text:
# Changeset 787:c2230649a493 in lemon-1.2 for lemon/full_graph.h

Ignore:
Timestamp:
11/15/09 19:57:02 (10 years ago)
Branch:
default
Phase:
public
Message:

Various doc improvements (#331)

• Add notes to the graph classes about the time of item counting.
• Clarify the doc for run() in BFS and DFS.
• Other improvements.
File:
1 edited

Unmodified
Added
Removed
• ## lemon/full_graph.h

 r780 /// only in the concept class. /// /// This class provides constant time counting for nodes and arcs. /// /// \note FullDigraph and FullGraph classes are very similar, /// but there are two differences. While this class conforms only /// completely static, the nodes can be indexed with integers from /// the range [0..nodeNum()-1]. /// The index of a node is the same as its ID. /// \sa index() Node operator()(int ix) const { return Parent::operator()(ix); } /// completely static, the nodes can be indexed with integers from /// the range [0..nodeNum()-1]. /// The index of a node is the same as its ID. /// \sa operator()() static int index(const Node& node) { return Parent::index(node); } /// only in the concept class. /// /// This class provides constant time counting for nodes, edges and arcs. /// /// \note FullDigraph and FullGraph classes are very similar, /// but there are two differences. While FullDigraph /// completely static, the nodes can be indexed with integers from /// the range [0..nodeNum()-1]. /// The index of a node is the same as its ID. /// \sa index() Node operator()(int ix) const { return Parent::operator()(ix); } /// completely static, the nodes can be indexed with integers from /// the range [0..nodeNum()-1]. /// The index of a node is the same as its ID. /// \sa operator()() static int index(const Node& node) { return Parent::index(node); }
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sin thetacos thetagtan theta

Solution : When the block slides down the plane with a constant speed then the inclination of the plane is equal to angle of repose (theta). <br> Coefficient of friction=tan of the angle of reposed =tan theta.

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# 10.2 Permutations and applications

 Page 1 / 2

## Permutations

The concept of a combination did not consider the order of the elements of the subset to be important. A permutation is a combination with the order of a selection from a group being important. For example, for the set $\left\{1,2,3,4,5,6\right\}$ , the combination $\left\{1,2,3\right\}$ would be identical to the combination $\left\{3,2,1\right\}$ , but these two combinations are different permutations, because the elements in the set are ordered differently.

More formally, a permutation is an ordered list without repetitions, perhaps missing some elements.

This means that $\left\{1,2,2,3,4,5,6\right\}$ and $\left\{1,2,4,5,5,6\right\}$ are not permutations of the set $\left\{1,2,3,4,5,6\right\}$ .

Now suppose you have these objects:

1, 2, 3

Here is a list of all permutations of all three objects:

1 2 3; $\phantom{\rule{0.277778em}{0ex}}$ 1 3 2; $\phantom{\rule{0.277778em}{0ex}}$ 2 1 3; $\phantom{\rule{0.277778em}{0ex}}$ 2 3 1; $\phantom{\rule{0.277778em}{0ex}}$ 3 1 2; $\phantom{\rule{0.277778em}{0ex}}$ 3 2 1. $\phantom{\rule{0.277778em}{0ex}}$

## Counting permutations

Let $S$ be a set with $n$ objects. Permutations of $r$ objects from this set $S$ refer to sequences of $r$ different elements of $S$ (where two sequences are considered different if they contain the same elements but in a different order). Formulas for the number of permutations and combinations are readily available and important throughout combinatorics.

It is easy to count the number of permutations of size $r$ when chosen from a set of size $n$ (with $r\le n$ ).

1. Select the first member of the permutation out of $n$ choices, because there are $n$ distinct elements in the set.
2. Next, since one of the $n$ elements has already been used, the second member of the permutation has $\left(n-1\right)$ elements to choose from the remaining set.
3. The third member of the permutation can be filled in $\left(n-2\right)$ ways since 2 have been used already.
4. This pattern continues until there are $r$ members on the permutation. This means that the last member can be filled in $\left(n-\left(r-1\right)\right)=\left(n-r+1\right)$ ways.
5. Summarizing, we find that there is a total of
$n\left(n-1\right)\left(n-2\right)...\left(n-r+1\right)$
different permutations of $r$ objects, taken from a pool of $n$ objects. This number is denoted by $P\left(n,r\right)$ and can be written in factorial notation as:
$P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}.$

For example, if we have a total of 5 elements, the integers $\left\{1,2,3,4,5\right\}$ , how many ways are there for a permutation of three elements to be selected from this set? In this case, $n=5$ and $r=3$ . Then, $P\left(5,3\right)=5!/7!=60!$ .

Show that a collection of $n$ objects has $n!$ permutations.

1. Proof: Constructing an ordered sequence of $n$ objects is equivalent to choosing the position occupied by the first object, then choosing the position of the second object, and so on, until we have chosen the position of each of our $n$ objects.

2. There are n ways to choose a position for the first object. Once its position is fixed, we can choose from $\left(n-1\right)$ possible positions for the second object. With the first two placed, there are $\left(n-2\right)$ remaining possible positions for the third object; and so on. There are only two positions to choose from for the penultimate object, and the $nth$ object will occupy the last remaining position.

3. Therefore, according to the fundamental counting principle, there are

$n\left(n-1\right)\left(n-2\right)...2×1=n!$

ways of constructing an ordered sequence of $n$ objects.

how can chip be made from sand
is this allso about nanoscale material
Almas
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where is the latest information on a no technology how can I find it
William
currently
William
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
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text:
# Why is a simply connected homology sphere a topological sphere?

I post this for a friend who currently doesn’t have access to this site.

It is about an implication in the last paragraph of the following paper:

The question is about the last sentence:

Since $$\tilde{M}$$ is simply connected, it is also a topological sphere.

Why is this true? Is it easy to work out a proof of this by hand ? If not may I ask a reference from which the question simply follows?

Sorry if it is a well-known fact in algebraic topology (if so may you point out a reference?)

Thanks a lot !

• I have not looked at the paper, but this may be the key point: en.wikipedia.org/wiki/Generalized_Poincar%C3%A9_conjecture Nov 25 '21 at 9:30
• It is more or less topological Poincare conjecture Nov 25 '21 at 18:01
• I am curious why someone doesn't have access to this site. Even Valery Fabrikant visits. (Привет!) Nov 28 '21 at 10:03
• @BenMcKay: maybe en.wikipedia.org/wiki/Great_Firewall Nov 28 '21 at 14:36

There are few steps.

The first one. A simply connected homology sphere is a homotopy sphere actually. It follows from the combination of the Whitehead and Hurewicz theorems. By the Hurewicz theorem, $$\pi_n(X) \cong H_n(X) \cong \mathbb Z$$. Therefore, there is a map inducing homology isomorphism. And by the Whitehead theorem it is a homotopy equivalence.

The second one. Any homotopy sphere is actually topological sphere. It follows from classification theorem in dimension 2. It follows from Poincare conjecture in dimension 3. And finally, in dimension n $$\geqslant 4$$ it follows from the generalized Poincaré conjecture, proved by Smale 1960-61 for n ≥ 5 and ca. 1982 by Michael Freedman for n = 4.

