text:
# symplectic structure of tangent bundle of $\mathbb{S}^{n-1}$

It is well known that $T\mathbb{S}^{n-1}$ is diffeomorphic to $M= f^{-1}(1)$ where $f:\mathbb{C}^n\rightarrow \mathbb{C}$ is $f(z):=\sum_{i=1}^{n} z_{i}^{2}$. Two questions:

1) Is $M$ a symplectic submanifold of $\mathbb{C}^n\sim \mathbb{R}^{2n}$ (with the standard symplectic structure)?

If the answer is affirmative, we can consider two symplectic structures for $T\mathbb{S}^{n-1}$. The first is the original structure of the tangent or cotangent bundle, the second one is the pull back structure of $M$.

2) Are these structures equivalent?

Here is a formula for an explicit symplectomorphism $F$ from $T^*S^{n-1}$ to the affine quadric $\{\sum z_{j}^{2}=1\}$ in $\mathbb{C}^n$:

$$F(p,q) = \left(\frac{1+\sqrt{1+4|p|^2}}{2}\right)^{1/2} q - i\left(\frac{1+\sqrt{1+4|p|^2}}{2}\right)^{-1/2}p$$

Here I view $T^*S^{n-1}$ as consisting of pairs $(p,q)\in \mathbb{R}^n\times\mathbb{R}^n$ with $p\cdot q=0$ and $|q|=1$. To check that $F$ is a symplectomorphism one can just check that $F$ pulls back the one-form $\alpha=-\sum y_jdx_j$ on $\mathbb{C}^n$ (which is a primitive for the standard symplectic form) to the canonical one-form on $T^*S^{n-1}$. The coefficients involving $\sqrt{1+4|p|^2}$ are designed to cause the map to have image in the quadric.

To indicate how I came up with this formula, the reasoning was that a suitably natural symplectomorphism ought to be equivariant with respect to the natural Hamiltonian $O(n)$-actions on both spaces. The easiest way to do this seems to be to take $F(p,q)=f(|p|)q-ig(|p|)p$ for some real functions $f$ and $g$. For the image to be in the quadric one needs $$f(|p|)^2- |p|^2g(|p|)^2=1.$$ Meanwhile if this is to be a symplectomorphism it should pull back the moment map for the $O(n)$-action on $\mathbb{C}^n$ to the moment map for the $O(n)$-action on $T^*S^{n-1}$. The norms of these moment maps are, respectively $(x+iy)\mapsto |x||y|$ and $(p,q)\mapsto |p|$, giving an equation $|f||g||p|=|p|$, i.e., $$|g|=\frac{1}{|f|}.$$ Solving these for $f$ and $g$ yields the formula at the top, which can then be directly confirmed to have the required properties.

• I didn't check your computation, but that's not the point. I just want to congratulate you for using the wording "Moment Map" :-}) – Patrick I-Z Nov 30 '13 at 17:58
• Mike thank you for your beutiful answer. But do you mean that an arbitrary symplectomorphism is necessarily equivariant with respect to O(n) action? – Ali Taghavi Dec 1 '13 at 18:21
• No, that certainly wouldn't be the case (for instance if you composed the symplectomorphism in the answer with a non-equivariant symplectomorphism of the domain--of which there are many--the result wouldn't be equivariant). The idea of asking for O(n)-equivariance was just to narrow down the search, and also to take advantage of the fact that an equivariant symplectomorphism will respect the moment maps (thus giving a function that the map should preserve, which is usually easier to arrange than preserving a form). – Mike Usher Dec 1 '13 at 21:13

$\def\C{{\mathbf C}} \def\d{\delta} \def\e{{\mathbf e}} \def\r{{\mathbf r}} \def\u{{\mathbf u}} \def\x{{\mathbf x}} \def\y{{\mathbf y}} \def\z{{\mathbf z}} \def\<{\langle} \def\>{\rangle}$ If I understand Now that I understand your questions correctly, the answers are 1) Yes 2) No Yes.

1) Writing $\z=\x+i\y$, we have $f(\z) = \|\x\|^2 - \|\y\|^2+2i\<\x,\y\>$. So $M=f^{-1}(1)$ (resp. $\mathrm T_\z M$) consists of all $\z$ (resp. $\d\z=\d\x+i\d\y$) such that $$\|\x\|^2 - \|\y\|^2= 1, \qquad \<\x,\y\>=0, \tag1$$ $$\llap{(\text{resp.}\qquad}\<\x,\d\x\>-\<\y,\d\y\>=0,\qquad\<\d\x,\y\>+\<\x,\d\y\>=0.) \tag2$$ Given $\z$ in (1), we can choose an orthonormal basis $(\e_1,\dots,\e_n)$ such that $\x = \|\x\|\e_1$ and $\mathbf y = \|\y\|\e_2$. Then (2) says that $(\d x_1,\d y_1)=\frac{\|\y\|}{\|\x\|}(\d y_2,-\d x_2)$ and hence \begin{align} \omega(\d\z,\d'\z) &= \sum_{i=1}^n(\d x_i\d'y_i-\d'x_i\d y_i)\\ &=\Bigl(1+\tfrac{\|\y\|^2}{\|\x\|^2}\Bigr)(\d x_2\d'y_2-\d'x_2\d y_2) + \sum_{i=3}^n(\d x_i\d'y_i-\d'x_i\d y_i) \end{align} which is obviously nondegenerate. (Or as Tim Perutz says, just note that $M$ is a complex submanifold.)

2) Consider (after Seidel's thesis pointed out by Daniel Pomerleano) the diffeo $\Phi:M\to\mathrm{TS}^{n-1}$ sending $\z=\x+i\y$ to the pair $(\u,\r)=(\frac{\x}{\|\x\|}, \|\x\|\y)$. Writing $\sigma$ for the canonical 2-form $d\<\u,d\r\>$ on $\mathrm{TS}^{n-1}$, we have \begin{align} \Phi^*\sigma &= d(\Phi^*\<\u,d\r\>)=d(\<\tfrac{\x}{\|\x\|},d(\|\x\|\y)\>)\\ &=d\<\x,d\y\> + \<\x,\y\>\tfrac{d\|\x\|}{\|\x\|} \end{align} and this equals $\omega$ since the second term vanishes by (1). Thus $\Phi$ is a symplectomorphism.

3) My initial (stupid and wrong) answer came from my thinking of another similar question. Namely $\mathrm{TS}^2$, thought of as the manifold of oriented affine lines in $\mathbf R^3$ (identify $(\u,\r)$ with the line $\r+\mathbf R\u$ with orientation $+\u$), carries a 2-parameter family of Euclid-invariant symplectic structures given by $$\omega_{k,s} = d\<k\u,d\r\> - s\Omega,\qquad k>0, s\in \mathbf R, \tag3$$ where the first term is the cotangent bundle 2-form, and the second ("magnetic") term is $-s$ times the area 2-form of the base $\mathrm S^2$.

On the other hand, identifying $\mathfrak{so}(3,\C)=\C^3$ with Lie bracket the vector product, the nonzero levels of your quadratic form $\<\z,\z\>=\sum_{i=1}^3 z_i^2$ are precisely the semisimple (co)adjoint orbits of $\mathrm{SO}(3,\C)$. As such they carry a (Kirillov-Kostant-Souriau) 2-form $\omega_{\mathrm{KKS}}(\d\z,\d'\z)=\mathrm{Re}\bigl(\frac1{s^2}\<\z,\d'\z\times\d\z\>\bigr)$ where $\<\z,\z\>=s^2$ is the equation of our orbit $\mathcal O$. So we can again ask if this is symplectomorphic with $\mathrm{TS}^2$ equipped with one of the forms (3). At least when $s^2$ is real positive, the answer is yes and the diffeomorphism $F:\mathrm{TS}^2\to\mathcal O$ I was considering (with $k=s=1$), $$F(\u,\r) = k\r\times\u + s\u + ik\r$$ is symplectic: $F^*\omega_{\mathrm{KKS}}=\omega_{k,s}$. Being familiar with this diffeomorphism, I then made the mistake of thinking it must be "the one" of which you were asking if it was symplectic relative to your (different) 2-forms. And it isn't.

