text:
# Tag Info

## Hot answers tagged terminology

5

A paper came out this week pointing to them having a banal (if amusing) origin: they are from two 27 year old microwave ovens. When people get impatient and open the door before the timer runs down, a short burst from the ovens' magnetron is released, which appears as a peryton if the telescope is pointed in the right direction. Figure 7. shows the perytons ...

4

The Weights and Measures Act (the origin of the Imperial Units) does not speak of temperature. It was intended to create a uniform system for trade. You don't sell temperature, in the way you sell a pint of milk or a yard of cloth. And frankly, when it was first conceived (before Magna Carta, which already stated: "There shall be but one Measure ...

4

I got a translation of the article from the German Wikipedia. Here's an excerpt: Perytons are in radio astronomy short radio signals having a length of a few milliseconds, which probably terrestrial are origin. The Perytons are named after mythical creatures . In radio astronomy, terrestrial are noise is always a problem. A well-known noise signal ...

3

cms and kgs are wrong. The SI units are abbreviations which are also used in the plural. You will write 2.6 m/s or 1 m/s, but say "2.6 meters per second" or "1 meter per second" respectively. Keep in mind the SI units are also used in tons of other languages that do not form the plural by attaching an -s. The units look the same in those languages. (e.g. ...

3

A good question, you are right the frequency remains constant (unless you have Doppler effects due to relative movement, but that's not your question). For visible light, refraction properties are quite often in question and as such it make sense to speak in terms of wavelength. As you go even higher in "frequency", physicists start talking in keV and MeV ...

3

Let $M$ be your spacetime, a smooth manifold equipped with (pseudo) Riemannian metric (for example $\mathbb{R}^{(1,3)}$ for special relativity). The set of reference frames is the frame bundle over $M$, usually denoted $FM$. Explicitly a frame at point $p$ in $M$ can be viewed as an ordered orthonormal basis (with respect to the the inner product defined ...

2

I think what he's saying is that $$F_{net} = F_{nc} + \nabla U,$$ which is pretty standard. $f^a$ is your net force, which is the sum of your conservative and nonconservative forces. Conservative forces can be written as the gradient of some potential, which is where you get your $\nabla U$ from. $f^e,$ then, are your nonconservative forces.

2

When we say something is conserved or that there is a conservation law for a given thing, we mean that the quantity of it does not change. You neither lose nor gain any of that thing. More specifically, conservation can come in two flavours. Something can be globally conserved. This means that the total amount of that something in the universe does not ...

2

When you say If something goes outside, then it will decrease inside! what you assume is exactly a conservation law. It may seem trivial, but it is not necessarily. Consider the population of a city, for example. At one point in time, you measure how many people are within the city borders; let's call this number $N_0$. Then, you observe all city ...

2

Static comes from the same root as stasis, meaning stop, immovable, To create static electricity, you have to rub two different materials. At the moment you rub them, the electrons already moved Note the word "create", creation is not static, and yes there are transient fields and currents during creation of a static field. The static describes the ...

2

To contract a tensor is to set two of the indices equal and sum over them, so given a tensor $A^i_j$ the contraction is $A=A^i_i=A^1_1+A^2_2+A^3_3+A^4_4$ The Bianchi identities you list have five indices. To contract them, you would set some pair equal and sum over them. Your second version is the same as the first, it just has the indices renamed. ...

2

The order of a quantity in general refers to the exponent of the quantity in an expression, ie $$x^3y^2$$ would be 3rd order in $x$ and 2nd order in $y$. According to the Feynman rules, each vertex in a Feynman diagrams contributes a factor of the coupling constant, so the order of each coupling constant is simply the number of vertices of that interaction. ...

2

Using elementary graph theory identities one can show that the number of loops in a connected diagram is related to the number of external lines and the number of vertices of type $i$ each of which has $n_i$ lines attached to it, is related by $$\sum \left(\frac{n_i}{2}-1\right) V_i -\tfrac{1}{2}E +1= L$$ So you can see that for a fixed process (fixed ...

2

As I recall, covariant refers to how an object transforms when you boost to another inertial frame. An example would be the relativistic 4-momentum $P^{\mu}$. Invariant refers to quantities which are unchanged under boosts to different frames. For example the product $P^{\mu}P_{\mu}=m$ has the same numerical value in any frame. Sometimes a relativistic ...

2

The kinetic term of the Lagrangian is proportional to $$g_{ij}v^iv^j$$ where the $v$s are the generalised velocities. Writing them as the time derivative of the generalised coordinates, i.e. $v^i\dot q^i$, taking the square root, and multiplying by a small time lapse $\epsilon$ you get $$\sqrt{g_{ij}\dot q^i\dot q^j}\epsilon,$$ which is a first order ...

2

According to the wiki page on Imperial and US customary units Fahrenheit is part of both the Imperial and US customary system. I can't think of any reason it wouldn't be included in the Imperial system. Note that in the wiki page on Imperial units it is mentioned that the weight's and measures act (which defined the Imperial system) explicitly used the ...

1

Dunno what book you're quoting, but you should realize that the index of refraction of air is $n = 1+ \epsilon$ (where I'm using the mathematics standard of $\epsilon$ being a tiny number). Thus the power in air is $1/F$

1

The equation you are quoting gives the power of a lens in terms of its geometry and refractive index. Simply rearranging the terms (dividing by $n$) gives you an expression for $\frac{1}{f}$ which is known as the power of the lens and is expressed in diopters. For the usual situation of a lens in air, we can put $n=1$ which leaves you with an even simpler ...

1

Positron IS an elementary particle, the anti-particle to the electron as you already know. But we do not get a "free positron" as a "free electron". They are usually generated through pair-production and get annihilated fast, or through radioactive decay (beta-decay) in weak interactions or in particle accelerators, and are present in cosmic rays too. A ...

1

The below seems to be a candidate for the first use of the term 'Majorana fermion'. (I'm not sure if it satisfies your other criteria.) Salam, Abdus, and J. Strathdee. Super-symmetry and non-Abelian gauges. International Centre for Theoretical Physics, Trieste (Italy), 1974.

1

Four component formalism is the "right" formalism, but it has negative energy eigenstates corresponding to the antiparticles. Most chemists and solid state physics are not interested in the antiparticles, and such negative energy solution causes trouble for conventional variational methods, where you might end up falling to negative infinity energy. It is ...

1

A site is just a place or location with given coordinates e.g. $(x_0,y_0,z_0)$

1

I would like to bring the ladder paradox here to explain simultaneity of events.A ladder (an inertial frame) is moving horizontally with a relatively high constant speed with respect to a garage (another inertial frame). The garage has an open door where the ladder can not actually enter if the ladder was at rest in the garage's frame but that is not ...

1

Regarding your assertions: Events $\varepsilon_{AJ}$ and $\varepsilon_{BK}$ were simultaneous in the inertial frame of participants $A$, $B$, $M$. This is a perfectly reasonable statement and it is the sort of language used in everyday physics. Participant $M$ was the middle between $J$ and $K$, in the inertial frame of participants $A$, $B$, $M$. ...

1

"Coincident" is defined in the Google online dictionary as (1) "occurring together in space OR time" (emphasis mine), and (2) "in agreement or harmony". "Simultaneous" is defined in the same dictionary as "occurring, operating, or done at the same time". Unfortunately, this dictionary lists "coincident" as a synonym of "simultaneous". I do not think this ...

1

This paper describes them. http://www.ursi.org/proceedings/procGA11/ursi/GP2-41.pdf They were apparently given a new name because their origin was uncertain.

1

i searched for the exact same problem recently after a debate with one of my colleagues. In my opinion, you already gave the answer to your question yourself. A source dipole is the flow field resulting from a sink and a source brought together. In a sink, all streamlines point radially inward to the singularity at the origin, in a source, all point ...

1

I think the magnon is a special case of the spin wave. Whereas spinon refers to the general quasiparticle that carries all spin of an electron, magnon refers to the limiting case of spin wave quantized in such a manner that it becomes part of an anti-magnetic cloud of quasiparticles. However, this may not be the complete story! Some usage of the terms: ...

1

UV stands for Ultraviolet and it is referring to a special kind of divergences in quantum field theory. In NLO loop diagrams, we often encounter divergences (infinite integrals) when we investigate what happens at $k \to \infty$, where $k$ is the internal momentum of a virtual particle in Feynman diagrams. These are precisely the divergences we call "UV ...

