search import H2OGridSearch

# select the values for k to grid over
# Note: this dataset is too small to see significant differences between these options
# the purpose of the example is to show how to use grid search with k if desired
hyper_params = {'k': [2,3,7]}

# this example uses cartesian grid search because the search space is small
# and we want to see the performance of all models. For a larger search space use
# random grid search instead: {'strategy': "RandomDiscrete"}
# initialize the estimator
seeds_kmeans = H2OKMeansEstimator(seed = 1234)

# build grid search with previously made Kmeans and hyperparameters
grid = H2OGridSearch(model = seeds_kmeans, hyper_params = hyper_params,
search_criteria = {'strategy': "Cartesian"})

# train using the grid
grid.train(x = predictors, training_frame = train, validation_frame = valid)

# sort the grid models by total within cluster sum-of-square error.
sorted_grid = grid.get_grid(sort_by='tot_withinss', decreasing=False)
print(sorted_grid)


text:
When adapting combinatorial metaheuristics to continuous problems, one sometimes use a sphere as an approximation of the "neighborhood". The idea is thus to draw the neighbours around a solution, for example in order to apply a simple mutation in a genetic algorithm.

Sometimes, one choose to use an uniform law, but how to draw random vectors uniformly in an hyper-sphere ?

The first idea that comes to mind is to use a polar coordinate system and draw the radius r and the angles a1...a2...ai...aN with a uniform law. Then, one convert the coordinates in the rectangular system, x1...x2...xi...xN.

The result is interesting for a metaheuristic design, but is not a uniform distribution:

The correct method is to draw each xi according to: xi=(ri1/Nai)/√(∑(ai))
(in LATEX : $x_{i}=\frac{r^{\frac{1}{N}}_i\cdot a_{i}}{\sqrt{{\displaystyle \sum _{j=1}^{N}a_{i}}}}$)

With ri uniformly drawn in U0,1 and a following a normal law NO,1

The result is then a true uniform distribution:

Credits goes to Maurice Clerc for detecting and solving the problem.


text:
# Anisotropic generalization of Buchdahl bound for specific stellar models

@inproceedings{Sharma2021AnisotropicGO,
title={Anisotropic generalization of Buchdahl bound for specific stellar models},
author={Ranjan Sharma and Arpita Ghosh and Soumik Bhattacharya and S. Das},
year={2021}
}
• Ranjan Sharma, +1 author S. Das
• Published 2021
• Physics
Anisotropy is one factor that appears to be significantly important in the studies of relativistic compact stars. In this paper, we make a generalization of the Buchdahl limit by incorporating an anisotropic effect for a selected class of exact solutions describing anisotropic stellar objects. In the isotropic case of a homogeneous distribution, we regain the Buchdahl limit 2M/R ≤ 8/9. Our investigation shows a direct link between the maximum allowed compactness and pressure anisotropy vi-a-vis… Expand

#### References

SHOWING 1-10 OF 19 REFERENCES
Effect of pressure anisotropy on Buchdahl-type relativistic compact stars
• Physics
• 2019
We consider exact models for dense relativistic stars with anisotropic pressures and containing Buchdahl-type spacetime geometry. The Buchdahl condition can be transformed to an Euler–Cauchy equationExpand
EXACT MODELS FOR ANISOTROPIC RELATIVISTIC STARS
• Physics
• 2002
We present a class of exact solutions of the Einstein gravitational field equations describing spherically symmetric and static anisotropic stellar type configurations. The solution is represented inExpand
A new exact anisotropic solution of embedding class one
• Physics
• 2016
Abstract.We have presented a new anisotropic solution of Einstein’s field equations for compact-star models. Einstein’s field equations are solved by using the class-one condition (S.N. Pandey, S.P.Expand
MAXIMUM MASS OF A CLASS OF COLD COMPACT STARS
• Physics
• 2005
We calculate the maximum mass of the class of compact stars described by the Vaidya–Tikekar27 model. The model permits a simple method of systematically fixing bounds on the maximum possible mass ofExpand
Exact relativistic model for a superdense star
• Physics
• 1982
Assuming that the physical 3-spacet = const in a superdense star is spheroidal, a static spherically symmetric model based on an exact solution of Einstein’s equations is given which will permitExpand
Local anisotropy in self-gravitating systems
• Physics
• 1997
Abstract We review and discuss possible causes for the appearance of local anisotropy (principal stresses unequal) in self-gravitating systems and present its main consequences. We consider bothExpand
Anisotropic compact stellar model of embedding class-I satisfying Karmarkar’s condition in Vaidya and Tikekar spheroidal geometry
• Physics
• 2020
We present a class of solutions for a spherically symmetric anisotropic matter distribution in Vaidya and Tikekar spheroidal geometry. Making use of the Vaidya and Tikekar (VT) metric ansatz (JExpand
Minimum mass–radius ratio for charged gravitational objects
• Physics
• 2007
We rigorously prove that for compact charged general relativistic objects there is a lower bound for the mass–radius ratio. This result follows from the same Buchdahl type inequality for chargedExpand
Absolute stability limit for relativistic charged spheres
• Physics
• 2008
We find an exact solution for the stability limit of relativistic charged spheres for the case of constant gravitational mass density and constant charge density. We argue that this provides anExpand
Buchdahl compactness limit and gravitational field energy
The main aim of this paper is essentially to point out that the Buchdahl compactness limit of a static object is given by \it{gravitational field energy being less than or equal to half of itsExpand


