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_

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.

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.

ADAPTIVE STEPSIZE RUNGE-KUTTA ALGORITHM We implemented the 4th order Runge-Kutta method to solve chaotic ODE system and it turned out to be a power solver. However, one of its drawback is that the step size is fixed. When the solution happens to be smooth and change slightly, the number of steps could be reduced for efficiency. On the other hand, if the solution is changing sharply, fixed step size could fail to account for the abrupt variations and the errors might accumulate over time. As a result, for the purpose of higher efficiency and accuracy, adaptive step size Runge-Kutta method would be a better ODE solver and could handle more equations at the same time. As mentioned by the author in the Numerical Recipes for c++, “many small steps should tiptoe through treacherous terrain while a few great strides through smooth uninteresting countryside.” The step size is controlled by the truncation errors between the fifth order and the fourth order values. If the error exceeds the tolerance, we should reduce the step size until the error is below the tolerance. On the other hand, if the error is small, it will be more efficient to increase the step size so that the error is about the same order of magnitude as the tolerance. The general form of the adaptive step size Runge-Kutta method is as follows. The tn and yn are the time and the vector containing N ODE values at the nth step. k_1 = hf(t_n, y_n)\\ k2 = hf(x_n+c_2h, y_n+a_{21}k_1)\\ ...\\ k6 = hf(t_n+c_6h, y_n+a_{61}h+...+a_{65}k5)\\ y_{n+1} = y_n+^{6} b_ik_i \\ y^*_{n+1} = y_n+^{6} b^*_ik_i where a’s, b′s, b*’s and c’s are the various constants found by Dormand and Prince and all the values are tabulated in Chapter 17.2 of Numerical Recipes. The yn + 1* is alculated to the fifth order so the error term can be written as \Delta = \left| y_{n+1} - y^*_{n+1} \right| Note that here the || is the absolute value of each element rather than the norm of the N-element vector. By picking the maximum value in the Δ vector, _err_, we can compare _err_ to our tolerance _err0_. First of all, if _err_ is larger than _err0_, we have to scale down the step size and recalculate again until the err < err0. Secondly, for scaling the step size, whichever the case is, we can always apply the same scaling factor, scale = \left( {err} \right)^{1/4} \times S The S here is the safe factor with the requirement |S|<1. We chose S = 0.8 in the program. The ${4}$ power was used instead of ${5}$ since it scales up/down the step size more efficiently and the function _sqrt_ in c++ is easier and more accurate. Thirdly, we cannot scale the step size abruptly, like from h₀ to 0.01h₀ or 50h₀. In such case, the adaptive step size Runge-Kutta method would have ignored some features in the 50h₀ step or become inefficient using 0.01h₀ step size. We have to put extreme scaling values _maxscale_ and _minscale_. The re-scaled step size would be passed to the next run and so on. Finally, in order to avoid infinite _while_ loops, we added a _counter_ to keep track of the number of loops. If _counter_ exceeds certain number (in the program 100), we will break the while loop and print the message that the err was never reduced below err0 at this step. However, this didn’t happen even though we’ve tried tens of cases with billions of steps. Usually, it only took less than 5 loops to reduce the error within our tolerance. Let’s talk about the specified error tolerance err0 and the number of steps to achieve certain final t value. The initial step size doesn’t make a difference because the step size will get scaled up/down depending upon the solution and system. Given the same number of steps, the final t value reached depends mainly on err0. The smaller the tolerance is, the smaller the step size would be for each step. Therefore, the final t would end up far below the expected value. We can either increase the number of steps or the error tolerance which is a more efficient way than the former one. In the following testing and computations, this issue took place most of the time and it’s necessary to choose a good and aacceptable err0 and number of steps to solve the ODE system. We found that err0 = 10−5 of the initial value would work well for most of the cases.

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.