1

Of course they both give informations about the motion of bodies. The kinematic equations tell us simply what are the valors of the variables of the specific motion , that because kinematic studies only the variables of the motion and their changing. kinematic equations give us indications about : Velocities (the most simple and known equation of kinematic ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 28 Nov 2015, 23:27

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# If k, m, and t are positive integers and k/6 + m/4 = t/12

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Manager
Joined: 16 Feb 2012
Posts: 237
Concentration: Finance, Economics
Followers: 5

Kudos [?]: 153 [3] , given: 121

If k, m, and t are positive integers and k/6 + m/4 = t/12 [#permalink]  23 Feb 2012, 01:28
3
KUDOS
3
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

53% (02:25) correct 47% (01:25) wrong based on 247 sessions
If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1?

(1) k is a multiple of 3.
(2) m is a multiple of 3.

In the explanation of this question they say that the sum of two multiples of 3 give the number that is also a multiple of 3.
Is that a general rule for any number? If someone can elaborate I would be grateful!
[Reveal] Spoiler: OA

_________________

Kudos if you like the post!

Failing to plan is planning to fail.

Math Expert
Joined: 02 Sep 2009
Posts: 30392
Followers: 5094

Kudos [?]: 57432 [8] , given: 8811

Re: General Math Question [#permalink]  23 Feb 2012, 01:40
8
KUDOS
Expert's post
2
This post was
BOOKMARKED
Stiv wrote:
If k, m, and t are positive integers and $$\frac {k}{6} + \frac {m}{4} = \frac {t}{12}$$ , do t and 12 have a common factor greater than 1?
(1) k is a multiple of 3.
(2) m is a multiple of 3.

In the explanation of this question they say that the sum of two multiples of 3 give the number that is also a multiple of 3.
Is that a general rule for any number? If someone can elaborate I would be grateful!

If k, m, and t are positive integers and $$\frac{k}{6} + \frac{m}{4} = \frac{t}{12}$$, do t and 12 have a common factor greater than 1 ?

$$\frac{k}{6} + \frac{m}{4} = \frac{t}{12}$$ --> $$2k+3m=t$$.

(1) k is a multiple of 3 --> $$k=3x$$, where $$x$$ is a positive integer --> $$2k+3m=6x+3m=3(2x+m)=t$$ --> $$t$$ is multiple of 3, hence $$t$$ and 12 have a common factor of 3>1. Sufficient.

(2) m is a multiple of 3 --> $$m=3y$$, where $$y$$ is a positive integer --> $$2k+3m=2k+9y=t$$ --> $$t$$ and 12 may or may not have a common factor greater than 1. Not sufficient.

As for your question:
If integers $$a$$ and $$b$$ are both multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference will also be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$ and $$b=9$$, both divisible by 3 ---> $$a+b=15$$ and $$a-b=-3$$, again both divisible by 3.

If out of integers $$a$$ and $$b$$ one is a multiple of some integer $$k>1$$ and another is not, then their sum and difference will NOT be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=6$$, divisible by 3 and $$b=5$$, not divisible by 3 ---> $$a+b=11$$ and $$a-b=1$$, neither is divisible by 3.

If integers $$a$$ and $$b$$ both are NOT multiples of some integer $$k>1$$ (divisible by $$k$$), then their sum and difference may or may not be a multiple of $$k$$ (divisible by $$k$$):
Example: $$a=5$$ and $$b=4$$, neither is divisible by 3 ---> $$a+b=9$$, is divisible by 3 and $$a-b=1$$, is not divisible by 3;
OR: $$a=6$$ and $$b=3$$, neither is divisible by 5 ---> $$a+b=9$$ and $$a-b=3$$, neither is divisible by 5;
OR: $$a=2$$ and $$b=2$$, neither is divisible by 4 ---> $$a+b=4$$ and $$a-b=0$$, both are divisible by 4.

Hope it's clear.
_________________
Math Expert
Joined: 02 Sep 2009
Posts: 30392
Followers: 5094

Kudos [?]: 57432 [0], given: 8811

Re: If k, m, and t are positive integers and k/6 + m/4 = t/12 [#permalink]  26 Jun 2013, 00:39
Expert's post
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Divisibility/Multiples/Factors questions to practice: search.php?search_id=tag&tag_id=354
All PS Divisibility/Multiples/Factors questions to practice: search.php?search_id=tag&tag_id=185

_________________
Moderator
Joined: 25 Apr 2012
Posts: 729
Location: India
GPA: 3.21
WE: Business Development (Other)
Followers: 37

Kudos [?]: 514 [1] , given: 723

Re: If k, m, and t are positive integers and k/6 + m/4 = t/12 [#permalink]  26 Jun 2013, 01:09
1
KUDOS
Stiv wrote:
If k, m, and t are positive integers and k/6 + m/4 = t/12, do t and 12 have a common factor greater than 1?

(1) k is a multiple of 3.
(2) m is a multiple of 3.

In the explanation of this question they say that the sum of two multiples of 3 give the number that is also a multiple of 3.
Is that a general rule for any number? If someone can elaborate I would be grateful!

We can solve the given expression and get the following

(2k+3m)/12= t/12 ------> this implies t= 2k +3 m

From St 1 we have k is a multiple of 3 so the above equation is of the form t= 2*3*a+ 3m i.e t= 6a +3m where a is a positive integer (since K is a positive integer "a" cannot be zero)

thus t = 3( 2a+m)
if a =1, m=1 then t= 9 ; an 9 and 12 have 3 as common factor other than 1
similarly if a=2, m=1 we have t=15, and both 15 and 12 have 3 as common factor
since t has 3 as one of its factors and 12 also has 3 as one of its factor and therefore "t" and 12 will always have 3 as a factor other than 1

from St2 we have t= 2k+ 3*3b -----> t= 2k+9b where b is a positive integer

Here if k=1 and b =1, then t= 11; 11 and 12 do not have any common factor other than 1
but if k=3 and b=3 then we have t= 24 ; 24 and 12 have many common factor

therefore ans should be A
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Manager
Joined: 28 Apr 2013
Posts: 159
Location: India
GPA: 4
WE: Medicine and Health (Health Care)
Followers: 1

Kudos [?]: 46 [0], given: 84

if k, m, and t are positive integers [#permalink]  22 Jan 2014, 18:21
If k, m, and t are positive integers and k/6 + m/4 = t/12 do t and 12 have a common factor greater than 1?

1. k is a multiple of 3.

2. m is a multiple of 3.

Source- Private notes
_________________

Thanks for Posting

LEARN TO ANALYSE

+1 kudos if you like

Math Expert
Joined: 02 Sep 2009
Posts: 30392
Followers: 5094

Kudos [?]: 57432 [0], given: 8811

Re: if k, m, and t are positive integers [#permalink]  22 Jan 2014, 18:25
Expert's post
rango wrote:
If k, m, and t are positive integers and k/6 + m/4 = t/12 do t and 12 have a common factor greater than 1?

1. k is a multiple of 3.

2. m is a multiple of 3.

Source- Private notes

Merging similar topics. Refer to the solutions above.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 7304
Followers: 384

Kudos [?]: 93 [0], given: 0

Re: If k, m, and t are positive integers and k/6 + m/4 = t/12 [#permalink]  01 Feb 2015, 04:33
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Re: If k, m, and t are positive integers and k/6 + m/4 = t/12   [#permalink] 01 Feb 2015, 04:33
Similar topics Replies Last post
Similar
Topics:
What is the value of [3(k + m) + 4k]/[2k + m]? 1 08 Apr 2015, 13:16
4 If m and k are positive integers, is m!+8k a multiple of k? 2 23 Jun 2014, 06:06
11 m = 4n + 9, where n is a positive integer. What is the greatest common 8 14 Feb 2012, 04:16
1 If k is a positive integer and m is the product of the fi 7 19 Jun 2011, 06:01
9 If a, b, k, and m are positive integers, is a^k a factor of 8 28 Dec 2009, 11:04
Display posts from previous: Sort by

# If k, m, and t are positive integers and k/6 + m/4 = t/12

 Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

format: plain


text:
Theorem

The unit ball $\mathbb D^3 \subset \R^3$ is equidecomposable to the union of two unit balls.

Proof

Let $\mathbb D^3$ be centered at the origin, and $D^3$ be some other unit ball in $\R^3$ such that $\mathbb D^3 \cap D^3 = \varnothing$.

Let $\mathbb S^2 = \partial \mathbb D^3$.

By the Hausdorff Paradox, there exists a decomposition of $\mathbb S^2$ into four sets $A, B, C, D$ such that $A, B, C,$ and $B \cup C$ are congruent, and $D$ is countable.

