INTRODUCTION: Topological Insulator has drawn a lot of attention recently in condensed matter physics. It describes the phase of matter in a different way, and gives us a new perspective toward materials. What is Topological Insulator(TI)? It is a material with bulk bandgap. However, at the surface of TI, it has edge states that propagate like a metal. We can imagine this like plastic tube wrapped with a aluminum foil around the tube. And it is proposed that it might be a potential candidate as fault tolerant quantum computation because the edge states are protected by time reversal symmetry. Protected state means that it is robust against impurity or imperfections in the crystal. The first 3D TI was observed in semiconducting alloy Bi1 − xSbx with angle resolved photoemission spectroscopy(ARPES) . The reason to choose Bisumth is due to its strong spin orbit interaction which is essential to see TI edge states. But the Bi1 − xSbx surface states are complicated, so it invokes other materials such as Bi₂Se₃ . INTEGER QUANTUM HALL EFFECT: To discuss TI surface edge states, we begin with the very similar cousin of TI edge states, which is quantum hall effect(QHE), or more accurately integer quantum hall effect. First observation of quantum hall effect was conducted in 1980 in MOSFET system in the low temperature and high magnetic field environment. The astonishing experiment result is that the conductivity(or resistivity) of the system has plateau with increasing magnetic field as following.

Research: the process by which we understand the world, ourselves, and other phenomena ranging from the alpha helix of a protein to the societal movements in politics is what life is.  That may be a bit too grandiose, but research is an integral part of advancing humankind forward.Said differently, research is the engine by which progress occurs.  Yet, performing successful research is something that is not easy. Unlocking the secrets of life is a laborious task and it requires cooperation amongst people in real time as well as those that came before us.  Key to this progress is clarity, completeness, and access.  One might guess that this is the what defines the research publishing community.  One might be wrong.Scholarly research is for the most part locked behind expensive paywalls in forms that resemble more of something like a trophy than a document used for conveying important of pieces of research in the best manner possible.  While there are lots of things occurring in scholarly research that don't make sense, one blatant one stands out.  We charge researchers to access other researchers documents-- not in the name of sustainability, but in the name of profit.  In fact, we had to invent a word to describe a publication process that is conducive to research: open access.  What is #OpenAccess and what is it good for?  Absolutely everything, as far as research is concerned.

INTRODUCTION TO VERSION CONTROL Many scientists write code as part of their research. Just as experiments are logged in laboratory notebooks, it is important to document the code you use for analysis. However, a few key problems can arise when iteratively developing code that make it difficult to document and track which code version was used to create each result. First, you often need to experiment with new ideas, such as adding new features to a script or increasing the speed of a slow step, but you do not want to risk breaking the currently working code. One often utilized solution is to make a copy of the script before making new edits. However, this can quickly become a problem because it clutters your filesystem with uninformative filenames, e.g. analysis.sh, analysis_02.sh, analysis_03.sh, etc. It is difficult to remember the differences between the versions of the files, and more importantly which version you used to produce specific results, especially if you return to the code months later. Second, you will likely share your code with multiple lab mates or collaborators and they may have suggestions on how to improve it. If you email the code to multiple people, you will have to manually incorporate all the changes each of them sends. Fortunately, software engineers have already developed software to manage these issues: version control. A version control system (VCS) allows you to track the iterative changes you make to your code. Thus you can experiment with new ideas but always have the option to revert to a specific past version of the code you used to generate particular results. Furthermore, you can record messages as you save each successive version so that you (or anyone else) reviewing the development history of the code is able to understand the rationale for the given edits. Also, it facilitates collaboration. Using a VCS, your collaborators can make and save changes to the code, and you can automatically incorporate these changes to the main code base. The collaborative aspect is enhanced with the emergence of websites that host version controlled code. In this quick guide, we introduce you to one VCS, Git (git-scm.com), and one online hosting site, GitHub (github.com), both of which are currently popular among scientists and programmers in general. More importantly, we hope to convince you that although mastering a given VCS takes time, you can already achieve great benefits by getting started using a few simple commands. Furthermore, not only does using a VCS solve many common problems when writing code, it can also improve the scientific process. By tracking your code development with a VCS and hosting it online, you are performing science that is more transparent, reproducible, and open to collaboration . There is no reason this framework needs to be limited only to code; a VCS is well-suited for tracking any plain-text files: manuscripts, electronic lab notebooks, protocols, etc.