PLOTTING OF BESSEL FUNCTIONS The Bessel Functions of the first kind Jν(x) and the second kind Yν(x) are two functions of interest in this program. The program incorporates two header files “NR3.H” and “BESSEL.H” and outputs the function values between x = 0 to x = 20. The first two Bessel functions J₀, J₁, Y₀ and Y₁ are built in within “BESSEL.H” and we can use the recurrence formula to construct higher order Bessel functions. J_{\nu+1}(x) = {x}J_\nu(x) - J_{\nu-1}(x) Y_{\nu+1}(x) = {x}Y_\nu(x) - Y_{\nu-1}(x) The first three Jν and Yν are plotted by MATLAB as shown in Fig. 1 and Fig. 2.

CALCULATION OF I/I₀ WITH VARYING n The analytical Fresnel Diffraction Theory gives the light intensity at x₀ on a viewing plane with distance z behind the edge the following form. I(x_0) = {\lambda z} \left| \int^{x_0}_{-\infty} \exp \left( {\lambda z} \right) dx \right| ^2 where λ and I₀ are the light wavelength and intensity respectively. It can be transformed into the following form by substituting $u_0 = x_0 {\lambda z}}$: {I_0} = {2}\left[ (C(u_0)-C(-\infty))^2 + (S(u_0)-S(-\infty))^2 \right] where C(u_0) = \int^{u_0}_0 \cos \left( {2} \right) du\\ S(u_0) = \int^{u_0}_0 \sin \left( {2} \right) du\\ As a result, instead of calculating the complex integral in Eq. (1), we simplify the task by evaluating C(u₀) and S(u₀). First, I’ll start with calculating both C(u₀) and S(u₀) by the trapezoid rule. The inputs of the program are the number of iterations n and the integral upper limit u₀. For mth iteration, the interval will be cut into 2m equally spaced trapezoids and the intensity ratio in Eq. (2) as well as the errors in C(u₀) and S(u₀) will be printed out. The output is shown in the following tables. As n increases, the errors in both functions are reduced dramatically, giving the convergence of the intensity ratio, shown as bold face numbers.

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.

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.

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,$$ x_1^2 d\sigma ={3} ||X||^2 d\sigma = {3}$$

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.

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.

Welcome to Authorea!

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)