For $r \in \R_{>0}$, define a function $r^{*} : \R^3 \to \R^3$ as $r^{*}(\mathbf x ) = r \mathbf x$, and define the sets:

$\displaystyle W = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(A)$
$\displaystyle X = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(B)$
$\displaystyle Y = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(C)$
$\displaystyle Z = \bigcup_{0 \mathop < r \mathop \le 1} r^{*}(D)$

Let $T = W \cup Z \cup \left\{{ \mathbf 0 }\right\}$.

$W$ and $X \cup Y$ are clearly congruent by the congruency of $A$ with $B \cup C$, hence $W$ and $X \cup Y$ are equidecomposable.

Since $X$ and $Y$ are congruent, and $W$ and $X$ are congruent, $X \cup Y$ and $W \cup X$ are equidecomposable.

$W$ and $X \cup Y$ as well as $X$ and $W$ are congruent, so $W \cup X$ and $W \cup X \cup Y$ are equidecomposable.

Hence $W$ and $W \cup X \cup Y$ are equidecomposable, by Equidecomposability is Equivalence Relation.

So $T$ and $\mathbb D^3$ are equidecomposable, from Equidecomposability Unaffected by Union.

Similarly we find $X$, $Y$, and $W \cup X \cup Y$ are equidecomposable.

Since $D$ is only countable, but $\mathbb{SO}(3)$ is not, we have:

$\exists \phi \in \mathbb{SO}(3): \phi \left({D}\right) \subset A \cup B \cup C$

so that $I = \phi \left({D}\right) \subset W \cup X \cup Y$.

Since $X$ and $W \cup X \cup Y$ are equidecomposable, by a theorem on equidecomposability and subsets, $\exists H \subseteq X$ such that $H$ and $I$ are equidecomposable.

Finally, let $p \in X - H$ be a point and define $S = Y \cup H \cup \left\{{p}\right\}$.

Since:

• $Y$ and $W \cup X \cup Y$
• $H$ and $Z$
• $\left\{{0}\right\}$ and $\left\{{p}\right\}$

are all equidecomposable in pairs, $S$ and $\mathbb B^3$ are equidecomposable by Equidecomposability Unaffected by Union.

Since $D^3$ and $\mathbb D^3$ are congruent, $D^3$ and $S$ are equidecomposable, from Equidecomposability is Equivalence Relation.

By Equidecomposability Unaffected by Union, $T \cup S$ and $\mathbb D^3 \cup D^3$ are equidecomposable.

Hence $T \cup S \subseteq \mathbb D^3 \subset \mathbb D^3 \cup D^3$ are equidecomposable and so, by the chain property of equidecomposability, $\mathbb D^3$ and $\mathbb D^3 \cup D^3$ are equidecomposable.

$\blacksquare$

Axiom of Choice

This theorem depends on the Axiom of Choice, by way of Hausdorff Paradox.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.

Conclusion

Whether this is a veridical paradox or an antinomy is being hotly discussed to the present day. There are even Facebook groups anti and pro.

Pick your personal choice of philosophy and start ranting.

If you feel really adventurous, check out Rudy Rucker's novel White Light.

Source of Name

This entry was named for Stefan Banach and Alfred Tarski.

They raised this question in a collaborative paper in 1924.


text:
4 Syntax highlighting

Python 3, 79 chars

## Python 3, 79 chars

a,b,c=eval(input())
d=(b*b-4*a*c)**.5/2/a
x=-b/2/a
print('r1=',x+d,'\nr2=',x-d)

Python's imaginary unity is j and not i. I used Python 3, because the power operator works also on negative numbers.

BTW, is it really needed to write r1= ... and r2= ... ?

Python 3, 79 chars

a,b,c=eval(input())
d=(b*b-4*a*c)**.5/2/a
x=-b/2/a
print('r1=',x+d,'\nr2=',x-d)

Python's imaginary unity is j and not i. I used Python 3, because the power operator works also on negative numbers.

BTW, is it really needed to write r1= ... and r2= ... ?

## Python 3, 79 chars

a,b,c=eval(input())
d=(b*b-4*a*c)**.5/2/a
x=-b/2/a
print('r1=',x+d,'\nr2=',x-d)

Python's imaginary unity is j and not i. I used Python 3 because the power operator works also on negative numbers.

BTW, is it really needed to write r1= ... and r2= ... ?

3 Syntax highlighting

## Python 3, 79 chars

Python 3, 79 chars

a,b,c=eval(input())
d=(b*b-4*a*c)**.5/2/a
x=-b/2/a
print('r1=',x+d,'\nr2=',x-d)

a,b,c=eval(input())
d=(b*b-4*a*c)**.5/2/a
x=-b/2/a
print('r1=',x+d,'\nr2=',x-d)

Python's imaginary unity is j and not i. I used Python 3, because the power operator works also on negative numbers.

BTW, is it really needed to write r1= ...r1= ... and r2= ...r2= ... ?

## Python 3, 79 chars

a,b,c=eval(input())
d=(b*b-4*a*c)**.5/2/a
x=-b/2/a
print('r1=',x+d,'\nr2=',x-d)

Python's imaginary unity is j and not i. I used Python 3 because the power operator works also on negative numbers.

BTW, is it really needed to write r1= ... and r2= ... ?

Python 3, 79 chars

a,b,c=eval(input())
d=(b*b-4*a*c)**.5/2/a
x=-b/2/a
print('r1=',x+d,'\nr2=',x-d)

Python's imaginary unity is j and not i. I used Python 3, because the power operator works also on negative numbers.

BTW, is it really needed to write r1= ... and r2= ... ?

2 added 61 characters in body

## Python 3, 79 chars

a,b,c=eval(input())
d=(b*b-4*a*c)**.5/2/a
x=-b/2/a
print('r1=',x+d,'\nr2=',x-d)

Python's imaginary unity is j and not i. I used Python 3 because the power operator works also on negative numbers.

BTW, is it really needed to write r1= ... and r2= ... ?

## Python 3, 79 chars

a,b,c=eval(input())
d=(b*b-4*a*c)**.5/2/a
x=-b/2/a
print('r1=',x+d,'\nr2=',x-d)

Python's imaginary unity is j and not i. I used Python 3 because the power operator works also on negative numbers.

## Python 3, 79 chars

a,b,c=eval(input())
d=(b*b-4*a*c)**.5/2/a
x=-b/2/a
print('r1=',x+d,'\nr2=',x-d)

Python's imaginary unity is j and not i. I used Python 3 because the power operator works also on negative numbers.

BTW, is it really needed to write r1= ... and r2= ... ?

1


text:
# Entire function such that $\lim_{ z \to \infty} \frac{f(z)}{z^N} = 0$ then $f$ is a polynomial of degree at most $N$.

I am tried to prove this:

Let a entire function $$f$$ such that $$\lim_{ z \to \infty} \frac{f(z)}{z^N} = 0.$$ Show that $$f$$ is a polynomial of degree at most $$N$$.

I find this results

Entire function bounded by a polynomial is a polynomial

So, I start my proof asuming that if $$f$$ is entire, it is equal to a power series centered at 0 with radius of convergence $$\infty$$, and its Taylor series is

$$f(z) = \sum_{n=0}^{\infty} \frac{f^{(N)}(0)}{N!}z^N$$

and I have the Cauchy formula says:

$$\frac{f^{n}(z_0)}{n!} = \frac{1}{2\pi i}\int_{|z-z_0|=r} \frac{f(z)}{(z-z_0)^{n+1}}dz$$

so if I consider $$z_0 = 0$$ and $$\frac{f^{n}(z_0)}{n!} {2\pi i} = \int_{|z|=r} \frac{f(z)}{z^{n+1}}dz$$

I tried to following the results of that, but I don't shure if I am in right way.

Can you help me to do it, please?