• Perhaps it's worth noting that the result that "any homotopy sphere is actually a topological sphere" is often called "the generalised Poincaré conjecture" (en.wikipedia.org/wiki/Generalized_Poincaré_conjecture), and its proof in dimension 4 is quite different from the proof in dimensions $\geq 5$.
– HJRW
Nov 28 '21 at 16:49
• @HJRW I thought that the only difference in dimension 4 is an application of the Whitney trick. Nov 30 '21 at 17:54
• No: see this recent Quanta article to get an idea of how much more difficult the 4d case is. quantamagazine.org/…
– HJRW
Nov 30 '21 at 20:50


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# Carbaryl과 Chlorothalonil의 공존이 Carassius auratus(goldfish)를 이용한 생물농축계수에 미치는 영향

• 민경진 (계명대학교 자연과학대학 공중보건학과) ;
• 김근배 (계명대학교 자연과학대학 공중보건학과) ;
• 차춘근 (계명대학교 자연과학대학 공중보건학과) ;
• 박천만 (계명대학교 자연과학대학 공중보건학과) ;
• 강회양 (계명대학교 자연과학대학 공중보건학과)
• Published : 1996.12.01

#### Abstract

This study was performed to investigate the effect of co-existence of carbaryl and chlorothalonil on the short-term bioconcentration factor in Carassius auratus(goldfish). The fishes were exposed to the combined treatment of carbaryl and chlorothalonil(0.05 ppm+0.005 ppm, 0.05 ppm+0.010 ppm, 0.10 ppm+0.005 ppm) for 1, 3 and 5 days, respectively. Carbaryl and chlorothalonil in fish and in test water were extracted with n-hexane and acetonitrile. GC-ECD was used to detect and quantitate carbaryl and chlorothalonil. 1-day, 3-day and 5-day bioconcentration factors($BCF_1, BCF_3$ and $BCF_5$) of each pesticide were calculated from the quantitation results. The depuration rate of each pesticide from the whole body of fish was determined over the 72-h period after combined treatment. The results were as follows: $BCF_1$ values of carbaryl were 3.521, 3.802 and 3.587, respectively, when the concentration of carbaryl and chlorothalonil in combined treatment were 0.05+0.005, 0.05+0.010 and 0.10+0.005 ppm. BCF3 values of carbaryl were 4.825, 4.556 and 3.828, respectively, and $BCF_5$ values of carbaryl were 3.974, 3.921 and 4.186, respectively, under the conditions. While $BCF_1$ of chlorothalonil were 0.829, 0.829 and 1.540, respectively, under the same condition of pesticide concentrations $BCF_3$ of chlorothalonil were 2.040, 2.208 and 3.633, respectively, and $BCF_5$ of chlorothalonil were 6.222, 6.667 and 7.095, respectively, under the conditions. Depuration rate constants of carbaryl were 0.022, 0.022 and 0.152, respectively, when the concentration of carbaryl and chlorothalonil in combined treatment were 0.05+0.005, 0.05+0.010 and 0.10+0.005 ppm. While depuration rate constants of chlorothalonil were 0.004, 0.004 and 0.006, respectively, under the same condition of pesticide concentrations. It was observed that no significant differences of carbaryl and chlorothalonil concentration in fish extracts, test water and $BCF_s$ of carbaryl and chlorothalonil between combined treatment and single treatment. It was considered that no appreciable interaction at experimental concentrations was due to low concentrations, 0.005~0.1 ppm. Co-existence of carbaryl and chlorothalonil had no effect on excretion of each pesticide and depuration rate of chlorothalonil was investigated 1/8 slower than that of carbaryl in combined treatment. Therefore, it is considered that the persistence of chlorothalonil in fish body would be higher than that of carbaryl.


text:
MathSciNet bibliographic data MR663889 58E99 (28D05 54F20 58F12) Handel, Michael A pathological area preserving $C\sp{\infty }$$C\sp{\infty }$ diffeomorphism of the plane. Proc. Amer. Math. Soc. 86 (1982), no. 1, 163–168. Article

For users without a MathSciNet license , Relay Station allows linking from MR numbers in online mathematical literature directly to electronic journals and original articles. Subscribers receive the added value of full MathSciNet reviews.


text:
# MacOS – Can I put /tmp and /var/log in a ramdisk on OS X

macmacosramdisktmpfs

For non-critical Linux systems, I often move things like /tmp and /var/log to tmpfs to save on some disk writing. I've been doing this for a year or so and if I ever need the logs across reboots, I just comment out a line in /etc/fstab and then start debugging.

In any case, I would like to do the same thing on OS X. I've seen posts on creating a ramdisk for OS X, but I'm looking for a more permanent solution that works on every boot. I always want /tmp and /var/log mounted in a ramdisk, with the ability to turn that off with a bit of command-line editing in vi if I have to.

#!/bin/bash
ramfs_size_sectors=$((${ramfs_size_mb}*1024*1024/512))
ramdisk_dev=hdid -nomount ram://${ramfs_size_sectors} newfs_hfs -v 'Volatile'${ramdisk_dev}
mkdir -p ${mount_point} mount -o noatime -t hfs${ramdisk_dev} ${mount_point} echo "remove with:" echo "umount${mount_point}"


text:
# Closedness in topological product

• Feb 24th 2009, 08:18 AM
Amanda1990
Closedness in topological product
Let $X$ and $Y$be topological spaces with $Y$compact and let $p_X : X \times Y \rightarrow X$denote the projection of the topological product on $X$, $p_X (x,y) = x$. Given that $W \subseteq X \times Y$is closed in $X \times Y$, prove that $p_X (W)$ is closed in X.
• Feb 24th 2009, 06:58 PM
aliceinwonderland
Quote:

Originally Posted by Amanda1990
Let $X$ and $Y$be topological spaces with $Y$compact and let $p_X : X \times Y \rightarrow X$denote the projection of the topological product on $X$, $p_X (x,y) = x$. Given that $W \subseteq X \times Y$is closed in $X \times Y$, prove that $p_X (W)$ is closed in X.

Tube lemma: Consider the product space $X \times Y$, where Y is compact. If N is an open set containing the slice $x_0 \times Y$ of $X \times Y$, then N contains some tube $V \times Y$ about $x_0 \times Y$, where V is a neighborhood of $x_0$ in $X$.

We shall show that $X \setminus p_X(W)$ is open in X using a tube lemma, which is equivalent to showing that $p_X(W)$ is closed in X.

Let $x_0$ be any element in $X \setminus p_X(W)$. By the tube lemma, there exists a tube $V \times Y$ such that

$x_0 \times Y \subset V \times Y \subset (X\setminus p_X(W))\times Y$.