• Not many people use $\delta$ to denote tangent vectors... – Patrick I-Z Nov 29 '13 at 22:42
• Only perverts like us! – Francois Ziegler Nov 29 '13 at 22:46
• For 1), there's no need to compute: $M$ is a complex submanifold of $\mathbb{C}^n$, hence a symplectic submanifold. – Tim Perutz Nov 30 '13 at 15:14
• I'm very confused: I think the answer to question 2 is "yes" if we mean "symplectomorphic". See e.g. Lemma 18.1 of Seidel's thesis or lemma 6.11 of arxiv.org/pdf/math/0006056v2.pdf. – Daniel Pomerleano Nov 30 '13 at 15:29
• Part 2 of the answer successfully shows that a certain specific diffeomorphism between the spaces is not a symplectomorphism...of course this is not the same as showing that they are not symplectomorphic. – Mike Usher Nov 30 '13 at 17:39


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A microphone array, if properly designed, has the capability to provide varying gains in different directions in space. We call this feature as spatially selectiveness of a microphone array.

With the spatial selective feature, a microphone array can enhance a desired target sound that travels in a direction, or suppress an interfering source from a different direction. Of course, if both the desired sound and the interfering sound come from the same direction, then the microphone array does not have the ability to differentiate.

As discussed in many of the published literatures, the direction of the desired target sound source is usually assumed known as a priori. However this condition is really satisfied in practice. We discuss in this short article a solution when the target sound source is not known.

As shown in the following diagram, the desired sound is travelling from the 0 degree angle. Ideally we would want to form a heart shaped beam pattern as shown in blue. It maximizes the directional gain at the incoming direction of the desired sound.

Since we assume no knowledge of the desired sound, we have to formulate it into a power optimization problem with a constraint on the beam forming vector w as below.

${maxE}\left[\left|{y}\left(t\right)\right|^2\right]$

$subject\ to:\left|w\right|^2=1$

where $y\left(t\right)=\sum_{m=1}^{M}{w_m\ast x_m\left(t\right)}$.

The constraint forces the solution to be within a finite range. When a single sound a sound is present, the solution yields the optimum beam pattern with maximum gain towards the target direction. If the target dynamically changes locations, such as, in the case that the sound source is a talker who walks around the microphone array in a circle, the algorithm will automatically follow and steer the maximum gain toward the talk location.

The main drawback with the algorithm is that it depends on a reliable double talk detector. It will not work correctly when interference or competing sound sources are present. We will address this problem in a separate article.

format: plain


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# The number of states in which $N$ binary strings with strictly adjacent $1$ bits have a specific number of strings with any $1$ bits

I have $N$ binary strings of length $L$. If a string has more than one $1$ bit, these bits must be adjacent to one-another. For example, for a length $L = 4$ string, the possible string states (of which there are $11$) would be: $\{"0000", "0001", "0010", "0100", "1000","0011","0110","1100","0111","1110","1111"\}$.

What function, $f(N,L,M,B)$, gives the number of "system" states where there are exactly $M \leq N$ strings with any $1$ bits, and the count of $1$ bits in all of the strings is $B$? What if we drop the constraint that $1$ bits need to be adjacent?

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To make sure that I understand: is $B$ the total number of $1$ bits in all $N$ strings? And are there $N-M$ copies of the $L$-bit zero string? –  Brian M. Scott Jan 2 '13 at 19:01
@Brian M. Scott Yes and yes. $B$ is the total number of $1$ bits in all of the $N$ strings. And there are $N-M$ strings with only $0$ bits. –  Egami Jan 2 '13 at 19:09

There are $\binom{N}{M}$ ways to select a subset of $M$ strings to populate. After that, we can attach the generating function $\sum_{n=1}^L (L-n+1) x^n = \frac{x}{(1-x)^2} (L - (L+1)x + x^{L+1})$ to each of the $M$ chosen strings, where the power of $x$ encodes the number of bits used in that string. The function you are looking for is then $\binom{N}{M}$ times the coefficient of $x^{B-M}$ in the Taylor series for $$\left[\frac{(L - (L+1)x + x^{L+1})}{(1-x)^2} \right]^M.$$ I don't see a good way to extract a general explicit formula, but for small fixed values of $L$ or $M$ it should be possible.

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For L=4 and M=4, using Series[((L - (L + 1) x + x^(L + 1))/(1 - x)^2)^M, {x, 0, 12}] in Mathematica, I get: 256 + 768 x + 1376 x^2 + 1840 x^3 + 1905 x^4 + 1608 x^5 + 1124 x^6 + 648 x^7 + 310 x^8 + 120 x^9 + 36 x^10 + 8 x^11 + x^12 + O[x]^13 –  Egami Jan 2 '13 at 22:08
How do I interpret the coefficient 256 for x^0, for example? I don't quite understand how this could make sense? –  Egami Jan 2 '13 at 22:11
@Egami The coefficient $x^0$ would be the number of ways to have $B=M=4$. In this case each of the 4 chosen strings can only get a single $1$, but there are $4$ places to put it. Therefore there are $4^4 = 256$ combinations. –  Erick Wong Jan 3 '13 at 0:04
Playing around in Mathematica and Maple, it looks like, for fixed $L$ or $M$, that you can get closed formed solutions for the coefficients, but they're a mess. –  Egami Jan 3 '13 at 3:32

For a string of length $L$, you can have any number of $1$ bits from $0$ to $L$. Given that there are $n \ \ 1$ bits, there are $L-n+1$ valid strings, as you can have from $0$ to $L-n \ \ 0$ bits at the start and the rest must be at the end. The one failure of this is having $0 \ \ 1$ bit, where there is only one string instead of $L+1$. So $$f(n,L)=\begin {cases}1 & n=0 \\ L-n+1 & n \gt 0 \end {cases}$$

If you drop the requirement that the $1$'s be adjacent, you just select the positions of the $1$ bits, so $f(n,L)={L \choose n}$

Added: The first step for what you want is to look for the number of partitions of $B$ things into parts of at most $N$. If order matters, if a set of strings $(0011, 0001,1111)$ is different from $(1111,0001,0011)$ you want compositions instead. Given the partition or composition, each element corresponds to a single string. For each element in it, you have the $f(n,L)$ above ways to select the bit string, so you multiply them together to see the number of possibilities for that partition. Unfortunately, for partitions you will overcount in case there are two strings with the same number of bits on because swapping them gets you back to the same place.

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Your answer makes perfect sense, though it wasn't exactly what I was asking? –  Egami Jan 2 '13 at 19:02
$N$ is the number of binary strings, and with the constraints, I'm trying to count the total number of ways we can have both exactly $M \leq N$ strings with no $1$ bits, and also exactly $B$ total $1$ bits. –  Egami Jan 2 '13 at 19:04
We can write the number of states for any given string as just: $(\sum_{i=1}^{L} i)+1$. –  Egami Jan 2 '13 at 19:05
@Egami: I didn't understand fully what you were asking and hoped it would help. The number of strings with no $1$ bits is $1$ of each length. My $n$ may be your $B$, but I am not sure. Your last comment is correct, and can be simplified to $1+\frac 12L(L+1)$ –  Ross Millikan Jan 2 '13 at 20:24


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# Algorithms to compute largest gap between smallest nonzero eigenvalues of sparse symmetric matrix

I am looking mainly for c/c++ implementations but also for theoretical algorithms to compute gaps between smallest positive eigenvalues of symmetric, singular matrix or real numbers. To be precise, I want to find index n of the eigenvalue such that between $\lambda_n$ and $\lambda_{n+1}$ (both are nonzero) there is the largest gap (I assume that eigenvalues are sorted). The matrix I am dealing with is huge, so I am allowed to use only sparse representation. I cannot store all the zero entries due to memory constraints. Can you recommend any c/c++ software which can compute it to me and uses sparse matrix representations? I have tried to install fortran ARPACK, but with no success. Maybe you know some papers with algorithms solving this problem. Thanks in advance!

• Hi. Why did you fail to install fortran Arpack? which operating system are you using (install it in ubuntu is very easy, for example)? have you tried with Arpack++? Actually I dont think there is any algorithm to get the largest gap, so it seems that you have to calculate all of them, or maybe you can use some information that you have about your matrix (where is the matrix coming from?). – sebas Feb 26 '14 at 22:40
• When you say largest gap do you mean the maximum $\lambda_n - \lambda_{n+1}$? – meawoppl Mar 2 '14 at 19:24


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## Determine the probability of finding the particle

A particle in an infinitely deep square well has a wave function given by the equation below, for 0 = x = L and zero otherwise.
psi_2(x) = sqrt(2/L)sin((2pix)/L)

(b) Determine the probability of finding the particle near L / 2 by calculating the probability that the particle lies in the range 0.480L = x = 0.520L.

(c) Determine the probability of finding the particle near L / 4 by calculating the probability that the particle lies in the range 0.230L = x = 0.270L.


