text:
which digit doesn't occur in this number?

I came across this question on a site and I didn't find any solution to this.....(except to use calculator)

The question was

Which digit doesn't occur in the number 2^29 ?

Plz help.....

Note by Poonayu Sharma
4 years, 1 month ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

• bulleted
• list

1. numbered
2. list

1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

If 2^29=n (mod 9),then 9-n is the answer. Noticing that 2^3=-1(mod 9),2^29=(2^3)^9x2^2=(-1)^9x2^2=-4=5 (mod 9). Therefore the answer is 9-5=4. This can be easily derived using mod. Hope you got it :)

- 4 years, 1 month ago

Are there other ways too?

- 4 years, 1 month ago

Sorry, I can't fathom any. :(... I'm really bad at intuition :-/

- 4 years, 1 month ago

Ohh...use of mod didn't strike me :P

I was trying to solve it using logs etc. . :P

Btw...thanks you sir

- 4 years, 1 month ago

You're welcome :D....but you're 3 yrs elder, so you must refrain from calling me sir, how could you? And this solution is also inspired by an experiment I did with the cyclicity of digital roots of powers of 2. You could check out my easy-peasy problem on that though :)

- 4 years, 1 month ago

U deserved "sir" :D.....and I solved the problem u talking about....but not with the (mod 9) method But with cyclic sum of digits of powers of 2

- 4 years, 1 month ago

Oh. Well at first ( when I encountered this problem in class 7) I solved this using cyclicity too :)

- 4 years, 1 month ago

Ohh...well dont u think it becomes a herculean task for a 7 grade guy to do such sums :P

- 4 years, 1 month ago

@Poonayu Sharma - Woah! I'm mind -blown!!!! How did you level up in Algebra and Number Theory so soon? What resources did you use to gain here?

- 4 years ago

I saved the challenging questions for later. ..then I read all the posts or notes related to them..For some.I used Google to search theories related to sum..and then tried to solve them....though I couldn't solve all(surprisingly I was able to solve 3 lvl 5 question) ;P ..I guess you can also do it ...faster than me :D

Next target ....JEE ,CALCULUS AND MECHANICS :P

- 4 years ago

What did you use to learn all those algebra question- inequalities , Symmetric bounding,,all that. And sorry, I really dont think I can do it faster :(.

- 4 years ago

Frankly speaking. ...The symmetric bounding question was a hit and trial by me..(I think one should not hesitate to use hit and trial if there are no other means at times.).I got it in the second try..

.whenever u see a problem. ..try to find a note about it .(That's the only thing I used to crack sums..also me and my 2 friends did few sums together)..I remember daniel liu once gave a link to a note which could be used to solve a problem...It was useful then ..If u don't find any such notes...Google it...believe me problems do become easier then.. :D Hoping ur level soars high :P

- 4 years ago

:D

- 4 years ago

Btw...u planning to do iit? :P

- 4 years ago

Nope. I do math only as a hobby. I'd wan't to take up medicine as my mother wants me to do it. U wanna go to IIT or somewhere else Mr. Future Cosmologist?

- 4 years ago

I am bad at bio and chemistry right from birth :P..I will do cosmology (wanna go to NASA) whether I get iit or not..currently I'm preparing for iit ...let's see if luck shines upon me and if i get a good rank then ill do iit ...well if not then I'm thinking for PhD in mathematics and then cosmology ( possibly astronaut too) :P

But the way medical ppl will have a tough competition because of u :P

- 4 years ago

Nice :). I also loved the way you ended it with a joke :). You want to become an astronaut or cosmologist? I must say there is a great difference between the two! But, whatever it be I'm sure you'd do well in it :P. You are preparing for IIT on your own? I feel ICSE curriculum is much better than CBSE. What do you feel?

- 4 years ago

I want to become both actually :P....I have joined classes here called pace for preparation. .I myself have studied from icse and I feel it's standard is better than other boardr...But while you prepare for iit ..It doesn't matter which board ur from ...All become more or less same

Thanks for ur compliment though ...

And good luck. .. (remember to enjoy life as much as u can now...After 10 ...its hell :( :P)

- 4 years ago

YUP. Why do you say life after 10th is hell? In fact, I would consider that heaven. I would have to study only my favorite two subjects. English is passable too. In fact, life now is hell for me!.

- 4 years ago

I used to think the same way bro....You will realise it after u leave ur school and join college or classes ...everything suddenly gets screwed up :P
but if u manage ur time and other things (which I failed to)...u wont have troubles .. Hope you enjoy it even after school :D

- 4 years ago

Oh! Thanks for your well wishes :)

- 4 years ago

You can view the question here with a detailed solution discussion.

Staff - 4 years, 1 month ago

Except there you are given that it's a 9 digit number with distinct digits. The first step in this problem is to show that it is a 9 digit number (using log) and then obviously by the way the problem is stated implies that the digits are distinct

- 4 years, 1 month ago

That rises an interesting question. Which powers (or 2) are missing exactly 1 digit (but could have repeats of others)? Unfortunately, I don't think there is an easy way to answer this question.

Staff - 4 years, 1 month ago

Yeah I forgot to mention it because I thought that its understood :P

- 4 years, 1 month ago

Thank you sir :D

- 4 years, 1 month ago


text:
• CHAOUCHE YASSINE

Articles written in Pramana – Journal of Physics

• Structural, elastic and dynamical properties of ${\rm YP}_{1-x}{\rm As}_{x}$ alloys from first principles calculations

The structural, elastic and lattice vibration properties of the ternary alloys ${\rm YP}_{1-x}{\rm As}_{x}$ at various As concentrations ($x$) from 0 to 1, $x=0, 0.25 ,0.5, 0.75$ and 1, are presented. The calculations were performed using the density functional perturbation theory (DFPT) within the generalised gradient approximation (GGA) and employing virtual crystal approximation (VCA). We studied the effect of arsenic composition on structural parameters, the phase transition pressure, the elastic constants, the optical and acoustic phonon frequencies at high symmetry points $\Gamma$, $X$ and $L$, the static and electronic dielectric constants, and the Born effective charge. It is established that all these properties follow a quadratic law in the concentration ($x$). There is a good agreement between our results and the available data for binary compounds YP and YAs. Our results are predictions of ${\rm YP}_{1-x}{\rm As}_{x}$.

• Theoretical prediction of lanthanum composition effects on structural, electronic and thermal properties of La$_x$Sc$_{1−x}$N alloys

In this study, we have explored the structural, electronic and thermal properties of La$_x$Sc$_{1−x}$N ternary alloys in rock-salt structure using the full-potential linearised augmented plane wave (FP-LAPW) method based on the density functional theory (DFT). To calculate the exchange-correlation energy and potential, we have used the Wu–Cohen generalised gradient approximation and the modified Becke–Johnson. We investigated the effect of composition on lattice parameters, bulk modulus and band gap. The variation of the calculated lattice constant with lanthanum composition is practically linear and shows a small deviation of the obtained results from Vegard’s law. Moreover, the effect of thermal macroscopic properties was also investigated employing the quasiharmonic Debye model, which takes into account the lattice vibrations. We have found a good accord between our results, the experimental data and previous theoretical results available in the literature for the binary compounds which can be a support for the ternary alloys in the future.

• # Pramana – Journal of Physics

Volume 96, 2022
All articles
Continuous Article Publishing mode

• # Editorial Note on Continuous Article Publication

Posted on July 25, 2019

Click here for Editorial Note on CAP Mode

© 2021-2022 Indian Academy of Sciences, Bengaluru.

format: plain


text:
🎄

# Tree Recursion

Many, many function calls.
We've now seen what standard recursion looks like — a function that calls itself. We'll now make our lives slightly more complicated by introducing tree recursion — a function that calls itself not once but multiple times.
A classic example of tree recursion is a Fibonacci Sequence. In this sequence, a given number is the sum of the previous two numbers. The sequence goes as follows:
$1, 1, 2, 3, 5, 8, 13, 21, 35, 56...$
Let's write this in code!
def fib(n):
'''
Accepts any value greater than or equal to 1 and returns that term in the
fibonacci sequence.
'''
if n == 0 or n == 1:
return n
else:
return fib(n-1) + fib(n-2)
An important observation here: Why do we have n==0 as a base case, even when that is not acceptable as an input. It's because while the user might never give us n as the value for n, our recursive calls might. Consider the case fib(2) — which makes the recursive calls fib(1) and fib(0)  — 1 + 0 — 1, as required!
We could have written something like n==1 or n==2 as our base case as well, but using this structure instead allows us to introduce the idea of base cases for inputs invalid for the user but valid for the recursive calls.
Regardless, something like fib(4) looks like the following in terms of recursive calls:
= fib(4)
= fib(3) + fib(2)
= (fib(2) + fib(1)) + (fib(1) + fib(0))
= ((fib(1) + fib(0)) + fib(1)) + (fib(1) + fib(0))
= ((1 + 0) + 1) + (1 + 0)
= (1 + 1) + 1
= 3
It is easier to visualize by representing function calls as nodes in a tree:
Recursive Tree for fib(4)
As you can see, this looks like an upside down tree (kinda... I wouldn't think too hard about it, most computer scientists don't go outside very often). The big idea here is that, at every stage, we make two recursive calls and then combine them in some way. This is tree recursion!


text:
PDF (letter size)

solving the spin 1 electron problem. Finding $$S_x,S_y,S_z$$

May 25, 2022   Compiled on May 25, 2022 at 9:36pm

Problem statement

Determine $$S_{x},S_{y},S_{z}$$ angular momentum spin matrices for the electron using spin 1. The given is that experiments show that $$S_{z}$$ has three possible values (eigenvalues). These are $$1,0,-1$$.

Solution

Using eigenbasis for $$S_{z}$$ as the following\begin {align} |S_{z=1}\rangle & =|1\rangle =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \tag {1}\\ |S_{z=0}\rangle & =|2\rangle =\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \nonumber \\ |S_{z=-1}\rangle & =|3\rangle =\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \nonumber \end {align}

Then the following three equations result from writing $$S_{z}|S_{z=\omega _{i}}\rangle =\omega _{i}|S_{z=\omega _{i}}\rangle$$ where $$\omega _{i}$$ is the eigenvalue. These three equation are solved to determine $$S_{z}$$. Let $$S_{z}=\begin {bmatrix} a & b & c\\ d & e & f\\ g & h & m \end {bmatrix}$$. Therefore\begin {align*} \begin {bmatrix} a & b & c\\ d & e & f\\ g & h & m \end {bmatrix}\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} & =1\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \qquad \\\begin {bmatrix} a & b & c\\ d & e & f\\ g & h & m \end {bmatrix}\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} & =0\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} \\\begin {bmatrix} a & b & c\\ d & e & f\\ g & h & m \end {bmatrix}\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} & =-1\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} \end {align*}

Or\begin {align*} \begin {bmatrix} a\\ d\\ g \end {bmatrix} & =\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} \qquad \\\begin {bmatrix} b\\ e\\ h \end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ 0 \end {bmatrix} \\\begin {bmatrix} c\\ f\\ m \end {bmatrix} & =\begin {bmatrix} 0\\ 0\\ -1 \end {bmatrix} \end {align*}

Which gives\begin {align} S_{z} & =\begin {bmatrix} a & b & c\\ d & e & f\\ g & h & m \end {bmatrix} \nonumber \\ & =\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \tag {1A} \end {align}

Now that $$S_{z}$$ is found, the goal is to determine $$S_{x},S_{y}.\$$Let $$S_{-}=S_{x}-iS_{y}$$ and $$S_{+}=S_{x}+iS_{y}$$. We start with $$S_{+}$$ (starting with $$S_{-}$$ will not work, as it will not be possible to determine $$S^{2}$$ that way. So we have to start with $$S_{+}$$). We always start with commutator $$\left [ S_{z},S_{+}\right ]$$ \begin {align*} \left [ S_{z},S_{+}\right ] & =\left [ S_{z},S_{x}+iS_{y}\right ] \\ & =\left [ S_{z},S_{x}\right ] +i\left [ S_{z},S_{y}\right ] \end {align*}

But $$\left [ S_{z},S_{x}\right ] =i\sum _{k}\epsilon _{ijk}S_{k}$$. Here $$i=3,j=1$$ since $$z=3,x=1$$. Then $$\left [ S_{z},S_{x}\right ] =i\epsilon _{312}S_{2}=iS_{y}$$ and similarly $$\left [ S_{z}S_{y}\right ] =i\sum _{k}\epsilon _{ijk}S_{k}$$. Here$$\ i=3,j=2$$ since $$z=3,y=1.$$ Then$$\ \left [ S_{z},S_{y}\right ] =i\epsilon _{321}S_{1}=-iS_{x}$$. The above becomes\begin {align*} \left [ S_{z},S_{+}\right ] & =iS_{y}+i\left ( -iS_{x}\right ) \\ & =iS_{y}+S_{x}\\ & =S_{+} \end {align*}