Now, for any $x_0$ in $X \setminus p_X(W)$, we have an open set V containing $x_0$ such that $V \subset (X \setminus p_X(W))$. Thus, $X \setminus p_X(W)$ is open in $X$.


text:
nested-sequence-0.2: List-like data structures with O(log(n)) random access

nested-sequence-0.2: List-like data structures with O(log(n)) random access

List-like data structures implemented using nested data types and polymorphic recursion. Also called "n-ary random access lists". They supports O(log(n)) lookup while still having amortized O(1) access to the left end of the sequence. Somewhat similar to finger trees, but much simpler, and the ternary and quaternary versions are also more memory efficient; however, modifying the right end of the sequence is still slow. See Data.Nested.Seq for general comments and Data.Nested.Seq.Binary.Lazy for an explanation of the data structure.


text:
University Calculus: Early Transcendentals (3rd Edition)

$-13$ and $-2$
Since, $f_x(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0+h,y_0)-f(x_0,y_0)}{h}$ and $f_y(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0+h)-f(x_0,y_0)}{h}$ $f_x(1,2)=\lim\limits_{h \to 0} \dfrac{f(1+h,2)-f(1,2)}{h}=\lim\limits_{h \to 0} \dfrac{-6h^2-13h}{h}=-13$ $f_y(1,2)=\lim\limits_{h \to 0} \dfrac{f(1,2+h)-f(1,2)}{h}$ or, $\lim\limits_{h \to 0} \dfrac{-4-2h+4}{h}=-2$


text:
# Mathematical notation#

Bokeh supports mathematical notations expressed in the LaTeX and MathML markup languages with a growing number of elements. Currently, you can use LaTeX and MathML notations with the following elements:

Bokeh uses the MathJax library to handle LaTeX and MathML. See the official MathJax documentation for more information on MathJax.

Note

If you use the components() function, make sure to include the bokeh-mathjax- resource in your html template.

## LaTeX#

To use LaTeX notation, you can pass a string directly to any supported element. This string needs to begin and end with one of the MathJax default delimiters. These delimiters are $$...$$, $...$, and $$...$$. For example: r"$$\sin(x)$$".

LaTeX on axis labels, titles, and labels

To use LaTeX notation as an axis label, title, or label, pass a raw string literal beginning and ending with MathJax default delimiters and containing LaTeX notation. For example:

from numpy import arange, pi, sin

from bokeh.models.annotations.labels import Label
from bokeh.plotting import figure, show

x = arange(-2*pi, 2*pi, 0.1)
y = sin(x)

p = figure(height=250, title=r"$\sin(x)\text{ for }x\text{ between }-2\pi\text{ and }2\pi$")
p.circle(x, y, alpha=0.6, size=7)

label = Label(
text="$$y = \sin(x)\$$",
x=150, y=130,
x_units="screen", y_units="screen",
)

p.yaxis.axis_label = r"$\sin(x)$"
p.xaxis.axis_label = r"$x\pi$"

show(p)

LaTeX and tick labels

To add LaTeX notations to your tick labels, use the major_label_overrides() function with an axis.

This function is used to replace values for existing tick labels with custom text. It accepts a dictionary with the tick label’s original value as the key and your custom value as the dict’s value.

Use this function to replace any plain text tick labels with LaTeX notation:

from numpy import arange

from bokeh.plotting import figure, show

x = arange(1, 4.5, 0.25)
y = 1 / x

plot = figure(height=200)
plot.circle(x, y, fill_color="blue", size=5)
plot.line(x, y, color="darkgrey")

plot.xaxis.axis_label = "Resistance"
plot.xaxis.ticker = [1, 2, 3, 4]
plot.yaxis.axis_label = "Current at 1 V"

plot.xaxis.major_label_overrides = {
1: r"$$1\ \Omega$$",
2: r"$$2\ \Omega$$",
3: r"$$3\ \Omega$$",
4: r"$$4\ \Omega$$",
}

show(plot)

LaTeX on RangeSlider and Slider widget titles

To use LaTeX notation in the title of a RangeSlider or Slider widget, pass a raw string literal beginning and ending with MathJax default delimiters and containing LaTeX notation as the title parameter. For example:

from bokeh.io import show
from bokeh.models import Slider

slider = Slider(start=0, end=10, value=1, step=.1, title=r"$$\delta \text{ (damping factor, 1/s)}$$")

show(slider)

LaTeX with div and paragraph widgets

To include LaTeX notation in the text of a div widget or paragraph widget, use the standard MathJax default delimiters anywhere within your string:

from bokeh.io import show
from bokeh.models import Div

div = Div(
width=400, height=100, background="#fafafa",
text="The Pythagorean identity is $$\sin^2(x) + \cos^2(x) = 1$$"
)

show(div)

To disable LaTeX rendering for a div or paragraph widget, set the widget’s disable_math property to True.

You can use some of Bokeh’s standard text properties to change the appearance of rendered math text. Use text_font_size to change the font size, use text_color to change the color. For example:

p.xaxis.axis_label = r"$$\nu \:(10^{15} s^{-1})$$"
p.xaxis.axis_label_text_color = "green"
p.xaxis.axis_label_text_font_size = "50px"

Text color and sizes defined in a Bokeh theme also work.

Additionally, you have the option to use the LaTeX extensions included in MathJax. For example, use \text{} to combine literal text with a math expression. Or use the color extension to change the color of the rendered LaTeX notation: \color{white} \sin(x). Text properties set with a LaTeX extension override any text properties set elsewhere in your code or in a theme.

Note

There are limitations to how much of LaTeX MathJax supports. See Differences from Actual TeX in the MathJax documentation for more details.

## MathML#

To add mathematical notations written in MathML, use Bokeh’s MathML model directly. This model has a text property that accepts a string containing MathML. For example:

from numpy import arange

from bokeh.models import MathML
from bokeh.plotting import figure, show

x = arange(-10, 10, 0.1)
y = (x * 0.5) ** 2

mathml = """
$<mrow> <mfrac> <mn>1</mn> <mn>4</mn> </mfrac> <msup> <mi>x</mi> <mn>2</mn> </msup> </mrow>$
"""

plot = figure(height=200)
plot.line(x, y)

plot.xaxis.axis_label = MathML(text=mathml)

show(plot)

Similar to LaTeX, you can also use Bokeh’s standard text properties text_font_size and text_color to change font size and color for MathML notations. For example:

plot.xaxis.axis_label = MathML(text=mathml)
plot.xaxis.axis_label_text_color = "green"
plot.xaxis.axis_label_text_font_size = "50px"

For more information, see MathML in the reference guide.


text:
## BdMO National 2013: Junior 9

BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### BdMO National 2013: Junior 9

The ratio of GCD and LCM of two integers is $1: 36$ and sum of the integers is $5460$. What is the difference between these two integers?