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# graph.famous

0th

Percentile

##### Creating named graphs

There are some famous, named graphs, sometimes counterexamples to some conjecture or unique graphs with given features. These can be created with this function

Keywords
graphs
##### Usage
graph.famous(name)
##### Arguments
name
Character constant giving the name of the graph. It is case insensitive.
##### Details

graph.famous knows the following graphs:

• Bull
{The bull graph, 5 vertices, 5 edges, resembles to the head of a bull if drawn properly.} Chvatal{This is the smallest triangle-free graph that is both 4-chromatic and 4-regular. According to the Grunbaum conjecture there exists an m-regular, m-chromatic graph with n vertices for every m>1 and n>2. The Chvatal graph is an example for m=4 and n=12. It has 24 edges.} Coxeter{A non-Hamiltonian cubic symmetric graph with 28 vertices and 42 edges.} Cubical{The Platonic graph of the cube. A convex regular polyhedron with 8 vertices and 12 edges.} Diamond{A graph with 4 vertices and 5 edges, resembles to a schematic diamond if drawn properly.} Dodecahedral, Dodecahedron{Another Platonic solid with 20 vertices and 30 edges.} Folkman{The semisymmetric graph with minimum number of vertices, 20 and 40 edges. A semisymmetric graph is regular, edge transitive and not vertex transitive.} Franklin{This is a graph whose embedding to the Klein bottle can be colored with six colors, it is a counterexample to the neccessity of the Heawood conjecture on a Klein bottle. It has 12 vertices and 18 edges.} Frucht{The Frucht Graph is the smallest cubical graph whose automorphism group consists only of the identity element. It has 12 vertices and 18 edges.} Grotzsch{The Grötzsch graph is a triangle-free graph with 11 vertices, 20 edges, and chromatic number 4. It is named after German mathematician Herbert Grötzsch, and its existence demonstrates that the assumption of planarity is necessary in Grötzsch's theorem that every triangle-free planar graph is 3-colorable.} Heawood{The Heawood graph is an undirected graph with 14 vertices and 21 edges. The graph is cubic, and all cycles in the graph have six or more edges. Every smaller cubic graph has shorter cycles, so this graph is the 6-cage, the smallest cubic graph of girth 6.} Herschel{The Herschel graph is the smallest nonhamiltonian polyhedral graph. It is the unique such graph on 11 nodes, and has 18 edges.} House{The house graph is a 5-vertex, 6-edge graph, the schematic draw of a house if drawn properly, basicly a triangle of the top of a square.} HouseX{The same as the house graph with an X in the square. 5 vertices and 8 edges.} Icosahedral, Icosahedron{A Platonic solid with 12 vertices and 30 edges.} Krackhardt_Kite{A social network with 10 vertices and 18 edges. Krackhardt, D. Assessing the Political Landscape: Structure, Cognition, and Power in Organizations. Admin. Sci. Quart. 35, 342-369, 1990.} Levi{The graph is a 4-arc transitive cubic graph, it has 30 vertices and 45 edges.} McGee{The McGee graph is the unique 3-regular 7-cage graph, it has 24 vertices and 36 edges.} Meredith{The Meredith graph is a quartic graph on 70 nodes and 140 edges that is a counterexample to the conjecture that every 4-regular 4-connected graph is Hamiltonian.} Noperfectmatching{A connected graph with 16 vertices and 27 edges containing no perfect matching. A matching in a graph is a set of pairwise non-adjacent edges; that is, no two edges share a common vertex. A perfect matching is a matching which covers all vertices of the graph.} Nonline{A graph whose connected components are the 9 graphs whose presence as a vertex-induced subgraph in a graph makes a nonline graph. It has 50 vertices and 72 edges.} Octahedral, Octahedron{Platonic solid with 6 vertices and 12 edges.} Petersen{A 3-regular graph with 10 vertices and 15 edges. It is the smallest hypohamiltonian graph, ie. it is non-hamiltonian but removing any single vertex from it makes it Hamiltonian.} Robertson{The unique (4,5)-cage graph, ie. a 4-regular graph of girth 5. It has 19 vertices and 38 edges.} Smallestcyclicgroup{A smallest nontrivial graph whose automorphism group is cyclic. It has 9 vertices and 15 edges.} Tetrahedral, Tetrahedron{Platonic solid with 4 vertices and 6 edges.} Thomassen{The smallest hypotraceable graph, on 34 vertices and 52 edges. A hypotracable graph does not contain a Hamiltonian path but after removing any single vertex from it the remainder always contains a Hamiltonian path. A graph containing a Hamiltonian path is called tracable.} Tutte{Tait's Hamiltonian graph conjecture states that every 3-connected 3-regular planar graph is Hamiltonian. This graph is a counterexample. It has 46 vertices and 69 edges.} Uniquely3colorable{Returns a 12-vertex, triangle-free graph with chromatic number 3 that is uniquely 3-colorable.} Walther{An identity graph with 25 vertices and 31 edges. An identity graph has a single graph automorphism, the trivial one.}

##### Value

• A graph object.

##### encoding

UTF-8

graph can create arbitrary graphs, see also the other functions on the its manual page for creating special graphs.

• graph.famous
##### Examples
solids <- list(graph.famous("Tetrahedron"),
graph.famous("Cubical"),
graph.famous("Octahedron"),
graph.famous("Dodecahedron"),
graph.famous("Icosahedron"))
Documentation reproduced from package igraph, version 0.5.1, License: GPL (>= 2)

### Community examples

Looks like there are no examples yet.


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1. ## Telescoping series

Dont quite understand how i would solve the following problem:

Repersent the series as a telescopic series: $\sum ln(n(n+2)/(n+1)^2)$ n=1 to infiniti

2. ## Re: Telescoping series

$\sum_{n=1}^N\ln\frac{n(n+2)}{(n+1)^2}=\ln1-\ln2-\ln(N-1)+\ln N$

using properties of ln.

Edit: The right-hand side should be $\ln1-\ln2-\ln(N+1)+\ln(N+2)$.

3. ## Re: Telescoping series

but what is the series represented as a partial fraction? I get:

-1/(n+1)^2

substituting in n=1 and so on i get ln-1/4, ln-1/9...which surely cannot be right

4. ## Re: Telescoping series

Originally Posted by NFS1
but what is the series represented as a partial fraction?
Hmm, I could not parse this sentence. Edit: OK. I parsed it, but I am not sure how to represent this series as a partial fraction.

Originally Posted by NFS1
I get:

-1/(n+1)^2
What do you claim about this expression? Obviously, $\frac{n(n+2)}{(n+1)^2}\ne-\frac{1}{(n+1)^2}$.

I am not sure partial fractions are necessary here. Simplify the expression under the sum using the formulas for logarithm of product and quotient.

5. ## Re: Telescoping series

Originally Posted by emakarov
Hmm, I could not parse this sentence. Edit: OK. I parsed it, but I am not sure how to represent this series as a partial fraction.

What do you claim about this expression? Obviously, $\frac{n(n+2)}{(n+1)^2}\ne-\frac{1}{(n+1)^2}$.

I am not sure partial fractions are necessary here. Simplify the expression under the sum using the formulas for logarithm of product and quotient.
Don't quite understand how to do that. I thought to express the series as telescopic series you should express the series as partial sums. If you can show a step by step solution it will be greatly appreciated.

6. ## Re: Telescoping series

The point is that $ln\left(\frac{n(n+2)}{(n+1)^2}\right)= ln(n+2)+ ln(n)- 2ln(n+1)= \left(ln(n+2)- ln(n+1)\right)- \left(ln(n+1)- ln(n)\right)$

7. ## Re: Telescoping series

Originally Posted by HallsofIvy
The point is that $ln\left(\frac{n(n+2)}{(n+1)^2}\right)= ln(n+2)+ ln(n)- 2ln(n+1)= \left(ln(n+2)- ln(n+1)\right)- \left(ln(n+1)- ln(n)\right)$
Yes, this is the best representation.

I was doing it as follows.

$\sum_{n=1}^N\ln\left(\frac{n(n+2)}{(n+1)^2}\right) =$
$\sum_{n=1}^N(\ln n+\ln(n+2)-2\ln(n+1))=$
$(\ln1+\ln3-2\ln2)+(\ln2+\ln4-2\ln3)+(\ln3+\ln5-2\ln4)+$ ... $+(\ln(N-1)+\ln(N+1)-2\ln N)+(\ln N+ \ln(N+2)-2\ln(N+1))=$
$\ln1-\ln2-\ln(N+1)+\ln(N+2)$

Attachment 24517

The horizontal axis corresponds to arguments of ln and the vertical axis corresponds to terms of the series. Blue dots (k, n) denote ln(k) in the nth term, and red dots (k, n) denote -2ln(k) in the nth term. Then three dots on the same vertical line correspond to ln(k) - 2ln(k) + ln(k) = 0, so only dots surrounded by the dashed line remain.