• For $n>N$, prove that $f^{(n)}(0)=0$. – Lord Shark the Unknown Apr 7 at 4:58
• Hint: show that $f^{(n)}(z)$ is a bounded entire function. Conclude using Liouville's. – Moya Apr 7 at 5:00
• There exists $R$ such that $|f(z)| <|z|^{N}$ for $|z| \geq R$. So $f$ is a polynomial of degree at most $N$ (by the result you have quoted). – Kavi Rama Murthy Apr 7 at 5:15

Hint following your idea: write $$f^{(n)}(0)$$ as line integral $$f^{(n)}(0) = \frac{n!}{2\pi i}\int_{|z|=r} \frac{f(z)}{z^{n+1}}\,dz$$ and apply the bound: $$\left|\int_{|z|=r}\frac{f(z)}{z^{n+1}}\,dz\right|\le \left(\sup_{|z|=r}\frac{f(z)}{z^N}\right)\frac1{r^{n+1-N}} \,2\pi r= \cdots$$ What happens when $$n$$ is large (how large?) enough?
• The integral is equals to 0 and $f^{(n)}(0) = 0$ but why I say that $f$ is a polynomial of grande at most N? It's because in the $n-th$ derivative is 0 and that's the limit? and where I use the hipotesis of my $lim_{z \to \infty} \frac{f(z)}{z^N} = 0$ – Ops Apr 7 at 17:10
• @Ops, the integral is $= 0$ for $n$ large enough. What is a Taylor series with coefficients $= 0$ for $n$ large enough? – Martín-Blas Pérez Pinilla Apr 7 at 17:12


text:
Volume 5, issue 2 (2005)

 Recent Issues
 The Journal About the Journal Subscriptions Editorial Board Editorial Interests Editorial Procedure Submission Guidelines Submission Page Author Index To Appear ISSN (electronic): 1472-2739 ISSN (print): 1472-2747
Some analogs of Zariski's Theorem on nodal line arrangements

A D Raza Choudary, Alexandru Dimca and Ştefan Papadima

Algebraic & Geometric Topology 5 (2005) 691–711
 arXiv: math.AT/0410363
Abstract

For line arrangements in ${ℙ}^{2}$ with nice combinatorics (in particular, for those which are nodal away the line at infinity), we prove that the combinatorics contains the same information as the fundamental group together with the meridianal basis of the abelianization. We consider higher dimensional analogs of the above situation. For these analogs, we give purely combinatorial complete descriptions of the following topological invariants (over an arbitrary field): the twisted homology of the complement, with arbitrary rank one coefficients; the homology of the associated Milnor fiber and Alexander cover, including monodromy actions; the coinvariants of the first higher non-trivial homotopy group of the Alexander cover, with the induced monodromy action.

Keywords
hyperplane arrangement, oriented topological type, 1–marked group, intersection lattice, local system, Milnor fiber, Alexander cover
Mathematical Subject Classification 2000
Primary: 32S22, 55N25
Secondary: 14F35, 52C35, 55Q52


text:
# Grow matrix to certain dimensions

Grow matrix to certain dimensions

grow(M, nrow, ncol, just = "tl", fillWith = "")

## Arguments

M matrix. number of requested rows of the output. number of requested columns of the output. string to control where "M" is placed within the larger matrix. Possible values are "t" (top), "b" (bottom), "l" (left) and "r" (right) which can be combined, e.g. "tl", "br", etc.. character or number which is used for filling the empty spots

## Value

Matrix with dimensions "nrow" and "ncol" accomodating the original matrix "M".


text:
# 129: compute the mean transit time In SoilR: Models of Soil Organic Matter Decomposition

## Description

This method computes the mean transit time for the linear time invariant system that can be constructed from the given operator and input distribution.

It relies on the mehtod getTransitTimeDistributionDensity using the same arguments.

## Arguments

 object inputDistribution

## Details

To compute the mean transit time for the distribution we have to compute the integral

\bar{T} = \int_0^∞ T \cdot S_r≤ft( \frac{\vec{I}}{I},0,T\right)\; dT

for the numerically computed density. To avoid issues with numerical integration we dont use as upper limit but cut off the integragion interval prematurely. For this purpose we calculate a maximum response time of the system as Lasaga

τ_{cycle} = \frac{1}{|\min(λ_i)|}

where λ_i are non-zero eigenvalues of the matrix {\bf A}.

## Author(s)

Carlos A. Sierra, Markus Mueller

## References

Lasaga, A.: The kinetic treatment of geochemical cycles, Geochimica et Cosmochimica Acta, 44, 815 – 828, doi10.1016/0016-7037(80)90263-X, 1980.

SoilR documentation built on May 4, 2017, 9:08 p.m.

Search within the SoilR package
Search all R packages, documentation and source code


text:
Indico search will be reestablished in the next version upgrade of the software: https://getindico.io/roadmap/

# Preparatory Joint Sessions on "Open questions and News Ideas"

7-8 July 2020
US/Eastern timezone

## Quantifying the effects of beam polarisation on charged triple gauge couplings

Not scheduled
20m
Short Talks

### Speaker

Jakob Beyer (DESY)

### Description

Charged triple gauge couplings (cTGCs) describe potential deviations from the Standard Model in the coupling of two $W$ bosons to a $Z$ boson or a photon. These couplings may interfere with the extraction of Higgs properties. Future $e^+e^-$ colliders need to measure these cTGCs with high precision to fully exploit the potential of their Higgs program.

Current proposals for $e^+e^-$ colliders vary in many aspects, most notably in their energy, luminosities and beam polarisations. Of those, the energy determines the relevant physics and the luminosity trivially scales the statistical precision.

Beam polarisation refers to the preferred direction of the spin in the beam particles. A dedicated magnet setup can flip the direction of a spin with non-zero polarisation.

Electroweak physics depends on the chirality and therefore the spin of the colliding particles. This flipping of the spin changes the allowed interactions. Both reducible and irreducible backgrounds can be suppressed by choosing the appropriate polarisation combination. The cTGC measurement is performed in the production of W bosons and highly sensitive to this effect.

While the physics changes when flipping the polarisations, the systematic effects remain the same. Beam polarisation may therefore provide a direct handle on systematic uncertainties in any measurement.

This study aims to quantify these effects of that a polarised beam may offer.

An extraction of cTGCs is performed together with a measurement of beam polarisations and 2-fermion parameters on generator level differential distributions. The fit is planned to be extended to include systematic effects that are correlated between the different processes.

A collider with polarised beams may achieve a qualitatively different measurement of cTGCs and other electroweak parameters. The extent to which polarisation may reduce systematics and provide access to electroweak physics needs to be clarified.

### Presentation Materials

There are no materials yet.


text:
## Getting Better

One of my mathy friends recently said, “I thought that when I finished graduate school I would have learned pretty much all the math I’d ever know.” When I was getting ready to graduate, I felt the same way. Which was a bit scary, because I felt like I didn’t know very much. But I figured that my future research life would be applying the techniques I used in my thesis to similar problems. I imagined that I would pick up some new ideas and go deeper into the area, but that my baseline math knowledge was pretty much set. This has been entirely wrong.

In fact, I have found myself working more on problems that I have absolutely no idea how to solve, and have realized that my interests refuse to stay bounded by the math that I know best. And problems don’t care! They may seem like problems in number theory, but oh, suddenly here comes representation theory, and a whole bunch of group theory! So I find myself trying to learn whole new areas, or relearn things I saw in courses but couldn’t really absorb at the time. It can be really hard, with work and so many other things going on.

I have this sense that if I had been a better graduate student, everything would be so much easier now, and that I should probably just take a few years off and go back to school. It would be great to be in a structured environment, designed by experts to help me learn more math. In my post-thesis research life, I find that so many topics from my classes really did turn out to be important, and those that seemed irrelevant at the time (I mean, when was I going to need measure theory?) have turned up over and over again. And I constantly wish I knew stuff from classes that I couldn’t bring myself to take (classical groups, linear programming… the wish-I-knew-list grows every day). But I can’t go back to school. I have a job now! I wasted my chance! Nooooo!

Of course, I have to remind myself that I actually learned a ton in graduate school. And at the time I was stressed out, worried about the very immediate problems presented by my thesis and teaching, as well as eventually finding a job. I was the best graduate student I could be, given who I was and the time I had. So what do I do with this wistful sense of what I should have been? I just keep doing the best I can, trying to solve problems, learning things from books, going to talks, and asking people questions when I get the chance. And, slowly, I do keep learning new things and solving new problems

The regretful sense that I should have been a better graduate student is increasingly countered by a feeling of excitement in the realization that I am actually still growing as a mathematician. In fact, as I (through great struggle) learn more, it becomes a little easier to see bigger pictures, and I even learn faster. Graduate school was just a jumping off point, and I keep gaining new perspectives. I am so glad that I was wrong about my math trajectory! It is actually thrilling to realize that, far from being done learning, I can just keep getting better at this.