INTRODUCTION Biofilms are diverse and complex microbial consortia, and, the biofilm lifestyle is the rule rather than the exception for microbes in many environments. Large and small-scale biofilm architectural features play an important role in their ecology and influence their role in biogeochemical cycles . Fluid mechanics impact biofilm structure and assembly , but it is less clear how other abiotic factors such as resource availability affect biofilm assembly. Aquatic biofilms initiate with seed propagules from the planktonic community . Thus, resource amendments that influence planktonic communities may also influence the recruitment of microbial populations during biofilm community assembly. In a crude sense, biofilm and planktonic microbial communities divide into two key groups: oxygenic phototrophs including eukaryotes and cyanobacteria (hereafter “photoautotrophs”), and heterotrophic bacteria and archaea. This dichotomy, admittedly an abstraction (e.g. non-phototrophs can also be autotrophs), can be a powerful paradigm for understanding community shifts across ecosystems of varying trophic state . Heterotrophs meet some to all of their organic carbon (C) requirements from photoautotroph produced C while simultaneously competing with photoautotrophs for limiting nutrients such as phosphorous (P) . The presence of external C inputs, such as terrigenous C leaching from the watershed or C exudates derived from macrophytes , can alleviate heterotroph reliance on photoautotroph derived C and shift the heterotroph-photoautotroph relationship from commensal and competitive to strictly competitive . Therefore, increased C supply should increase the resource space available to heterotrophs and increase competition for mineral nutrients decreasing nutrients available for photoautotrophs (assuming that heterotrophs are superior competitors for limiting nutrients as has been observed ). These dynamics should result in the increase in heterotroph biomass relative to the photoautotroph biomass along a gradient of increasing labile C inputs. We refer to this differential allocation of limiting resources among components of the microbial community as niche partitioning. While these gross level dynamics have been discussed conceptually and to some extent demonstrated empirically , the effects of biomass dynamics on photoautotroph and heterotroph membership and structure has not been directly evaluated in plankton or biofilms. In addition, how changes in planktonic communities propagate to biofilms during community assembly is not well understood. We designed this study to test if C subsidies shift the biomass balance between autotrophs and heterotrophs within the biofilm or its seed pool (i.e. the plankton), and, to measure how changes in biomass pool size alter composition of the plankton and biofilm communities. Specifically, we amended marine mesocosms with varying levels of labile C input and evaluated differences in photoautotroph and heterotrophic bacterial biomass in plankton and biofilm samples along the C gradient. In each treatment we characterized plankton and biofilm community composition by PCR amplifying and DNA sequencing 16S rRNA genes and plastid 23S rRNA genes.