This all means that I was wrong about the partial sum in post #2.

8. ## Re: Telescoping series

Ok so the series represented as telescopic series is ln1 - ln2 + ln(n+2) - ln (n+1).

How would i go about finding the sum of the telescopic series? Would i use log laws again so that the telescopic series is ln1/2 + ln(n+2)/(n+1)?? If so what would the answer to the sum be??

9. ## Re: Telescoping series

Now that you have what the sum is up to n, take the limit as n goes to infinity.

10. ## Re: Telescoping series

Originally Posted by Prove It
Now that you have what the sum is up to n, take the limit as n goes to infinity.
What is the sum?? is it: ln1 - ln2 + ln(n+2) - ln (n+1) or using log laws is it ln(1/2 + (n+2)/(n+1))?

11. ## Re: Telescoping series

Originally Posted by NFS1
What is the sum?? is it: ln1 - ln2 + ln(n+2) - ln (n+1) or using log laws is it ln(1/2 + (n+2)/(n+1))?
The first is correct, but the second is not. It should be \displaystyle \begin{align*} \ln{\left[ \frac{1(n+2)}{2(n+1)} \right]} \end{align*}, and now take the limit.


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# Set Definition by Predicate/Examples/Integers Greater Than or Equal to 2

Jump to navigation Jump to search

## Example of Set Definition by Predicate

Let $S$ be the set defined as:

$S := \set {x \in \Z: 2 \le x}$

Then $S$ is the set of all integers greater than or equal to $2$:

$S = \set {2, 3, 4, \ldots}$


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# GM going green

1. Nov 29, 2006

### Ivan Seeking

Staff Emeritus
http://www.detnews.com/apps/pbcs.dll/article?AID=/20061129/UPDATE/611290443

2. Nov 29, 2006

### Skyhunter

You realize of course that this means when the Democrats pass higher CAFE standards that the auto industry will not object. Not going to look good for the Republicans.

To some extent this is GM finally interpreting the handwriting written on the wall. With Toyata kicking their butt they had to respond. The irony is that had they stood behind and marketed the EV1, instead of pushing the Hummer they would be pwning the industry right now. A good example of short sightedness leading to dead ends.

I wonder where GM plans to build these cars?

It would sure be nice to see GM employ more Americans.

3. Nov 30, 2006

### turbo

It might not matter if the government keeps letting the auto industry cheat. Look at the PT Cruiser. It's a little wagon built on a Neon chassis. Somehow, Dodge/Chrysler was allowed to call it a truck, and average its fuel economy in with the Ram truck series to avoid having to produce more fuel-efficient trucks. It's one thing to pass regulations, and another entirely to make them difficult to circumvent.

4. Nov 30, 2006

### Staff: Mentor

Since GM's future economic viability is very much in doubt, "nice" cannot be a consideration.

5. Nov 30, 2006

### Staff: Mentor

Heh, I didn't know that (though it doesn't surprise me). As with gun regulations (what is an "assault weapon"?), it is often easy to circumvent regulations. Keeping them in line requires vigilance, which requires a willingness to act, which is something beaurocratic governments do not posess.

Perhaps the answer would have to be some sort of inspector general in charge of making such determinations (similar to the way building permits are issued). Such a person is empowered to make judgement calls to make the reality of a design fit with the intent of a law. There are obvious downsides to that, though.

6. Nov 30, 2006

### edward

Somehow they convinced the government that the PT Cruiser was a crossover vehicle and for some mysterious reason that made it a truck.

Ford did the same thing with the Freestar.

7. Nov 30, 2006

### Futobingoro

How does that follow? If competition is already forcing GM into the production of fuel-efficient cars, why are higher CAFE standards even necessary?

Environmental and economic concerns can align. And when an industry shifts to a strategy more favorable to the environment, it is not necessarily a concession by capitalists and Republicans.

With the auto industry's hybrid/electric cars and Boeing's new 787, the top priority was the almighty dollar, make no mistake. But by improving fuel efficiency to produce the most marketable products, these corporations also decreased the impact on the environment. Yes, these corporations might not object to tougher fuel regulations, as the market is already regulating them, but this isn't a setback for Republicans.

For instance: most shipping companies have drastically cut paper consumption by switching to electronic tracking. So this somehow means that if there were a movement in Congress to pass stricter regulations governing the consumption of paper products, the shipping industry would not object; and it would not look good for Republicans?!

There have been extreme cases of nitpicking over current events of late. When there is even the slightest indicator of a failure of the Republican party, it is sure to show up here.

What's the scoop on the economic sanctions placed on North Korea? What does Trent Lott's ascendancy to minority whip mean? There are more, but I think you get the point.

It is interesting to note that there has been no mention of Murtha's flop in his run for House majority leader or Alcee Hasting's failed bid for chairman of the House Intelligence Committee. The search terms for Murtha and Hastings are telling. The cynic in me wanted to start a thread speculating that these events showed a lethal divide in the Democratic majority, but I decided against it.

8. Nov 30, 2006

### Skyhunter

That is a good point.

With Barbara Boxer as Environmental & Public Works Committee Chair, I think that regulations will be somewhat more robust than in the past. She tried and failed to add a simple provision to exclude luxury SUV's from the small business tax credit. (Remember the $2000 hybrid tax incentive -vs- the$100,000 Hummer incentive.) Now with the Republicans in the minority, and the neo-nuts in the extreme minority, we just might be able to get good CAFE legislation passed.

The major source, over half, of the pollution in the Bay Area is caused by automobiles. One of the side benefits of fuel economy is that when a vehicle burns less fuel, it generates less pollution.


text:
De Brogile Application

$\lambda=\frac{h}{p}$

Shirley Liu 2I
Posts: 64
Joined: Fri Sep 28, 2018 12:16 am

De Brogile Application

I know you can use the De Brogile equation for matter, such as balls, cars, etc., but can you also use the De Brogile equation for finding the wavelengths of electrons and photons or can you only use the c= wavelength x frequency for that ?

Sierra Cheslick 2B
Posts: 61
Joined: Fri Sep 28, 2018 12:27 am

Re: De Brogile Application

You use de Broglie to find the wavelengths of subatomic particles (electrons, protons, and neutrons) but you use lambda = hc/energy for photons since they do not mass.

Jake Gordon 1A
Posts: 61
Joined: Fri Sep 28, 2018 12:15 am

Re: De Brogile Application

Equations using the speed of light are only applicable to electromagnetic radiation (photons) they lack rest mass and therefore do not work for the debroglie relation

On the other hand, everything with rest mass can not be related by using the speed of light and the debroglie relation must be used

Sapna Ramappa 1J
Posts: 63
Joined: Fri Sep 28, 2018 12:24 am

Re: De Brogile Application

You can use de Broglie for anything with mass, so that would include electrons. However, you can't use the equation for photons, because photons don't have mass. In this case, you would use an equation with the speed of light.

305127455
Posts: 30
Joined: Fri Sep 28, 2018 12:16 am

Re: De Brogile Application

Technically De Brogile equation can be used on anything with rest mass; however, the equation is not so helpful on macro objects since their wave property is not obvious.


text:
Author Topic: Problem 3  (Read 15377 times)

Thomas Nutz

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Problem 3
« on: November 12, 2012, 04:40:07 PM »
Dear Professor,

do we need to find a closed expression for the solution in problem 3, or can we express it as a sum over n?

Victor Ivrii

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Re: Problem 3
« Reply #1 on: November 12, 2012, 04:45:03 PM »
Dear Professor,

do we need to find a closed expression for the solution in problem 3, or can we express it as a sum over n?

Yes, either as a series or as an integral (or present two forms)--there is no "closed" expression

Thomas Nutz

• Full Member
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Re: Problem 3
« Reply #2 on: November 17, 2012, 03:21:05 PM »
Are we supposed to derive eq. 19 in lecture notes 24 (laplacian outside the disk) for problem 3b) or can we start from this formula (it given in the context of an exercise)?

Victor Ivrii

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Re: Problem 3
« Reply #3 on: November 17, 2012, 04:25:27 PM »
Are we supposed to derive eq. 19 in lecture notes 24 (laplacian outside the disk) for problem 3b) or can we start from this formula (it given in the context of an exercise)?

Better forget about this formula, separate variables and use the boundary value provided and not try to derive formula at all.

Ian Kivlichan

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Re: Problem 3
« Reply #4 on: November 19, 2012, 09:30:20 PM »
Hopeful solutions attached!