Mar Cortesa,b, Alejandra Climenta, Laura Dubreuil Vallc, Giulio Ruffinic  Douglas Labara, Dylan Edwardsa a. Non invasive Brain Stimulation and Human Motor Control Laboratory, Burke Medical Research Institute, Weill Medical College of Cornell University, 785 Mamaroneck Avenue, 10605, White Plains, NY, USA.b. Universitat de Barcelona, Barcelona, Spain. c. Neuroelectrics Corporation, Cambridge (MA), USA. Corresponding author:Mar CortesNon invasive Brain Stimulation and Human Motor Control Laboratory, Burke Medical Research Institute, Weill Medical College of Cornell University, 785 Mamaroneck Avenue, 10605, White Plains, NY, [email protected]: +1 914 368 3181 Abstract: (250)[MC1]  Background: Existing strategies to enhance motor function following Spinal Cord Injury (SCI) are suboptimal leaving patients with considerable disability. Available evidence suggests that transcranial direct current stimulation (tDCS) is a promising method to improve motor dysfunction. How tDCS affects resting brain activity monitored by EEG is little explored. Objective: Investigate the effects of anodal tDCS on brain signaling (EEG) and neurophysiology (TMS) when targeting forearm muscles below the level of the lesion in chronic SCI subjects. Methods: We conducted a randomized, single blind, sham-controlled, cross-over study in seven chronic SCI subjects with cervical lesions. We investigated the effects of 20-minute anodal tDCS applied over the left primary motor cortex (M1) on electroencephalography (EEG) power spectrum density, coherence and frequency band power. Subjects were randomized to receive either 1mA or sham stimulation.  The EEG data acquisition pre and post stimulation comprised 5-minute takes of 24 bit, 500 S/s 8-channel EEG using StarStim Ag/AgCl EEG electrodes (at F3, F4, Cz, C4, P3 and P4; and Pi Ag/AgCl electrodes at C3, anode, AF8, return). Results: In comparison to sham stimulation, 20-minutes of active 1mA tDCS induced a pattern of faster activity around the anodal stimulating electrode, and slowing activity near the return electrode in the frequency (full band) and mean power domain (gamma band). In addition, tDCS increased coherence in the fastest bands (gamma, beta 2) and decreased coherence in slower frequency bands (theta, SMR), with no relation with brain topography or the stimulation electrode polarity. Conclusions: These findings show that tDCS is capable of inducing modulation of ongoing oscillatory brain rhythms captured by EEG, in spinal cord injury patients. The combined use of EEG and tDCS sets the stage for optimizing tDCS protocols targeting motor cortex and may have application in treatment of motor dysfunction and chronic pain.  Key words: spinal cord injury, EEG, tDCS, motor cortex, TMS Abbreviations:    Introduction[JC2]  Transcranial direct current stimulation (TDCS) delivered over primary motor cortex (M1) can increase or decrease corticomotor excitability as determined by the amplitude of the motor evoked potential (MEP) from stimulating M1 with supra-threshold transcranial magnetic stimulation (TMS). The effect on MEP amplitude depends on the duration and intensity of stimulation and the polarity and spatial arrangement of the TDCS electrodes, and these effects can persist for up to an hour after TDCS is ceased. It is assumed that the TDCS after-effect results from some form(s) of neuromodulation at the cortical level such as synaptic potentiation or depression, or effects on ion channels. However, further insight into the cortical mechanisms in the human is limited experimentally. The electroencephalogram (EEG) measures far-field potentials from synchronized neural populations over cortical regions that extend beyond M1, and so offers a method for measuring the effect of TDCS on neural activity across the cortex.  In particular, such measurements do not depend on spinal cord conduction and alpha-motoneuron activation, which is the case for MEP recordings. This could have advantages for determining the effect of TDCS at the cortical level in spinal cord injury (SCI). While most SCI studies focus on the level of injury, there is increasing interest in the compensatory changes that might occur in the cortex, and whether they might facilitate or impede recovery. In the present study we measured the effect of EEG after 20 min of anodal or sham TDCS applied to the dominant M1 in chronic cervical SCI. We show that there are changes in EEG power, frequency and coherence that are spatially-related to the TDCS electrode configuration, but that spread widely across the hemispheres. [MC3]  Materials and methods Participants and study designThe randomized, single blind, sham controlled, cross over study was conducted on seven chronic SCI subjects, all males with an average age of 51.14 ± 10.57 years (mean ±SD, range 34 to 65 years). Participants presented traumatic SCI at the cervical level (C4-C8); some degree of motor function