1. - 1.71. Our proof of the Cauchy-Schwarz inequality, Theorem 1.13, used that when U is a unit vector, $$0 \leq ||V−(U·V)U||^2 = ||V||2 −(U·V)^2$$. Therefore if U is a unit vector and equality holds, then V = (U · V)U. Show that equality occurs in the Cauchy Schwarz inequality for two arbitrary vectors V and W only if one of the vectors is a multiple (perhaps zero) of the other vector. Answer: In the first case, when W = 0, the W is a multiple of V; In the second case, when W is nonzero, then consider the unit vector $U = {||W||}$. Then, by the result in the question, it follows that V = (U · V)U. Therefore: $$V = (U · V)U = ({||W||}· V)·{||W||}= ({||W||^2})·W$$ As $({||W||^2})$ is a constant, so V is a multiple of W. 2. -2.19. Suppose C is an n by n matrix with orthonormal columns. Use Theorem 2.2 to show that $$||CX|| \leq ||X||$$ Use the Pythagorean theorem and the result of Problem 2.17 to show that in fact $$||CX|| = ||X||$$ for such a matrix. Answer: (1) First we calculate C. Let Cj denote the jth column of C. Since C has orthonormal columns, each Cj has norm 1. Then $$||C|| = ^n^n}C_{ij}^2}$$ $$=^n(^n}C_{ij}^2)}$$ $$=^n|| C_j||^2}$$ Now by Theorem 2.2, $$||CX|| \leq ||C||||X|| = ||X||$$ as desired. (2)By Problem 2.17, CX = x₁C₁ + x₂C₂ + … + xnCn. To find the norm of RHS, we need to apply the Pythagorean theorem (we need that C has orthonormal columns) to get $$||x_1 C_1 + x_2 C_2 + \dots + x_n C_n||^2 = ||x_1 C_1||^2 + ||x_2 C_2||^2 + \dots + ||x_n C_n||^2$$ Now we can put the pieces together: $$||CX||^2 = ||x_1 C_1 + x_2 C_2 + \dots + x_n C_n||^2$$ $$=||x_1 C_1||^2 + ||x_2 C_2||^2+\dots+ ||x_n C_n||^n$$ $$=x_1^2||C_1||^2 + x_2^2|| C_2||^2+\dots+ x_n ^2||C_n||^n$$ $$=x_1^2 + x_2^2+\dots+ x_n ^2$$ $$= ||X||^2$$ Since norms are nonnegative, we can conclude that ||CX|| = ||X||. 3. -2.44. Use the Cauchy-Schwarz inequality $$|A · B| \leq||A||||B||$$ to prove: (a) the function f(X)=C · Xis uniformly continuous, (b) the function g(X, Y)=X · Y is continuous. Answer: (a) In case 1, If C = 0 then f(X)=0 for all X,so |f(X)−f(Y)| = 0 < ε for all ε, X, Y. In case 2, where C = (c1, c2, ..., cn)≠(0, 0, ..., 0). By the definition of f and properties of the dot product, |f(X)−f(Y)| = |C · Y − C · Y|=|C · (X − Y)|. By the Cauchy-Schwartz inequality we get $$|f(X)− f(Y)|=|C·(X−Y)|\leq||C||||X−Y||$$ Let ε > 0 and take $δ=\frac {ε}{||C||}$ If $||X−Y||<δ= \frac {ε}{||C||}$ Then we have |f(X)−f(Y)| ≤ ||C||||X − Y|| < ε for all X and Y in Rn. Therefore for any C, f is uniformly continuous. (b) Fix V(A, B) in R2n, to show that g(V)=g(X, Y)=XY is continuous at (a, b) Given ϵ > 0, we need to find δ > o If $||U-V||=\sqrt {(X-A)^2+(Y-B)^2}<\delta$ then ||g(U)−g(V)||=|XY − AB|<ϵ By the triangle inequality, we know that $$|XY-AB|= |[(X-A)+A][(Y-B)+B]-AB|$$ $$=|(X-A)(Y-B)+B(X-A)+A(Y-B)|$$ $$\leq||X-A||+||B|| ||X-A||+||A|| ||Y-B||$$ If $\sqrt {(X-A)^2+(Y-B)^2}<\delta$ Then ||X − A|| + ||B||||X − A|| + ||A||||Y − B|| $$\leq (1+||A||+||B||)\sqrt {(X-A)^2+(Y-B)^2}$$ $$\leq(1+||A||+||B||)\epsilon$$ $$\leq(1+||A||+||B||) \frac {\delta}{1+||A||+||B||}$$ $$=\delta$$ So, given ϵ > 0, set $\delta= min(1,\frac {\delta}{1+||A||+||B||})$ we have $\sqrt {(X-A)^2+(Y-B)^2}<\delta$ Therefore, for any (A, B) in R2n, g(X, Y)=X · Y is continuous. 4. -2.45. In the triangle inequality ||A + B|| ≤ ||A|| + ||B|| put A = X − Y and B = Y. Deduce ||Y|| − ||X|| ≤ ||Y − X||. Show that if two points are within one unit distance of each other, then the difference of their norms is less than or equal to one. Answer: Let A and Bbe in Rn. Apply the triangle inequality we have $$||A + B|| \leq ||A|| + ||B||.$$ LetX = A + B, Y = B,so A = X − Y. so we have $$||X|| \leq ||X − Y|| + ||Y||, ||X|| − ||Y|| \leq ||X − Y||.$$ Exchange the symbol X and Y we get $$ ||Y|| − ||X|| \leq ||Y − X||.$$ So when X and Y are within one unit of distance of each other which means ||Y − X|| ≤ 1, by the inequality we can conclude $$||Y|| − ||X|| \leq ||Y − X||\leq 1$$ 5. -3.9. Suppose a function F from Rn to Rm is differentiable at A. Justify the following statements that prove $$L_AH = DF(A)H,$$ that is, the linear function LA in Definition 3.4 is given by the matrix of partial derivativesDF(A). (a) There is a matrixC such that LA(H)=CH for all H. (b) Denote by Ci the i − th row of C. The fraction $${||H||} = {||H||}$$ tends to zero as ||H|| tends to zero if and only if each component $$\frac {f_i(A+H)− f_i(A)−C_iH}{ ||H||}$$ tends to zero as ||H|| tends to zero. (c) Set H = hej in the i − thcomponent of the numerator to show that the partial derivative fi, xj(A) exists and is equal to the (i, j) entry of C. Answer: (a) F from Rn to Rm is differentiable at A. By the definition of differentiability there is a linear function LA(H) such that $${||H||}$$ tends to 0 as ||H|| tends to zero. By Theorem 2.1, every linear function from Rn to Rm can be written as LA(H)=CH for all H, where C is some m × n matrix. (b) The absolute value of each component of a vector is less than or equal to the norm of the vector so for each H and i we have $$0 \leq| f_i(A + H) − f_i(A) − C_i · H| \leq ||F(A + H) − F(A) − CH||$$ where Ci is the i − th row of C. Since $${||H||}$$ tends to 0, by the squeeze theorem, both${||H|| }$ and${||H|| }$ (c) Let H = hej = (0, 0, ..., 1, 0, ..., 0), the 1 in thej − th place. By part (b), tend to zero as ||H|| tends to zero. $$ {|h|} =0$$ Therefore $$ {h} =0$$ Since Ci · ej = cij and by the definition of partial derivatives,we have $$ {h} = \frac {\partial f_i}{\partial x_j} (A)$$ So we can conclude that cij = c(A) 6. 