text:
Generally, a file’s type is identified by server side to check if this file is legal for storing. For instance, a cloud document editor can only allow users to import .doc/.docx/.pages files.

However, only detecting the expanded-name is not enough, some of users can change the file expanded-name to avoid this detection. Therefore, we have to find a solution to address this unsafe issue.

## Could a file’s content can be modified ?

If we just modify the file’s expanded-name, the file’s content won’t be changed. We can do an experiment. When we try to play a video file with a .pdf extension, it can still play well. In other words, the video player has another method to detect whether it is a video file or not. After searching on Google, I found that every file has a File Signature (or Magic Number), which represents for the real type of a file. Fortunately, it is constantly embedded in the header of a file (first 4~8 bytes).

However, most solutions provided online are implementation on server side. It is not a good way to use. Only the file that has been uploaded to the server can be detected. A large file has to be waited for a long time to upload, which might make users mad. Therefore, I intended to find a front-end solution.

## How to detect file type through HTML5 ?

Thanks to File API of HTML5, we can easily get the file signature of a file.

Here is the source code:

## In Practice

However, when I use it in practice, the file signature’s length is not constant. Some of files only need first 4 bytes to detect, but others need first 8 bytes. Even worse, some files’ signatures begin from the 512 bytes. How to use a universal solution to detect these different signatures ?

My current solution is that establishing a signatures library at first, and then using the expanded-name of files to match the signature of such kind of extension. If the signature can’t match with the expanded-name, it will be considered illegally.

This is the signatures library I create:

Implementation:

## Summary

Detecting file type through front-end may be the fastest way, but there are also some cons.

### 1. Part of files has the same signatures

For instance, the signatures of files that Microsoft Office create (.xlsx,.docx,.pptx, etc.) are equal to that of zip files.

Test:

File Type Successfully Intercepted
mp4 to pdf
zip to docx
zip to jpg
docx to zip
docx to xlsx

### 2. Compatibility

Unfortunately, HTML5’s FileReader is not supported by all browsers.

Feature Firefox (Gecko) Chrome Internet Explorer* Opera* Safari
Basic support 3.6 (1.9.2) 7 10 🚫 🚫

### 3. File Signature Library

The library of file signatures can be found on File Signature Database, but not all the files has a signature on it, even though this website keeps updating them.


text:
# Difference between revisions of "2015 AMC 8 Problems/Problem 9"

## Problem

On her first day of work, Janabel sold one widget. On day two, she sold three widgets. On day three, she sold five widgets, and on each succeeding day, she sold two more widgets than she had sold on the previous day. How many widgets in total had Janabel sold after working $20$ days?

$\textbf{(A) }39\qquad\textbf{(B) }40\qquad\textbf{(C) }210\qquad\textbf{(D) }400\qquad \textbf{(E) }401$

## Solutions

### Solution 1

First, we have to find how many widgets she makes on Day $20$. We can write the linear equation $y=-1+2x$ to represent this situation. Then, we can plug in $20$ for $x$: $y=-1+2(20)$ -- $y=-1+40$ -- $y=39$ The sum of $1,3,5, ........ 39$ is $\dfrac{(1 + 39)(20)}{2}= \boxed{\textbf{(D)}~400}$

### Solution 2

The sum is just the sum of the first $20$ odd integers, which is $20^2=\boxed{\textbf{(D)}~400}$