I don’t know why I didn’t realize it sooner. I mean, I guess I thought my professors just started out good at math; that they were of a different species of mathematician. I definitely didn’t think of them as still learning. Perhaps I had, without even knowing it, absorbed the idea of mathematics as a sort of inborn talent, that you could fulfill as a young person but which fades as time goes on. Thank G. H. Hardy, with his incredibly annoying statement in A Mathematician’s Apology, “No mathematician should ever allow himself to forget that mathematics, more than any other art or science, is a young man’s game.” Perhaps Hardy’s words applied to his life. After all, he was a Tripos star, at the top of his field from the very beginning. He may have just crammed in a huge amount of math early on, and never really felt that he was growing in the same way later in life. However, I definitely followed a different path, and many others have as well. Richard Guy, anyone? (Happy 100th birthday!) In fact, I am not the first to point out that this whole idea is suspect. I reject the notion that these early years are the peak of my career, and that people’s mathematical value or potential is determinable by their 40th birthday. Hooray for early achievers! But also, hooray for people who start later and move slower but keep getting better all the time.

Realizing that I can keep getting better as a mathematician has made me really empathize with my students, who often think that they are already basically good or bad at math, and are afraid that they have hit their limit and are not capable of understanding some difficult concepts. This takes me back to growth mindset, (which I just explained to my linear algebra class on the first day of class!) and how easy it is even for me to fall into the trap of thinking that I can’t grow in some way.

On the level of the profession, it is important to reach out to talented young people from all backgrounds, both to maximize mathematical progress and to make the mathematical world a more diverse place. However, I think that it is also important to keep the door open for those who are not early talents, who may not choose mathematics until much later but still have a lot to offer. This is vital to broadening the profession, since despite our best efforts, students from underrepresented groups in STEM or less affluent school districts may not come into contact with interesting mathematics until later in life, and may have far fewer opportunities to pursue mathematics young even if they are interested.

So that’s what I’m thinking about as I start the new semester. Good luck to everyone with the challenges of fall! I would love to hear your thoughts on all this in the comments.

Okay, this illustration is a stretch. But I’m going for it:  I am totally getting better at math. Like, I used to know a flamingo of math, and now I know many flamingos of math. And someday…

… someday I will know a giraffe of math! Okay, sorry about that. I wanted to illustrate with a picture or a video of the Beatles singing “Getting Better,” but do you know how copyrighted that stuff is?!? These are actually pictures from my summer trip to the San Diego zoo.

Posted in research, Uncategorized | Tagged , | 2 Comments

## Summer Research Wrap-up

We’re already back to school here at Hood, but I wanted to take a minute to reflect on my summer undergraduate research experience. Overall it was very positive. My students met all the goals I had for them, got to present at MathFest, and we all had a pretty great time.

I’d mentioned in a previous post that we used Slack extensively to coordinate our research. It ended up being a collaborative lab notebook of a sort, as well as an email replacement. I loved not having to search through email threads to find relevant bits of conversation or documented work. I loved it so much I’m trying to think of how I can force my classes and my department to use it with me.

When one of my students had to leave town for a couple of weeks, we also used an app called Baiboard in conjunction with Google Hangouts to keep in touch. Baiboard served as our collaborative whiteboard while we were chatting on Google, so that we could see what the others were writing without having to try to type things on the fly, or hold paper up to the camera. I will definitely use this app again for things like online office hours.

This summer was also my first experience teaching students to use LaTeX, and it went much better than I expected. They collaborated on Overleaf to write up their results and make their Beamer presentation for MathFest. They even got comfortable with the tikz package for their graphics. The only real help I gave them was a couple old documents of mine to modify, and they picked the rest up incredibly quickly (somehow). One student has a lot of CS experience so I’m not entirely surprised that he adapted well, but even my normally-computer-phobic student became a LaTeX whiz almost immediately. I think at that point in the summer they might have just been excited to have problems whose solutions were Googleable.

They wrote their presentation and practiced it for a couple of my department members, who had excellent advice. I haven’t even seen that many undergraduate talks, much less helped anyone write one, so I was very grateful for their experience. We ended up scaling down the scope of the talk a little and let them focus on a few main aspects of their results, which helped the cohesiveness a lot. The students rehearsed more than I ever have for anything in my life, and when the time came to present, they got over their nerves and just killed it. I was beaming.

Bringing students to a conference did present some interesting logistical challenges: spending three days – including about 12 hours in the car – with two rising sophomores meant I definitely overheard more juicy student gossip than I ever needed to. And I just assumed that they’d know how to get around a city on their own, even though they haven’t spent much time away from their parents. This proved false when I got a call asking me to rescue them from across town when they’d gotten on the wrong bus by accident.

But we all had a really good time at MathFest. I enjoyed getting to interact with these students outside of the classroom, and getting to know them outside of their mathematical interests. Turns out they’re pretty cool people! And they loved getting to see aspects of mathematics they’d never been exposed to, though I think they found a lot of the conference just plain overwhelming.

I probably won’t try to apply for this research grant next summer, as I’d rather take a break from managing a group and work exclusively on my own stuff for awhile. But I will definitely be doing this again in the future. I think this was about the best first experience I could have hoped for, and I think my students felt the same.

## National Service through Math

Original design for the “Be Patriotic” poster by Paul Stahr, 1917-18. Public Domain, from Wikipedia

Math-ional Service? Two thoughts lead to one blog today. First: the recent political conventions (and non-stop political coverage for the last year before the conventions) have got me thinking about government and public service. The ideal of government is to serve the public. We all participate in government in a basic way when we vote, but surely there are other ways that citizens (like, say, mathematicians) can serve each other and the world through involvement in government. Second thought: once in a while, I want to do something else. I mean, do a job other than math professor. Occasionally, I want to do ANYTHING but be a math professor. Generally I am just looking for a temporary change, so I can use the skills that make me a good professor in a different way, or towards different goals… like maybe serving the public through working with the US government! Luckily, there are actually opportunities to do this as a professor. This blog is devoted to some government service opportunities for academic PhD mathematicians.

You may have heard of the American Association for the Advancement of Science (AAAS) Science and Technology Policy Fellowships. These are year-long fellowships open to professional researchers and educators with PhDs from any STEM field (master’s degrees in Engineering are also ok). The program accepts applicants to work in seven different areas to help with government policy related to science in technology. Fellows are based in Washington DC and receive a $75,000-$100,000 stipend plus benefits—this is a viable job for a full year. There are both congressional and executive branch fellowships available. The AMS sponsors one congressional fellowship each year specifically for a mathematician.

People do this program at many different stages of math life—early career, sabbatical, or transitioning out of academic mathematics. This work is right in line with my interests, so I applied for the Science and Technology Policy Fellowship two years ago. The application process is outlined on the website above. If you are chosen as a semi-finalist (I was), you will do a video-conference interview. Before my interview I was asked to write a briefing memo, explaining an area in which my expertise would be helpful to a policy maker. The people on the interview panel were very engaging. We talked about the memo I’d written and my interests in math and public policy. I have to say that the interview was a positive experience in itself. Though I was eventually became a finalist, I received an offer from Villanova at the same time, so I didn’t actually get to participate in the program. However, I think this fellowship is fantastic and I hope I get the chance to do this at some point in my career.

Another opportunity that gets less press is the Institute for Defense Analyses (IDA) Centers for Communication and Computing summer program for faculty, known as SCAMP. The Center for Communications Research has two offices, one in Princeton, NJ, the other in La Jolla, CA. Each summer these centers run 10-week summer workshops in which visiting and permanent researchers work on problems of interest to the government, including “cryptography, cryptanalysis, algorithms, high-performance computing, information processing, signal processing, and network security, as well as related areas of pure and applied mathematics,” according the CCR website.  I talked to several people who have gone to SCAMPs, and they reported really enjoying the problems and collaborative atmosphere. One thing to consider is that attending these workshops requires obtaining a security clearance, which is not a quick (or necessarily easy) process. It probably makes sense to start working on this early in the fall if you are interested in attending the following summer. If you are interested in joining a SCAMP, email CCR La Jolla or CCR Princeton for more information.

Finally, an opportunity that gets incredibly little press: the Defense Science Study Group (DSSG). I recently heard about it from an alumnus who had a great experience in the program. An internet search made the DSSG seem very mysterious—I found a few references from alumni (including this and this), but very little official information. (Apparently there is a real website, which was down but will hopefully be up again soon.) Here’s the general idea: the DSSG is a group of science and engineering professors that meet/go on adventures for approximately 20 days per year for two years. A new class is chosen every two years from a field of faculty nominated by their home universities. The participants learn about national security issues, from active US government security personnel. As their brochure explains, “Group members interact with top-level officials from the Department of Defense (DoD), and other Government organizations, various Intelligence agencies, The White House, and Congress. Visits to military bases throughout the United States provide members with a unique perspective of operating forces and allow DSSG members to meet with senior commanders responsible for our nation’s defense. Tours of defense laboratories and industrial facilities provide further insight into the technical dimensions of national security.”