6.14. Justify the following items which prove: If f is continuous on R₂ and ∫RfdA = 0 for all smoothly bounded sets R, then is identically zero. (a) If f(a, b)=p > 0 then there is a disc D of radius r > 0 centered at (a, b)in which $f(x,y)> \frac {1}{2}p$ (b) If f is continuous and f(x, y)≥p₁ > 0 on a disk R then ∫RfdA ≥ p₁(Area(R)). ∫RfdA = 0for all smoothly bounded regions R, then f cannot be positive at any point. (d) f is not negative at any point either. (e) f = 0 at all points. Answer: First, we can assume that f is continuous on R2 and ∫RfdA = 0 for all smoothly bounded sets R (a) Because f is continuous at (a, b), by the definition of continuity, there is r > 0 such that for all (x, y) such that||(x, y)−(a, b)||<r, we have |f(x, y)−f(a, b)| < p/2.Then we assume p > 0, so p − p/2 < f(x, y)<p + p/2 In particular, f(x, y)>p/2 (b) As R is bounded, the closure of R is closed and bounded. So we can apply the extreme value theorem which means f is bounded on the closure of R. In particular, f is bounded on R. f is also integrable on R; in fact ∫RfdA = 0. Apply the lower bound property, ∫RfdA ≥ p₁(Area(R)) holds. (c) Suppose f is positive at (a, b). From (a), there is a disc Rof nonzero radius on which f(x, y)>f(a, b)/2 > 0. From (b), ∫RfdA ≥ (f(a, b)/2)·area(R)>0 But we assumed that ∫RfdA = 0 for all smoothly bounded sets R, it comes to a contradiction. Therefore f cannot be positive at any point. (d) As we know that −f is continuous, and that for all smoothly bounded regions R, by linearity, we have −fdA = −fdA = −0 = 0 . From(c),we know that −f cannot be positive at any point. Thus, we conclude that f cannot be negative at any point. (e) Therefore, for any (a, b), f(a, b) is defined and is neither positive nor negative, so it must be 0. 7. -6.44. Justify the following steps to prove that if f is integrable on R₂ and g is a continuous function with 0 ≤ g ≤ f then g is integrable on R₂. (a) ∫D(n)gdA exsits (b) 0 ≤ ∫D(n)gdA ≤ ∫D(n)fdA (c) The numbers ∫D(n)gdA are an increasing sequence bounded above. (d) limn → ∞∫D(n)gdA exsits Answer: Check D : D = R² unbounded, g 0, continuous, so we need to prove limn → ∞∫D(n)gdA exsits. (a) g ≥ 0 is continuous on R₂ and D(n) is bounded for each n so g is integrable over D(n) (b) By theorem 6.9 Larea(D)≤I(f, D) and the fact 0g, we know that 0area(D)≤∫D(n)gdA so if 0 ≤ f(x, y)−g(x, y) then $$ 0=0 area(D)\leq f(x,y)-g(x,y) dA \leq f dA - g dA $$ Therefore, 0 ≤ ∫D(n)gdA ≤ ∫D(n)fdA (c) Let Cn = ∫D(n)gdA. Because g ≥ 0, D(n)≤D(n + 1). Then C₁, C₂, C₃...Cn is an increasing sequence. Since 0 ≤ ∫D(n)gdA ≤ ∫D(n)fdA and $$ f dA = f dA$$ exists, We got $$ g dA \leq f dA$$ (e) By the Monotone Convergence Theorem for sequences, ∫D(n)gdA increasing and bounded above is convergent so limn → ∞∫D(n)gdA = limn → ∞Cn exists 8. 6.50. Justify steps (a)–(d) to prove that if a continuous function f is integrable on an unbounded set D then |∫DfdA| ≤ ∫D|f|dA (a)∫DfdA = ∫Df+dA − ∫Df−dA ≤ ∫Df+dA + ∫Df−dA = ∫D|f|dA (b)∫D(−f)dA ≤ ∫D|f|dA (c)−∫DfdA ≤ ∫D|f|dA (d)|∫DfdA| ≤ ∫D|f|dA (a) By Definition 6.9, if f is continuous and integrable on an unbounded set D, then |f| is integrable on D. Rewrite f(x, y)=f+(x, y)−f−(x, y) where f+(x, y)=f(x, y) if f(x, y)≥0 and 0 otherwise, and f−(x, y)= − f(x, y)if f(x, y)≤0 and 0 otherwise. So, by the definition of ∫DfdA, $$\int_ {D} f dA=\int_ {D} f_+ dA - \int_ {D} f_- dA $$ Since ∫Df−dA is nonnegtive $$\int_ {D} f_+ dA - \int_ {D} f_- dA \leq \int_ {D} f_+ dA + \int_ {D} f_- dA$$ Since f+ ≥ 0 and f− ≥ 0 are integrable over D $$\int_ {D(n)} f_+ dA + \int_ {D(n)} f_- dA =\int_ {D(n)} (f_+ + f_-) dA$$ By the properties of limits of increasing sequence D(n), we know ∫D(n)(f+ + f−)dA converges so $$\int_ {D} f_+ dA + \int_ {D} f_- dA =\int_ {D} (f_+ + f_-) dA$$ By the equation f(x, y)=f+(x, y)−f−(x, y), we got $$\int_ {D} f dA \leq \int_ {D} \left|f \right|dA$$ (b) In the same way, we apply (a) to the functions − f to get $$\int_ {D} -f dA \leq \int_ {D} \left|-f \right|dA= \int_ {D} \left|f \right|dA$$ (c)By the properties of limits and the equation ∫D(n) − fdA=_ D(n) f dA ,weget$$- \int_ {D} f dA \leq \int_ {D} \left|f \right|dA$$(d)Ifbaand −b ≤ a then|b|≤a. From (a), we got $$\int_ {D} f dA \leq \int_ {D} \left|f \right|dA$$, From (b) and (c), we got $$- \int_ {D} f dA \leq \int_ {D} \left|f \right|dA$$ Therefore, we can conclude that $$\left| \int_ {D} f dA\right| \leq \int_ {D} \left|f \right|dA$$ 9. -4.21. Find the point on the plane $$z = x − 2y + 3$$ that is closest to the origin, by finding where the square of the distance between (0, 0) and a point (x, y) of the plane is at a minimum. Use the matrix of second partial derivatives to show that the point is a local minimum. Let $$ D=d^2 = f(x,y)= x^2+y^2+(x-2y+3)^2 $$, to find the local extrema we let $$\triangledown f = (4x−4y+6,−4x+10y−12)=0$$ at ( − 0.5, 1). so $$ H(-0.5,1)= \left[ { c c } 4 & -4 \\ -4 & 10 \right] $$ Because 4 > 0 and (4)(10) − (−4)2 = 24 > 0. So by the Theorem 4.3, it is positive definite. By theorem 4.8, If ▿f(A)=0 and the Hessian matrix [fxixj(A]) is positive definite at A, then f(A) is a local minimum. Therefore, f has a local minimum at point ( − 0.5, 1) 10. -7.32. Let S be the unit sphere centered at the origin in R³. Evaluate the following items, using as little calculation as possible (a)∫S1dσ (b)∫S||X||²dσ (c) Verify that ∫Sx₁²dσ = ∫Sx₂²dσ = ∫Sx₃²dσ using either a symmetric argument or parametrizations. Can you do this without evaluating them? (d) Use the result of parts (b) and (c) to deduce the value of ∫Sx₁²dσ Answer: (a) In geometry, ∫S1dσ means the area of the unit sphere in R³ So ∫S1dσ = π · 1³ = 4π (b) For all X S we have ||X||² = 1, therefore ∫S||X||²dσ = ∫S1dσ = 4π (c) Rotation by /2 about the x₃-axis corresponds to some transformation on the domain of the parametrization of S. We know that x₁ comes to the same position as x₂, Therefore ∫Sx₁²dσ = ∫Sx₂²dσ In the same way, make a rotation by π/2 about the x₂ , we got ∫Sx₁²dσ = ∫Sx₃²dσ Therefore, ∫Sx₁²dσ = ∫Sx₂²dσ = ∫Sx₃²dσ (d) By the definition of norm ||X||, we know that ||X|| = x₁² + x₂² + x₃² So, $$ ||X||^2 d\sigma= x_1^2 +x_2^2 +x_3^2 d\sigma= 3 x_1^2 d\sigma = 4\pi$$ Therefore,$$ x_1^2 d\sigma ={3} ||X||^2 d\sigma = {3}$$