This implies\begin {align} \left [ S_{z},S_{+}\right ] & =S_{+}\nonumber \\ S_{z}S_{+}-S_{+}S_{z} & =S_{+}\nonumber \\ S_{z}S_{+} & =S_{+}S_{z}+S_{+}\tag {2} \end {align}

Picking $$|1\rangle$$ to start with, as it lead to ﬁnding $$S^{2}$$, which we must ﬁnd before making any progress. $$S^{2}$$ is proportional for the identity matrix $$I$$. From the above we obtain \begin {align} S_{z}S_{+}|1\rangle & =\left ( S_{+}S_{z}+S_{+}\right ) |1\rangle \nonumber \\ & =S_{+}S_{z}|1\rangle +S_{+}|1\rangle \tag {3} \end {align}

But $$S_{z}|1\rangle =|1\rangle$$ since the eigenvalue is $$1$$ associated with $$|1\rangle$$ eigenvector. The above becomes\begin {align*} S_{z}S_{+}|1\rangle & =S_{+}|1\rangle +S_{+}|1\rangle \\ & =2S_{+}|1\rangle \end {align*}

The above shows that $$S_{+}|1\rangle$$ is eigenvector of $$S_{z}$$ associated with the eigenvalue $$2$$ which is not compatible with experiments. Therefore the only logical result is that\begin {equation} S_{+}|1\rangle =0|1\rangle \tag {4} \end {equation} Taking the adjoint gives\begin {align*} \left ( S_{+}|1\rangle \right ) ^{\dag } & =0|1\rangle ^{\dag }\\ \langle 1|S_{+}^{\dag } & =0\langle 1| \end {align*}

Hence\begin {align} \langle 1|S_{+}^{\dag }S_{+}|1\rangle & =0\langle 1|1\rangle \nonumber \\ \langle 1|S_{+}^{\dag }S_{+}|1\rangle & =0\tag {4A} \end {align}

The above is used to ﬁnd $$S^{2}$$. Since $$S_{+}=S_{x}+iS_{y}$$ then the above becomes\begin {align*} \langle 1|\left ( S_{x}^{\dag }-iS_{y}^{\dag }\right ) \left ( S_{x}+iS_{y}\right ) |1\rangle & =0\\ \langle 1|S_{x}^{\dag }S_{x}+iS_{x}^{\dag }S_{y}-iS_{y}^{\dag }S_{x}+S_{y}^{\dag }S_{y}|1\rangle & =0 \end {align*}

But $$S_{x},S_{y}$$ are Hermitian. Therefore $$S_{x}^{\dag }=S_{x},S_{y}^{\dag }=S_{y}$$ and the above reduces to\begin {align*} \langle 1|S_{x}^{2}+iS_{x}S_{y}-iS_{y}S_{x}+S_{y}^{2}|1\rangle & =0\\ \langle 1|S_{x}^{2}+i\left ( S_{x}S_{y}-S_{y}S_{x}\right ) +S_{y}^{2}|1\rangle & =0\\ \langle 1|S_{x}^{2}+i\left [ S_{x},S_{y}\right ] +S_{y}^{2}|1\rangle & =0 \end {align*}

But $$\left [ S_{x},S_{y}\right ] =iS_{z}$$, therefore the above becomes$\langle 1|S_{x}^{2}-S_{z}+S_{y}^{2}|1\rangle =0$ And $$S^{2}=S_{x}^{2}+S_{y}^{2}+S_{z}^{2}\$$, therefore $$S_{x}^{2}+S_{y}^{2}=S^{2}-S_{z}^{2}$$. Hence \begin {equation} S_{+}^{\dag }S_{+}=S^{2}-S_{z}^{2}-S_{z}\tag {4B} \end {equation} Therefore (4A) becomes$\langle 1|S^{2}-S_{z}^{2}-S_{z}|1\rangle =0$ Expanding the above gives\begin {align*} \langle 1|S^{2}|1\rangle -\langle 1|S_{z}^{2}|1\rangle -\langle 1|S_{z}|1\rangle & =0\\ \langle 1|S^{2}|1\rangle & =\langle 1|S_{z}^{2}|1\rangle +\langle 1|S_{z}|1\rangle \end {align*}

But $$\langle 1|S_{z}|1\rangle =1$$ and $$\langle 1|S_{z}^{2}|1\rangle =1$$, therefore the above becomes\begin {equation} \langle 1|S^{2}|1\rangle =2\tag {5} \end {equation} It is not possible to use the above to solve for a general $$S^{2}$$ which is $$3\times 3$$ matrix.  But since $$S^{2}$$ must be proportional to the Identity matrix for all spin numbers, then it must diagonal matrix with same element on the diagonal, then let $$S$$ be$\begin {bmatrix} a & 0 & 0\\ 0 & a & 0\\ 0 & 0 & a \end {bmatrix}$ Substituting this in (5) gives\begin {align*} \langle 1|\begin {bmatrix} a & 0 & 0\\ 0 & a & 0\\ 0 & 0 & a \end {bmatrix} ^{2}|1\rangle & =2\\ \langle 1|\begin {bmatrix} a^{2} & 0 & 0\\ 0 & a^{2} & 0\\ 0 & 0 & a^{2}\end {bmatrix} |1\rangle & =2=\\ a^{2} & =2 \end {align*}

Hence $$a=\sqrt {2}$$. Therefore\begin {equation} S=\sqrt {2}\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix} \tag {6} \end {equation} Now that $$S$$ is found, the next step is to ﬁnd $$S_{+}|2\rangle$$ and $$S_{+}|3\rangle$$.  Starting with (2), but now applying it to $$|2\rangle$$ gives$S_{z}S_{+}|2\rangle =S_{+}S_{z}|2\rangle +S_{+}|2\rangle$ But $$S_{z}|2\rangle =0|2\rangle$$ since $$|2\rangle$$ is the eigenvector associated with $$0\,$$ eigenvalue, the above becomes$S_{z}S_{+}|2\rangle =S_{+}|2\rangle$ Which means $$S_{+}|2\rangle$$ is eigenvector of $$S_{z}$$ associated with eigenvalue $$+1$$ which is compatible with experiment. Hence $S_{+}|2\rangle =c|1\rangle$ Where we used $$|1\rangle$$ since that is the eigenvector of $$S_{z}$$ associated with $$+1$$ eigenvalue. Now we need to ﬁnd $$c$$. Taking adjoint of both sides of the above gives\begin {align*} \left ( S_{+}|2\rangle \right ) ^{\dag } & =\left ( c|1\rangle \right ) ^{\dag }\\ \langle 2|S_{+}^{\dag } & =c^{\ast }\langle 1| \end {align*}

Therefore\begin {align} \langle 2|S_{+}^{\dag }S_{+}|2\rangle & =c^{\ast }c\langle 1|1\rangle \tag {7}\\ & =\left \vert c\right \vert ^{2}\nonumber \end {align}

But $$S_{+}^{\dag }S_{+}=S^{2}-S_{z}^{2}-S_{z}$$  which was found earlier above in (4B). Therefore the above equation becomes$\langle 2|\left ( S^{2}-S_{z}^{2}-S_{z}\right ) |2\rangle =\left \vert c\right \vert ^{2}$ Using $$S^{2}$$ found in (6) the above becomes\begin {align*} \langle 2|\left ( 2\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \right ) |2\rangle & =\left \vert c\right \vert ^{2}\\ \langle 2|\begin {bmatrix} 0 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} |2\rangle & =\left \vert c\right \vert ^{2}\\\begin {bmatrix} 0 & 1 & 0 \end {bmatrix}\begin {bmatrix} 0 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix}\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} & =\left \vert c\right \vert ^{2}\\ 2 & =\left \vert c\right \vert ^{2} \end {align*}

which implies $$c=\sqrt {2}$$. Therefore \begin {equation} S_{+}|2\rangle =\sqrt {2}|1\rangle \tag {8} \end {equation} Finally, to ﬁnd $$S_{+}|3\rangle$$,  starting again with (2), but now applying it to $$|3\rangle$$ gives$S_{z}S_{+}|3\rangle =S_{+}S_{z}|3\rangle +S_{+}|3\rangle$ But $$S_{z}|3\rangle =-|3\rangle$$ since $$|3\rangle$$ is the eigenvector associated with $$-1\,$$ eigenvalue, the above becomes\begin {align*} S_{z}S_{+}|3\rangle & =-S_{+}|3\rangle +S_{+}|3\rangle \\ & =0S_{+}|3\rangle \end {align*}

Which means $$S_{+}|3\rangle$$ is eigenvector of $$S_{z}$$ associated with eigenvalue $$0$$ which is compatible with experiment. Hence $S_{+}|3\rangle =b|2\rangle$ Where we used $$|2\rangle$$ since that is the eigenvector of $$S_{z}$$ associated with $$0$$ eigenvalue. Now we need to ﬁnd $$b$$. Taking adjoint of both sides of the above gives\begin {align*} \left ( S_{+}|3\rangle \right ) ^{\dag } & =\left ( b|2\rangle \right ) ^{\dag }\\ \langle 3|S_{+}^{\dag } & =b^{\ast }\langle 2| \end {align*}

Therefore\begin {align} \langle 3|S_{+}^{\dag }S_{+}|3\rangle & =b^{\ast }b\langle 2|2\rangle \tag {9}\\ & =\left \vert b\right \vert ^{2}\nonumber \end {align}

But $$S_{+}^{\dag }S_{+}=S^{2}-S_{z}^{2}-S_{z}$$  which was found earlier in (4B). Therefore the above equation becomes$\langle 3|\left ( S^{2}-S_{z}^{2}-S_{z}\right ) |3\rangle =\left \vert b\right \vert ^{2}$ Using $$S^{2}$$ from eq (6) the above becomes\begin {align*} \langle 3|\left ( 2\begin {bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} \right ) |3\rangle & =\left \vert b\right \vert ^{2}\\ \langle 3|\begin {bmatrix} 0 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} |3\rangle & =\left \vert b\right \vert ^{2}\\\begin {bmatrix} 0 & 0 & 1 \end {bmatrix}\begin {bmatrix} 0 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix}\begin {bmatrix} 0\\ 0\\ 1 \end {bmatrix} & =\left \vert b\right \vert ^{2}\\ 2 & =\left \vert b\right \vert ^{2} \end {align*}

which implies $$b=\sqrt {2}$$. Therefore \begin {equation} S_{+}|3\rangle =\sqrt {2}|2\rangle \tag {10} \end {equation} Now that $$S_{+}|1\rangle ,S_{+}|2\rangle ,S_{+}|3\rangle$$ are found, $$S_{+}$$ can be calculated. From (4,6,8) the result is\begin {align*} S_{+}|1\rangle & =0|1\rangle \\ S_{+}|2\rangle & =\sqrt {2}|1\rangle \\ S_{+}|3\rangle & =\sqrt {2}|2\rangle \end {align*}

Hence\begin {align*} S_{+} & =\begin {bmatrix} \langle 1|S_{+}|1\rangle & \langle 1|S_{+}|2\rangle & \langle 1|S_{+}|3\rangle \\ \langle 2|S_{+}|1\rangle & \langle 2|S_{+}|2\rangle & \langle 2|S_{+}|3\rangle \\ \langle 3|S_{+}|1\rangle & \langle 3|S_{+}|2\rangle & \langle 3|S_{+}|3\rangle \end {bmatrix} \\ & =\begin {bmatrix} 0 & \langle 1|\sqrt {2}|1\rangle & \langle 1|\sqrt {2}|2\rangle \\ 0 & \langle 2|\sqrt {2}|1\rangle & \langle 2|\sqrt {2}|2\rangle \\ 0 & \langle 3|\sqrt {2}|1\rangle & \langle 3|\sqrt {2}|2\rangle \end {bmatrix} \\ & =\sqrt {2}\begin {bmatrix} 0 & \langle 1|1\rangle & \langle 1|2\rangle \\ 0 & \langle 2|1\rangle & \langle 2|2\rangle \\ 0 & \langle 3|1\rangle & \langle 3|2\rangle \end {bmatrix} \end {align*}