### Solution 3

We can easily find out she makes $2*20-1 = 39$ widgets on Day $20$. Then, we make the sum of $1,3, 5,......,35,37,39$ by adding $(1+39)+(3+37)+(5+35)+...+(19+21)$ which have 10 pairs $40$. So The sum of $1,3,5, ........ 39$ is $(40*10)$ = \boxed{\textbf{(D)}~400}\$ ----LarryFlora


text:
de.mpg.escidoc.pubman.appbase.FacesBean
English

# Item

ITEM ACTIONSEXPORT
Matrix factorisations for rational boundary conditions by defect fusion

Behr, N., & Fredenhagen, S. (2015). Matrix factorisations for rational boundary conditions by defect fusion. Journal of high energy physics: JHEP, 2015(05): 055. doi:10.1007/JHEP05(2015)055.

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Behr, Nicolas, Author
Fredenhagen, Stefan1, Author
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1Quantum Gravity & Unified Theories, AEI-Golm, MPI for Gravitational Physics, Max Planck Society, escidoc:24014

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Free keywords: High Energy Physics - Theory, hep-th,Mathematical Physics, math-ph,Mathematics, Mathematical Physics, math.MP
Abstract: A large class of two-dimensional $\mathcal{N}=(2,2)$ superconformal field theories can be understood as IR fixed-points of Landau-Ginzburg models. In particular, there are rational conformal field theories that also have a Landau-Ginzburg description. To understand better the relation between the structures in the rational conformal field theory and in the Landau-Ginzburg theory, we investigate how rational B-type boundary conditions are realised as matrix factorisations in the $SU(3)/U(2)$ Grassmannian Kazama-Suzuki model. As a tool to generate the matrix factorisations we make use of a particular interface between the Kazama-Suzuki model and products of minimal models, whose fusion can be realised as a simple functor on ring modules. This allows us to formulate a proposal for all matrix factorisations corresponding to rational boundary conditions in the $SU(3)/U(2)$ model.

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Dates: 2014-07-2720152015
Publication Status: Published in print
Pages: 33+23 pages, 3 figures
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Identifiers: arXiv: 1407.7254
DOI: 10.1007/JHEP05(2015)055
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Title: Journal of high energy physics : JHEP
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Publ. Info: Bologna, Italy : Società italiana di fisica
Pages: - Volume / Issue: 2015 (05) Sequence Number: 055 Start / End Page: - Identifier: ISSN: 1126-6708
CoNE: http://pubman.mpdl.mpg.de/cone/journals/resource/111021927548002


text:
Ligand compatibility with SEWING is in its early stages, but is currently supported via the RosettaScripts interface via a Ligands subtag to any SEWING mover. If the Ligands tag is not provided, any ligand in the input structure will be stripped from the structure and ignored during the alignment/design process. Using this tag ensures that the SEWING Assembly will be aware of the ligand atoms and that any residue contacts with the ligand will be preserved throughout the SEWING protocol. However, to verify that there won't be any clashes with the ligand during the assembly process, you need to provide a LigandClashRequirement.

## Ligand Subtag

The Ligands tag takes a list of Ligand subtags. The Ligand tag has several options:

• which will be listed here


text:
# Equivalence relations on S given no relation?

On my assignment, one of the questions ask to list all equivalence relations on S and count how many are of partial orders.

1. Let S = {u, v, w}. List all equivalence relations on S. How many of these are also partial orders?

Am I to assume the equivalence relations are the possible combinations of each pair S x S ({}, {{u, u}}, {{u, u}, {u, v}} ...) or should their always be a given relation?

HINT: Since $S$ has $3$ elements, $S\times S$ has $3^2=9$ elements. Every subset of $S\times S$ is a relation on $S$, so there are $2^9=512$ relations on $S$. Fortunately, most of them are not equivalence relations, and you can ignore those. For instance, an equivalence relation must by definition be reflexive, so if $E$ is an equivalence relation on $S$, then you know that $E$ must contain each of the ordered pairs $\langle u,u\rangle,\langle v,v\rangle$, and $\langle w,w\rangle$. That leaves only $6$ ordered pairs that might or might not be in $E$. You could then take into account the restrictions on $E$ imposed by the requirements that $E$ be symmetric and transitive. However, this is doing it the hard way.
In fact, you know that every equivalence relation on $S$ determines and is determined by the partition of $S$ whose parts are its equivalence classes. What are the partitions of $S$? One of them is $\big\{\{u,v\},\{w\}\big\}$; what are the others? There aren’t many at all.
• @COMP232: I’m not sure what you mean by multiplication here, but those are indeed the four partitions. As an example, the second one induces the equivalence relation $$\{\langle u,u\rangle,\langle u,v\rangle,\langle v,v\rangle,\langle w,w\rangle\}\;,$$ and the first one induces the relation of equality on $S$. – Brian M. Scott Nov 29 '13 at 3:21