(parts 1,2)

Ian Kivlichan

• Sr. Member
• Posts: 51
• Karma: 17
Re: Problem 3
« Reply #5 on: November 19, 2012, 09:31:27 PM »
Hopeful solutions attached!

(part 3)

Chen Ge Qu

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Re: Problem 3
« Reply #6 on: November 19, 2012, 09:33:01 PM »
a) Part 1 of 2

Chen Ge Qu

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Re: Problem 3
« Reply #7 on: November 19, 2012, 09:33:48 PM »
a) Part 2 of 2

Chen Ge Qu

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Re: Problem 3
« Reply #8 on: November 19, 2012, 09:34:25 PM »
Part b) attached

Calvin Arnott

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Re: Problem 3
« Reply #9 on: November 19, 2012, 09:38:03 PM »
Problem 3

Part a. Solve:

$$\Delta u := u_{xx} + u_{yy} = 0, \phantom{\ }r < a$$
$$u \bigr|_{r=a} = f\left(\theta\right)$$
\text{in polar coordiantes } \left(r,\theta\right) \text{, with: } f(x) = \left\{\begin{aligned} &\phantom{-\ }1 &&: 0 < \theta < \pi\\ &-1 &&: \pi < \theta < 2\pi \end{aligned} \right.

If $a \le 0$, clearly the problem is not well-posed or yields only a singular point-solution, as for polar coordinates we take only $r \ge 0$. Let $a > 0$. Then this is the Dirichlet problem on the disk of radius $a$, with BC $h\left(\theta\right) = f\left(\theta\right)$. Because we have $u = f$ on the circumference of the disk, we have that $|u| \le |f| = 1$ on the disk by the maximum principle, and $|u|$ is bounded. Then, it was derived that by changing to polar coordinates $\left(x,y\right) \mapsto \left(r,\theta\right)$ with Dirichlet BC: $\{u = h\left(\theta\right) : r = a\}$, $u\left(r,\theta\right)$ has a solution of the form:

$$u\left(r,\theta\right) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^n \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)$$

$$A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\cos\left(n\phi\right)d\phi$$

$$B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\sin\left(n\phi\right)d\phi$$

Substituting in $f\left(\theta\right)$ and splitting the integral between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$:

$$\implies A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}f\left(\phi\right)\cos\left(n\phi\right)d\phi = \frac{1}{\pi a^n}\left( \int_{0}^{\pi}\cos\left(n\phi\right)d\phi - \int_{\pi}^{2\pi}\cos\left(n\phi\right)d\phi\right)$$

$$= \frac{1}{\pi a^n}\left( \frac{1}{n}\sin\left(n\phi\right) \bigr|_{0}^{\pi} - \frac{1}{n}\sin\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = 0 \text{, as } n \in \mathbb{N}$$

$$\implies B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}f\left(\phi\right)\sin\left(n\phi\right)d\phi = \frac{1}{\pi a^n}\left( \int_{0}^{\pi}\sin\left(n\phi\right)d\phi - \int_{\pi}^{2\pi}\sin\left(n\phi\right)d\phi\right)$$

$$= \frac{1}{\pi a^n}\left( -\frac{1}{n}\cos\left(n\phi\right) \bigr|_{0}^{\pi} + \frac{1}{n}\cos\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = \frac{1}{n \pi a^n}2\left(1-\left(-1\right)^n\right) \text{, as } n \in \mathbb{N}$$

$$\implies u\left(r,\theta\right) = \sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n \frac{2\left(1-\left(-1\right)^n\right)}{n \pi } \sin\left(n \theta\right) \phantom{\ } \square$$

If instead we approached the Dirichlet problem on the disk from Poisson's formula, we have that $u\left(r,\theta\right)$ has a solution of the form:

$$u\left(r,\theta\right) = \left(a^2 - r^2\right)\int_{0}^{2\pi}\frac{h\left(\phi\right)}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}$$

Substituting in $f\left(\theta\right)$ for $h\left(\theta\right)$, and splitting between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$ gives the irreducible integrals:

$$= \left(a^2 - r^2\right)\left(\int_{0}^{\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi} - \int_{\pi}^{2\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}\right) \phantom{\ } \blacksquare$$

Part b. Solve:

$$\Delta u := u_{xx} + u_{yy} = 0, \phantom{\ }r > a$$
$$u \bigr|_{r=a} = f\left(\theta\right)$$
$$\max |u| < \infty$$
\text{in polar coordiantes } \left(r,\theta\right) \text{, with: } f(x) = \left\{\begin{aligned} &\phantom{-\ }1 &&: 0 < \theta < \pi\\ &-1 &&: \pi < \theta < 2\pi \end{aligned} \right.

Answer:If $a \le 0$, clearly the problem is not well-posed. For $r < 0$ our BC $u \bigr|_{r=a} = f\left(\theta\right)$ is undefined, as $r \ge 0$, and for $r = 0$ the singular point-BC does not specify a solution. Let $a > 0$. Then this is the Dirichlet problem on the exterior of the disk of radius $a$, with BC $h\left(\theta\right) = f\left(\theta\right)$. We can view this as an equivalent problem on the interior of the disk by considering the rotationally invariant projection of the real line of the Riemann sphere to itself with the rule: $r\mapsto \frac{1}{r}$ : $\{ z : z \in \mathbb{R}_\infty , z \ge a \} \mapsto \{ \frac{1}{z} : \frac{1}{z} \in \mathbb{R}_{\infty} , \frac{1}{z} \le \frac{1}{a}\}$. This gives a mapping the Dirichlet problem on the exterior of the disk in $r$ with BC at $a$ to the Dirichlet problem on the interior of the disk in $\frac{1}{r}$ with BC at $\frac{1}{a}$. This map exists as $|u|$ bounded as $r \rightarrow \infty$ which implies that we have a removable
singularity of $u$ at our point at infinity, so $u$ can be analytically extended there, and so may the equivalent mapped point at $0$ when $\frac{1}{r} \ \rightarrow 0$ as $r \rightarrow \infty$.

We derived that the Dirichlet problem on the interior of the disk with BC: $\{u = h\left(\theta\right) : r = a\}$, $u\left(r,\theta\right)$ has a solution of the form:

$$u\left(r,\theta\right) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^n \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)$$

$$A_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\cos\left(n\phi\right)d\phi$$

$$B_n = \frac{1}{\pi a^n} \int_{0}^{2\pi}h\left(\phi\right)\sin\left(n\phi\right)d\phi$$

Substituting in $f\left(\theta\right)$, mapping $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$, and splitting the integral between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$, gives us the solution on the exterior:

$$\implies u\left(r,\theta\right) = \frac{1}{2} A_0 + \sum_{n=1}^{\infty} r^{-n} \left(A_n \cos\left(n \theta\right) + B_n \sin\left(n \theta\right)\right)$$

$$\implies A_n = \frac{1}{\pi a^{-n}} \int_{0}^{2\pi}f\left(\phi\right)\cos\left(n\phi\right)d\phi = \frac{a^{n}}{\pi }\left( \int_{0}^{\pi}\cos\left(n\phi\right)d\phi - \int_{\pi}^{2\pi}\cos\left(n\phi\right)d\phi\right)$$

$$= \frac{a^{n}}{\pi}\left( \frac{1}{n}\sin\left(n\phi\right) \bigr|_{0}^{\pi} - \frac{1}{n}\sin\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = 0 \text{, as } n \in \mathbb{N}$$

$$\implies B_n = \frac{1}{\pi a^{-n}} \int_{0}^{2\pi}f\left(\phi\right)\sin\left(n\phi\right)d\phi = \frac{a^{n}}{\pi }\left( \int_{0}^{\pi}\sin\left(n\phi\right)d\phi - \int_{\pi}^{2\pi}\sin\left(n\phi\right)d\phi\right)$$

$$= \frac{a^{n}}{\pi }\left( -\frac{1}{n}\cos\left(n\phi\right) \bigr|_{0}^{\pi} + \frac{1}{n}\cos\left(n\phi\right) \bigr|_{\pi}^{2\pi}\right) = \frac{a^n}{n \pi }2\left(1-\left(-1\right)^n\right) \text{, as } n \in \mathbb{N}$$

$$\implies u\left(r,\theta\right) = \sum_{n=1}^{\infty} \left(\frac{a}{r}\right)^n \frac{2\left(1-\left(-1\right)^n\right)}{n \pi } \sin\left(n \theta\right) \phantom{\ } \square$$

If instead we approached the Dirichlet problem on the interior disk from Poisson's formula, we have that $u\left(r,\theta\right)$ has a solution of the form:

$$u\left(r,\theta\right) = \left(a^2 - r^2\right)\int_{0}^{2\pi}\frac{h\left(\phi\right)}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}$$