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

### Re: BdMO National 2013: Junior 9

let two numbers are a and b.
and let $\left ( a,b \right )=x \Leftrightarrow a=xa_{1}$ and $b=xb_{1}$ where $\left ( a_{1},b_{1} \right )=1$
so,$\left [ a,b \right ]=xa_{1}b_{1}$

but given that, $\frac{(a,b)}{[a,b]}=\frac{1}{36}$
$\frac{x}{xa_{1}b_{1}}=\frac{1}{36}\Leftrightarrow \frac{1}{a_{1}b_{1}}=\frac{1}{36}\Leftrightarrow a_{1}b_{1}=36$

so now we get,$a_{1}b_{1}=36$ and $(a_{1},b_{1})=1$
the pair which satisfy these two property of $a_{1}$ and $b_{1}$ is $(4,9)$; that means $a_{1}=4$ and $b_{1}=9$

$\therefore a=4x$ and $b=9x$
given that $a+b=5460$
so, $a+b=5460\Leftrightarrow 4x+9x=5460\Leftrightarrow 13x=5460\Leftrightarrow x=420$

finally $x=420;a_{1}=4;b_{1}=9$

so $a=420*4=1680$ and $b=420*9=3780$
and difference $=3780-1680=2100$......(ans:)....

Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm

### Re: BdMO National 2013: Junior 9

Another possible pair satisfying the conditions for $(a_1, b_1)$ is $(1,36)$.
Although it does not lead to any significant discovery, ignoring it might mean you losing out on some easy points for the problem.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.

"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

### Re: BdMO National 2013: Junior 9

i noticed that. but for pair (1,36), our next equation comes 37x=5460. but 37 does not divide 5460. then x becomes fraction. but x must be a integer. so (1,36) is not possible. i forget to include these steps in my solution

Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm

### Re: BdMO National 2013: Junior 9

Yes, that pair does not lead us to a solution. But it is a good practice to include these small details in your solution.
Otherwise you will lose 1 or 2 points in BdMO/BdMC/IMO. It was just a heads up.
I request, if it's still possible, please hide the solutions. It helps others.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.

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Swapnil Barua
Posts: 14
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### Re: BdMO National 2013: Junior 9

In 2 numbers GCD their common factor remains multiple of those numbers and in LCM it remains with the common factors and other factors remains multiple. So in the ratio of GCD and LCM common factors cancel out and remains the others factors without those common factors.
Now, 36=2×2×3×3 these 4 factors can be of similar numbers [case 1] or 2×2 factor is of 1 number and 3×3 is of another.[case 2]
First take these 4 factors are of similar numbers, If 1 number is 36x another is x
According to the question
36x+x= 5460
or, 37x= 5460
or,x= 147.57 [ Not an integer]
Now take the 2nd case,
If 1 number is 4x another is 6x
According to the question
4x+6x=5460
or, 10x= 5460
or, x= 546
Difference=6x-4x=2x= 2× 546= 1092


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# In a population of 100,000 males, 80% can be expected to live to age 6

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In a population of 100,000 males, 80% can be expected to live to age 6 [#permalink]

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In a population of 100,000 males, 80% can be expected to live to age 60 and 60% can be expected to live to 80. Given that a male in this group is 60, what is the probability that he lives to 80?

A. 48%
B. 60%
C. 75%
D. 78%
E. 80%
[Reveal] Spoiler: OA

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Re: In a population of 100,000 males, 80% can be expected to live to age 6 [#permalink]

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13 May 2017, 12:26
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Of the 100000 males, there are 80000 males who make it to age 60.
Similarly, there are 60000 males who make it to age 80.

Since we are asked the probability of how many in the age of 60, make it to 80 :
$$\frac{60000}{80000}*100$$= 75%(Option C)
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Re: In a population of 100,000 males, 80% can be expected to live to age 6 [#permalink]

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15 May 2017, 05:19
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pushpitkc wrote:
Of the 100000 males, there are 80000 males who make it to age 60.
Similarly, there are 60000 males who make it to age 80.

Since we are asked the probability of how many in the age of 60, make it to 80 :
$$\frac{60000}{80000}*100$$= 75%(Option C)

Bunuel wrote:
In a population of 100,000 males, 80% can be expected to live to age 60 and 60% can be expected to live to 80. Given that a male in this group is 60, what is the probability that he lives to 80?

A. 48%
B. 60%
C. 75%
D. 78%
E. 80%

https://www.dartmouth.edu/~chance/teach ... apter4.pdf Refer to example 4.2

P(a) = 80 percent of the 100000 males, 80,000 males, can be expected to live to 60- but

P(b) = 60 percent of 100000 males, 60,000, males can be expected to live to

b is a subset of a so 60000/80000 * 100 = 75

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Re: In a population of 100,000 males, 80% can be expected to live to age 6 [#permalink]

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06 Jul 2017, 03:15
Bunuel wrote:
In a population of 100,000 males, 80% can be expected to live to age 60 and 60% can be expected to live to 80. Given that a male in this group is 60, what is the probability that he lives to 80?

A. 48%
B. 60%
C. 75%
D. 78%
E. 80%

Let me first write down the data given in the question, and for simplicity, I'm assuming tat there are 100 people.

By age 0
Alive = 100

By age 60
Alive = 80

By age 80
Alive = 60

Now, consider the following questions and my understanding

What is the percentage of people who are already 60 will be alive at 80?

Answer = 60/80 = 75%

In how many ways can we select a person who is 60 years old and will be alive at 80?

$$\frac{60C1}{80C1}$$

75%

Now my understanding of the original question

If a person is randomly selected from people who are 60, what is the probability that he will live to 80?

In this case. we have first already selected a person, now this particular person can be a part of the 60 people who will live to 80, or of the 20 people who will not. We have to calculate the possibility that this particular person is a part of the 60 people who live and not of the 20 people who don't. How do we do that?
I know that I'm getting way too much confused here, but your answer to these questions will really help me understand probability 100%.

Thanks
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Re: In a population of 100,000 males, 80% can be expected to live to age 6 [#permalink]

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06 Jul 2017, 03:27
No. of males that can live to age 60 = 80,000
No. of males that can live to age 80 = 60,000

Probability that a 60-year old male lives to 80 = 60,000/80,000 = 3/4. Ans - C.
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Re: In a population of 100,000 males, 80% can be expected to live to age 6 [#permalink]

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06 Jul 2017, 06:08
ShashankDave wrote:
Bunuel wrote:
In a population of 100,000 males, 80% can be expected to live to age 60 and 60% can be expected to live to 80. Given that a male in this group is 60, what is the probability that he lives to 80?

A. 48%
B. 60%
C. 75%
D. 78%
E. 80%

Let me first write down the data given in the question, and for simplicity, I'm assuming tat there are 100 people.

By age 0
Alive = 100

By age 60
Alive = 80

By age 80
Alive = 60

Now, consider the following questions and my understanding

What is the percentage of people who are already 60 will be alive at 80?

Answer = 60/80 = 75%

In how many ways can we select a person who is 60 years old and will be alive at 80?

$$\frac{60C1}{80C1}$$

75%

Now my understanding of the original question

If a person is randomly selected from people who are 60, what is the probability that he will live to 80?

In this case. we have first already selected a person, now this particular person can be a part of the 60 people who will live to 80, or of the 20 people who will not. We have to calculate the possibility that this particular person is a part of the 60 people who live and not of the 20 people who don't. How do we do that?
I know that I'm getting way too much confused here, but your answer to these questions will really help me understand probability 100%.