ABSTRACT Biological soil crusts (BSC) are key components of ecosystem productivity in arid lands and they cover a substantial fraction of the terrestrial surface. In particular, BSC N₂-fixation contributes significantly to the nitrogen (N) budget of arid land ecosystems. In mature crusts, N₂-fixation is largely attributed to heterocystous cyanobacteria, however, early successional crusts possess few N₂-fixing cyanobacteria and this suggests that microorganisms other than cyanobacteria mediate N₂-fixation during the critical early stages of BSC development. DNA stable isotope probing (DNA-SIP) with ¹⁵N₂ revealed that _Clostridiaceae_ and _Proteobacteria_ are the most common microorganisms that assimilate ¹⁵N₂ in early successional crusts. The _Clostridiaceae_ identified are divergent from previously characterized isolates, though N₂-fixation has previously been observed in this family. The Proteobacteria identified share >98.5 %SSU rRNA gene sequence identity with isolates from genera known to possess diazotrophs (e.g. _Pseudomonas_, _Klebsiella_, _Shigella_, and _Ideonella_). The low abundance of these heterotrophic diazotrophs in BSC may explain why they have not been characterized previously. Diazotrophs play a critical role in BSC formation and characterization of these organisms represents a crucial step towards understanding how anthropogenic change will affect the formation and ecological function of BSC in arid ecosystems. KEYWORDS: microbial ecology / stable isotope probing / nitrogen fixation / biological soil crusts