Substituting in $f\left(\theta\right)$ for $h\left(\theta\right)$, taking $\{r \mapsto \frac{1}{r}, a \mapsto \frac{1}{a}\}$, and splitting the integral between $\left(0,\pi\right)$ and $\left(\pi,2\pi\right)$, gives us the solution on the exterior in the form of two irreducible integrals:

$$= \left(a^{-2} - r^{-2}\right) \left(\int_{0}^{2\pi}\frac{f\left(\phi\right)}{a^{-2}-2 a^{-1} r^{-1} \cos\left(\theta-\phi\right)+r^{-2}}\frac{d \phi}{2\pi}\right)$$

$$= \frac{a^2 r^2}{a^2 r^2}\left(a^{-2} - r^{-2}\right)\left(\int_{0}^{2\pi}\frac{f\left(\phi\right)}{a^{-2}-2 a^{-1} r^{-1} \cos\left(\theta-\phi\right)+r^{-2}}\frac{d \phi}{2\pi}\right)$$

$$=\left(r^2 - a^2\right)\int_{0}^{2\pi}\frac{f\left(\phi\right)}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}$$

$$= \left(r^2 - a^2\right)\left(\int_{0}^{\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi} - \int_{\pi}^{2\pi}\frac{1}{a^2-2 a r \cos\left(\theta-\phi\right)+r^2}\frac{d \phi}{2\pi}\right) \phantom{\ } \blacksquare$$
« Last Edit: November 19, 2012, 09:41:00 PM by Calvin Arnott »

Ian Kivlichan

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Re: Problem 3
« Reply #10 on: November 19, 2012, 09:40:30 PM »
Calvin: it's a bit simpler to write out ((-1)^(n+1) +1) as 0 or 2!

p.s. I am envious of your skills in LaTeX

Zarak Mahmud

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Re: Problem 3
« Reply #11 on: November 19, 2012, 09:55:51 PM »
Calvin: it's a bit simpler to write out ((-1)^(n+1) +1) as 0 or 2!

p.s. I am envious of your skills in LaTeX

Me too. Pretty impressive.

Victor Ivrii

Yes, simpler definitely to write $n=2m+1$ with $m=0,1,\ldots$ as "even" coefficients are $0$.


text:
# What is the correct mathematical expression for the assumption ‘consolidation is occurring under small changes in effective stress’ made in arriving the differential equation for transient flow during one-dimensional consolidation?

This question was previously asked in
UPPSC AE Civil 2019 Official Paper I (Held on 13 Dec 2020)
View all UPPSC AE Papers >
1. av = constant
2. $$\frac{1}{{\left( {1 + {e_0}} \right)}}$$.dx.dy.dz = constant
3. γw = constant
4. $$h = \frac{u}{{{\gamma _W}}}$$

 S. No. Assumption Mathematical Expression 1 One dimensional flow in the z-direction $${a_v} = constant$$ 2 Darcy’s Law is valid $${V_z} = {k_z}\frac{{\partial h}}{{\partial z}}$$ 3 Soil is homogenous $${k_z} \ne f\left( z \right)$$ 4 Flow occurs on account of excess pore water pressure $$h = \;\frac{u}{{{\gamma _w}}}$$ 5 Water is incompressible $${\gamma _w} = constant$$ 6 The volume of soil solids is constant $$\frac{1}{{1 + {e_0}}}{d_x}{d_y}{d_z} = constant$$ 7 Consolidation is occurring under small changes in effective stress $$\frac{{\partial {v_x}}}{{\partial x}} = \;\frac{{\partial {v_y}}}{{\partial y}} = 0$$


text:
Environ Eng Res > Volume 25(2); 2020 > Article
Macedonio, Frappa, Brunetti, Barbieri, and Drioli: Recovery of water and contaminants from cooling tower plume

### Abstract

Membrane assisted condenser is an innovative membrane operation that exploits the hydrophobic nature of microporous membranes to promote water vapor condensation and recovery. It can be used for water and chemicals recovery from waste gaseous streams. In this work, the testing of membrane condenser for water and ammonia recovery from synthetic streams (i.e., a saturated air stream with ammonia) simulating the plume of cooling tower is illustrated. The modeling of the process was carried out for predicting the membrane-based process performance and for identifying the minimum operating conditions for effectively recovering liquid water. The experimental data were compared with the results achieved through the simulations showing good agreement and confirming the validity of the model. It was found that the recovery of water can be increased growing the temperature difference between the plume and the membrane module (DT), the relative humidity of the plume (RHplume) and the feed flow rate on membrane area ratio. Moreover, the concentration of NH3 in the recovered liquid water increased with the growing DT, at increasing NH3 concentration in the fed gaseous stream and at growing relative humidity of the feed.