text:
# Null space for $\mathcal{N}(A)$ given SVD of $A$

Let $A \in \mathbb{R}^{m \times n}$ be a matrix with singular value decomposition

$$A = U \tilde{\Sigma} V^{T}.$$

Let $\text{rank}(A) = r \leq \min(m, n)$. My textbook notes that the first $r$ columns of $U$ then constitute a basis for $\mathcal{R}(A)$ (the range of $A$). That makes sense, since $Ax = U \tilde{\Sigma} V^{T} x = U \tilde{x}$ where $\tilde{u} = \tilde{\Sigma}V^{T} x$.

However, my book also states that the first $n - r$ columns of $V$ constitute a basis for $\mathcal{N}(A)$ (the null space of $A$). Shouldn't it instead be that the last $n - r$ rows of $V$ are a basis for $\mathcal{N}(A)$? If we want $V^{T} x = 0$, then multiplication by $x$ must produce a linear combination of the rows of $V$ that sum to the zero vector.

Every matrix $$\mathbf{A} \in \mathbb{C}^{m\times n}_{\rho}$$ has a singular value decomposition

\begin{align} \mathbf{A} &= \mathbf{U} \, \Sigma \, \mathbf{V}^{*} \\ % &= % U \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S}_{\rho\times \rho} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] % V* \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ \color{red}{\mathbf{V}_{\mathcal{N}}}^{*} \end{array} \right] \\ % % &= % U \left[ \begin{array}{cc} \color{blue}{\mathbf{U}_{\mathcal{R}}} & \color{red}{\mathbf{U}_{\mathcal{N}}} \end{array} \right] % Sigma \left[ \begin{array}{cccc|cc} \sigma_{1} & 0 & \dots & & & \dots & 0 \\ 0 & \sigma_{2} \\ \vdots && \ddots \\ & & & \sigma_{\rho} \\\hline & & & & 0 & \\ \vdots &&&&&\ddots \\ 0 & & & & & & 0 \\ \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{\mathbf{V}_{\mathcal{R}}}^{*} \\ \color{red}{\mathbf{V}_{\mathcal{N}}}^{*} \end{array} \right] \\ % & = % U \left[ \begin{array}{cccccccc} \color{blue}{u_{1}} & \dots & \color{blue}{u_{\rho}} & \color{red}{u_{\rho+1}} & \dots & \color{red}{u_{n}} \end{array} \right] % Sigma \left[ \begin{array}{cc} \mathbf{S} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} \end{array} \right] % V \left[ \begin{array}{c} \color{blue}{v_{1}^{*}} \\ \vdots \\ \color{blue}{v_{\rho}^{*}} \\ \color{red}{v_{\rho+1}^{*}} \\ \vdots \\ \color{red}{v_{n}^{*}} \end{array} \right] % \end{align}

The $\rho$ singular values are ordered and satisfy $$\sigma_{1} \ge \sigma_{2} \ge \dots \sigma_{1} > 0$$

The column vectors are orthonormal basis vectors: \begin{align} % R A \color{blue}{\mathcal{R} \left( \mathbf{A} \right)} &= \text{span} \left\{ \color{blue}{u_{1}}, \dots , \color{blue}{u_{\rho}} \right\} \\ % R A* \color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} &= \text{span} \left\{ \color{blue}{v_{1}}, \dots , \color{blue}{v_{\rho}} \right\} \\ % N A* \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)} &= \text{span} \left\{ \color{red}{u_{\rho+1}}, \dots , \color{red}{u_{m}} \right\} \\ % N A \color{red}{\mathcal{N} \left( \mathbf{A} \right)} &= \text{span} \left\{ \color{red}{v_{\rho+1}}, \dots , \color{red}{v_{n}} \right\} \\ % \end{align} The dimensions of the subspace matrices are \begin{align} % \color{blue}{\mathbf{U}_{\mathcal{R}}} &\in \mathbb{C}^{m\times\rho} \\ % \color{blue}{\mathbf{V}_{\mathcal{R}}} &\in \mathbb{C}^{n\times\rho} \\ % \color{red}{\mathbf{U}_{\mathcal{N}}} &\in \mathbb{C}^{m\times (m-\rho)} \\ % \color{red}{\mathbf{V}_{\mathcal{N}}} &\in \mathbb{C}^{n\times (n-\rho)} \\ % \end{align} To answer your question: Yes, the final $n-\rho$ columns of $\mathbf{V}$ are a span for $\color{red}{\mathbf{V}_{\mathcal{N}}}$.