Thanks

Hi Shashank,

You have clearly understood the Q correctly even while answering and rewording the Q.
Now you pick up a person A who is part of gang of 80 person reaching 60years.
Now he can get into 60 person who do reach 80 years..
So probability is 60/80..
And this is exactly you have mentioned earlier
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Re: In a population of 100,000 males, 80% can be expected to live to age 6 [#permalink]

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14 Jul 2017, 10:36
Bunuel wrote:
In a population of 100,000 males, 80% can be expected to live to age 60 and 60% can be expected to live to 80. Given that a male in this group is 60, what is the probability that he lives to 80?

A. 48%
B. 60%
C. 75%
D. 78%
E. 80%

Since 80% and 60% are expected to live to ages 60 and 80, respectively, 80,000 and 60,000 males are expected to live to ages 60 and 80, respectively. Thus, the probability that a male lives to age 80, given that he is 60, is:

60,000/80,000 = 3/4 = 75%

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Re: In a population of 100,000 males, 80% can be expected to live to age 6 [#permalink]

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11 Sep 2017, 14:28
JeffTargetTestPrep wrote:
Bunuel wrote:
In a population of 100,000 males, 80% can be expected to live to age 60 and 60% can be expected to live to 80. Given that a male in this group is 60, what is the probability that he lives to 80?

A. 48%
B. 60%
C. 75%
D. 78%
E. 80%

Since 80% and 60% are expected to live to ages 60 and 80, respectively, 80,000 and 60,000 males are expected to live to ages 60 and 80, respectively. Thus, the probability that a male lives to age 80, given that he is 60, is:

60,000/80,000 = 3/4 = 75%

what if he wasnt 60. what if he was 30, would the probability change?
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Re: In a population of 100,000 males, 80% can be expected to live to age 6   [#permalink] 11 Sep 2017, 14:28
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# In a population of 100,000 males, 80% can be expected to live to age 6

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text:
If α<!-- α --> , β<!-- β --> , γ<!-- γ --> , δ<!-- δ

If $\alpha$, $\beta$, $\gamma$, $\delta$ are smallest positive angle in ascending order of magnitude which have their sines equal to positive quantity $\lambda$ , then if
$4\mathrm{sin}\frac{\alpha }{2}+3\mathrm{sin}\frac{\beta }{2}+2\mathrm{sin}\frac{\gamma }{2}+\mathrm{sin}\frac{\delta }{2}=2\sqrt{1+\lambda }$
then $\lambda$ = ?
Options-
(1)$\mathrm{sin}\alpha$
(2)$\mathrm{sin}\beta$
(3)$\mathrm{sin}\gamma$
(4)$\mathrm{sin}\delta$
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percolarse2rzd
See we have ascending order for positive sines as $a,\pi -a,2\pi +a,3\pi -a$ as sine is positive only in first two quadrants .Now these are values of $\alpha ,\beta ,\gamma ,\delta$ plugging in and simplifying we get $\mathrm{sin}\left(\alpha /2\right)+\mathrm{cos}\left(\alpha /2\right)=\sqrt{1+\lambda }$ thus squaring we get $\lambda =\mathrm{sin}\left(\alpha \right)$


text:
# Minkowski Inequality for $p \le 1$

I've been trying to prove the concavity of a particular function which I reduced to proving the reverse Minkowski Inequality for $p \le 1$, $p \ne 0$ for arguments in $\mathbb{R}^{n}_{+}$. That is,

$$\|x \|_p + \|y\|_p \le \|x+y\|_p$$

I've found a proof for the case of $0 < p \le 1$, but the case when $p < 0$ still evades me. All proofs of Minkowski's Inequality (in the proper direction) usually rely on Hölder's Inequality, which in turn relies on Young's Inequality. However, Young's does not apply for exponents below $0$, and I am rather jammed up finding another way. Can anyone offer a little direction?

-
Isn't that false, at the very least for $p=1$? Together with the usual Minkowski inequality you deduce that $$\| x+y \| _p = |\ x\|_p + \| y \|_p ,$$ which is false generally. – Ragib Zaman Oct 17 '11 at 9:12
Quite right you are if either $x,y < 0$. I've amended it to only consider values with elements greater than or equal to 0. – duckworthd Oct 17 '11 at 21:28

Hölder is reversed when $p<0$. The proof is a simple change of variables. See Bullen's book "Handbook of Means and Their Inequalities" p.202-3 for details. Hence Minkowski is also reversed.
Using the concavity of the function $f$ defined by $f(x)=x^p$, $0<p<1$, we get that for $t\in (0,1)$, $$(x_k+y_k)^p=\bigg(t\dfrac{x_k}{t}+(1-t)\dfrac{y_k}{1-t}\bigg)^p\geq t\dfrac{x_k^p}{t^p}+(1-t)\dfrac{y_k^p}{(1-t)^p}.$$ Taking the sum, we get that $$||x+y||_p^p\geq t\dfrac{||x||_p^p}{t^p}+(1-t)\dfrac{||y||_p^p}{(1-t)^p}.$$ By choosing $t=\dfrac{||x||_p}{||x||_p+||y||_p}$ and $1-t=\dfrac{||y||_p}{||x||_p+||y||_p}$, we get that


text:
## Cylinder Candy ZOJ - 3866 (求定积分)

Cylinder Candy
ZOJ - 3866

 Edward the confectioner is making a new batch of chocolate covered candy. Each candy center is shaped as a cylinder with radius r mm and height h mm.

The candy center needs to be covered with a uniform coat of chocolate. The uniform coat of chocolate is d mm thick.

You are asked to calcualte the volume and the surface of the chocolate covered candy.

Input

There are multiple test cases. The first line of input contains an integer T(1≤ T≤ 1000) indicating the number of test cases. For each test case:

There are three integers r, h, d in one line. (1≤ r, h, d ≤ 100)

Output

For each case, print the volume and surface area of the candy in one line. The relative error should be less than 10-8.

Sample Input

2
1 1 1
1 3 5

Sample Output

32.907950527415 51.155135338077
1141.046818749128 532.235830206285

(xr)2+y2=d2

x=f(y)=d2y2+r

V=πd0πx2dy

V1=πd0(d2y2+r)2dy

=πd0(d2y2+r2+2rd2y2)dy

=πd3+πr2d13πd3+πd02rd2y2dy

=πr2d+23πd3+πd02rd2y2dy

y=dsinθ

d0d2y2=π20dcosθddsinθ=π20d2cos2θdθ

=d2π20cos2θ+14d2θ

=d2π0cost+14dt

=14πd2

V1=πr2d+23πd3+14πd2

V2=π(r+d)2h

V=2V1+V2

y=f(x)

S=2πbaf(x)1+f(x)2dx

S1=2πd0(d2y2+r)1+y2d2y2dy

=2πd2+π2dr

S2=2πr2

S3=2π(r+d)h

S=S1+S2+S3

code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
const double pi = acos(-1.0);
int main(){
int t;
scanf("%d",&t);
while(t--){
double r,d,h;
scanf("%lf%lf%lf",&r,&h,&d);
printf("%.10f ",2.0*(2.0/3.0*pi*d*d*d+pi*r*r*d+1.0/2.0*pi*pi*d*d*r)+pi*(r+d)*(r+d)*h);
printf("%.10f\n",2.0*(2.0*pi*d*d+pi*pi*d*r)+2.0*pi*r*r+2.0*pi*(r+d)*h);
}
return 0;
}


text:
# Using Adobe Minion Pro with LaTeX

## Tags: howtos, latex

Published on
« Previous post: Wine + CUPS — Next post: Using gnokii to manage a Nokia 6300 … »

This short note explains how to use the font "Adobe Minion Pro" with the LaTeX typesetting system.