INTRODUCTORY PARAGRAPH We explored the microbial contributions to decomposition using a sophisticated approach to DNA Stable Isotope Probing (SIP). Our experiment evaluated the dynamics and ecological characteristics of functionally defined microbial groups that metabolize labile and structural C in soils. We added to soil a complex amendment representing plant derived organic matter substituted with either ¹³C-xylose or ¹³C-cellulose to represent labile and structural C pools derived from abundant components of plant biomass. We found evidence for ¹³C-incorporation into DNA from ¹³C-xylose and ¹³C-cellulose in 49 and 63 operational taxonomic units (OTUs), respectively. The types of microorganisms that assimilated ¹³C in the ¹³C-xylose treatment changed over time being predominantly _Firmicutes_ at day 1 followed by _Bacteroidetes_ at day 3 and then _Actinobacteria_ at day 7. These ¹³C-labeling dynamics suggest labile C traveled through different trophic levels. In contrast, microorganisms generally metabolized cellulose-C after 14 days and did not change to the same extent in phylogenetic composition over time. Microorganisms that metabolized cellulose-C belonged to poorly characterized but cosmopolitan soil lineages including _Verrucomicrobia_, _Chloroflexi_ and _Planctomycetes_.

MODEL Transistor The transistor is modeled generically by a heavily simplified virtual-source (short-channel) MOSFET model . Although this model was first defined for Silicon transistors, it has been successfully adapted to numerous other contexts, including Graphene and Gallium Nitride devices, both HEMTs and MOSHEMT+VO₂ HyperFETs . Following Khakifirooz , the drain current ID is expressed {W}=Q_{ix_0}v_{x_0}F_s where Qiz₀ is the charge at the virtual source point, vx₀ is the virtual source saturation velocity, and Fs is an empirically fitted “saturation function” which smoothly transitions between linear (Fs ∝ VDS/VDSSAT) and saturation (Fs ≈ 1) regimes. The charge in the channel is described via the following semi-empirical form first proposed for CMOS-VLSI modeling and employed frequently since (often with modifications, eg ): Q_{ix_0}=C_nV_\ln\left[1+\exp\left\{-V_T}{nV_}\right\}\right] where Cinv is an effective inversion capacitance for the gate, nVthln10 is the subthreshold swing of the transistor, VGSi is the transistor gate-to-source voltage, VT is the threshold voltage, and Vth is the thermal voltage kT/q. For precise modeling, Khakifirooz includes further adjustments of VT due to the drain voltage (DIBL parameter) and the gate voltage (strong vs weak inversion shift), as well as a functional form of Fs. For a first-pass, we will ignore these effects, employ a constant VT, and assume the supply voltage is maintained above the gate overdrive such that Fs ≈ 1. However, we will add on a leakage floor with conductance Gleak. Altogether, the final current expression (for the analytical part of this analysis) is {W}=nv_{x_0}C_V_{th}\ln\left[1+\exp\left\{-V_}{nV_{th}}\right\}\right]+}{W}V_

This document describes the design for the VLA Sky Survey (VLASS) Pilot Survey which will be carried out in the 27 May – 5 Sep 2016 time-frame.

Loaded dice, and sticky spheres (a) For discrete system, the entropy S_{discrete} = -k_B \sum p_i\log p_i S_{discrete} = {2}k_B \log 2 (b) Entropy for being hit by a comet with $\rho_c(\theta,\phi) = {4\pi}$: S_c = -k_B \iint \rho_s \log (\rho_s) d\Omega = k_B \log (4\pi) Entropy for being hit by an asteroid with $\rho_a(\theta,\phi) = {\pi^2}$: S_a = -k_B \iint \rho_a \log (\rho_a) d\Omega = -k_B \int \left( {\pi}\log(\cos\theta) - {\pi}\log\pi \right)d\theta S_a = -k_B \left({2}(1-\log 4) - 2\log(\pi) \right) = k_B \log \left({} \right) The entropy difference \Delta S = S_c - S_a = k_B\log\left(}{\pi} \right)

Entropy of socks For a child’s room containing 100 things in the room, we only need to consider the configuration space since those things are stationary in the room. The room before tightening up, the number of configurations is \Omega = V^N = 5^{100} Let’s consider the center-of-mass location accuracy of 1 cm. First of all, the center-of-mass can be located itself at any point of the room. Therefore, = 5 Then we can assume that the center of mass in x direction is at 0. For each randomly located thing A1 at x, there would be the another thing A1′ located at the −x. We now have only 50 things we can throw at will. For the A50′, since the deviation of 1cm is acceptable, A50′ can have roughly 1m of freedom in the x direction. Therefore, = (L^{50}*1)^3 = 5^{50} where L = 51/3. \Omega' = \times = 5^{51} Entropy decrease can be found to be \Delta S = k_B - k_B = -49k_B \Delta S_{total} = 10^9\times\Delta S = 1.09\times10^{-12} For 1 liter baloon contracting by 1% in volume, \Delta S = Nk_B - Nk_B = {T}\log(0.99) where P = 1atm, V = 1L and T = 300K. \Delta S = -0.0034 (J/K)

This project paper mainly focuses on the physical models and the current characteristics of charge density waves (CDW) in the niobium triselenide (NbSe₃) with anisotropic band structures. The models discussed in this paper are classical single particle transport model (a.k.a. sliding model) and quantum tunneling model. The classical model gives a threshold electric field beyond which the CDW material is a conductor. Finally, CDW effects could be modulated by applying gate voltage, resulting in field-effect CDW transistors.

Hello Thesis 1 -- (Anezka's Section) This is a long "to do" list...Read them all first and pick your favorites. These should be enough to keep you busy all summer! Make sure you give yourself and your ideas "simmer" or relaxation time! That is when your ideas have a chance to cook and mature.