Therefore\begin {equation} S_{+}=\begin {bmatrix} 0 & \sqrt {2} & 0\\ 0 & 0 & \sqrt {2}\\ 0 & 0 & 0 \end {bmatrix} \tag {11} \end {equation} Now that we know $$S_{+}$$ we turn our attention to ﬁnding $$S_{-}$$. Considering the commutator $$\left [ S_{z},S_{-}\right ]$$ \begin {align*} \left [ S_{z},S_{-}\right ] & =\left [ S_{z},S_{x}-iS_{y}\right ] \\ & =\left [ S_{z},S_{x}\right ] -i\left [ S_{z},S_{y}\right ] \end {align*}

But $$\left [ S_{z},S_{x}\right ] =iS_{y}$$ and $$\left [ S_{z}S_{y}\right ] =-iS_{x}$$. The above becomes\begin {align*} \left [ S_{z},S_{-}\right ] & =iS_{y}-i\left ( -iS_{x}\right ) \\ & =iS_{y}-S_{x}\\ & =-S_{-} \end {align*}

This implies\begin {align} \left [ S_{z},S_{-}\right ] & =-S_{-}\nonumber \\ S_{z}S_{-}-S_{-}S_{z} & =-S_{-}\nonumber \\ S_{z}S_{-} & =S_{-}S_{z}-S_{-}\tag {12} \end {align}

Picking $$|1\rangle$$ to start with, gives \begin {align} S_{z}S_{-}|1\rangle & =\left ( S_{-}S_{z}-S_{-}\right ) |1\rangle \nonumber \\ & =S_{-}S_{z}|1\rangle -S_{-}|1\rangle \tag {13} \end {align}

But $$S_{z}|1\rangle =|1\rangle$$ since the eigenvalue is $$1$$ associated with $$|1\rangle$$ eigenvector from (1). The above now becomes\begin {align*} S_{z}S_{-}|1\rangle & =S_{-}|1\rangle -S_{-}|1\rangle \\ & =0S_{-}|1\rangle \end {align*}

The above shows that $$S_{+}|1\rangle$$ is eigenvector of $$S_{z}$$ associated with the eigenvalue $$0$$ which is compatible with experiments. This implies \begin {equation} S_{-}|1\rangle =c|2\rangle \tag {14} \end {equation} Where $$|2\rangle$$ was used above, since that is the eigenvector with $$0$$ eigenvalue. $$c$$ is constant to be found. Taking adjoint of both sides of the above gives\begin {align*} \left ( S_{-}|1\rangle \right ) ^{\dag } & =\left ( c|2\rangle \right ) ^{\dag }\\ \langle 1|S_{-}^{\dag } & =c^{\ast }\langle 2| \end {align*}

Therefore\begin {align} \langle 1|S_{-}^{\dag }S_{-}|1\rangle & =c^{\ast }c\langle 2|2\rangle \tag {15}\\ & =\left \vert c\right \vert ^{2}\nonumber \end {align}

But \begin {align*} S_{-}^{\dag }S_{-} & =\left ( S_{x}-iS_{y}\right ) ^{\dag }\left ( S_{x}-iS_{y}\right ) \\ & =\left ( S_{x}^{\dag }+iS_{y}^{\dag }\right ) \left ( S_{x}-iS_{y}\right ) \end {align*}

Since $$S_{x},S_{y}$$ are Hermitian, then $$S_{x}^{\dag }=S_{x}$$ and $$S_{y}^{\dag }=S_{y}$$. The above becomes\begin {align*} S_{-}^{\dag }S_{-} & =\left ( S_{x}+iS_{y}\right ) \left ( S_{x}-iS_{y}\right ) \\ & =S_{x}^{2}-iS_{x}S_{y}+iS_{y}S_{x}+S_{y}^{2}\\ & =S_{x}^{2}+i\left ( S_{y}S_{x}-S_{x}S_{y}\right ) +S_{y}^{2}\\ & =S_{x}^{2}+i\left [ S_{y},S_{x}\right ] +S_{y}^{2} \end {align*}

But $$\left [ S_{y},S_{x}\right ] =-iS_{z}$$. The above becomes\begin {align} S_{-}^{\dag }S_{-} & =S_{x}^{2}+i\left ( -iS_{z}\right ) +S_{y}^{2}\nonumber \\ & =S_{x}^{2}+S_{y}^{2}+S_{z}\tag {16} \end {align}

Since $$S^{2}=S_{x}^{2}+S_{y}^{2}+S_{z}^{2}$$, then $$S_{x}^{2}+S_{y}^{2}=S^{2}-S_{z}^{2}$$. This implies\begin {equation} S_{-}^{\dag }S_{-}=S^{2}-S_{z}^{2}+S_{z}\tag {16A} \end {equation} Substituting (16A) in (15) gives\begin {equation} \langle 1|\left ( S^{2}-S_{z}^{2}+S_{z}\right ) |1\rangle =\left \vert c\right \vert ^{2}\tag {17} \end {equation} But \begin {equation} S_{z}^{2}=\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix}\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} =\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} \tag {18} \end {equation} And $$S^{2}$$ was found in (6). Substituting (6,18) back in (17) gives an equation to solve for $$c$$\begin {align*} \langle 1|\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} +\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} |1\rangle & =\left \vert c\right \vert ^{2}\\ \langle 1|\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 0 \end {bmatrix} |1\rangle & =\left \vert c\right \vert ^{2}\\\begin {bmatrix} 1 & 0 & 0 \end {bmatrix}\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} 1\\ 0\\ 0 \end {bmatrix} & =\left \vert c\right \vert ^{2}\\ 2 & =\left \vert c\right \vert ^{2} \end {align*}

Hence $$c=\sqrt {2}$$. From (14) this implies\begin {equation} S_{-}|1\rangle =\sqrt {2}|2\rangle \tag {19} \end {equation} Now we pick $$|2\rangle$$ and using (12) gives \begin {align} S_{z}S_{-}|2\rangle & =\left ( S_{-}S_{z}-S_{-}\right ) |2\rangle \nonumber \\ & =S_{-}S_{z}|2\rangle -S_{-}|2\rangle \tag {20} \end {align}

But $$S_{z}|2\rangle =0|2\rangle$$ since the eigenvalue is $$0$$ associated with $$|2\rangle$$ eigenvector. The above now becomes$S_{z}S_{-}|2\rangle =-S_{-}|2\rangle$ The above shows that $$S_{-}|2\rangle$$ is eigenvector of $$S_{z}$$ associated with the eigenvalue $$-1$$ which is compatible with experiments. This implies \begin {equation} S_{-}|2\rangle =b|3\rangle \tag {21} \end {equation} Where $$|3\rangle$$ was used above, since that is the eigenvector with $$-1$$ eigenvalue. $$b$$ is constant to be found. Taking adjoint of both sides of the above gives\begin {align*} \left ( S_{-}|2\rangle \right ) ^{\dag } & =\left ( b|3\rangle \right ) ^{\dag }\\ \langle 2|S_{-}^{\dag } & =b^{\ast }\langle 3| \end {align*}

Therefore\begin {align*} \langle 2|S_{-}^{\dag }S_{-}|2\rangle & =b^{\ast }b\langle 3|3\rangle \\ & =\left \vert b\right \vert ^{2} \end {align*}

But $$S_{-}^{\dag }S_{-}=S^{2}-S_{z}^{2}+S_{z}$$ as calculated earlier in (16A). Hence the above becomes $\langle 2|\left ( S^{2}-S_{z}^{2}+S_{z}\right ) |2\rangle =\left \vert b\right \vert ^{2}$ Using $$S_{z}^{2},S^{2}$$ calculated earlier in the above gives\begin {align*} \langle 2|\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end {bmatrix} -\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 1 \end {bmatrix} +\begin {bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end {bmatrix} |2\rangle & =\left \vert b\right \vert ^{2}\\ \langle 2|\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 0 \end {bmatrix} |2\rangle & =\left \vert b\right \vert ^{2}\\\begin {bmatrix} 0 & 1 & 0 \end {bmatrix}\begin {bmatrix} 2 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 0 \end {bmatrix}\begin {bmatrix} 0\\ 1\\ 0 \end {bmatrix} & =\left \vert b\right \vert ^{2}\\ 2 & =\left \vert b\right \vert ^{2} \end {align*}

Hence $$b=\sqrt {2}$$. From (21) this implies\begin {equation} S_{-}|2\rangle =\sqrt {2}|3\rangle \tag {22} \end {equation} And ﬁnally using $$|3\rangle$$ in (12) results in\begin {align*} S_{z}S_{-}|3\rangle & =\left ( S_{-}S_{z}-S_{-}\right ) |3\rangle \\ & =S_{-}S_{z}|3\rangle -S_{-}|3\rangle \end {align*}

But $$S_{z}|3\rangle =-|3\rangle$$ since the eigenvalue is $$-1$$ associated with $$|3\rangle$$ eigenvector. The above now becomes$S_{z}S_{-}|3\rangle =-2S_{-}|3\rangle$ The above shows that $$S_{-}|3\rangle$$ is eigenvector of $$S_{z}$$ associated with the eigenvalue $$-2$$ which is not compatible with experiments. This implies \begin {equation} S_{-}|3\rangle =0|1\rangle \tag {23} \end {equation} Now that $$S_{-}|1\rangle ,S_{-}|2\rangle ,S_{-}|3\rangle$$ are all known, we are ready to determine $$S_{-}$$. From (19,22,23)\begin {align*} S_{-}|1\rangle & =\sqrt {2}|2\rangle \\ S_{-}|2\rangle & =\sqrt {2}|3\rangle \\ S_{-}|3\rangle & =0|1\rangle \end {align*}

Therefore\begin {align*} S_{-} & =\begin {bmatrix} \langle 1|S_{-}|1\rangle & \langle 1|S_{-}|2\rangle & \langle 1|S_{-}|3\rangle \\ \langle 2|S_{-}|1\rangle & \langle 2|S_{-}|2\rangle & \langle 2|S_{-}|3\rangle \\ \langle 3|S_{-}|1\rangle & \langle 3|S_{-}|2\rangle & \langle 3|S_{-}|3\rangle \end {bmatrix} \\ & =\begin {bmatrix} \langle 1|\sqrt {2}|2\rangle & \langle 1|\sqrt {2}|3\rangle & \langle 1|0|1\rangle \\ \langle 2|\sqrt {2}|2\rangle & \langle 2|\sqrt {2}|3\rangle & \langle 2|0|1\rangle \\ \langle 3|\sqrt {2}|2\rangle & \langle 3|\sqrt {2}|3\rangle & \langle 3|0|1\rangle \end {bmatrix} \\ & =\sqrt {2}\begin {bmatrix} 0 & 0 & 0\\ 1 & 0 & 0\\ 0 & 1 & 0 \end {bmatrix} \end {align*}

Therefore\begin {equation} S_{-}=\begin {bmatrix} 0 & 0 & 0\\ \sqrt {2} & 0 & 0\\ 0 & \sqrt {2} & 0 \end {bmatrix} \tag {24} \end {equation} Now that $$S_{+},S_{-}$$ are known, $$S_{x},S_{y}$$ can be found. Using\begin {align} S_{-} & =S_{x}-iS_{y}\tag {25}\\ S_{+} & =S_{x}+iS_{y}\tag {26} \end {align}

Adding the above two equations gives, and using (11,24)\begin {align*} S_{-}+S_{+} & =2S_{x}\\\begin {bmatrix} 0 & 0 & 0\\ \sqrt {2} & 0 & 0\\ 0 & \sqrt {2} & 0 \end {bmatrix} +\begin {bmatrix} 0 & \sqrt {2} & 0\\ 0 & 0 & \sqrt {2}\\ 0 & 0 & 0 \end {bmatrix} & =2S_{x}\\ S_{x} & =\frac {1}{2}\begin {bmatrix} 0 & \sqrt {2} & 0\\ \sqrt {2} & 0 & \sqrt {2}\\ 0 & \sqrt {2} & 0 \end {bmatrix} \\ & =\frac {\sqrt {2}}{2}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} \end {align*}

Hence\begin {equation} S_{x}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end {bmatrix} \tag {27} \end {equation} And subtracting (25,26) gives\begin {align*} S_{-}-S_{x} & =-2iS_{y}\\\begin {bmatrix} 0 & 0 & 0\\ \sqrt {2} & 0 & 0\\ 0 & \sqrt {2} & 0 \end {bmatrix} -\begin {bmatrix} 0 & \sqrt {2} & 0\\ 0 & 0 & \sqrt {2}\\ 0 & 0 & 0 \end {bmatrix} & =-2iS_{y}\\\begin {bmatrix} 0 & -\sqrt {2} & 0\\ \sqrt {2} & 0 & -\sqrt {2}\\ 0 & \sqrt {2} & 0 \end {bmatrix} & =-2iS_{y}\\ S_{y} & =\frac {-1}{2i}\begin {bmatrix} 0 & -\sqrt {2} & 0\\ \sqrt {2} & 0 & -\sqrt {2}\\ 0 & \sqrt {2} & 0 \end {bmatrix} \\ & =\frac {i}{2}\begin {bmatrix} 0 & -\sqrt {2} & 0\\ \sqrt {2} & 0 & -\sqrt {2}\\ 0 & \sqrt {2} & 0 \end {bmatrix} \\ & =\frac {\sqrt {2}}{2}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} \end {align*}

Hence \begin {equation} S_{y}=\frac {1}{\sqrt {2}}\begin {bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end {bmatrix} \tag {28} \end {equation} This completes the solution. We have found $$S_{x},S_{y}$$, starting from just knowing the eigenvalues of $$S_{z}$$.


text:
# Gray Level Image Processing

## histogramme

### Method

To make the histogram of an image, it's count how many pixels exist for each possible value that a pixel can take. If we have a gray level image where the values ??are represented to 8 bits, we will have a histogram of 255 values. 8 bits gives 28 possible values .
To make this processing, we create a vector hist whose size is defined by the number of possible different value of pixel (ie size = 2k where k is the number of bits for gray level ) and we put all of them to zero, then we must scan the entire picture, line by line and column by column. For each value of a pixel, we increase the value of the histogram associated with that value. for example, if we find the pixel value 200 then hist (200) = hist (200) + 1.