The goal of this program is to forge connections between scientists and engineers and the national security community, and to encourage academic scientists and engineers to contribute their knowledge to addressing national security issues. It is administered by IDA and sponsored by the Defense Advanced Projects Research Agency (DARPA). For a math professor with strong national security interests, who is interested in doing things like flying in military planes and visiting aircraft carriers, this program would be a perfect fit. However, individuals can’t directly apply for this program. To get into the program, you need to be nominated by someone, usually an official at your university (like a provost). While some institutions have internal application processes, many do not. Some provosts may not know about this opportunity or may not have considered nominating a mathematician for it. With that in mind, the alum I talked to suggested that if you are interested you should definitely contact your provost about a nomination. Recruitment for the next two-year class begins this September/October.

Of course, programs like these are not the only math-related ways to serve/effect change in government. You could always just go to Congress, Mr. Smith style. The Association for Women in Mathematics is doing this (okay, visiting congressional offices, not addressing Congress). In 2015, the AWM leadership and student chapter members visited Capitol Hill offices to meet with congressional staffers. Georgia College students, from a chapter led by Dr. Marcela Chiorescu, describe their group’s experience in DC in this press release. Villanova’s AWM student chapter is hoping to visit congressional offices this December—so excited about this!

Are there more programs out there? Other ideas on how to serve or get involved? Let me know in the comments!


text:
Guides How To Lose Weight Reviewing the 2020 Research About Vitality Code Diet

Reviewing the 2020 Research About Vitality Code Diet

The Vitality Code Diet uses a ketogenic model to help users lose weight and feel their best while continuing to eat the foods they love.

It is no secret that a number of people across the United States today lead unhealthy, sedentary lifestyles which eventually leads them to suffer from a number of physical and mental ailments. For example, as per data released by the US Center for Disease Control (CDC), obesity-related issues were found to affect more than 93 million adults in the country (through 2015~2016.) In this regard, it also bears mentioning that people suffering from excess weight-related issues have also been found to suffer from a number of other problems such as heart disease, stroke, type 2 diabetes and certain types of cancer.

If that wasn’t shocking enough, a research study released a few years ago has shown that the approximate medical cost of treating obesity-related issues in the United States lays somewhere around the $147 billion threshold. The study also states that the average cost of treating people suffering from obesity is nearly$1,500 more than people with normal weight levels.

Lastly, a survey conducted by a reputable fitness media outlet in 2013 showed that of a select group of individuals who claimed to be dieting regularly, two out of five quit within the first week itself. Not only that, it is estimated that only 20% of these individuals stuck with their goals and made it past the three-month mark. Clearly, consumers often need more than just simple diet and exercise to guarantee success in their weight loss and wellness goals.

Let's review Vitality Code Diet and see if the new health plan is worthy of a second look or if sticking with a keto shake, keto diet pill or ketogenic diet challenge may be optimal.

Check out the Vitality Code Diet Here

What is the Vitality Code Diet?

As the name seems to suggest, Vitality Code is a ketogenic diet plan that seeks to help users maintain their vitality and wellness through a proper diet. According to Vitality Code Diet creator Amanda S., most weight loss programs get key aspects of the weight loss equation completely wrong. She had previously tried out a number of different weight loss plans, including Paleo, Whole 30, Elimination Diets, Carb Cycling, HCG, Alkaline Diet and Mediterranean Diet. For a number of reasons, though, all of these weight loss plans fell short of helping her achieve her goals.

The Vitality Code Diet outlines that most diets fail because of three main reasons. These include:

• Forced Starvation. Most weight loss systems impose insane eating restrictions on their users for extended periods of time. And while such methods may work initially, they usually end up doing more harm than good to one’s body in the long run.
• Nutritional Imbalance. Another reason why many diet plans fail to deliver long-lasting results is because they compromise on the nutritional intake of certain key ingredients, minerals, and vitamins that our body requires for optimal functionality.
• Sudden Shift in Hormonal Levels. Many diets (such as High-Fat Low-Carb) force people to give up certain foods that they are used to consuming on a regular basis. This can cause one’s hormone levels to fluctuate wildly. This causes malfunctions in the thyroid and pituitary glands, which can negate the effects of the weight loss routine and even cause weight gain.

Users probably have a number of questions about this interesting new take on the traditional ketogenic diet. While our information is limited because we have not personally purchased a subscription to the Code, this section will include answers to some of the most commonly asked questions about The Vitality Code and the company behind it.

Q: Who created The Vitality Code?

A: This weight loss regimen was created by Amanda S. Little information is given about her background; she merely explains that she was a mother of four and became disenchanted with other weight loss programs that were available to her. This is a bit troubling, as consumers cannot be sure that the program was created by someone with a clinical understanding of human nutrition.

A: The Vitality Code relies on the principle of ketosis. When the body enters a state of ketosis, it begins to burn fat instead of carbohydrates for the energy it needs. This helps to quickly burn through fat and lose weight for the user. For The Vitality Code, this state of ketosis is attained without the use of calorie counters or the gym. It is unclear how, precisely, the program helps users enter this state; the site simply explains that most keto diets “do it wrong.”

Q: Who can use The Vitality Code?

A: The Vitality Code is not made specifically for any demographic; all adults can use the program. We would advise against children beginning any type of ketogenic diet without the prior approval of their primary care physician.

Q: Can you eat whatever you want on The Vitality Code Diet?

A: Yes! Generally, the only restriction applied to ketogenic diets is that users should curb their intake of carbohydrates, mainly found in breads and grains. It is unclear what specific restrictions are placed on users by The Vitality Code, but it is likely that the program asks users to gradually decrease their carb intake.

Vitality Code Diet Benefits

To start off with, the Vitality Code Diet seeks to address many of the common health issues that a lot of people around the world deal with on a day to day basis. For example, when followed diligently, the system can help allay a host of issues related to dizziness, digestive malfunctioning, respiratory difficulties, and chronic fatigue.

According to its creators, The Vitality Code:

• Optimizes Gastrointestinal Health. By gradually eliminating the unhealthy fat and grease deposits that tend to line up inside our intestines, the Vitality Code diet helps rejuvenate one’s gastrointestinal system, thereby making digestion smoother and much less taxing on our circulatory system.
• Promotes physical and mental balance. One of the key facets of Vitality Code is that it helps restore the hormonal balance in our bodies. Not only that, the diet also helps maximize the efficiency of our neural network, thus allowing for increased mental clarity, enhanced focus, improved spatial awareness, etc.
• Helps reduce anxiety. People who have followed the Vitality Code diet diligently have seen a drastic reduction in their anxiety and nervousness levels. This is because the program helps maximize the efficiency of our circulatory system in a highly streamlined manner.
• Mitigates health problems. When followed on a regular basis, the official company website claims that the Vitality Code diet can help prevent a number of major physical ailments such as appendicitis, arthritis, bladder pain, candida, carpal tunnel syndrome, colitis, etc.
• Achieves ketogenic state. As mentioned in an earlier section, this diet is largely based on the keto weight loss system – with a few minor changes here and there. In this regard, followers of this diet are required to refrain from consuming substantial quantities of carbohydrates — since they are known to metabolize into sugars and fats quite quickly.
• Sustainable in the long run. Research studies have shown that the Vitality Code diet plan can be followed by users for extended periods of time without any harmful side effects.
• Optimizes metabolism. One of the most important aspects of Vitality Code is that it helps enhance the metabolic activity of our bodies. This not allows for faster weight loss but also allows individuals to feel more vital and fresh through the course of their daily activities.
• Is easy-to-follow. The plan clearly lays out all the foods that are good for one’s body and those that aren’t. Not only that, the program also helps educate users about a number of so-called “health foods” that have been clinically shown to sabotage one’s weight loss efforts.
• Accelerates fat loss. Thanks to the ketogenic principles that this diet is based upon, it helps users lose large amounts of weight within short periods of time. In this regard, it should be pointed out that when followed diligently, users can lose around 3-5 kgs within the first few weeks of following the Vitality Code diet.
• Spurs muscular development. Traditionally, high carb-low fat diets have been shown to reduce one’s muscle mass levels. However, the Vitality Code system not only maintains our muscle mass but also helps increase the strength of our tissues and tendons so as to enhance our overall strength and endurance capacity.