### 1. Introduction

International awareness has been increasing in recent years, in particular for what concerns the importance of minimizing the scarcity of potable water and the impact of air, water, and solid waste pollutants. One area that has received a considerable amount of attention is the problem of industrial wastewater treatment [1, 2]. In industrial processes, the recycling and reusing of process streams and, in particular, of water, is necessary for minimizing fresh water requirements. For example, in power plants the two main sources of emissions are from the stack and the cooling towers. Streams emitting from a stack become saturated in the desulfurization step, and the streams from cooling towers are typically river or seawater evaporated to cool the steam cycle stream. If the evaporated water can be effectively recovered, it can represent a real new source of high quality water, to be utilized both at industrial (as make-up water) and at human level (as drinkable water). The main traditional technologies for the capture of evaporated water by gaseous streams are: heat exchangers, absorption, adsorption, and membranes. Implementation of heat exchangers to condense out the water vapor is an obvious option that has been considered for a long time [3, 4]. An example can be found in the project “Recovery of water from boiler flue gas” funded by the U.S. Department of Energy, under Award No. DE-FC2606NT42727 [5, 6]. The condensing heat exchanger apparatus, developed and tested, consisted of a long rectangular duct containing water-cooled heat exchangers arranged in series. Hot flue gas entered the apparatus from the back end of the boiler and was cooled as it passed through the heat exchangers. Condensation in flue gas is a complicated phenomenon since heat and mass transfer of water vapor and various acid vapors simultaneously occur in the presence of noncondensable gases. As a consequence, the condensed water is relatively dirty and corrosive. Therefore, the recovered water is typically of low quality that requires further treatment, and the infrastructure is inherently prone to corrosion over time, appending additional maintenance cost.
Another technique available to remove water vapor from gas streams is using a desiccant drying system [7, 8]. Siemens Power Generation, Inc. and the Energy & Environmental Research Center, at the University of North Dakota, performed a U.S. Department of Energy supported pilot-scale program that checked the technical viability of a liquid desiccant-based dehumidification technology to efficiently extract water from the power plant flue gas [9]. In particular, Copen et al. [10] designed a pilot scale Water Extraction from Turbine Exhaust system to recover water vapor, and declare that the system can remove 23–63% water vapor from flue gas by volume. However, a desiccant system has as disadvantages the regeneration of the desiccant and the low quality of the produced water.
Membrane technology has been researched as a promising option to recover evaporated water. Many works exist that employ dense membranes that selectively permeate water vapor over other gases with excellent selectivity and permeability, reaching above 107 (H2O/N2) and 105 barrer [11, 12]. However, the evaporated water through the cooling tower and stack leaves at near atmospheric pressure, requiring additional compression or vacuum to apply the necessary driving force for separation. Hence, current vapor separation membranes are completely limited by pressure ratio [13] and mass transfer resistance [11].
Supported liquid membranes (thus combining advantages of absorbent and membranes) [14] were also employed but the work is largely in the research stage without accurate economic analysis.
A hydrophilic porous membrane device was developed in one of the projects supported by U.S. Department of Energy’s (DOE) National Energy Technology Laboratory. The developed nanoporous ceramic membrane device (called Transport Membrane Condenser (TMC) [1518]) condenses water and recovers heat from flue gas. Water vapor contained in the flue gas condenses inside the membrane nanopores and passes through by direct contact with low temperature water from the permeate side. In this way, the transported water is recovered along with virtually its whole latent heat. Non-condensable gases such as CO2, O2, NOx, and SO2 are supposed to be inhibited from passing through the membrane by the condensed water clogging the tube membrane pores. The recovered water is of high quality and mineral free and therefore can be used directly as boiler makeup water, as well as for other processes.
Not only tubes of alumina coated with a final coating of Zirconium dioxide but also other membrane materials (such as polyethersulphone, mixed cellulose ester (e.g., cellulose triacetate) and polyvinylchloride) were used in hydrophilic membrane-based dehumidifiers [19]. The advantage in the use of these materials for separating water vapor from the vapor/gas mixture is their strong affinity to the water molecules. More recent works by Zhao et al. [20, 21], Chen et al. [22, 23] and Kim et al. [24] further investigated the efficacy of TMC configurations. In particular, Kim et al. [24] found that TMC is completely limited by the rate of condensation, and proper membranes and an effective module design must be developed for enhancing the vapor pressure gradient.
Another technology currently investigated for water recovery from waste gaseous stream is membrane-assisted condenser (MC). The latter belongs to the class of membrane contactors where microporous hydrophobic membranes are used not as selective barriers but for promoting water vapor condensation and recovery from humidified gaseous streams (e.g., flue gas, plume, etc.). Its working principle was recently introduced by Drioli and co-workers [2529] (Fig. S1). In details, membrane condenser feed is a gaseous stream, at a certain temperature and, in most of the cases, water saturated. It is fed to the membrane condenser module whose temperature is equal or lower than the one of the feed. In doing this, water droplets already present in the feed together with the amount that condenses onto the membrane surface (and in the membrane module) are blocked by the hydrophobic nature of the membrane. Therefore, the liquid water is recovered at the retentate side, whereas the other gases at the permeate side of the membrane unit. The hydrophobic nature of the membranes not only avoids water droplets dragging, but promotes vapour condensation exploiting the principle of dropwise condensation where, when condensation takes place on a surface that is not wet by the condensate, water beads up into droplets and rolls off the surface. Water vapor preferentially condenses on solid surfaces rather than directly from the vapor because of the reduced activation energy of heterogeneous nucleation in comparison to homogeneous nucleation [30, 31]. While the excess energy of a surface controls the heterogeneous nucleation process, it also determines the wetting behavior of the condensate, which has a significant impact on the overall performance. However, the effect of surface-structure geometry on condensation behavior remains under debate. What it is sure is that the use of hydrophobic surfaces improves heterogeneous condensation and recently, the use of hydrophobic surfaces structured at capillary length scales was proposed to enhance heterogeneous condensation [30, 32, 33].
Moreover, in a membrane condenser unit, the modulation of contact time between saturated stream and membrane, as well as the control of temperature and/or pressure difference between membrane sides, allow controlling the fraction of components present in the feed gaseous stream that will be retained in the condensed water. Membrane condenser technology was initially developed in the FP7 CAPWA project (Capture of evaporated water with novel membranes) [34], and it is now further studied and scaled up in the framework of the EU Horizon 2020 project MATChING (Materials & Technologies for Performance Improvement of Cooling Systems performance in Power Plants) [35]. In both projects, membrane condenser was and is utilized for the recovery of evaporated water from humidified waste gaseous streams. In particular, membrane condenser was used in CAPWA for water recovery from flue gas (i.e., a waste gaseous stream at high temperature containing SOx, NOx and particles) whereas, currently, in MATChING project it is developed for water recovery from plume of cooling tower (i.e., the stream of saturated exhaust air leaving the cooling tower). Plume temperature is relatively low with respect to that of flue gas. Moreover, the drift of fine droplets emitted from the cooling towers can contain salts (in particular when seawater is utilized as make-up water), biocides and algaecides. These are dosed to the circulating cooling water to prevent growths of micro-organisms such as bacteria, fungi and algae that could interfere with the continuous flow of the water and reduce the heat transfer efficiency of the cooling tower. A normal industrial practice is to use two biocides, such as oxidizing and non-oxidizing types to complement each other’s strengths and weaknesses, and to ensure a broader spectrum of attack. Another very important reason for using biocides in cooling towers is to prevent the growth of Legionella. Various studies have shown that from 40 to 60% of cooling towers tested contained Legionella [36]. Biocidal substances and products used as anti-fouling agents are, for example chlorine (used as a short-life biocide in industrial water treatment) and certain types of quaternary ammonium compounds (added to industrial water systems to act as an algicide, protecting the water from infestation and growth of algae). The emission of these compounds in the atmosphere significantly affects the air quality in the areas near the plant; therefore, membrane condenser can be considered as an efficient pre-treatment unit able to both separate water vapour contained in waste gaseous streams, also retaining some of the most effective contaminants that, when discharged in the atmosphere, affect air quality.
In this manuscript, we analyzed, both theoretically and experimentally, the use of membrane condenser for the recovery of water vapor from the plume of a cooling tower. A membrane condenser prototype with 0.0258 m2 of active membrane area was used for the treatment of a gaseous stream simulating the plume (that is, for the reasons listed above, a water saturated air stream containing ammonia). The results of the experiments were utilized for the validation of the simulation study of the process. The latter allows predicting the membrane-based process performance in a wide range of operating conditions and identifying the minimum operating conditions for effectively recovering liquid water.

### 2.1. Membrane Condenser Prototype

Fig. 1 shows the scheme of the membrane condenser prototype built up for performing the measurements.
The membrane module made in stainless steel with four inlet/outlet lines, is placed in a chamber set at a temperature equal or lower than the one of the gaseous stream for maintaining the required temperature difference between the feed (i.e., the plume) and the membrane module. Nineteen microporous hydrophobic polypropylene commercial membranes are assembled in the module by only one fixed end for an active membrane area of about 258 cm2. According to manufacturer’s data (Membrana), each membrane used (3M Capillary Membrane MF-PP Series, Type S6/2) has inner diameter equal to 1.8 mm, wall thickness of 450 micron and nominal pore size of 0.20 micron. The humidified gaseous stream, consisting of saturated or super saturated air with ammonia, is fed to the upper part of the module by a mass flow controller; the membrane allows the liquid water to be retained in the retentate side and collected at the bottom of the module, opportunely designed with a conic shape to avoid stagnant zones, whereas the dehydrated stream is recovered in the permeate side. In dependence on the operating conditions used during the experiments, the permeate stream could contain a certain level of relative humidity, lower than the feed one, and a part of the gaseous stream fed to the module could not pass through the membrane remaining in the retentate side. Indeed, for all the streams, relative humidity and temperature sensors are placed at the inlet and at the outlet of the permeate and retentate streams. A back-pressure controller is placed on the retentate side and the pressure is measured by means of a manometer. All the lines connecting the module with the rest of the instrumentation placed out of thermostatic chamber are thermally insulated in order to avoid any undesired condensation.
The composition of the recovered liquid water is analysed by an ionic chromatography Thermo Scientific Dionex Integrion System.
Table S1 summarizes the operating conditions used in the experiments.
The experimental results are reported in terms of water recovery, defined as the mass of liquid water retained in the retentate side with respect to the total water contained as vapor in the gaseous feed stream (Eq. (1)):
##### (1)
$Water recovery=Liquid water retained in the retentate sideTotal water contained in the feed stream×100, %$

### 2.2. Membrane Condenser Simulations

Modeling and simulations of membrane condensers were carried out to explore the effect of some process parameters (such as temperatures of plume and module, temperature difference (T) between plume and module, feed flow rate (Q) on membrane area (A) ratio (Q/A) on membrane condenser performance, and considering a wider range of operating conditions with respect to that used for experiments. For set cases, they were also compared with the experimental results. Detailed relations and models for transport phenomena through the membrane condenser are outside the scope of this work and details can be found in cited literature [25, 27]. The results of the simulation give indications on the amount of recovered liquid water, on its composition and on the energy consumption of the process.
Fig. 2(a) shows the effect of cooling on the amount of recovered water. It can be observed that lower temperatures of the feed gaseous stream require larger T (temperature difference between the feed and the gas before entering the membrane module) for achieving a positive amount of recovered water.
The recovery of water can be improved acting on the ratio Q/A, as it can be seen in Fig. 2(b) showing the amount of water that can be recovered from a plume at 25°C when Q/A ranges from 1 to 15 m h−1.
Simulations indicate that for obtaining positive amount of recovered water, either a Q/A ratio ≥ 2 m/h or T ≥ 2°C (if Q/A = 1 m/h) are necessary when Tplume = 25°C. Less stringent conditions are instead required at Tplume equal to 30°C where a Q/A ratio ≥ 2 or T ≥ 1 (if Q/A = 1 m/h) have to be utilized.
For what concenrs the energy consumption of the membrane condenser, Fig. 3 shows the condenser heat duty at different temperature (Tplume) of a supersatured (RH = 110%) plume. It can be seen that the power needed for achieving the same percentage of recovered water increases with initial feed temperature, although the temperature reduction required for obtaining the desired percentage of water recovery decreases with Tplume (Fig. 2(a)). This is due to the fact that, at increasing temperature and constant RH, the amount of water vapor in the gaseous mixture is higher. Therefore, for achieving the same percentage of water recovery, a higher volume of water has to be condensed requiring a higher heat duty.