# Preparation

• Download an older version of Adobe Acrobat Reader. Choose the .tar.gz for Linux. It is vital that you do not download the most recent version. Unfortunately, the font metrics (which we will install later) do not work with the most recent version of the font.
• Extract the archive; the font is in the file COMMON.TAR in the directory Adobe/Reader8/Resource/Font.
• Install otfinfo from /usr/ports/print/typetools (for FreeBSD users) or using the package manager of your choice.
• Use otfinfo -v to inspect the fonts. For MinionPro-Bold.otf, I get Version 2.015;PS 002.000;Core 1.0.38;makeotf.lib1.7.9032. This version works for me.
• If not already done, create a local texmf-structure. For the venerable teTeX-distribution, I had to set the environment variable TEXMFHOME to $HOME/.texmf. I also created the directories .texmf, .texmf-config, .texmf-var. • Copy updmap.cfg to $HOME/.texmf-config/web2c/. This file needs to be modified for the new font.
• Run texhash

# Font installation

Follow the excellent documentation from CTAN. You will be provided with detailed step-by-step instructions and some helpful scripts that convert the fonts. After placing everything in the correct directory (as the tutorial suggest), you are good to go: A simple \usepackage{MinionPro} should do the trick.

My gratitude goes to Achim Blumensath, Andreas Bühmann, and Michael Zedler for creating and maintaining the aforementioned package. Otherwise, I would have been hopelessly lost in the mazes of my own incompetence.


text:
## Calculus: Early Transcendentals 8th Edition

(a) $$\sec^2 x=\frac{1}{\cos^2 x}$$ (b) $$\sec x\tan x=\frac{\sin x}{\cos^2 x}$$ (c) $$\cos x-\sin x=\frac{\cot x-1}{\csc x}$$
What we must do here is to differentiate both sides of the equation. (a) $$\tan x=\frac{\sin x}{\cos x}$$ First consider the left side. Its derivative would be $$A=(\tan x)'=\sec^2 x$$ Then consider the right side, using the Quotient Rule to differentiate. $$B=\Bigg(\frac{\sin x}{\cos x}\Bigg)'=\frac{(\sin x)'\cos x-\sin x(\cos x)'}{\cos^2 x}$$ $$B=\frac{\cos x\cos x-\sin x(-\sin x)}{\cos^2 x}$$ $$B=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}$$ $$B=\frac{1}{\cos^2 x}$$ Therefore, the result would be $$\sec^2 x=\frac{1}{\cos^2 x}$$ (b) $$\sec x=\frac{1}{\cos x}$$ The left side: $$A=(\sec x)'=\sec x\tan x$$ The right side: we would use Reciprocal Rule. In detail, $$B=\Bigg(\frac{1}{\cos x}\Bigg)'=-\frac{(\cos x)'}{\cos^2 x}$$ $$B=\frac{\sin x}{\cos^2 x}$$ So, the result would be $$\sec x\tan x=\frac{\sin x}{\cos^2 x}$$ (c) $$\sin x+\cos x=\frac{1+\cot x}{\csc x}$$ The left side: $$A=(\sin x+\cos x)'=\cos x-\sin x$$ The right side (use Quotient Rule): $$B=(\frac{1+\cot x}{\csc x})'=\frac{(1+\cot x)'\csc x-(1+\cot x)(\csc x)'}{\csc^2 x}$$ $$B=\frac{-\csc^2 x\csc x-(1+\cot x)(-\csc x\cot x)}{\csc^2 x}$$ $$B=\frac{-\csc^3 x+\csc x\cot x+\csc x\cot^2 x}{\csc^2 x}$$ $$B=\frac{-\csc^2 x+\cot x+\cot^2 x}{\csc x}$$ $$B=\frac{-\csc^2+\cot x+(\csc^2-1)}{\csc x}$$ (since $\cot^2 x=\csc^2-1$)$$B=\frac{\cot x-1}{\csc x}$$ Eventually, we have $$\cos x-\sin x=\frac{\cot x-1}{\csc x}$$


text:
MATH3401

Lecture 20

Contour integrals

Further examples

Example 4: Evaluate   $$I = \ds \int_C \frac{dz}{z}$$, where $$C= 2e^{i\theta}$$, $$0\leq \theta \leq 2 \pi$$.

Can't repeat argument from Example 3 directly.

 $$I = I_1 + I_2$$ where $$I_j = \ds \int_{C_j}\frac{dz}{z}$$

Contour integrals

Further examples

Example 4 (cont): On $D = \C \setminus \{\text{negative real axis}\cup \{0\}\},$ $$\Log z$$ is a primitive for $$\dfrac{1}{z}$$. $$C_1\subset D$$, so Theorem in Lecture 19 $$\Rightarrow$$

$$I_1 = \Log(2i)-\Log(-2i)$$ $$= \ln|2i| + i\dfrac{\pi}{2}-\left(\ln|-2i| + i\left(-\dfrac{\pi}{2}\right)\right)$$

$$\quad = \pi i$$.

Remark: Agrees with calculation from Lecture 19, Example 1.

Contour integrals

Further examples

Example 4 (cont): For $$I_2$$, on $D' = \C \setminus \{\text{positive real axis}\cup \{0\}\},$ $$1/z$$ has a primitive, e. g. $$\mathcal \Log z = \ln|z| + i \,{\Large\mathcal a}\mathcal rg(z)$$, $$0 \leq {\Large\mathcal a}\mathcal rg(z) < 2 \pi$$.

Note $$C_2 \subset D'$$. So Theorem in Lecture 19 $$\Rightarrow$$

$$I_2 = \mathcal \Log(-2i)-\mathcal \Log(2i)$$ $$= \ln 2 + i\dfrac{3\pi}{2}-\left(\ln 2 + i\left(\dfrac{\pi}{2}\right)\right)$$

$$\quad = \pi i$$.

Thus $$I= I_1 + I_2 = 2 \pi i$$.

Contour integrals

Further examples

Example 4 (cont): So $\int_C z^n dz = \left\{ \begin{array}{lr} 0 & n\in \Z\setminus \{-1\}\\ 2 \pi i & n=-1 \end{array} \right.$

for any circle $$C$$ centred at the origin, positively oriented.