Here is the program that performs the histogram.

clear all; close all;

[filename, pathname] = uigetfile({'*.jpg;*.tif;*.PNG;*.gif;*.bmp','All Image Files';...
'*.*','All Files' },'mytitle',...
'C:\Work\myfile.jpg') % récupération du chemin du fichier

x = imread(filename); % récupération des données

if (length(size(x))>2)% si l'image en niveau de gris est sur 3 composantes couleurs
x=x(:,:,1);% on prend un seul plan
end

% nombre de bit de la valeur des pixels
k = whos('x');
if k.class == 'uint8'
k=8;
elseif k.class == 'int16'
k=16;
elseif k.class == 'int32'
k=32;
elseif k.class == 'int64'
k=64
end

%taille de la matrice
[ligne colonne]=size(x);
hist = zeros(2^k,1);% création du vecteur de résultat
for i=1:ligne %on parcourt toutes les lignes
for j=1:colonne %on parcourt toutes les colonnes
hist(x(i,j)+1) = hist(x(i,j)+1)+1;% on compte les valeurs
end
end
% affichage de l'image et de l'histogramme
figure(1)
subplot(222)
imshow(x)
subplot(221)
plot([0:(2^k)-1],hist)

The program have been applyed on left image, the result is shown to the right.

The histogram gives us information on the distribution of pixel values ??in the image. An histogram with many values ??near zero mean a picture under exposed (dark image) while an histogram which have high values lead an over exposed image (image very bright). The average of this histogram shows the value which separates the histogram into two identical parts. It shows where is the majority of pixels. With the previous example we obtain a mean value of 116.9947, it is called Average Optical Density (AOD). It is calculated as .

aod=sum(hist.*[0:(2^k)-1]')/(ligne*colonne)

### Linear correction

On certain image, the histogram is used to apply a correction. Imagine an image where the large majority (see all) of pixels are located on a small part of possible values. Example only from 30 to 100 for an image of 8 bits (255 values at all). In this case the color (gray level) are distributed over a very short range. It difficult to differentiate them with the eye. The correction will help to spread the colors over the entire range. With the above example the pixels with ??30 values will be put to a zero value and the others will be spread over all the 255 values possible.

for i=1:ligne
for j=1:colonne
img(i,j) = round((img(i,j)-bas)*(255/haut));
end
end

Where bas is the value which will go to zero and haut the value which will go to 255 (or the maximal)

let'see the result on an under exposed image.

### Logarithmic Correction

This type of correction is applied to the image with a high concentration of point either in the dark (close to zero) in the light (close to 255). The goal is to spread the pixels in a nonlinear way. If it's a dark image, we will space the pixel which are low values and close in them when they are high. the same methodology applies if the picture is bright but the reasoning in reverse ( inverse the logaritm fonction ).
How do it? We will make the correction as linear but instead of applying a proportional transformation, it is logarithmic. for example an image or k = 8, the logarithm is 0 to 255, as the next equation:

for i= 1:255
hist(i)=round(log(i)*((2^k)/log(2^k)));
end
plot(hist)

We find the following curve:

We will again go through all pixels and each pixel value found (abscyss) we associate the value log (ordinate). In this way dark pixels are spaced in the color space and the eye see them better. When the pixel are white, which are less many, they are condensed on very little range value (they are less important because fewer many).
The result:

### Correction by cumulative sum of histogram

This time, we will use the histogram of the original image to construct a nonlinear function. For this function we will follow several steps.

• normalize the histogram function: the sum of the vector histogram is 1. we have so:

$hist=\genfrac{}{}{0.1ex}{}{hist}{ligne×colonne}$

• do the cumulative sum of the histogram:

$histsum\left(i\right)=\sum _{a=1}^{i}hist\left(a\right)$

• denormalization of the nonlinear function set for a max = 2*k.

$histsum=histsum\cdot {2}^{k}$

Then, as for Previous corrections, we apply the function for each pixel.
Matlab program:

clear all; close all;

[filename, pathname] = uigetfile({'*.jpg;*.tif;*.PNG;*.gif;*.bmp','All Image Files';...
'*.*','All Files' },'mytitle',...
'C:\Work\myfile.jpg') % récuperation du chemin du fichier

x = imread(filename); % récuperation des données

if (length(size(x))>2)% si l'image en niveau de gris est sur 3 composantes couleurs
x=x(:,:,1);% on prend un seul plan
end

% nombre de bit de la valeur des pixel
info = whos('x');
if strcmp(info.class,'uint8')
k=8;
elseif strcmp(info.class,'uint16')
k=16;
elseif strcmp(info.class,'uint32')
k=32;
elseif strcmp(info.class,'uint64')
k=64;
end

%taille de la matrice
[ligne colonne]=size(x);
hist = zeros(2^k,1);% création du vecteur de résultat
for i=1:ligne %on parcourt toutes les lignes
for j=1:colonne %on parcourt toutes les colonnes
hist(x(i,j)+1) = hist(x(i,j)+1)+1;% on compte les valeurs
end
end
%normalisation d'histogramme
hist=hist/(ligne*colonne);
%somme cumulative de l'histogramme normalisé
histsum = cumsum(hist);
% dénormalisation
histsum=round(histsum*2^k);

% affichage de l'image et de l'histogramme
figure(1)
subplot(222)
imshow(x)
subplot(221)
plot([0:(2^k)-1],hist)

for i=1:ligne
for j=1:colonne
x(i,j) = histsum(x(i,j)+1);
end
end

hist = zeros(2^k,1);
for i=1:ligne
for j=1:colonne
hist(x(i,j)+1) = hist(x(i,j)+1)+1;
end
end

subplot(224)
imshow(x)
subplot(223)
plot([0:(2^k)-1],hist)

And the result


text:
## Study shows two-stroke scooters dominant source of air pollution in many cities; asymmetric polluters

##### 23 May 2014

A study by European researchers has found that two-stroke (2S) scooters, although constituting a small fraction of the fleet, can dominate urban vehicular pollution through organic aerosol and aromatic emission factors up to thousands of times higher than from other vehicle classes. The study by the team led by researchers from the Laboratory of Atmospheric Chemistry, Paul Scherrer Institute, Switzerland, appears in the journal Nature Communications.

The team calls 2S scooters “asymmetric polluters” as their emission factors (EFs) and evidence from air quality measurements before and after bans on scooters in Asian cities suggest they may dominate vehicular pollution despite their relatively small numbers.

Despite being high emitters of primary PM, regulations for scooters are generally less stringent than for other vehicles, for example, in Europe having reached Euro 5/V (a fifth tranche of regulations), for passenger cars and trucks, versus only Euro 2 for scooters… Accordingly, a scientific report to the European Commission suggests that scooters will emit more volatile organic compounds (VOCs) than all other vehicles combined in Europe by 2020. Furthermore, high PM levels and toxic aromatic hydrocarbons, important SOA precursors, have been observed in many cities, especially in Asia. Globally, organic aerosol (OA) dominates PM, with SOA accounting for the largest fraction.

Here we show that 2S scooters emit significant amounts of primary organic aerosol (POA), aromatic VOCs and also produce significant SOA [secondary organic aerosol]. … Chemical analysis of the emissions shows that SOA is mainly produced via photo-oxidation of aromatic VOCs, present in gasoline, from the exhaust. This shows that the known issue of incomplete fuel combustion during the 2S cycle is also responsible for SOA formation. Finally, we present the first online measurements of aged exhaust showing that SOA formation also produces reactive oxygen species (ROS) with potentially detrimental effects on our lungs.

—Platt et al.

For the study, the team combined results from two measurement campaigns where 2S scooter exhaust was injected through a heated inlet into smog chambers to produce SOA via photochemistry. During the first study, an in-use Euro 1 (E1) and a new Euro 2 (E2a) 2S scooter were run in idle or simulated low power. During the second campaign, emissions from a different Euro 2 2S scooter (E2b) were sampled during ECE47 driving cycles.

They also used ambient data are from roadside/tunnel measurements in the US, EU and Asia for light-duty and heavy-duty vehicles, and data from Indian in-use 2S autorickshaws for comparison to the European scooters of this study.

 Emission factors (EFs) from scooters and other vehicles. EFs plotted as box-and-whiskers (median line, red; 25th and 75th percentile, box; 10th and 90th percentile, whiskers) of (a) POA, (b) aged OA (POA + SOA formation), (c) benzene and (d) light aromatics (benzene, toluene and C2–C4 alkylated benzenes). Points shown next to the box-and-whiskers are the individual data points, colored depending on measurement region for ambient data. 2S scooters were run in idle or during driving cycles (ECE47). Data on the other vehicles shown are from the literature for light-duty and heavy-duty vehicles (LDVs and HDVs).LDVs data are further divided between vehicles meeting Euro 5 and those not meeting Euro 5, labelled

They found that the oxidation of VOCs in 2S scooter emissions produces significant SOA (g carbon (C) kg-1 fuel), with total OA on average 2.9 and 2.4 times higher than POA after aging for idling and driving 2S scooters, respectively. In addition, substantial toxic aromatic emissions (up to ~40% of emitted VOC volume for the scooters of this study) of benzene, toluene and C2–C4 alkylated benzenes, which are recognized SOA precursors, were present in the exhaust.

Among the aromatics, benzene is of particular concern due to its carcinogenicity, they noted. Levels in the raw 2S scooter exhaust were as high as 300,000 mg m-3 or 146 ppm (v) from idling. The EU annual mean limit for the protection of human health is 5 mg m-3 (ref. 13), while the US National Institute for Occupational Safety and Health (NIOSH) recommends that workers wear special breathing equipment when exposed to benzene at levels exceeding 1 ppm for 15 min. Waiting in traffic behind an idling 2S scooter may therefore be highly deleterious to health, the authors said.

In general, ambient EFs [emission factors] from Asian vehicles are in the same range as European and US vehicles, while emissions from in-use 2S rickshaws are slightly higher than from the European scooters of this study. POA emissions from 2S scooters are on average around 20 (maximum 2,780) times higher than ambient (light-duty dominant) values, and aged OA emissions on an average 53–771 times higher than laboratory studies on other vehicle types.

It should be noted that absolute aerosol concentrations can influence EFs: higher measurement concentrations would lead to higher EFs. SOA formation is most significant from idling scooter emissions, while smaller at higher engine loads. However, POA emissions are higher under the latter conditions, and the aggregate POA + SOA emission at high load is comparable with that from idling.

—Platt et al.

The team suggested several reasons for these relatively large OA and aromatic emissions from 2S scooters.:

1. 2S engines, unlike four-stroke (4S), require addition of lubricant oils to the fuel, some of which is emitted in the exhaust.

2. During the 2S engine cycle some of the fresh fuel/air mixture passes directly through the engine, increasing VOC emissions, which may explain the high SOA formation.

3. Scooters generally utilize rich combustion (low-air/fuel ratio), improving drivability while producing higher CO, VOC and PM emissions (but lower NOx). Accordingly, the VOC emissions measured—in particular aromatics as found in raw gasoline—are also on average 124 and 11 times higher from idling and driving 2S scooters, respectively, compared with those from other vehicles.

4. Scooter after-treatment systems are inherently inefficient due to their relatively small size and longer light-off times.