La inminente creación de un Ministerio de Ciencia y Tecnología pretende potenciar el rol público en la investigación y desarrollo de conocimiento. Un desafío clave para la comunidad científica en este escenario es justificar de cara a la ciudadanía la cantidad de energía y recursos públicos que se le brindarán. Disciplinas como _Education and Public Outreach_ ponen foco en las acciones que la comunidad científica debe desempeñar. Este documento propone una estrategia municipal complementaria, anclada en la disciplina _Citizen Science_, para facilitar las acciones los ciudadanos y ciudadanas interesados en influir en la creación de conocimiento, cuyos impuestos estan siendo utilizados. Se presenta una estrategia que aprovecha el ámbito municipal y su territorialidad para reflejar las necesidades, anhelos y requerimientos de las vecinas y vecinos con respecto a la investigación y desarrollo científico vinculada a su territorio. Además, se propone un sistema que refleja el caracter emergente de estos intereses, necesidades y anhelos, y al mismo tiempo brinda la oportunidad de vincular a los investigadores a través de sus proyectos, ya sea en etapa de propuesta o existentes, que satisfacen esos intereses, necesidades y anhelos, y así facilitar un vínculo de las vecinas y vecinos con proyectos que se hacen cargo de sus inquietudes.

The purpose of this homework is to explore the Monte Carlo Algorithm and apply it to the simplified protein folding model in 2D. MONTE CARLO METHOD Monte Carlo Method uses the randomly generated possible solutions to a certain problem in a solution space and test its degree of goodness based on certain physical requirements. The ways of generating the possible solutions are usually two. First, we can generate the possible solutions totally at random. For example, we use random number generator to do Monte Carlo Integration. Second, we can generate the possible solutions from the previous step by randomly changing some parameters of the previous one. We will use the later one to generate our 2D protein structures in this homework. The general flow of Monte Carlo Method is shown as follows. Note that we use the term “conformation space” instead of “solution space” since we are talking about protein structures here. 1. Start from a initial state in the conformation space. 2. Randomly change the previous state, subjecting to requirement 1. 3. Determine the degree of goodness by criterion 2. 4. Accept/ reject this state by physical rule 3. 5. If it is accepted, pass this state and repeat 2 through 5 for certain number of steps. 6. If it is rejected, do not pass the state and repeat 2 through 4 until the new state is accepted. The requirement 1, criterion 2 and rule 3 are problem-specific and we will mention these in our protein folding problem. 2D PROTEIN FOLDING Proteins are composed of 20 different amino acids (AAs) in a polypeptide chain and due to the mutual interactions between those AAs, proteins will favor some folded states to lower the Gibbs free energy. The interactions are mostly negative because of hydrophobic effects or ion-ion interactions. In order for proteins to perform certain biological functions, their unique structures are essential. We can use a simple bead-and-chain model for a 2D protein chain, assuming that all the AAs are of the same size and the peptide bond between two AAs is rigid, being only one unit and unstretchable. Each AA occupies one grid point of the 2D space and cannot be in the same point of any other AAs. When protein folds, the non-covalent interactions apply to the two non-bonding AAs separate by one unit. An we can calculate the relative Gibbs free energy ΔG by summing all the interactions of non-bonding neighbors. \Delta G = E_0 = E_{t(i)t(j)} where (i, j) are the indices of two neighboring AAs of types (t(i),t(j)) and E is an 20 × 20 interaction matrix. With this model in mind, we can determine the requirements mentioned in the previous section. 1. Requirement 1: 1. The modified AA cannot occupy other’s positions. 2. The modified AA must be one unit away from its neighbor(s). 3. The best way to modify an AA’s position is to move (1, 0), (1, 1), (0, 1), ( − 1, 1), ( − 1, 0), ( − 1, −1), (0, −1) and (1, −1), eight possible changes. 2. Criterion 2: Evaluate the interaction energy, E₀ and use this number to determine the goodness of the state. The lower, the better. 3. Rule 3: 1. If the new state has lower E₀, the protein will adopt this state in order to reach the minimum of the folding landscape. 2. If the new state has higher E₀, the protein does not favor such state. However, there is still some probability to jump from lower energy state to higher energy ones, P = e−(Enew − E₀)/kT. Once the model is set and the steps are clear, we can start to do the simulation. PROGRAM CODES First, we need a general random number generator, _myrand(seed)_. We will test its validity and then apply it to alter the position of a randomly selected AA. Given different _seed_, the function will give different random number sequences. Our seeds for generating interaction matrix E and the AA sequence in the protein are two distinct yet fixed value. So we will guarantee that we use exactly the same protein and interactions throughout the calculation. Other than those, the seed will be set by _time(NULL)_ and independent of our bias. Second, there are several subroutines to do the Monte Carlo calculations and to make sure the protein is subject to some requirements. Note that the information of proteins is stored in a 45 × 3 matrix with the first column being AA types, second the x positions and third the y positions. 1. _neighbor()_: inputs a Protein Vector and outputs the pairs of indices of two non-bonding AAs. 2. _Energy()_: input pairs of neighbor indices, Protein Vector, interaction matrix E and outputs the energy E₀. 3. _n2ndistance()_: inputs a Protein Vector and outputs the end-to-end distance of the protein. 4. _pcheck()_: inputs Protein Vector and the index of certain AA and check if that AA occupies others’ positions. The function outputs _true_ if the protein is not allowed, _false_ otherwise. 5. _conformationchange()_: input Protein Vector and make a position change to one of its AA and outputs a modified new Protein Vector. RESULTS First we tested the random number generator _myrand()_. The random number generator gives a uniform distribution of numbers between 0 and 1. And the points (xn + 1, xn) cover the 1 × 1 square without noticeable patterns as shown in Fig. 1. We further test it by estimating π. {4} = }{N} where Nin is the number of points in the quarter circle and N is the total number of points.As the total number of points increases, the RHS will reach ${4}$ asympotically, as shown in Fig. 2.