Cauchy-Goursat

Let $$C$$ be a simple closed curve in $$\C$$. If $$f$$ is analytic on $$C$$ and its interior, then $\int_C f\left(z\right)dz = 0.$

Cauchy-Goursat

Remark: $$\ds \int_C f\left(z\right)dz = 0$$ $$\nRightarrow$$ $$f$$ is analytic in and on $$C$$.

e. g. $$\ds \int_C z^n dz = 0$$, $$n= -2, -3, -4, \ldots$$

Proof of C-G: Do it for

Cauchy-Goursat

Key step in proof: $$M$$-$$\ell$$ estimate

Suppose $$f$$ is continuous on a contour $$C$$ given by $$z = z(t)$$, $$a\leq t \leq b$$. Then exists $$M$$ such that $|f\left(z\right)| \leq M \quad \forall z\in\C,$

via extreme value theorem in $$\R$$.

Cauchy-Goursat

So $$\ds \int_C f \left(z\right) dz =\left| \ds \int_a^b f \left(z \left(t\right)\right) z' \left(t\right) dt\right|$$

$$\qquad \qquad \quad \leq \ds \int_a^b \left| f \left(z \left(t\right)\right) \right| \cdot \left| z' \left(t\right)\right|dt$$

$$\qquad \qquad \quad \leq M \ds \int_a^b \left| z' \left(t\right)\right|dt$$ $$= M \ell$$.

where $$\ell = \ell\left(C\right)=\text{length}$$ of $$C$$.

Cauchy-Goursat

Recall: a domain $$D$$ is simply connected if for every simple closed contour $$C$$ in $$D$$, there holds $\text{Int}\, C \subset D.$

i. e. "no holes". In other words, all simple closed contours are null homotopic.

If $$D$$ is not simply connected, it is multiply connected.

Cauchy-Goursat extension

Multiply connected domains

If $$f$$ is analytic on simple closed contours $$C$$, $$C_1, C_2, \ldots, C_n\subset \text{Int}\, C$$, and the interior of the domain bounded by $$C_1, C_2, \ldots, C_n$$, with $$C$$ positively oriented and $$C_k$$ negatively oriented, then $\int_C f + \sum_{j=1}^{n} \int_{C_j} f = 0.$


text:
## Intermediate Algebra for College Students (7th Edition)

$\sum_{i=1}^{12}(i^4)$
Since, we have $1^4+2^4+....+12^4$ This can be summed up with the summation notation as: $\sum_{i=1}^{12}(i^4)$


text:
Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Hi there!

Over the last few months, my lab-mate Irene Villa (see more of her work here!)
and I, have been discussing ecological niche overlap. The niche concept dates back to ideas first proposed by ornithologist J. Grinnell (1917). Later on, G.E. Hutchinson
(1957) defined the ecological niche of a species as the n-dimensional hyper-volume of ecological factors that define a space where the species can exist indefinitely. These ecological factors can be either abiotic (fundamental niche) or biotic variables, and the intersection of both defines the realized niche, or potential ecological range where the species can survive. The last necessary element for the existence of a species in a particular place (actual presence or distribution) is related to the “movement” ability or dispersal capacity of the species to arrive to those places where ecological factors are suitable. This theory is nicely summarized in the BAM diagram (Biotic-Abiotic-Movement; Soberón, 2007).

In the last decades, with the development of cartographic information about climatic variables (mainly temperature and precipitation) and mathematical algorithms, many research efforts have been directed towards modeling the ecological niche of species from presence records and bioclimatic information (Elith & Leathwick, 2009). As these techniques were developed, many researchers focused on the comparison of ecological niches of two species, and the degree of overlap between them. A wide range of literature has been published about this topic (Peterson, 2011) but, in the last years, a general consensus has been reached about the methods to measure this overlap. Briefly, the method consist of two steps: 1) building an environmental space using bioclimatic values in the presence records and background points with any multivariate technique (usually Principal Component Analysis), and 2) representing both species niches in this space via kernel density, comparing them using D or I indexes and applying similarity or identity tests to assess significance (Warren et al., 2008).

Following this general approach, the R package ecospat (Di Cola et al., 2016) provides an easy way to perform a Principal Component Analysis and to compare two species niches using D overlapping index, identity tests, etc. However, Irene and I, realized that, when using this approach, in many cases for which niche overlap was not obtained, this was due to the lack of overlap for just one of the two PC axes, while a complete overlap was retrieved for the other axis. Taking into account that both axes do not usually contribute equally to explain environmental space (the first axis usually explains much more variability than the following ones), we worked on a function using the ggplot2 R package to create a nice plot for visual exploration of the Principal Component scores in both axes. This plot allows us to examine the factors that are or are not producing niche overlap, since it shows three plots in one: each axis separately and the joint distribution.

Here is the function code (too large to print here). You can download and run it in an R console. The following is an example with two wall lizards from a Mediterranean area. You can download ecological values for the 19 bioclimatic variables at presence points of these two species here.

##########################
#######   EXAMPLE   ######
##########################
library(ecospat)
library(ggplot2)
library(grid)
library(gridExtra)
library(gtable)
library(RColorBrewer)

# Read wall lizards ecological data

# Note: This data represent the values for the 19 bioclimatic variables at occurrence
# points for two Mediterranean wall lizards, Iberian wall lizard (Podarcis hispanicus,
# mainly in Iberian Peninsula) and Lilford's wall lizard (Podarcis lilfordi, from the
# The first 1125 rows are the bioclimatic values for P.hispanicus presences, while the
# next 44 rows represent bioclimatic values for P.lilifordi. The following rows represent
# available bioclimatic conditions (background) obtained from 10000 random points
# generated in a buffer of 400 km approximatively around presence points. Some inaccuracies
# (such as 2 Lilford's wall lizard presences in Iberian Peninsula) were not removed (this is
# just an example!!).
# *Data searches:

# From this data, we perform a Principal Component Analysis using "dudi.pca" function from "ecospat" package.
# We will choose 2 axes for the environmental space representation.

pca_Ph_Pl <-dudi.pca(na.omit(data_Ph_Pl, center = T, scale = T, scannf = F, nf = 2))
2

# Now, we can use directly this result with the niceOverPlot function to represent a
# 2D environmental space in a central plot, and each environmental gradient represented
# by each axis at top and right. We must provide to the function with the number of presences
# of each species in the same order as in the input data (nº of presences
# for Sp1 and nº of presences for Sp2)

niceOverPlot(pca_Ph_Pl, n1=1125 , n2= 44)

Results from niceOverPlot (Blue area: P. lilifordi; Pink area: P. hispanicus)

This plot could help us to interpret the results obtained from niche overlap analysis for these two species. In this example, we can see that, in 2D environmental space, the bioclimatic preferences of the two wall lizard species do not overlap. But you can notice that this lack of overlap is due to Axis2 of the PCA, which represents around 18% of contribution in environmental space. Now, we can focus on the bioclimatic variables that are involved in this axis to understand which factors are preventing the overlap.
In upcoming posts, we will discuss some alternative ways to address the D overlap index to take into account cases like our example.

Have fun!

Podarcis lilifordi, from the Balearic Islands

References

-Soberón, J. (2007). Grinnellian and Eltonian niches and geographic distributions of species. Ecology letters, 10(12), 1115-1123.