Despite the difficulty of making a precise estimation of a relative contribution to vehicular PM and aromatics from 2S scooters, the researchers suggested that many 2S scooters likely fall into the super-polluting category, especially as a considerable number of scooters are in operation in some regions without any form of emissions control. In contrast, the scooters presented in the study all were equipped with two-way oxidation catalysts, which reduce emissions of carbon monoxide and VOCs). Further, real-world emissions may be further exacerbated by poor maintenance and tampering, rife for scooters.

Using the average 2S scooter EF (ECE47 driving cycle) … suggests that 2S scooters contribute to around 60% of roadside POA in Bangkok, where they account for 10% of fuel consumption. In a more extreme case (comparing the 75th percentile for scooters and 25th percentile for ambient light-duty dominated), 2S scooters would contribute over 96% to roadside POA. Note that these values are based on the European scooters of this study. …Emissions from some in-use Asian 2S vehicles may be higher, by a factor of three.

… Our data suggest that 2S scooters are a significant, and in many cities the largest, source of vehicular PM and toxic SOA and aromatic hydrocarbons, despite being a relatively small fraction of the total fleet. Therefore, given the alternative technologies available, restrictions on 2S scooters, already implemented in China, could improve air quality in many cities around the globe.

—Platt et al.

Resources

• S.M. Platt et al. (2014) “Two-stroke scooters are a dominant source of air pollution in many cities,” Nature Communications 5, Article number: 3749 doi: 10.1038/ncomms4749

Yes, Kuala Lumpur City is a perfect example.

OK, I can well believe that.
Question is - what do you replace them with ?

Bikes, E-Bikes, Electric scooters, 4 stroke scooters ?

My recommendation would be e-scooters, but my guess would be 4 stroke scooters - at least the same companies can make and sell them.
I had (have) a 4 stroke Vespa Et4 (125cc) and it is a very nice vehicle (though I mainly prefer to cycle).

They tend to put the whole family on these vehicles, so bikes and e-bikes wouldn't be powerful enough, and a powerful enough e-scooter would be too expensive, hence the 4 stroke (or a small 4 stroke motorbike).

There have been few examples of two-stroke engines with a separate closed loop lubrication system. These are just as clean as four-stroke engines but they are more expensive to build. So what! It's cheaper to pollute, so let's just keep on polluting with conventional two-strokes.

We have a vintage Vespa from Italy. It is a work of art and has not been driven in over 15 years (kept only for artistic purposes).
Honda has a simple answer, the SH125i and SH150i micro hybrid 4 stroke, gets 47.4 km/liter of gasoline. Honda the world's largest scooter manufacturer gave up on 2 strokes years ago as well.

Yup, 4 stroke scooters. 2 stroke might survive if they can pull a rabbit trick on emissions. Though I wouldn't hold my breath....rather I *would* hold my breath.

Just switch to E-bike and be done with it!

Now, battery powered lawnmower and trimmer and chainsaw can all be powered by lithium battery and are quite affordable, costing perhaps 50% more than gasoline version, but one can recoup this in a short time given the much higher cost of gasoline vs cost of electricity per unit of work done.

In an E-bike, the cost differential is perhaps no more than 30% vs the gasoline version because the engine makes up smaller proportion of cost of the entire vehicle than in a lawn equipment. With rapid charging and charged twice daily at work and at home, daily riding range would be more than sufficient. THus, charged during the day to capture solar energy and charged at night to capture excess wind energy. Pollution level will drop precipitously with E-bikes mass adoption.

Yes Roger. China and India (and many other countries) could actively support the construction, sales and usage of adapted low cost E-bikes and E-Scouters for their local market and export.

For example, license plates could be $100+ depending on emissions for the gasoline units and as low as$1 for the electrified versions.

E-bikes are the ultimate answer. They have already started to dominate the high end motorcycle models. A company from Texas called KLD Energy Technologies is selling their oneDrive electric system to Eclimo of Malaysia that has an electric scooter with a 53 mile range and gets an impressive 234 kme per liter (or 6 kWh/100 miles).
Electric scooters have enormous torque so they can also handle typical double duty as pickup truck and rickshaw.

In USA, we're starting to see 4 strokers everywhere, including weed-whackers, chainsaws, snowthrowers, etc.... (and might I add, about time!)

2 stokers are a thing of the past.... (with regards to 2-stroke gasoline of course)

O TOLMON NIKA,

Exactly. As far as I know, the only 2 stroke scooter left on the market in Europe and North America is an expensive direct-injection Aprilia. They are gone from the mainstream market. Any other country can legislate them away easily, four-stroke alternatives are just as cheap.

I love the two stroke. alas only direct injection can save it... or can it? Help us please Orbital and Rotax!

Take a look at the PatATi, a simple two-stroke (only three moving parts, no reed valves, no rotary valves) with asymmetric transfer and asymmetric intake, at http://www.pattakon.com/pattakonPatAT.htm#PatATi
Thanks

The comments to this entry are closed.


text:
The Chebyshev Functions
 Definition: The First Chebyshev Function or Theta Function is the function $\theta : X \to \mathbb{R}$ defined for all $x \in \mathbb{R}$ by $\displaystyle{\theta(x) = \sum_{p \leq x} \ln p}$.
 Definition: The Second Chebyshev Function or Psi Function is the function $\psi : X \to \mathbb{R}$ defined for all $x \in \mathbb{R}$ by $\displaystyle{\psi(x) = \sum_{p^k \leq x} \ln p}$.
Note that if $p^k \leq x$ then $\Lambda (p^k) = \ln p$ and otherwise, $\Lambda (n) = 0$. Therefore the psi function can be expressed alternatively by the formula:


text:
It is currently 22 Jan 2018, 10:30

### 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

# A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43361
A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

17 Jun 2015, 04:55
Expert's post
20
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

53% (02:26) correct 47% (02:14) wrong based on 178 sessions

### HideShow timer Statistics

A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If the dowel is painted on all surfaces and then cut into 1/2 inch cubes, what fraction of the resulting cube faces are painted?

(A) 1/3
(B) 3/8
(C) 7/16
(D) 1/2
(E) 9/16

Kudos for a correct solution.
[Reveal] Spoiler: OA

_________________
SVP
Joined: 08 Jul 2010
Posts: 1920
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

17 Jun 2015, 05:21
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
Bunuel wrote:
A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If the dowel is painted on all surfaces and then cut into 1/2 inch cubes, what fraction of the resulting cube faces are painted?

(A) 1/3
(B) 3/8
(C) 7/16
(D) 1/2
(E) 9/16

Kudos for a correct solution.

Number of pieces in which First Side of Box of Dimension 4 will be cut in cubes of Dimension (1/2)inch = 4/(1/2) = 8
Number of pieces in which Second Side of Box of Dimension 1 will be cut in cubes of Dimension (1/2)inch = 1/(1/2) = 2
Number of pieces in which Third Side of Box of Dimension 1 will be cut in cubes of Dimension (1/2)inch = 1/(1/2) = 2

i.e. Total Number of smaller cubes of Dimension (1/2 each) = 8 x 2 x 2 = 32

METHOD-1
Total Cubes painted on 3 faces = Cubes available on the vertices of Box = No. of Vertics of Box = 8
Total Cubes painted on 2 faces = Cubes available on the edges of Box x 4 = 6 x 4 = 24
Total Cubes painted on 1 faces = Cubes available on the Faces but not on the edges of Vertices
Total Cubes painted on 0 faces = Cubes NOT available on anyone of the Faces

i.e. Total Painted faces = (8*3) + (24*2) + 0 + 0 = 72
Total faces = No. of Cubes x Faces on each cube = 32x6 = 192

Fraction = 72/192 = 3/8

[Reveal] Spoiler:
B

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

SVP
Joined: 08 Jul 2010
Posts: 1920
Location: India
GMAT: INSIGHT
WE: Education (Education)
Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

17 Jun 2015, 05:22
3
KUDOS
Expert's post
2
This post was
BOOKMARKED
Bunuel wrote:
A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If the dowel is painted on all surfaces and then cut into 1/2 inch cubes, what fraction of the resulting cube faces are painted?

(A) 1/3
(B) 3/8
(C) 7/16
(D) 1/2
(E) 9/16

Kudos for a correct solution.

Number of pieces in which First Side of Box of Dimension 4 will be cut in cubes of Dimension (1/2)inch = 4/(1/2) = 8
Number of pieces in which Second Side of Box of Dimension 1 will be cut in cubes of Dimension (1/2)inch = 1/(1/2) = 2
Number of pieces in which Third Side of Box of Dimension 1 will be cut in cubes of Dimension (1/2)inch = 1/(1/2) = 2

i.e. Total Number of smaller cubes of Dimension (1/2 each) = 8 x 2 x 2 = 32

METHOD-2
Total painted Area = Total Surface Area of Bigger Box = 2 (lb+bh+lh) = 2(4x1 + 1x1 + 4x1) = 18

Total Surface Area of all the smaller cubes = No. of Smaller Cubes x Surface Area of Each Cube

Please Note: Surface Area of Each Cube = $$6a^2$$ where $$a$$ is the each dimension of Cube

i.e. Total Surface Area of all the smaller cubes = $$32 * [6x(1/2)^2]$$ = 48

Painted faces as fraction of total faces = Painted Area / Total area of smaller Cubes = 18/48 = 3/8

[Reveal] Spoiler:
B

_________________

Prosper!!!
GMATinsight
Bhoopendra Singh and Dr.Sushma Jha
e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772
Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi
http://www.GMATinsight.com/testimonials.html

22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION

Math Expert
Joined: 02 Aug 2009
Posts: 5538
Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

17 Jun 2015, 05:45
Bunuel wrote:
A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If the dowel is painted on all surfaces and then cut into 1/2 inch cubes, what fraction of the resulting cube faces are painted?

(A) 1/3
(B) 3/8
(C) 7/16
(D) 1/2
(E) 9/16

Kudos for a correct solution.

Hi,
we have to find the painted faces and total faces..

1)painted faces..
we have four faces of size 4*1 sq inches...when cut at 1/2 inches , it will give us 8*2 external (painted) faces.. total faces 8*2*4=64faces
remaining two faces of size 1*1sq inches...when cut at 1/2 inches, it will give us 2*2 external (painted) faces.. total faces 2*2*2=8faces
total 64+8=72 faces

2) total number of faces: number of cubes=4*1*1/(0.5*0.5*0.5)=8*2*2=32..
each cube has 6 faces.. so total faces=32*6=192

fraction of faces painted=painted faces/total faces=72/192=3/8
ans B
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

Manager
Joined: 26 Dec 2011
Posts: 120
Schools: HBS '18, IIMA
Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

17 Jun 2015, 06:53
1
KUDOS
Bunuel wrote:
A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If the dowel is painted on all surfaces and then cut into 1/2 inch cubes, what fraction of the resulting cube faces are painted?

(A) 1/3
(B) 3/8
(C) 7/16
(D) 1/2
(E) 9/16

Kudos for a correct solution.

Solution -
Total area of the Wooden dowel(Painted) = 2*(4*1+1*4+1*1) = 18 inch^2
No of cubes after cut into 1/2 inch = (4*1*1)/(1/2) = 32
Area of 1/2 inch cube = 6*(1/2)*(1/2) = 3/2 inch^2

Total area of all the 1/2 inch cubes = 32*(3/2) = 48 inch^2*

Fraction of the resulting cube faces are painted = 18/48 = 3/8. ANS B.

Thanks

_________________

Thanks,

Current Student
Joined: 29 Mar 2015
Posts: 44
Location: United States
Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

18 Jun 2015, 23:45
Bunuel wrote:
A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If the dowel is painted on all surfaces and then cut into 1/2 inch cubes, what fraction of the resulting cube faces are painted?

(A) 1/3
(B) 3/8
(C) 7/16
(D) 1/2
(E) 9/16

Kudos for a correct solution.

The surface area of the wooden dowel is the area of each of the 6 faces. There is are four 4x1 faces and two 1x1 faces, for a total of 4*4+2=18 square inches that are painted.
If it is cut into 1/2 inch cubes, then that means that it will be cut into 8x2x2 1/2 inch cubes. The surface area of each of these cubes is 6*1/2*1/2=6/4=1.5 square inches. There are a total of 8x2x2=32 cubes, so the total surface area is 32*1.5=32+16=48 square inches. 18/48 square inches are painted. 18/48=3/8 so answer is B.
Senior Manager
Status: Math is psycho-logical
Joined: 07 Apr 2014
Posts: 432
Location: Netherlands
GMAT Date: 02-11-2015
WE: Psychology and Counseling (Other)
Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

19 Jun 2015, 00:14
Hello,

I am a little bit confused here. Are we looking for the fraction of the resulting, painted surface areas, or for the fraction of the number of resulting faces that are painted?