The assertion is true if the singular values are ordered in a way that the first $n-r$ diagonal entries of $\tilde \Sigma$ are zero.

Let $v_1, v_2, \dots$ be the columns of $V$. Let $$v= \sum_{i=1}^{n-r}x_i v_i$$ be a linear combination of the first $n-r$ columns. This means that in the first row of $v$, there's a linear combination of the first row of the $v_i$s, in the second row the second and so on.

Thus we can write $v=Vx$, where $x= \begin{pmatrix}x_1\\ x_2\\ \vdots \\ x_{n-r}\\ 0 \\ \vdots \\0 \end{pmatrix}.$

Then we have that $V^T V = I$, so that

$$Av = U \tilde \Sigma V^T V x = U \tilde \Sigma x.$$

Now, since we required $\tilde \Sigma$ to be of a shape that the left-upper $(n-r)\times (n-r)$ block consists of zeros and this is precisely where $x$ doesn't have any zeros, so we get $$\tilde \Sigma x = 0, \mbox{ hence } Ax = 0.$$

If the zeros wouldn't be on the top of the diagonal of $\tilde \Sigma$, but on the bottom, then indeed, we'd need to look at the last $n-r$ columns of $V$.

Addendum: I'm pretty sure that you should be able to proof this with some kind of nice duality argument (since by $A= U\tilde \Sigma V^T$ we get $A^T = V \tilde \Sigma ^T U,$ which shows that the columns of $V$ span $\mathcal R(A^T) = (\mathcal N(A))^\perp$.)


text:
# data¶

Utilities to find files from NIPY data packages

 Bomber(name, msg) Class to raise an informative error when used BomberError Error when trying to access Bomber instance DataError Datasource(base_path) Simple class to add base path to relative path VersionedDatasource(base_path[, config_filename]) Datasource with version information in config file datasource_or_bomber(pkg_def, **options) Return a viable datasource or a Bomber find_data_dir(root_dirs, *names) Find relative path given path prefixes to search Return specified or guessed locations of NIPY data files make_datasource(pkg_def, **kwargs) Return datasource defined by pkg_def as found in data_path

## Bomber¶

class nibabel.data.Bomber(name, msg)

Bases: object

Class to raise an informative error when used

__init__(name, msg)

Initialize self. See help(type(self)) for accurate signature.

## BomberError¶

class nibabel.data.BomberError

Bases: nibabel.data.DataError, AttributeError

Error when trying to access Bomber instance

Should be instance of AttributeError to allow Python 3 inspect to do various hasattr checks without raising an error

__init__(*args, **kwargs)

Initialize self. See help(type(self)) for accurate signature.

## DataError¶

class nibabel.data.DataError

Bases: Exception

__init__(*args, **kwargs)

Initialize self. See help(type(self)) for accurate signature.

## Datasource¶

class nibabel.data.Datasource(base_path)

Bases: object

Simple class to add base path to relative path

Initialize datasource

Parameters
base_pathstr

path to prepend to all relative paths

Examples

>>> from os.path import join as pjoin
>>> repo = Datasource(pjoin('a', 'path'))
>>> fname = repo.get_filename('somedir', 'afile.txt')
>>> fname == pjoin('a', 'path', 'somedir', 'afile.txt')
True

__init__(base_path)

Initialize datasource

Parameters
base_pathstr

path to prepend to all relative paths

Examples

>>> from os.path import join as pjoin
>>> repo = Datasource(pjoin('a', 'path'))
>>> fname = repo.get_filename('somedir', 'afile.txt')
>>> fname == pjoin('a', 'path', 'somedir', 'afile.txt')
True

get_filename(*path_parts)

Prepend base path to *path_parts

We make no check whether the returned path exists.