-Elith, J., & Leathwick, J. R. (2009). Species distribution models: ecological explanation and prediction across space and time. Annual review of ecology, evolution, and systematics, 40, 677-697.

-Peterson, A. T. (2011). Ecological niche conservatism: A time‐structured review of evidence. Journal of Biogeography, 38(5), 817-827.

-Di Cola, V., Broennimann, O., Petitpierre, B., Breiner, F. T., D'Amen, M., Randin, C., ... & Pellissier, L. (2016). ecospat: an R package to support spatial analyses and modeling of species niches and distributions. Ecography.


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+0

# What is the value of a + b?

0
351
1
+78

Let a and b real numbers such that x4+2x3-x2+ax+b = (Q(x))2 for some polynomial Q(x). What is the value of a + b?

Oct 16, 2018

#1
+6193
+2

$$p(x)=x^4+2x^3-x^2+ax+b = (q(x))^2,~\text{for some polynomial }q(x)$$

$$\text{we know q(x) will be of order 2 so write it as }\\ q(x)=q_2 x^2 + q_1 x + q_0 \\ (q(x))^2 = q_2^2 x^4+2 q_1 q_2 x^3+\left(q_1^2+2 q_0 q_2\right) x^2+2 q_0 q_1 x+q_0^2$$

$$\text{and we have equations }\\ q_2^2 = 1\\ 2q_1q_2 = 2\\ (q_1^2+2q_0q_2)=-1\\ 2q_0q_1=a\\ q_0^2=b$$

$$\text{clearly }q_2 = \pm 1\\ \text{suppose }q_2=1 \\ 2q_1 (1)=2,~q_1=1\\ (1+2q_0(1))=-1 \\ q_0=-1 \\ b=q_0^2 = 1 \\ a=2q_0 = 2$$

$$\text{now suppose }q_2=-1 \\ 2q_1(-1) = 2,~q_1=-1 \\ (1+2q_0(-1))=-1,~q_0=1 \\ b=q_0^2 = 1 \\ a = 2(-1)(-1) = 2$$

$$\text{so in both cases }a=2,~b=1 \\ \text{and }a+b = 3$$

I kind of suspect there is a simpler way to solve this.

.
Oct 17, 2018
edited by Rom  Oct 17, 2018


text:
My Math Forum > Math 3 8's to make 24 -- my challenge puzzle

 Math General Math Forum - For general math related discussion and news

November 25th, 2014, 07:44 PM   #21
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Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
Thanks: 521

Math Focus: Calculus/ODEs
Quote:
 Originally Posted by Math Message Board tutor You have rocks in your head, and your a psychotic liar. You need those rocks kicked out of your head. I told you to stop posting on this thread, you punk. You're to do what I tell you to do, you persistent troll. Get off the thread!
Please post civilly and without painting violent pictures.

You cannot dictate who posts where, no matter how much you may dislike someone. I would advise you to revise your list of ignored users if you do not like what a particular person has to say.

But, you cannot be calling people names and telling them they need for their heads to be kicked. Be advised that this will not be tolerated.

 November 25th, 2014, 07:48 PM #22 Banned Camp   Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191 I'm advising you, MarkFL and all of the moderators, *take care of business* and ban him now and delete his posts off of this thread, because they're inappropriate. How many times must I send PMs to your moderators for his repeat harassment!? YOU take care of business on YOUR end now. Last edited by skipjack; November 27th, 2014 at 11:28 AM.
November 25th, 2014, 07:53 PM   #23
Senior Member

Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
Thanks: 521

Math Focus: Calculus/ODEs
Quote:
 Originally Posted by Math Message Board tutor I'm advising you, MarkFL and all of the moderators, *take care of business* and ban him now and delete his posts off of this thread, because they're inappropriate. How many times must I send PM's to your moderators for his repeat harassment!? YOU take care of business on YOUR end now.
We are taking care of business, thank you very much.

You are abusing the "Report Post" feature, when it is in fact you who are posting inappropriately. You apparently do not take it well when someone disagrees with you, and then you call for their posts to be deleted.

We are not going to delete posts because you don't like that someone disagreed with you.

So, when someone disagrees with you, refute them civilly, or ignore them, but do not call them names, or otherwise become aggressive.

Last edited by skipjack; November 27th, 2014 at 11:29 AM.

November 25th, 2014, 08:00 PM   #24
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Quote:
 Originally Posted by jks $\displaystyle \large \frac{8}{\left(\sqrt{\sqrt{8+8}}\right)!} \ !$ $\displaystyle \large \sqrt{8 \times \left(\sqrt{\sqrt{8+8}}\right)!} \quad !$ I think I can guess why the number of factorials is limited to 2.
These two expressions don't count, because the factorial signs next to each parenthetical expression
can be removed and the expression still equals 24. In other words,
they are not necessary. Only the outermost factorial sign on each one to make them four factorial, respectively, are required.

 November 25th, 2014, 08:01 PM #25 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs $\displaystyle \frac{8!!}{8+8}=24$
November 25th, 2014, 08:12 PM   #26
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Quote:
 Originally Posted by MarkFL $\displaystyle \frac{8!!}{8+8}=24$
The double factorial one is valid.

Then there are more than the five I referred to.

Last edited by Math Message Board tutor; November 25th, 2014 at 08:15 PM.

 November 26th, 2014, 07:39 PM #27 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications $\displaystyle \large \sqrt{\sqrt{8 \times 8}+8} \quad !$
November 26th, 2014, 07:53 PM   #28
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Quote:
 Originally Posted by jks $\displaystyle \large \sqrt{\sqrt{8 \times 8}+8} \quad !$
Yes.

November 26th, 2014, 08:02 PM   #29
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Joined: Jan 2014
From: The backwoods of Northern Ontario

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Quote:
 The double factorial one is valid.
I don't understand how it can be valid

8! = 40320
Thus 8!!= 40320!

This woud be a very large number. Divide it by 16 and you surely don't get 24.

 November 26th, 2014, 08:50 PM #30 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications $\displaystyle 8!! \ = \ 8 \times 6 \times 4 \times 2$ Look here. Thanks from MarkFL

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# Where to find datasheets in bulk?

Where is collection of data sheets of large table with parameters as columns and models as rows?

Something like:

Model        Breakdown_voltage On_resistance <other_values>
QWERTY122AS  120V              6_Ohm         ....
AWERTY123SD  30V               1_Ohm         ....
<all other popular models>

for transistors, diodes and other parts.

When searching Google it usually provides data for one/few models or even offer download a PDF file. Where to find it in bulk?

• In the late Paleozoic manufacturers would publish their collected datasheets on things called CD-ROMs. I still have a number of those. For instance the Motorola collection required 2 CDs. That was before DVDs. Aug 10 '11 at 8:54
• I see there is many disjoint databases per manufacturer, unlikely with standard interface. Is there a comprehensive database of all parts (both currently available and legacy) of all manufacturers with their basic characteristics?
– Vi0
Aug 11 '11 at 15:31