So, the ratio of "total inches of resulting painted surface areas"/"total inches of surface areas" or for the ratio of "total number of resulting faces that are painted"/"total number of resulting faces"?
Math Expert
Joined: 02 Aug 2009
Posts: 5538
Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

19 Jun 2015, 00:36
pacifist85 wrote:
Hello,

I am a little bit confused here. Are we looking for the fraction of the resulting, painted surface areas, or for the fraction of the number of resulting faces that are painted?

So, the ratio of "total inches of resulting painted surface areas"/"total inches of surface areas" or for the ratio of "total number of resulting faces that are painted"/"total number of resulting faces"?

Hi,
we are looking for the faces and that is mentioned in the Q..
A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If the dowel is painted on all surfaces and then cut into 1/2 inch cubes, what fraction of the resulting cube faces are painted
so the coloured portion in your query is the ratio you have to find
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

BANGALORE/-

Math Expert
Joined: 02 Sep 2009
Posts: 43361
Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

22 Jun 2015, 06:18
Expert's post
1
This post was
BOOKMARKED
Bunuel wrote:
A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If the dowel is painted on all surfaces and then cut into 1/2 inch cubes, what fraction of the resulting cube faces are painted?

(A) 1/3
(B) 3/8
(C) 7/16
(D) 1/2
(E) 9/16

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

If you draw a picture, this problem becomes a matter of counting:

Total Cubes = (4 inches × 2 cubes per inch) × (1 × 2) × (1 × 2) = 32 cubes

Total Cube Faces = 32 cubes × 6 faces per cube = 192 faces total

We now consider the faces that were painted on the front and back of the dowel, the top and bottom of the dowel, and the ends of the dowel. In the diagram above, we can see 16 faces on the front, 16 faces on the top, and 4 faces on the end shown. Of course, there are other sides: the back, the bottom, and the other end.

The fraction of faces that are painted = 72/192 = 24(3)/24(8) = 3/8.

Notice that there is no shortcut to solving this kind of problem, so don't waste time looking for one—just draw the diagram and count.

 ! Even if you can easily picture 3-D shapes and objects in your head, it is still better to draw a picture on your scrapboard.Wrong answer choices are often those you might get by losing track of your progress as you process the object in your mind.

This kind of process can also help you with questions that deal with the relative size of different objects.

[Reveal] Spoiler:
Attachment:

2015-06-22_1814.png [ 38.37 KiB | Viewed 2614 times ]

Attachment:

2015-06-22_1815.png [ 29.26 KiB | Viewed 2598 times ]

_________________
Manager
Joined: 01 Sep 2016
Posts: 208
GMAT 1: 690 Q49 V35
A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

17 Aug 2017, 14:53
1
KUDOS
Hey Experts- Bunuel mikemcgarry - Can you please validate my approach below. I used the volume method to find the number of cubes that would be contained in the cuboid instead of the counting technique. Is it alright to use this method? Are there any caveats to consider. For example- the volume approach often does not take into account the exact cuts and specifics of a figure. This approach might be working in this question, but does it work well with other scenarios as well?

BEFORE the cuts
There are 6 faces to the dowel, and each face is a rectangle.
4 of the faces have dimensions 4 by 1. So, the area of each rectangle = (4)(1) = 4
So, the total area of those 4 rectangles = (4)(4) = 16
The remaining 2 faces have dimensions 1 by 1. So, the area of each rectangle (square) = (1)(1) = 1
So, the total area of those 2 rectangles = (2)(1) = 2

So, BEFORE the cuts, the TOTAL surface area of the dowel = 16 + 2 = 18
In other words, there are 18 square inches of paint.

AFTER the cuts
Each cube has dimensions 1/2 by 1/2 by 1/2
So, the VOLUME of each cube = (1/2)(1/2)(1/2) = 1/8
BEFORE the cut, the VOLUME of the dowel = (4)(1)(1) = 4
So, the NUMBER of cubes = 4/(1/8) = 32
So, after the cuts, there are 32 mini cubes

Each individual mini-cube has 6 sides, and each side is a 1/2 by 1/2 square.
So, the area of ONE square = (1/2)(1/2) = 1/4
So, the total surface area of one mini cube = (6)(1/4) = 6/4 = 3/2
So, the TOTAL surface are of all 32 mini cubes = (32)(3/2) = 48 square inches

So, the 32 mini cubes have a TOTAL surface are of 48 square inches, and 18 square inches are painted.

So, the correct answer is 18/48
_________________

we shall fight on the beaches,
we shall fight on the landing grounds,
we shall fight in the fields and in the streets,
we shall fight in the hills;
we shall never surrender!

Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4648
Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

17 Aug 2017, 15:43
2
KUDOS
Expert's post
bkpolymers1617 wrote:
Hey Experts- Bunuel mikemcgarry - Can you please validate my approach below. I used the volume method to find the number of cubes that would be contained in the cuboid instead of the counting technique. Is it alright to use this method? Are there any caveats to consider. For example- the volume approach often does not take into account the exact cuts and specifics of a figure. This approach might be working in this question, but does it work well with other scenarios as well?

BEFORE the cuts
There are 6 faces to the dowel, and each face is a rectangle.
4 of the faces have dimensions 4 by 1. So, the area of each rectangle = (4)(1) = 4
So, the total area of those 4 rectangles = (4)(4) = 16
The remaining 2 faces have dimensions 1 by 1. So, the area of each rectangle (square) = (1)(1) = 1
So, the total area of those 2 rectangles = (2)(1) = 2

So, BEFORE the cuts, the TOTAL surface area of the dowel = 16 + 2 = 18
In other words, there are 18 square inches of paint.

AFTER the cuts
Each cube has dimensions 1/2 by 1/2 by 1/2
So, the VOLUME of each cube = (1/2)(1/2)(1/2) = 1/8
BEFORE the cut, the VOLUME of the dowel = (4)(1)(1) = 4
So, the NUMBER of cubes = 4/(1/8) = 32
So, after the cuts, there are 32 mini cubes

Each individual mini-cube has 6 sides, and each side is a 1/2 by 1/2 square.
So, the area of ONE square = (1/2)(1/2) = 1/4
So, the total surface area of one mini cube = (6)(1/4) = 6/4 = 3/2
So, the TOTAL surface are of all 32 mini cubes = (32)(3/2) = 48 square inches

So, the 32 mini cubes have a TOTAL surface are of 48 square inches, and 18 square inches are painted.

So, the correct answer is 18/48

Dear bkpolymers1617,

I'm happy to respond. :-)

What you did is fine. I would think that this approach might take slightly longer than the counting method, but whatever makes the most intuitive sense to you is often the easier and faster method. This is a relatively straightforward problem. As you move into harder problems, and you are wondering about the relative strengths of two different approaches, experiment: do them both, and see which one is faster. Ideally, you will find that you are confident in more than one way to solve a number of problems.

Does this make sense?

Mike :-)
_________________

Mike McGarry
Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)

EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 10734
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If [#permalink]

### Show Tags

12 Jan 2018, 12:07
Hi All,

This question can be approached in a couple of different ways (most of which involve lots of math steps). Based on the 'spread' of the answer choices though, you can do a little bit of math and use a bit of logic to get to the correct answer.

To start, it would help to physically draw the rectangular solid that is described (including the "cut lines"). Since a (1 in.) x (1 in.) x (1 in.) cube will contain eight 1/2 in. mini-cubes, the (4 in.) x (1 in.) x (1in.) dowel will end up being cut into 4(8) = 32 mini-cubes.

When the outside of the dowel is painted, you should recognize that each of the smaller cubes will have paint on either 3 faces (the 8 'corner' pieces) or 2 faces (all of the non-corner pieces - 24 in total). Put a different way - the 8 corner pieces have HALF of their faces painted and all of the other pieces have a THIRD of their faces painted. This then can be looked at as a Weighted Average. The AVERAGE fractional number of faces painted must be closer to 1/3 than to 1/2. There's only one answer that fits...

[Reveal] Spoiler:
B

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

Special Offer: Save \$75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

Re: A rectangular wooden dowel measures 4 inches by 1 inch by 1 inch. If   [#permalink] 12 Jan 2018, 12:07
Display posts from previous: Sort by


text:
# How to evaluate $\sum J_0(\alpha n) z^{-n}$ in closed form?

I need to evaluate $\sum_{n = -\infty}^{\infty} J_0(\alpha n) z^{-n}$ in closed form, where $z$ is complex variable and $J_0()$ is the zeroth order Bessel function of the first kind. How do I evaluate this summation?

-
Explanation. In "signal processing" they say "Z transform" but mathematicians don't use that terminology. The question may be translated to:evaluate $$\sum_{n=-\infty}^{\infty}J_0(\alpha n)z^{-n}$$ is closed form. –  GEdgar Mar 24 '12 at 16:27
Since $J_0(x)$ is asymptotically proportional to $x^{-1/2} \cos x$, wouldn't the series diverge for all $z$? –  Steven Stadnicki Mar 28 '12 at 16:56
@StevenStadnicki Since $Z = r e^{j\theta}$, for large enough value of $r$ the sum $\sum |J_0(\alpha n) r^{-n}|$converges? –  sauravrt Mar 28 '12 at 17:07
@sauravrt But you're summing over positive and negative $n$, so if the terms go to $0$ on one side (e.g., if $|r|< 1$ then the terms go to $0$ for $n\rightarrow\infty$) then they grow unboundedly on the other (e.g., $n\rightarrow -\infty$). The harmonic term tweaks this behavior slightly, but not enough to get any kind of convergence. –  Steven Stadnicki Mar 28 '12 at 17:24

You can prove (or maybe already know) that in fact $J_0(-z)=J_0(z)$. So we can split up the sum as $$\underset{n=-\infty}{\overset{\infty}{\sum}}J_0(an)z^{-n}=\underset{n=-\infty}{\overset{-1}{\sum}}J_0(an)z^{-n}+\underset{n=0}{\overset{\infty}{\sum}}J_0(an)z^{-n} = \underset{n=1}{\overset{\infty}{\sum}}J_0(an)z^{n}+\underset{n=0}{\overset{\infty}{\sum}}J_0(an)z^{-n}$$ $$= J_0(0)+\underset{n=1}{\overset{\infty}{\sum}}J_0(an)(z^n+z^{-n}) = 1+\underset{n=1}{\overset{\infty}{\sum}}J_0(an)(z^n+z^{-n})$$


text:
RUS  ENG JOURNALS   PEOPLE   ORGANISATIONS   CONFERENCES   SEMINARS   VIDEO LIBRARY   PERSONAL OFFICE
 General information Latest issue Forthcoming papers Archive Impact factor Subscription Guidelines for authors License agreement Submit a manuscript Search papers Search references RSS Latest issue Current issues Archive issues What is RSS

 Izv. RAN. Ser. Mat.: Year: Volume: Issue: Page: Find

 Izv. Akad. Nauk SSSR Ser. Mat., 1990, Volume 54, Issue 2, Pages 396–417 (Mi izv1100)

Triples of infinite iterates of metrizable functors

V. V. Fedorchuk

Abstract: The concepts of metrizable, uniformly metrizable, and perfectly metrizable functors are introduced. The triple $\mathscr F^+$, $\mathscr F^{++}$, $\mathscr F^\omega$ of infinite iterates of a perfectly metrizable functor $\mathscr F$ is defined. The geometric properties of such triples are investigated for the continuous hyperspace functor and the probability measure functor.