Parameters
*path_partssequence of strings
Returns
fnamestr

result of os.path.join(*path_parts), with self.base_path prepended

list_files(relative=True)

Recursively list the files in the data source directory.

Parameters
relative: bool, optional

If True, path returned are relative to the base paht of the data source.

Returns
file_list: list of strings

List of the paths of all the files in the data source.

## VersionedDatasource¶

class nibabel.data.VersionedDatasource(base_path, config_filename=None)

Datasource with version information in config file

Initialize versioned datasource

We assume that there is a configuration file with version information in datasource directory tree.

The configuration file contains an entry like:

[DEFAULT]
version = 0.3

The version should have at least a major and a minor version number in the form above.

Parameters
base_pathstr

path to prepend to all relative paths

config_filanameNone or str

relative path to configuration file containing version

__init__(base_path, config_filename=None)

Initialize versioned datasource

We assume that there is a configuration file with version information in datasource directory tree.

The configuration file contains an entry like:

[DEFAULT]
version = 0.3

The version should have at least a major and a minor version number in the form above.

Parameters
base_pathstr

path to prepend to all relative paths

config_filanameNone or str

relative path to configuration file containing version

## datasource_or_bomber¶

nibabel.data.datasource_or_bomber(pkg_def, **options)

Return a viable datasource or a Bomber

This is to allow module level creation of datasource objects. We create the objects, so that, if the data exist, and are the correct version, the objects are valid datasources, otherwise, they raise an error on access, warning about the lack of data or the version numbers.

The parameters are as for make_datasource in this module.

Parameters
pkg_defdict

dict containing at least key ‘relpath’. Can optioanlly have keys ‘name’ (package name), ‘install hint’ (for helpful error messages) and ‘min version’ giving the minimum necessary version string for the package.

data_pathsequence of strings or None, optional
Returns
dsdatasource or Bomber instance

## find_data_dir¶

nibabel.data.find_data_dir(root_dirs, *names)

Find relative path given path prefixes to search

We raise a DataError if we can’t find the relative path

Parameters

sequence of paths in which to search for data directory

*namessequence of strings

sequence of strings naming directory to find. The name to search for is given by os.path.join(*names)

Returns
data_dirstr

full path (root path added to *names above)

## get_data_path¶

nibabel.data.get_data_path()

Return specified or guessed locations of NIPY data files

The algorithm is to return paths, extracted from strings, where strings are found in the following order:

1. The contents of environment variable NIPY_DATA_PATH

2. Any section = DATA, key = path value in a config.ini file in your nipy user directory (found with get_nipy_user_dir())

3. Any section = DATA, key = path value in any files found with a sorted(glob.glob(os.path.join(sys_dir, '*.ini'))) search, where sys_dir is found with get_nipy_system_dir()

4. If sys.prefix is /usr, we add /usr/local/share/nipy. We need this because Python 2.6 in Debian / Ubuntu does default installs to /usr/local.

5. The result of get_nipy_user_dir()

Therefore, any paths found in NIPY_DATA_PATH will be searched before paths found in the user directory config.ini

Parameters
None
Returns
pathssequence of paths

Notes

We have to add /usr/local/share/nipy if sys.prefix is /usr, because Debian has patched distutils in Python 2.6 to do default distutils installs there:

Examples

>>> pth = get_data_path()

## make_datasource¶

nibabel.data.make_datasource(pkg_def, **kwargs)

Return datasource defined by pkg_def as found in data_path

data_path is the only allowed keyword argument.

pkg_def is a dictionary with at least one key - ‘relpath’. ‘relpath’ is a relative path with unix forward slash separators.

The relative path to the data is found with:

names = pkg_def['name'].split('/')
rel_path = os.path.join(names)

We search for this relative path in the list of paths given by data_path. By default data_path is given by get_data_path() in this module.

If we can’t find the relative path, raise a DataError

Parameters
pkg_defdict

dict containing at least the key ‘relpath’. ‘relpath’ is the data path of the package relative to data_path. It is in unix path format (using forward slashes as directory separators). pkg_def can also contain optional keys ‘name’ (the name of the package), and / or a key ‘install hint’ that we use in the returned error message from trying to use the resulting datasource

data_pathsequence of strings or None, optional

sequence of paths in which to search for data. If None (the default), then use get_data_path()

Returns
datasourceVersionedDatasource

An initialized VersionedDatasource instance