More Vitality Code Facts

In addition to the main benefits offered by the Vitality Code, users can expect to avoid some of the pitfalls that come with many other diets in the current dieting industry. These points are important; the types of trends avoided by the Vitality Code often deter successful adherence to the diet, reversing the effects of weight loss and preventing real progress for users.

By using the Vitality Code system, users will avoid:

• Counting calories. Unlike a number of other systems, Vitality Code does not require people to monitor their daily food intake incessantly or even count their calories meticulously. Instead, all that is required from the user is that they minimize the consumption of certain harmful nutrients from their regular meal plans.
• Food tracking. When following the Vitality Code dietary system, it is not necessary for users to keep an accurate tab of the food that they have consumed over the course of their day.
• Food intake restrictions. As mentioned earlier, people making use of Vitality Code do not have to starve themselves or keep their food intake at a bare minimum. Instead, they can consume large meal portions without having to compromise on the taste or quality of their food.
• The gym. More often than not, dieting plans require users to spend substantial amounts of time in the gym to see real results. However, when following this diet plan, no such thing is necessary. While the gym certainly won't hurt the process, users don't need to obsess over lifting weights or running to see results while using the Vitality Code.

To purchase Vitality Code, just follow the instructions on the official Vitality Code website.  Right now, the program is being offered at a discounted rate of $37, which is$30 less than its regular cost of \$67. Payments can be completed using a host of safe and secure methods including AMEX, Visa, Mastercard, etc.

For additional questions or support, contact customer support at this email address: customerservice@vitalitycodediet.com.

Last but not least, the product comes with a 60-day money-back guarantee in case users are not satisfied with its overall quality or effectiveness.

Final Thoughts

The main premise of the Vitality Code is simple; the weight loss plan allows users to take advantage of the ketogenic diet in a way that keeps them from excessive time in the gym, hours spent counting calories, and the inevitable pitfalls that come with diets that preach under-eating and self-deprivation. While consumers should still consult their physician before beginning any new radical diet program, The Vitality Code offers a simple solution to the problem of obesity.

Check out the Vitality Code Diet Here


text:
# Express the given limit as a definite integral

I'm supposed to convert this limit to a definite integral but I'm not sure how I'm supposed to do it.

$$\lim_{n\to \infty}\sum_{i=1}^n \left(\frac{1}{n}\right)\cdot\sqrt{ \frac i{n}}$$

Would appreciate some help/guidelines on how to tackle problems like this because I haven't found that many great sources to learn from when Googling it. My textbook is also really vague on this and doesn't really explain it or include examples.

• What is $\Delta x$? Once you recognize that, what is $f(x_i^*)$? From this, you can determine what $f(x)$ is and the bounds for the integral. – Clayton Feb 14 '18 at 21:02
• It's the limit of Riemann sums for $\int_0^1f.$ What is $f$ in this case? – zhw. Feb 14 '18 at 21:05
• I would assume that $f(x) = \sqrt x$ here? And $\Delta x$ is equal to $\frac 1 {n}$ ? – gbgult Feb 14 '18 at 21:08
• For any $f\in C^0([0,1])$ we have that $$\lim_{n\to +\infty}\frac{1}{n}\sum_{k=1}^{n}f\left(\frac{k}{n}\right) = \int_{0}^{1}f(x)\,dx.$$ In your case $f(x)=\sqrt{x}$ gives that the limit equals $\frac{2}{3}$. – Jack D'Aurizio Feb 14 '18 at 21:12
• @JackD'Aurizio is the $f$ in the sum necessarily the same $f$ of the integral? To me it seems that only on $\mathbb{Q}$ they have to give the same picture. Since for example for no $k,n\in \mathbb{N}\quad f(k/n)=f(1/\sqrt{2}).$ I accept that the result is the same but the functions don't have to be the same, right? – ty. Feb 14 '18 at 21:43

By inspection, this looks like a Riemann sum of some function where the partition consists of $n$ evenly spaced points between $0$ and $1$. That is, if $f(x)=\sqrt{x}$, and $P=\{0,1/n,2/n,\ldots, n/n\}$, then a (uppeR) Riemann sum of this function over $P$ is $$\sum_{i=1}^n\left(\frac{i+1}{n}-\frac{i}{n}\right)f\left(\frac{i}{n}\right)=\sum_{I=1}^n\frac{1}{n}\sqrt{\frac{i}{n}}.$$
Since $f$ is integrable on $0,1$, this converges to $\int_0^1 f$ as $n\to\infty$.
• Is there some "general things" that I should start looking for in problems like this to see what $\Delta x$ etc is? This was a pretty easy question so I guess it won't be as simple for me going forward to more advanced stuff. Also, I don't really get how you get that $( \frac {i+1} {n} - \frac i {n}) f( \frac i {n})$ is equal to the sigma notation in my post? – gbgult Feb 14 '18 at 22:19


text:
# Calculating angles when sides are known - Without Trignometric ratios.

The question is:

ABCD is a parallelogram and BFDE is a square . If AB is 20 and CF is 16 what is the perimeter of the parallelogram.

The question is fairly simple and I know how to solve it. However how would I get the remaining angles of triangle FDC (highlighted) if i know that CD=20 and FD=12. I also know that angle F is 90 however the triangle doesn't seem to be 30-60-90 or a 45-45-90. Also I need to solve this without Trigonometric ratios since I wont be using calculator

-
Your question is basically "What are the angles of a 3-4-5" triangle, which is pretty googlable. The short answer is that you need to use trig functions to express these numbers. –  Cam McLeman Jun 19 '12 at 0:08
I realize that but I could obtain these angles using trigonometric ratios. I was just curious if it was possible to obtain them without trigonometric ratios which i guess is not possible. –  MistyD Jun 19 '12 at 0:12

The triangles $\,ABE\,,\,CDF\,$are congruent, thus $\,CD=20\,$. Apply now Pythagoras and get $\,FD^2=20^2-16^2=144\Longrightarrow FD=12=FB=ED\,$ , so $\,AD=BC=16+12=32\,,\,AB=CD=20\,$ and now just add.
The angle is not at all pleasant. In particular (in radians) it is not a rational multiple of $\pi$. –  André Nicolas Jun 19 '12 at 0:55


text:
Finding the TFs that regulate a set of genes
4
0
Entering edit mode
6.3 years ago
kamula ▴ 10

I've Identified a list of co-expressed genes, and I'd like to figure out what is regulating their expression. Currently what I have done is run this group of genes through scope to identify motifs in their promoters. How do I then take those motifs and associate them with a TF? Or is there a better way of solving my problem that you know of?

motif tf expression • 2.9k views
0
Entering edit mode
6.3 years ago
TriS ★ 4.5k

A good starting point is the ENCODE CHiPSeq significance tools

starting from a list of genes it looks at published ENCODE data and tells you which TF bind to the region around your gene, you can define the distance you prefer from TSS or gene body

otherwise, to be a lil more fancy, you can use Homer

in this case you provide the region you want to analyze (i.e. 2kb up/downstream from your gene) and it will give you a list of TF that can bind the motifs in the regions you queried. it will do it for known and de-novo motifs.

0
Entering edit mode
6.3 years ago

Approach 1

-Collect immediate early promoters (1000 bp) --db like promoser
-Run the sequences through the TF databases
-Identify the over represented TFs

Approach 2

-Identify the TFs that are regulated (up or down) in your expression data set
-Correlate the expression of TFs with their target genes  (IPA might be helpful in this)

Approach 3

-Collect immediate early promoter elements (1000 bp) --db like promoser
-Submit the sequences to MEME
-See if you could over represented motifs that match with TF motifs (cumbersome)

0
Entering edit mode
6.3 years ago
Zhilong Jia ★ 2.0k

Another tool: Enrichr can make TF enrichment analysis using Transcription Factor PPIs

0
Entering edit mode
5.2 years ago
jin ▴ 80

You can retrieve the regulation among these genes at PlantRegMap for 132 plant species, which also provides a Tool to find enriched TFs in the gene set.


text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Mar 2019, 10:33

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# The average height of 40 students in a class is 172 cm. 30 students wh

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 53734
The average height of 40 students in a class is 172 cm. 30 students wh  [#permalink]

### Show Tags

20 Feb 2019, 05:11
00:00

Difficulty:

45% (medium)

Question Stats:

70% (02:10) correct 30% (02:40) wrong based on 23 sessions

### HideShow timer Statistics

The average height of 40 students in a class is 172 cm. 30 students whose average height is 172.5 cm left the class and 40 students whose average height is 170.5 cm joined the class. Find the average height of the present class.