Full text: PDF file (1113 kB)
References: PDF file   HTML file

English version:
Mathematics of the USSR-Izvestiya, 1991, 36:2, 411–433

Bibliographic databases:

UDC: 515.12
MSC: Primary 54B30, 54E45, 54E50; Secondary 54B20, 54C50, 54F25

Citation: V. V. Fedorchuk, “Triples of infinite iterates of metrizable functors”, Izv. Akad. Nauk SSSR Ser. Mat., 54:2 (1990), 396–417; Math. USSR-Izv., 36:2 (1991), 411–433

Citation in format AMSBIB
\Bibitem{Fed90} \by V.~V.~Fedorchuk \paper Triples of infinite iterates of metrizable functors \jour Izv. Akad. Nauk SSSR Ser. Mat. \yr 1990 \vol 54 \issue 2 \pages 396--417 \mathnet{http://mi.mathnet.ru/izv1100} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=1062520} \zmath{https://zbmath.org/?q=an:0715.54005} \adsnasa{http://adsabs.harvard.edu/cgi-bin/bib_query?1991IzMat..36..411F} \transl \jour Math. USSR-Izv. \yr 1991 \vol 36 \issue 2 \pages 411--433 \crossref{https://doi.org/10.1070/IM1991v036n02ABEH002028}

• http://mi.mathnet.ru/eng/izv1100
• http://mi.mathnet.ru/eng/izv/v54/i2/p396

 SHARE:

Citing articles on Google Scholar: Russian citations, English citations
Related articles on Google Scholar: Russian articles, English articles

This publication is cited in the following articles:
1. V. V. Fedorchuk, “Probability measures in topology”, Russian Math. Surveys, 46:1 (1991), 45–93
2. Vitaly V. Fedorchuk, Yury V. Sadovnichy, “On some categorical properties of uniform spaces of probability measures”, Topology and its Applications, 82:1-3 (1998), 131
3. V.V. Fedorchuk, “On a preservation of completeness of uniform spaces by the functor PscT”, Topology and its Applications, 91:1 (1999), 25
4. Yu. V. Sadovnichii, “Lifting the functors $U_\tau$ and $U_R$ to the categories of bounded metric spaces and uniform spaces”, Sb. Math., 191:11 (2000), 1667–1691
5. A. A. Zaitov, “Open Mapping Theorem for Spaces of Weakly Additive Homogeneous Functionals”, Math. Notes, 88:5 (2010), 655–660
•  Number of views: This page: 233 Full text: 103 References: 34 First page: 3


text:
# Elliptic complex

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Let $M$ be a differential manifold and $E_i$ be smooth vector bundles over $M$. Let $P_i:\Gamma(E_i) \to \Gamma(E_{i+1})$ form a differential complex (viz $P_i \circ P_{i-1} = 0$. Then this differential complex is said to be elliptic if the following sequence of symbols is exact:
$\ldots \pi^*(E_{i-1}) \to^{\sigma(P_{i-1})} \pi^*(E_i) \to^{\sigma(P_i)} \pi^*{E_{i+1}} \ldots$


text:
# Configure MPI for Distributed Computing

This section will cover how to configure ADCME for MPI functionalities.

## Configure the MPI backend

The first step is to configure your MPI backend. There are many choices depending on your operation system. For example, Windows have Microsoft MPI. There are also OpenMPI and Intel MPI available on most Linux distributions. If you want to use your own MPI backend, you need to locate the MPI libraries, header files, and executable (e.g., mpirun). You need to build ADCME with the following environment variable:

• MPI_C_LIBRARIES: the MPI shared library, for example, on Windows, it may be
C:\\Program Files (x86)\\Microsoft SDKs\\MPI\\Lib\\x64\\msmpi.lib

On Unix systems, it may be /opt/ohpc/pub/compiler/intel-18/compilers_and_libraries_2018.2.199/linux/mpi/intel64/lib/release/libmpi.so Note that you must include the shared library in the variable.

• MPI_INCLUDE_PATH: the directory where mpi.h is located, for example,

C:\\Program Files (x86)\\Microsoft SDKs\\MPI\\Include Or in a Unix system, we have /opt/ohpc/pub/compiler/intel-18/compilers_and_libraries_2018.2.199/linux/mpi/intel64/include/

The simplest way is to add these variables in the environment variables. For example, in Linux, we can add the following lines in the ~/.bashrc file.

export MPI_C_LIBRARIES=/opt/ohpc/pub/compiler/intel-18/compilers_and_libraries_2018.2.199/linux/mpi/intel64/lib/libmpi.so
export MPI_INCLUDE_PATH=/opt/ohpc/pub/compiler/intel-18/compilers_and_libraries_2018.2.199/linux/mpi/intel64/include/
alias mpirun=/opt/ohpc/pub/compiler/intel-18/compilers_and_libraries_2018.2.199/linux/mpi/intel64/bin/mpirun

In the case you do not have an MPI backend, ADCME provides you a convenient way to install MPI by compiling from source. Just run

using ADCME
install_openmpi()

This should install an OpenMPI library for you. Note this functionality does not work on Windows and is only tested on Linux.

## Build MPI Libraries

The MPI functionality of ADCME is not fulfilled at this point. To enable the MPI support, you need to recompile the built-in custom operators.

using ADCME
ADCME.precompile(true)

At this point, you will be able to use MPI features.

## Build MPI Custom Operators

You can also build MPI-enabled custom operators by calling

using ADCME
customop(with_mpi=true)

In this case, there will be extra lines in CMakeLists.txt to setup MPI dependencies.

IF(DEFINED ENV{MPI_C_LIBRARIES})
set(MPI_INCLUDE_PATH $ENV{MPI_INCLUDE_PATH}) set(MPI_C_LIBRARIES$ENV{MPI_C_LIBRARIES})
message("MPI_INCLUDE_PATH = ${MPI_INCLUDE_PATH}") message("MPI_C_LIBRARIES =${MPI_C_LIBRARIES}")
include_directories(\${MPI_INCLUDE_PATH})
ELSE()
message("MPI_INCLUDE_PATH and/or MPI_C_LIBRARIES is not set. MPI operators are not compiled.")
ENDIF()

## Running MPI Applications with Slurm

To run MPI applications with slurm, the following commands are useful

sbatch -n 4 -c 8 mpirun -n 4 julia app.jl

This specifies 4 tasks and each task uses 8 cores. You can also replace sbatch with salloc.

To diagonose the application, you can also let mpirun print out the rank information, e.g., in OpenMPI we have

sbatch -n 4 -c 8 mpirun --report-bindings -n 4 julia app.jl


text:
# Rotate doesn't work with BarChart and BoxWhiskerChart

Rotate@BarChart[RandomReal[1, {4, 5}], ChartLayout -> "Percentile"]

Even though the FullForm generated by BarChart is Graphics expression, Rotate has no effect on BarChart and many other packaged plot functions. However, it's frequently necessary to rotate, scale and align such plots to other graphics in composite data visualizations (eg, imagine a heatmap augmented with row and column descriptive statistics visualized with two corresponding BoxWhiskerChart)

How can rotatable, translatable, etc. graphics be extracted from functions like BarChart and BoxWhiskerChart?

-

You left off the angle to rotate through:

Rotate[BarChart[RandomReal[1, {4, 5}], ChartLayout -> "Percentile"], 90 Degree]

Rotate[BarChart[RandomReal[1, {4, 5}], ChartLayout -> "Percentile"], 45 Degree]

Strange that it didn't throw an error, though.

-
Indeed, it should give an error message. –  Jens Apr 3 '12 at 21:14
Indeed. Thank you @F'x, Eli. I'm so used to GraphicConverter and other apps that have default 90 deg Rotate. –  alancalvitti Apr 4 '12 at 0:43
@jens In cases where the inputs do not match the defined formats Mathematica just returns the function unevaluated. This is by design. –  Sjoerd C. de Vries Apr 4 '12 at 16:09
@SjoerdC Not really. Try Map[x] where you see the red pointer because of the wrong format, and you get a message ::argtu: The same holds here with Rotate[x] (you get a red pointer to warn you, but no message). –  Jens Apr 4 '12 at 16:22
@jens The OP used the function in prefix format. In that case there is no possibility for syntax coloring. I guess many functions have a reasonable output built-in given insufficient parameters . Take RotateLeft[{x}] for instance. Same for Rotate[x]. Simple user-defined functions return unevaluated if the argument patterns don't match the definition. –  Sjoerd C. de Vries Apr 4 '12 at 16:37

In addition to Rotate you could use the BarOrigin option to BarChart to change the orientation. In this case the graphic is still wider than it is tall:

BarChart[RandomReal[1, {4, 5}],
ChartLayout -> "Percentile", BarOrigin -> Left]

-
Thanks @Brett. These features and options (eg "Percentile") are barely mentioned in Documentation. If only Wolfram would link MMA.SE to help. –  alancalvitti Apr 4 '12 at 0:44


text:
Article

# F-Willmore submanifold in space forms

Frontiers of Mathematics in China (Impact Factor: 0.45). 10/2011; 6(5):871-886. DOI: 10.1007/s11464-011-0140-y

ABSTRACT We introduce an F-Willmore functional of submanifold in space forms, which generalizes the well-known Willmore functional. Its critical point
is called the F-Willmore submanifold, for which the variational equation and Simons’ type integral inequality are obtained.

KeywordsMean curvature–Willmore submanifold–Simons’ type integral inequality

0 Bookmarks
·
167 Views
• Source
##### Article: Some conformal invariants of submanifolds and their applications
Bollettino dell Unione Matematica Italiana 01/1974; 10:380-385.
• Source
##### Article: Willmore submanifolds of the Möbius space and a Bernstein-type theorem
[Hide abstract]
ABSTRACT: We study Willmore immersed submanifoldsf: M m →S n into then-Möbius space, withm≥2, as critical points of a conformally invariant functionalW. We compute the Euler-Lagrange equation and relate this functional with another one applied to the conformal Gauss map of immersions intoS n . We solve a Bernestein-type problem for compact Willmore hypersurfaces ofS n , namely, if ∃a ∈ℝ n+2 such that <γf, a > ≠ 0 onM, whereγ f is the hyperbolic conformal Gauss map and <, > is the Lorentz inner product ofℝ n+2, and iff satisfies an additional condition, thenf(M) is an (n−1)-sphere.
manuscripta mathematica 01/1993; 81(1):203-222. · 0.50 Impact Factor


text:
# On components of 2-factors in claw-free graphs

Haitze J. Broersma, Daniël Paulusma, K. Yoshimoto

Research output: Chapter in Book/Report/Conference proceedingConference contributionAcademicpeer-review

## Abstract

For a non-hamiltonian claw-free graph $G$ with order $n$ and minimum degree $\delta$ we show the following. If $\delta=4$, then $G$ has a 2-factor with at most $(5n-14)/18$ components, unless $G$ belongs to a finite class of exceptional graphs. If $\delta \ge 5$, then $G$ has a 2-factor with at most $(n-3)/(\delta -1)$ components. These bounds are sharp in the sense that we can replace nor 5/18 by a smaller quotient nor $\delta -1$ by $\delta$.
Original language Undefined EuroComb 2007: European Conference on Combinatorics, Graph Theory and Applications Amsterdam Elsevier 289-293 5 https://doi.org/10.1016/j.endm.2007.07.050 Published - Aug 2007

### Publication series

Name Electronic Notes in Discrete Mathematics Elsevier 1 29 1571-0653 1571-0653

• IR-62062
• METIS-245869
• EWI-11587


text:
# A little data munging

For the past couple of weeks, I’ve been doing analyses of other people’s data. Usually, the data comes in the form of a spreadsheet, which works out fine. I don’t like using spreadsheets myself, but it’s easy to extract the data and put it in a CSV or TSV file, which I can then manipulate in Python.

Sometimes, though, what I get is a PDF of someone else’s report, with the data in one or more tables. This is more challenging. If the PDF was generated directly from the writer’s word processor, the text in the tables is selectable, and I can copy and paste it into a plain text document. Unfortunately, the columns often aren’t delimited the way I’d like.

Here’s part of a multipage table I had to analyze. It has 21 columns and, after combining all the pages, about 100 rows. I opened the PDF in Preview, selected the text, and copied it.

I crossed my fingers and pasted it into a new BBEdit document. The columns were lost entirely; every table cell appeared on its own line. But luckily there was order to it, as the data came in row by row. In other words, something like this

Col 1 Col 2 Col 3 Col 4 Col 5
a b c d e
f g h i j
k l m n o

was turned into this (I didn’t select the column headings because I didn’t like the names given in the table).

a
b
c
d
e
f
g
h
i
j
k
l
m
n
o

At least it would have turned into that if it weren’t for the missing data. If you look back at the image of the table, you’ll see that lots of cells are empty. What that meant was the table looked more like this

Col 1 Col 2 Col 3 Col 4 Col 5
a b e
g h
k n o

and turned into this after copying and pasting into BBEdit.

a
b
e
g
h
k
n
o

I couldn’t think of an automated way to add the missing lines, so I did it by hand, adding empty lines to the BBEdit document where needed.

a
b

e

g
h

k

n
o

While I’m pretty sure pandas has a way of reading a file like this and putting it into a data frame, I’m not fluent enough in pandas to know how to do it without poring over the documentation. And because I’d just spent a good chunk of time adding blank lines to the file, I wanted to get the analysis going right away. So I wrote this little script to read in the text file and spit out a 21-column TSV.

python:
1:  #!/usr/bin/env python
2:
3:  from sys import stdout
4:
5:  with open('table.txt') as f:
7:
8:  for i, b in enumerate(a):
9:    if (i + 1) % 21 > 0:
10:      stdout.write(b.replace('\n', '\t'))
11:    else:
12:      stdout.write(b)

The rows are 21 items long, so we want to turn the line feeds into tab characters except every 21st one. The only tricky part is recognizing that the array of lines, a, is 0-indexed, and the modulo operator used to determine where a row stops must be applied to line numbers that start with 1, not 0. Hence the (i + 1) % 21 in Line 9. After running the script, I added a row of column names to the top of the file and had a nice TSV file for importing.