4TH ORDER RUNGE KUTTA ODE SOLVER Suppose we have the a set of N ODEs with the following form. }{dt} = (t, ) where $$ and $$ are N-dimensional vectors of functions. The 4th order Runge Kutta takes the following algorithm each step h forward and by calculating a large number of steps, we can have the numerical solution to the set of ODEs. } = (t_n) = h(t_n,)\\ = h(t_n+0.5h,+0.5)\\ = h(t_n+0.5h,+0.5)\\ = h(t_n+h,+) The value for $}$ one step away can be calculated. } = }+{6}\left( +2+2+ \right) Finally, by starting from $}$, the solution can be obtained.

FINITE DIFFERENCE ALGORITHM We have used Runge-Kutta method to solve ODEs with initial conditions; however, many of the physical systems cannot be described by ODEs. The partial differential equations with boundary conditions are inevitable for physical systems such as time-independent Schrodinger equations and electromagnetic fields. Those equations involve partial differential in spatial coordinates with specified boundaries and/or sources. When trying to solve such system numerically, the finite difference algorithm is a straightforward and easy-to-program method. We will use the 2D Laplace equation as an example to demonstrate the algorithm. \nabla^2\phi(x,y) = 0 The finite difference algorithm requires to mesh-grid the space of interest and the value on each grid point (i, j) represents the ϕ(x = ih₁, y = jh₂)=ϕij where h₁ and h₂ are the spacing in x and y axes. It’s always required to set up another “flag” grid which determines whether certain combination of (i, j) is on the boundary or source (electrodes). Next, based on the finite difference of adjacent values to ϕij and spacing, the second partial differential can be approximated. We can solve for ϕij in terms of its neighbors for the first iteration. ^{FD} = {4}\left( +++ \right) The flag grid is like a mask, leaving all the specified boundary values unchanged during iterations. Similar to Runge-Kutta method, we evaluate the maximum error in each iteration. \Delta = max | ^{FD}- | If the maximum error is larger than the specified tolerance Δϕij > ϵmax, we should pass the new values of ϕij to Eq. (2) and repeat the calculation until the error is below the tolerance or the number of iterations exceeds certain upper bound. The new ϕijnew can be the following form, which is the weighed average of ϕij and ϕijFD. ^{new} = (1-\omega) + \omega^{FD} where 0 < ω < 2. If 1 < ω < 2, the algorithm converges more quickly and this is known as the successive over-relaxation algorithm. The regime where 0 < ω < 1 corresponds to the under-relaxation where the solution converges relatively slowly. Here, we will solve the Laplace equation in a cylinder with three disks of inner and outer radius. Instead of utilizing Eq. (2), we will use the finite difference in cylindrical coordinates and assume azimuthal symmetry. ^{FD} = {4}\left( +++ \right)+{8j}\left( - \right) ,where z = ih and r = jh. The above equation applies to points where r ≠ 0. On the central axis where r = 0, we should have ^{FD} = {6}\left( 4++\right) In order to use finite difference algorithm, we need to set up matrices in c++. We used _arrayt.h_ to generate and do the calculations for matrices. OVER-RELAXATION First, we investigated the powerful role of ω here. We made the grid spacing h = 0.1 and error tolerance ϵmax = 2 and calculated the number of iterations for the error to go below the tolerance. The results are shown in the following table. ω 1.0 1.2 1.4 1.6 1.8 1.9 ---------------------- ----- ----- ----- ----- ----- ------ Number of iterations 655 511 431 371 359 1077 The best choice of ω seems to be around ω = 1.8 and we would use this over-relaxation value for the following calculation. SOLUTIONS For solving the electric potential inside the cylinder with three disk electrodes and boundaries, the grid spacing is the same h = 0.1 while the error tolerance is much smaller, ϵmax = 0.01. This one percent error would increase the number of iterations significantly to N = 4231. The contour plot of the numerical solution in space is shown in Fig. 1, where the red line represents the middle disk electrode held at ψ = 2000V and the black line and rectangle are the two grounded electrodes ψ = 0V. The electric potential contour is distorted by the two grounded electrodes, compressing the potential lines in the z direction. The potential at about r = 10mm is symmetric while that closer to the central axis is slightly asymmetric due to the mismatch in inner radii of the two grounded electrodes. The electric potential is mainly restricted among the disks and near the origin. Outside of those disks, the potential is negligibly zero everywhere. The configuration is seen in the electron guns of the electron microscopes to focus electrons by adjusting the fixed voltage of the center electrode.

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