A. 170 cm
B. 170.2 cm
C. 170.5 cm
D. 175
E. 177 cm

_________________
Manager
Joined: 10 Oct 2018
Posts: 144
Location: United States
GPA: 4
WE: Human Resources (Human Resources)
Re: The average height of 40 students in a class is 172 cm. 30 students wh  [#permalink]

### Show Tags

20 Feb 2019, 05:29
Average height of 40 students in a class is 172 cm
$$\frac{Σ40}{40}$$=172 => Σ40= 6880.......................(1)

30 students whose average height is 172.5 cm leave the class
$$\frac{Σ30}{30}$$=172.5 => Σ30= 5175.......................(2)

(1)-(2) will give Σ10= 1705.............(3)

40 students whose average height is 170.5 cm join the class
$$\frac{Σ40}{40}$$=170.5 => Σ40= 6820.......................(4)

The average height of the present class will be the average of 50 students.
Σ50=(3)+(4)=1705+6820=8525

Thus, $$\frac{Σ50}{50}$$=$$\frac{8525}{50}$$=170.5
_________________

SVP
Joined: 18 Aug 2017
Posts: 2400
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: The average height of 40 students in a class is 172 cm. 30 students wh  [#permalink]

### Show Tags

20 Feb 2019, 09:51
Bunuel wrote:
The average height of 40 students in a class is 172 cm. 30 students whose average height is 172.5 cm left the class and 40 students whose average height is 170.5 cm joined the class. Find the average height of the present class.

A. 170 cm
B. 170.2 cm
C. 170.5 cm
D. 175
E. 177 cm

sum of 40 students ; 172 * 40 ; 6880
and of 30 students who left class; 172.5 * 30; 5175
now left with 10 students in class and weight of class; 6880-5175; 1705
the weight of 40 new students; 40 * 170.5 ; 6820
so new total of 50 students; 6820 + 1705 ; 8525
avg ; 8525/50 ; 170.5 IMO C
_________________

If you liked my solution then please give Kudos. Kudos encourage active discussions.

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8997
Location: Pune, India
Re: The average height of 40 students in a class is 172 cm. 30 students wh  [#permalink]

### Show Tags

20 Feb 2019, 10:30
1
1
Bunuel wrote:
The average height of 40 students in a class is 172 cm. 30 students whose average height is 172.5 cm left the class and 40 students whose average height is 170.5 cm joined the class. Find the average height of the present class.

A. 170 cm
B. 170.2 cm
C. 170.5 cm
D. 175
E. 177 cm

Simply use the concept of aves discussed here to reduce calculations:

Avg of 40 students is 172 cm
30 students leave whose avg is 172.5 cm. So these 30 students had .5 cm extra i.e. total 15 extra.
When 15 cm is taken away from 10 students, this means 1.5 cm is taken away from each student so average goes down to 170.5 cm.

Now when 40 students with an average of 170.5 cm join, the average remains 170.5 cm.

_________________

Karishma
Veritas Prep GMAT Instructor

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 5375
Location: United States (CA)
Re: The average height of 40 students in a class is 172 cm. 30 students wh  [#permalink]

### Show Tags

22 Feb 2019, 16:39
Bunuel wrote:
The average height of 40 students in a class is 172 cm. 30 students whose average height is 172.5 cm left the class and 40 students whose average height is 170.5 cm joined the class. Find the average height of the present class.

A. 170 cm
B. 170.2 cm
C. 170.5 cm
D. 175
E. 177 cm

The sum of the heights of the original 40 students is 40 x 172 = 6,880 cm.

The sum of the 30 students who left is 30 x 172.5 = 5,175 cm.

The sum of the 40 students who joined is 40 x 170.5 = 6,820 cm

So the new sum is 6,880 - 5,175 + 6,820 = 8,525 cm, and the new number of students is 40 - 30 + 40 = 50. Therefore, the new average height is 8,525/50 = 170.5 cm.

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

Re: The average height of 40 students in a class is 172 cm. 30 students wh   [#permalink] 22 Feb 2019, 16:39
Display posts from previous: Sort by


text:
# Why upper/lower limit always has only one value?

I am a beginner to calculus, and I have a simple question on limits.

Consider the function $f(x)= 1/x$ for all real $x$. Then we know that upper limit of $x$ tends to infinity is $0$. This is because x can be arbitrarily large, and the larger $x$ is, the smaller $f(x)$ is. So $f(x)$ gets closer and closer to $0$, but it never is equal to $0$. So upper limit of $x$ tends to infinity is $0$. But we can also think of it this way that as $x$ increases, $f(x)$ decreases. As $f(x)$ tends to $0$, we can also say that it tends towards $-1$ because $x$ is getting larger and larger, thus $f(x)$ is getting smaller and smaller, and the smaller it gets, the closer it gets to both $0$ and $-1$. My question is: Why do we say upper limit of $x$ tends to infinity is $0$ and not $-1$?

• Because -1 is always at a distance of at least 1 from $f(x)$, whereas $0$ is at distance $\varepsilon$ if $x$ is big enough. – Andy Apr 23 '13 at 18:27
• The function must not just get closer and closer to the limit, but must be able to get arbitrarily close. $f(x)$ can get closer and closer to $-1$, but you cannot make $f(x)$ as close as you like to $-1$. However not only does $f(x)$ get closer and closer to $0$, you can make $f(x)$ as close as you like to $0$ by taking $x$ sufficiently large. – user50229 Apr 23 '13 at 18:58

## 2 Answers

I think you are confusing limit with lower bound. A lower bound is a number below all the values of the function. $-1$ is a fine lower bound for $\frac 1x$ as long as $x \gt 0$ because it is less than all the values of $\frac 1x$. $-\frac 12,-2,$ and $-\pi$ are also lower bounds for the same reason. A limit, if it exists, is unique. It is what the function "comes very close to". $\lim_{x \to \infty} \frac 1x = 0$ because, as you say, for $x$ very large $\frac 1x$ is very close to zero. It is not very close to $-1$, so $-1$ is not a (or the) limit of $\frac 1x$

$f(x)\underset{x\to+\infty}{\longrightarrow} l \underset{definition}{\Leftrightarrow} \forall \varepsilon >0, \exists A > 0, \forall x \ge A, \left|f(x)-l\right|\le\varepsilon$

For all $\varepsilon > 0$, we can take $A=\cfrac{1}{\varepsilon}$ and then, $\forall x \ge A,\left|\cfrac{1}{x}-0\right|=\cfrac{1}{x} \le \cfrac{1}{A}=\varepsilon$

So $\boxed{\cfrac{1}{x}\underset{x\to+\infty}{\longrightarrow}0}$

Now take $\varepsilon = \cfrac{1}{2}$

$\forall A \ge 0,$ for $x=2+A$, $\left|\cfrac{1}{x}-1\right|=\left|\cfrac{1}{2+A}-1\right|=\left|\cfrac{-1-A}{2+A}\right|=\cfrac{1+A}{2+A}\ge\cfrac{1}{2}=\varepsilon$

So since it doesn't work for that specific $\varepsilon$, it can not work $\forall \varepsilon$ so

$\boxed{\cfrac{1}{x}\not\underset{x\to+\infty}{\longrightarrow}1}$

(for some reason I can't get the strike on the arrow... If someone knows how to format it properly, please edit)

Also, the limit is always unique.

Suppose $f(x)\underset{x\to+\infty}{\longrightarrow} l_1$ and $f(x)\underset{x\to+\infty}{\longrightarrow} l_2$

$\forall \varepsilon >0, \exists A > 0, \forall x \ge A, \left|f(x)-l_1\right|\le\varepsilon$

$\forall \varepsilon >0, \exists A > 0, \forall x \ge A, \left|f(x)-l_2\right|\le\varepsilon$

Since $\forall x,\left|l_1-l_2\right|=\left|(l_1-f(x))-(l_2-f(x))\right|\le \left|f(x)-l_1\right|+\left|f(x)-l_2\right|$

$\forall \varepsilon >0, \exists A > 0, \forall x \ge A, \left|l_1-l_2\right|\le \varepsilon$

We can remove useless variables from the formula and we get

$\forall \varepsilon >0, \left|l_1-l_2\right|\le \varepsilon$

So $\left|l_1-l_2\right|=0$

That is $\boxed{l_1=l_2}$

• Could you please explain it? – abcd Apr 23 '13 at 18:28
• @abcd: I edited. – xavierm02 Apr 23 '13 at 18:50