in wrist extension (score 1-5 over 5 on the Medical Research Council (MRC) scale for motor strength in the right extensor carpi radialis (ECR) muscle); chronic injury (>9 months after injury); and tolerance to sitting upright for at least one hour (see Table 1 for baseline characteristics). Patients were excluded if they presented with: progressive neurodegenerative disorder; concomitant traumatic brain injury or stroke; clinically significant cognitive impairment; medically unstable; change in medication during the study; or presented contraindications to brain stimulation (history of seizures/epilepsy, presence of metallic implants in the brain, pacemaker, pregnancy). The subjects were randomized in two groups depending on whether they were receiving stimulation or not:  a) Sham or control group, b) 1mA or active group. We investigated the effects of 20-minute anodal t-DCS applied over the primary motor cortex (M1) on: a) quantitative electroencephalography (EEG) power spectrum density, coherence and frequency band power. At Baseline we collected: clinical and functional evaluations were performed prior the brain stimulation intervention, and included the upper extremity motor score (UEMS), American Spinal Injury Association impairment scale (AIS), spinal cord independence measure (SCIM III) and visual analog scale (VAS) pain questionnaires. [DE4]  The study was approved by the Burke Medical Rehabilitation Institutional Review Board and conformed to the standards set out by the 1964 Declaration of Helsinki.                  AddTable 1 Transcranial direct current stimulation (t-DCS) interventionParticipants remained seated in their own wheelchair or were provided with a comfortable chair. The StarStimNE noninvasive wireless t-DCS/EEG neurostimulator (NE Neuroelectrics®, Barcelona, Spain) was used to both record EEG data and deliver the direct current sequentially. The StarStimNE neurostimulator included a wireless neoprene cap, based on the International 10-20 system, which was placed on the participants’ heads by aligning the central CZ electrode position with the vertex (intersection of nasion-inion and inter-aural line mid-point).Small Ag/AgCl gelled electrodes, with a surface contact area of 3.14 cm2 specific to the StarStimNE device (Pi electrodes, Neuroelectrics®), were placed over the left M1 at C3 (anode) and contralateral supraorbital area, (AF8; cathode) (Figure 1). The electrodes were connected to a control box device, which was wirelessly connected to a computer and communicated with the NIC software (version 1.2, Neuroelectrics®).  Add Figure 1 During anodal stimulation, direct current was delivered from a current-control circuit in the battery-driven stimulator within the control box device. The current was set at 1mA intensity and applied for 20 minutes. [DE5] For the sham stimulation, electrodes were placed in the same position and participants received a short ramp (30sec total up / down[DE6] ) at the beginning and end of the stimulation period.  Electroencephalography (EEG)The StarStimNE multichannel wireless device (NE Neuroelectrics®, Barcelona, Spain), which allows for simultaneous electroencephalography (EEG) and tDCS[DE7] , was used to record EEG data. The EEG data consisted of 5-minute takes (pre and post stimulation) with 24 bit resolution, 500 S/s 8-channel EEG collected with StarStimNE Ag/AgCl EEG electrodes (geltrodes, NE022; positions at F3, F4, Cz, C4, P3 and P4; Pi stimulating electrodes at C3 and AF8), based on the International 10-20 system, with the cap aligned to the central CZ electrode position (vertex).[DE8]  The electrodes were connected to a control box device, which was wirelessly connected to a computer and communicated with the NIC software (version 1.2, Neuroelectrics®). Transcranial magnetic stimulation (TMS) Electromyography (EMG): A bipolar surface EMG electrode (1 cm diameter, 2 cm inter-pole distance; Biometrics Ltd, UK) was placed over the right ECR muscle, with the forearm relaxed in a pronated position and supported by a cushion. The EMG activity was amplified and filtered on site (x1000 gain, band-pass filter 20-460 Hz; SX230-1000), digitized at 2 kHz (CED 1401, Cambridge Electronic Design, Cambridge, UK) and stored for offline analysis using Spike 2.6 software. Measurements were performed at rest. During the experiment, free running EMG was continuously monitored with visual feedback of EMG to ensure complete muscle relaxation.The right ECR muscle was selected for clinical relevance; where restoration of motor function in this muscle can help increase independence in quadriplegic subjects with activities of daily living.  