This script is, I realize, both longer than necessary and uses more memory than necessary. Maybe I should’ve used a Perl one-liner:

perl -ne 'chomp; if ($.%21) {print "$_\t"} else {print "$_\n"}' table.txt  This doesn’t slurp in the entire file, and because the line number variable, $., starts counting at 1, there’s no off-by-one trickiness to worry about. An awk one-liner would be similar.

But I wasn’t in a Perl or awk frame of mind when I did this. I’d been working in Python, so that’s what I used. And there are better ways of writing this in Python, too. As I sit here drinking tea and not trying to finish a paying job, I imagine the fileinput module would be useful.

python:
1:  #!/usr/bin/env python
2:
3:  from fileinput import input, lineno
4:  from sys import stdout, argv
5:
6:  columns = int(argv[1])
7:  del argv[1]
8:  for line in input():
9:    if lineno() % columns:
10:      stdout.write(line.replace('\n', '\t'))
11:    else:
12:      stdout.write(line)

That gets rid of both the slurping and the off-by-one correction. And it doesn’t have the name of the file or the number of columns in the table hard-wired into the code. It can be called this way:

shape-table 21 table.txt

Now I have something I can use again the next time I need to extract a table of data from a PDF. Which pretty much guarantees I’ll never run into this kind of problem again.


text:
# When quotient map is open?

Quotient map from $X$ to $Y$ is continuous and surjective with a property : $f^{-1}(U)$ is open in $X$ iff $U$ is open in $Y$.

But when it is open map? what condition need?

-
A quotient map $f \colon X \to Y$ is open if and only if for every open subset $U \subseteq X$ the set $f^{-1} (f (U))$ is open in $X$. A sufficient condition is that $f$ is the projection under a group action. –  Andrea Sep 1 '11 at 13:40
anything that is locally a projection should be open –  yoyo Sep 1 '11 at 22:21
Another sufficient condition: that the map be a surjective submersion of smooth manifolds. –  Bruno Stonek May 17 '13 at 16:48
@Andrea: "A sufficient condition is that f is the projection under a group action" Why, please? –  Ian Mallett Oct 26 '13 at 21:45
If $\pi \colon X \to X/G$ is the projection under the action of $G$ and $U \subseteq X$, then $\pi^{-1} (\pi (U)) = \cup_{g \in G} g(U)$. –  Andrea Oct 27 '13 at 14:54


text:
# Integration by parts formula lagrange notation

• vincenthardwick
• 114
• 100% (3) 3 out of 3 people found this document helpful

This preview shows page 75 - 78 out of 114 pages.

Integration by parts formula (Lagrange notation)integraldisplayf(x)g(x) dx=f(x)G(x)integraldisplayfprime(x)G(x) dx.HereGis an antiderivative ofg.The integration by parts formula is useful for integrating some products ofthe formf(x)g(x). You can see that it changes the problem of integratingf(x)g(x) into the problem of integratingfprime(x)G(x), whereGis anantiderivative ofg.So it’s useful for integrating products of the formf(x)g(x) that have thefollowing characteristics.You can find an antiderivativeGofg; in other words, you canintegrate the expressiong(x).The productfprime(x)G(x) is easier to integrate than the productf(x)g(x). For example, this may be the case iffprime(x) is simpler thanf(x).When we integrate by using the integration by parts formula, we say thatwe’reintegrating by parts. Here’s an example.177
Unit 8Integration methodsExample 23Integrating by partsFind the integralintegraldisplayxsinxdx.SolutionThe integrand is a product of two expressions,xand sinx. Youcan integrate the second expression, sinx, and differentiating the firstexpression,x, makes it simpler. So try integration by parts.Letf(x) =xandg(x) = sinx.Thenfprime(x) = 1, and an antiderivative ofg(x) isG(x) =cosx.So the integration by parts formula givesintegraldisplayxsinxdx=integraldisplayf(x)g(x) dx=f(x)G(x)integraldisplayfprime(x)G(x) dx=x×(cosx)integraldisplay1×(cosx) dx=xcosx+integraldisplaycosxdxFind the integral in this expression.=xcosx+ sinx+c.You can see that in Example 23 the use of the integration by parts formulachanged the problem of integratingxsinxinto the problem of integratingcosx, which is easier. Here are two similar examples for you to try.Activity 34Integrating by partsFind the following integrals.(a)integraldisplayxcosxdx(b)integraldisplayxexdxAs you become more familiar with integration by parts, you’ll probablyfind that you can carry it out more quickly and conveniently if youremember the informal version of the formula stated below. You can recitethis version in your head as you apply integration by parts, in a similarway to the informal versions of the product and quotient rules fordifferentiation that you met in Unit 7. It’s useful for applying integration178
4Integration by partsby parts in fairly simple cases, like the ones that you’ve seen so far, whichis usually all that you need to do.Integration by parts formula (informal)integral of product = (first)×parenleftbiggantiderivativeof secondparenrightbiggintegral ofparenleftbiggparenleftbiggderivativeof firstparenrightbigg×parenleftbiggantiderivativeof secondparenrightbiggparenrightbigg.

Upload your study docs or become a

Course Hero member to access this document

Upload your study docs or become a

Course Hero member to access this document

End of preview. Want to read all 114 pages?

Upload your study docs or become a

Course Hero member to access this document

Term
Summer
Professor
NoProfessor


text:
# Math Help - Normal distribution problem

1. ## Normal distribution problem (can anyone help?)

Hi

Can someone help me understand the following question? Please look on this link.

http://i48.tinypic.com/20hk4z9.png

A hint on how to solve this would be much appreciated. I don't know where to begin with it. It's impossible to integrate right? so what other way can be approached?

Any help is welcome - many thanks in advance

2. ## Re: Normal distribution problem

Hi zzizi!

You're not supposed to integrate it (which is possible btw).
You're supposed to apply the rule "integration by parts" to find a recurrence relation.

Integration by parts is summarized as:

$\int u dv = uv - \int v du$

Do you know this rule?
And how to apply it?

3. ## Re: Normal distribution problem

Thank you, I thought I had to integrate it.


text:
## Signals of deconfinement at SPS energies

### Marcus Bleicher

We investigate hadron production as well as transverse hadron spectra from proton-proton, proton-nucleus and nucleus-nucleus collisions from 2 $A\cdot$GeV to 21.3 $A\cdot$TeV within two independent transport approaches (HSD and UrQMD) that are based on quark, diquark, string and hadronic degrees of freedom. The comparison to experimental data on transverse mass spectra from $pp$, $pA$ and C+C (or Si+Si) reactions shows the reliability of the transport models for light systems. For central Au+Au (Pb+Pb) collisions at bombarding energies above $\sim$ 5 A$\cdot$GeV, furthermore, the measured $K^{\pm}$ transverse mass spectra have a larger inverse slope parameter than expected from the default calculations. We investigate various scenarios to explore their potential effects on the $K^\pm$ spectra. In particular the initial state Cronin effect is found to play a substantial role at top SPS and RHIC energies. However, the maximum in the $K^+/\pi^+$ ratio at 20 to 30 A$\cdot$GeV is missed by ~40% and the approximately constant slope of the $K^\pm$ spectra at SPS energies is not reproduced either. Our systematic analysis suggests that the additional pressure - as expected from lattice QCD calculations at finite quark chemical potential $\mu_q$ and temperature $T$- should be generated by strong interactions in the early pre-hadronic/partonic phase of central Au+Au (Pb+Pb) collisions.


text:
Browse by: Author name - Classification - Keywords - Nature

XV: 24, 320-346, LNM 850 (1981)
DELLACHERIE, Claude; LENGLART, Érik
Sur des problèmes de régularisation, de recollement et d'interpolation en théorie des martingales (General theory of processes)
The optional section theorem implies that an optional process $X$ is completely determined by its values $X_T$ at all stopping times. Conversely, given random variables $X_T$, ${\cal F}_T$-measurable and such that $X_S=X_T$ a.s. on the set $\{S=T\}$, is it possible to aggregate'' them into an optional process $X$? This is the elementary form of the general problem discussed in the paper, in the case where the random variables $X_T$ satisfy a supermartingale inequality. The problem solved is more general: the optional $\sigma$-field is replaced by any of the $\sigma$-fields considered in 1449 (including previsible, accessible, etc), and the family of all stopping times is replaced by a suitable family (called a chronology)
Keywords: General filtrations, Strong supermartingales, Snell's envelope, Section theorems
Nature: Original
Retrieve article from Numdam


text:
# Math Help - Probability question

1. ## Probability question

Hey, lets say I have 5 probabilities of the chance that 5 different things are is "on", so the probabilities for each of the items would be something like:

.75, .81, .63, .9, .1

I'd like to calculate the probability that, given the individual probabilities, that 3 or more of these items are really "on".

I'm guessing the probability that all 5 are "on" is simply the 5 numbers multiplied together, which is ~.03.

I'd imagine I'd have to add this number to the probability that 4 are and on the probability that 3 of them are on to get the total probability for all possible outcomes? Not sure how to calculate these though...

2. Hello, paulig!

This is long, tedious problem . . .

There are 5 objects $\{A,B,C,D,E\}$ that could be "on".

The probabilities for each being "on" is: . $\begin{array}{ccc} P(A) &=& 0.75 \\ P(B) &=& 0.81 \\ P(C) &=& 0.63 \\ P(D) &=& 0.90 \\ P(E) &=& 0.10 \end{array}$

Find the probability that 3 or more of these items are "on".

All five are on: there is one case.

. . $P(A \cap B \cap C \cap D \cap E) \;=\;(0.75)(0.82)(0.63)(0.90)(0.10) \;=\;0.3444525$

Four are on: there are 5 cases.

. . $\begin{array}{ccccc} P(A \cap B \cap C \cap D) &=& (0.75)(0.81)(0.63)(0,90) &=& 0.334 452 5 \\ P(A \cap B \cap D \cap E) &=& (0.75)(0.81)(0.63)(0.10 &=& 0.038 2725 \\ \vdots && \vdots && \vdots \\ P(B \cap C \cap D \cap E) &=& (0.81)(0.63)(0.90)(0.10) &=& 0.045 9270 \end{array}$

Three are on: there are 10 cases.

. . $\begin{array}{ccccc}
P(A \cap B \cap C) &=& (0.75)(0.81)(0.63) &=& 0.382725 \\
P(A \cap B \cap D) &=& (0.75)(0.81_)0.90) &=& 0.546750 \\
P(A \cap B \cap E) &=& (0.75)(0.81)(0.10) &=& 0.060750 \\
P(A \cap C \cap D) &=& (0.75)(0.63)(0.90) &=& 0.425250 \\
\vdots && \vdots && \vdots \\
P(C \cap D \cap E) &=& (0.63)(0.90)(0.10) &=& 0.056700 \end{array}$

I'll wait in the car . . .
.

3. Thanks for your reply. Ok so that's it then, I guess I posted it cause I thought there might be an easier way to calculate this, some kind of mathematical tool to do the job easily, unfortunately that doesn't seem to be the case, thanks again!

P

4. Just a comment on Soroban's solution--

This solution assumes the events are independent. There is nothing in the problem statement to this effect. If it is *not* the case that the events are independent, then you do not have sufficient information to solve the problem.


text:
Math proof

1. Sep 27, 2016

kregg87

1. The problem statement, all variables and given/known data
Show that any vector in a vector space V can be written as a linear combination of a basis set for that same space V.

2. Relevant equations
http://linear.ups.edu/html/section-VS.html
We are suppose to use the 10 rules in the above link, plus the fact that if you have a lineraly independent set
{X1,X2,...,Xn} then -> c1X1+c2X2+...+cnXn = 0 vector implies that all the constants (c1,c2, etc) are zero.

Not looking for a complete solution, just not sure where to start. Ive tried proof by contradiction and a couple other ways and non have worked out for me.

3. The attempt at a solution

2. Sep 28, 2016

Ray Vickson

Define "basis". (I know the usual definition, but what is the one YOU are using?)

3. Oct 5, 2016

kregg87

My definition is a linearly independent set of N vectors, where N in the dimension of the space. My definition of dimension, N, is it the max number of mutually lineraly independant vectors possible.

4. Oct 5, 2016

Staff: Mentor

So what can you say about a given vector? Can it be linearly independent from your basis? Or otherwise, what does it mean it can't?