### 3. Experimental Measurements

As reported in Table S1, experimental measurements were carried out at various temperature of the fed gaseous stream (i.e., temperature of the plume), RH of the feed, at various temperature difference DT between the plume and the module, at two different Q/A values, and at various NH3 concentration in the feed.
Fig. 4 shows the recovered water at around 25°C for T ranging from 5 to 12°C. It can be seen that the amount of recovered water increases at growing T. In particular, it was found that considering a plume at 25°C and utilizing a Q/A = 1.2 m/h, water recovery ranges from 36.24% (T = 5.84°C and RH = 104.9%) to more than 55% (T ≈ 10°C, Tplume ≈ 25°C and RH > 140%). Moreover, Fig. 4 shows the comparison between the results obtained through the simulations and those of the experimental measurements. It can be observed a good agreement between the experimental data and the model. This confirms the validity of the simulation study done and its suitability for predicting the performance of the condenser.
Good agreement between theoretical and experimental results was achieved also in the experiments performed considering the same temperature of the plume (≈ 25°C) and at a higher Q/A ratio (Fig. 5 on the left). In this case, the amount of recovered water increases from 46.0% (T = 6°C) to more than 70% (T ≈ 13.9°C) when the RH of the plume is as high as ≈142% (Fig. 5 on the left). On the contrary, as shown in Fig. 6 on the right and accordingly to simulations, even if TPlume is higher (≈ 30°C), the recovery of water is reduced by lowering RHPlume to around 106%. In particular, in the latter case water recovery passes from 32.12% for a T = 6.0°C to more than 64% for T ≈ 15.95°C.
The influence of RH on the amount of recovered water, maintaining almost constant T, is clearly visible in Fig. 6 where it can be observed the increase of water recovery, at growing RH, at two different TPlume (40 and 45°C). The highest value was 48.3%, reached in the case of TPlume = 45°C when T = 11.3°C and RHPlume = 120%.
As observed with the modeling, also at experimental level was found the increase of water recovery at growing QFeed/AMembrane ratio when all the other parameters (T of the plume, RH and T) are kept constant (Table S2).
The quality of liquid water recovered by membrane condenser is strictly affected by various parameters which influence the final contaminants content. As mentioned, in this work we considered the presence of NH3, as it is a contaminant usually present in the plume. For what concerns the composition of the recovered liquid water, results reported in Fig. 7 show the increase of NH3 concentration with the growing temperature difference DT between the fed plume and the membrane module (when the temperature of the fed plume is constant whereas the temperature of the membrane module decreases). This trend is because Henry’s law constant of NH3 increases when the temperature of the membrane module decreases (T in Eq. (2)):
##### (2)
$kH=kH0×exp(-ΔsolnHR(1T-1T0))$
where
$ΔsolnHR=-dlnkHd(1/T)$
k0 and T0 refer to standard condition (298.15 K) and are equal to 61 mol l−1 atm−1 and 4,200 K, respectively.
As a consequence, NH3 solubility in aqueous solution rises when the membrane module temperature decreases.
Other parameters allowing controlling the concentration of NH3 are the initial feed concentration and the RHfeed. In fact, Fig. 7 and 8 show the increase of NH3 concentration in the recovered liquid water at growing NH3 concentration in the fed gaseous stream (Fig. 8) and RHplume (Fig. 7).
In the performed experiments, the lowest NH3 concentration in the recovered water was found equal to 0.59 ppm when the initial NH3 concentration in the feed was of about 100 ppm (Table S3).

### 4. Conclusions

In this work we analyzed, both theoretically and experimentally, the use of membrane condenser for the recovery of water vapor from the plume of a cooling tower. The simulation allowed predicting the process performance in a wide spectrum of operating conditions and identifying the minimum operating conditions for effectively recovering liquid water. It was found that the recovery of water can be increased with the grow of DT, RH and Q/A. Moreover, at Tplume low and equal to 25°C, either a Q/A ratio ≥ 2 mh or DT ≥ 2°C are necessary for obtaining positive amount of recovered water. It was also observed that the power needed for achieving the same percentage of recovered water increases with initial feed temperature.
The simulation study was validated through experiments performed on a membrane condenser prototype and a good agreement was found. Considering a plume at 25°C and utilizing Q/A = 1.2 m/h, water recovery ranged from 36.24% (DT = 5.84°C and RH = 104.9%) to more than 55% (DT≈10°C, RH > 140%) and these values increased at Q/A ratio of 2.7 m/h and RH ≈ 142%, passing from 46.04% (DT = 6.0°C) to more than 70% (DT≈13.9°C). As observed with the modeling, also at experimental level was found the increase of water recovery at growing Q/A ratio when all the other parameters (T of the plume, RH and DT) are kept constant.
The experimental analyses performed considering the presence of NH3 in the fed gaseous stream revealed that the concentration of NH3 in the recovered liquid water increased with the growing temperature difference DT between the fed plume and the membrane module, at increasing NH3 concentration in the fed gaseous stream and at growing relative humidity of the feed. Therefore, apart the recovery of water, another important advantage of membrane condenser is the possibility to recover the chemicals necessary in the plant and, at the same time, to reduce their emissions. These are fundamental aspect to be considered in highly industrialized realities where the problems connected to water shortage and air pollution (inducing health problems and climate concerns) are becoming more and more pressing.

### Acknowledgments

The H2020 is gratefully acknowledged for funding this work through the project “Materials & Technologies for Performance Improvement of Cooling Systems in Power Plants (acronym MATChING)” (GA 686031).

### References

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##### Fig. 1
Scheme of the MC prototype.
##### Fig. 2
Water recovery as a function of temperature difference between the plume and the membrane module. Relative humidity of the plume (RHPlume) = 110%. A = 0.026 m2. (a) Water recover at different values of TPlume and QFeed/AMembrane = 1. (b) Water recovery at different values of QFeed/AMembrane and TPlume = 25°C.
##### Fig. 3
Power needed to drive the process as a function of water recovery.
##### Fig. 4
Water recovery as a function of temperature difference between the plume and the membrane module, at different RHPlume. Symbols: experimental results; lines: simulation results.
##### Fig. 5
Water recovery as a function of temperature difference between the plume and the membrane module, at different TPlume. Symbols: experimental results; lines: simulation results.
##### Fig. 6
Water recovery as a function of temperature difference between the plume and the membrane module, at different RHPlume.
##### Fig. 7
NH3 in recovered water as a function of temperature difference between the plume and the membrane module, at different TPlume and RHPlume.
##### Fig. 8
NH3 in recovered water as a function of NH3 in the plume at different temperature difference between the plume and the membrane module.
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text:
# Determine the truth value of the statement if the universe of each variable consists of (i) all real numbers, (ii) all integers.

∀x (x > 0 → ∃y ((√x)/y = 3))

I've learned about different types of proofs, and I'm thinking here that I would first negate the statement to then prove the negation. Would the negation be: ∃x(x ≯ 0 → ∀y((√x)/y ≠ 3)) ?

If it's as easy as finding a number for x in which the statement is false, I could say:

let x = 25

(√25)/y = 5/y ≠ 3

Is this sufficient proof?

In either case, how would the proof be different with integers than with real numbers? Thanks ahead.

• In your counterexample, what if $y=\frac53$? You can't pass negation through implication like that; instead consider writing your implication $P\implies Q$ in the form $Q\vee \neg P$ and then negating that. – Steven Stadnicki Oct 9 '14 at 23:08

You shouldn't need to use a contradiction proof in either case. For the real number case, it is easier to use a constructive existence proof for the value of y. For the integer case, a counter example will suffice.

The field you choose changes the truth value of the proposition.

In the case of real numbers, you need to show that there exists some real number y such that ${\frac{\sqrt{x}}{y}=3}$ whenever x>0.

It is easy to see that letting y = ${\frac{\sqrt{x}}{3}}$ for any value of x>0 will yield:
${\frac{\sqrt{x}}{y} = \frac{3\sqrt{x}}{\sqrt{x}} = 3}$.

so the proposition is true in the real number field.

However, as your counter example shows it isn't true for the integers since letting ${x=25}$ causes ${\frac{\sqrt{25}}{y} = \frac{5}{y}}$ which does not equal 3 for any integer y.

Thus, the proposition is false among the integers.


text:
# Solve the proportion x/8 =4/12 A.1/24 B.1/3 C.3 D.8/3 E.24

###### Question:

Solve the proportion x/8 =4/12
A.1/24
B.1/3
C.3
D.8/3
E.24

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