Transcranial Magnetic Stimulation (TMS): A figure-of-eight coil (Model DB-80, Tonika Elektronik A/S, Farum, Denmark), connected to a MagPro X100 series (MagVenture A/S, Farum, Denmark) magnetic stimulator, was placed congruent with the head and the handle rotated 45° lateral from mid-sagittal so as to induce currents in the brain approximately perpendicular to the central sulcus. Resting motor threshold (RMT) was established at C3, and was defined as the minimum TMS intensity required to elicit a reliable MEP in the contralateral ECR amplitude of >50 µV in at least 50% of consecutive trials.[DE9]  EEG data processing/statistical analysisResting quantitative EEG and corticospinal excitability (transcranial magnetic stimulation; MEP) were recorded before (PRE) and at the end (POST) of each intervention. Quantitative EEG measures included: mean power, mean frequency and mean coherence. The three measures were normalized as the percentage change between PRE and POST stimulation, for each frequency band and electrode. The frequency bands were defined as: Theta= [4 8] Hz, Alpha-1=[8 10] Hz, Alpha-2=[10 12] Hz, SMR=[12 16] Hz, Beta-1=[16 25] Hz, Beta-2=[25 35] Hz, Gamma=[35 40] Hz.The mean power is defined as the average power in uV2 in a given frequency band. The mean frequency is defined as the average frequency in a given band, weighting each frequency value by their corresponding power at that frequency. The mean coherence of a given electrode is defined as the average coherence of this electrode with all the other ones (i.e., the similarity of that signal with all the other electrodes).--> Giulio to revise - DONEThe EEG data were referenced to the average of all 8 electrodes. An automated quality check of the data was then carried out using 8 sec epochs. Epochs have been rejected if they do not meet quality criteria (too high mean power at full band or line noise, or motion artifacts as detected by the built in accelerometer). PSD changes and coherence analysis was carried out for each quality passed epoch. Average PSD’s and coherence have then been computed for each subject.EEG data were discarded from analysis for channels presenting bad signal quality during the entire 5-minute recording. Those subjects that did not have more than 5 valid channels were completely discarded. After the discarding process, in order to have the same subjects in all the conditions, 7 subjects were analyzed for Sham and 1mA.Resting MEP amplitude (peak-to-peak; ECR muscle) was measured following single-pulse TMS (10/12) set at 120% [DE12] of the RMT over the C3[DE13] . Raw and normalized values were used for analysis[DE14] . Results are presented as mean ± standard deviation (SD), and standard error of the mean (SEM)[DE15] .After normalizing the EEG features as the percentage change between PRE and POST stimulation, statistical comparison between Sham and Active (1mA) stimulation has been done. The p value in the T tests has been calculated with a Wilcoxon one-tail T test with the assumption of paired samples. Changes are considered statistically significant when p =< 0.05. PLEASE ADD THE STUDY DATA ANALYSIS – GIULIO / LAURA --> Described above Results Effects of tDCS on EEGNormalized pre-post EEG power, frequency and coherence for those subjects who received 1mA tDCS compared to those who received sham stimulation shows: a) significant increase of the mean power in the 1mA group in the Gamma frequency band under C3 (p=0.0035), the anodal stimulating electrode; b) a significant increment in the active group of mean frequency around the anode (stimulating) electrode (C3, F3) (p=0.0047) with a decreased mean frequency in the Alpha band near the return electrode (P4) (p=0.035); and c) significantly increased mean coherence in the active group in the fastest frequency bands Beta2 under Cz and C4 (p=0.007), Gamma band under Cz electrode (p=0.0023), C3 electrode (p=0.0006) and C4 electrode (p=0.0006), and SMR band under C3 electrode (p=0.05). LAURA – ADD THE TYPE OF TEST USED FOR THE RESULTS AND THE P VALUES OBTAINED --> Described  above Effects of tDCS on Corticospinal excitabilityBaseline values for resting MEP amplitude were similar between interventions (average: 0.37±0.05 mV; mean±SEM). No significant changes were observed for 1 mA a-tDCS or sham. The stimulation intensities used to obtain the RMT were not significantly different between 1 mA (64±17% MSO) and sham (66±16% MSO). All participants presented MEP responses.   Discussion1. tdcs does modulate EEG in SCI population. 2. The effect of the TDCS in EEG is location specific and band specific. 3. The fastest bands are more affected. Augmented power and frequency of fast bands (Gamma and B2) under the anode electrode - there is more fast activity under the anodal electrode.  Why is this?3.1. Anodal stimulation leads to increased spontaneous neuronal firing, increase motor cortex excitability. So we would expect more fast activity and higher amplitudes under the anode electrode. 3.2. Brain activity changes (increase fast band) are related to motor learning – changes in Mu rhythm (preparation for movement, getting ready for movement). This is shown in the coherence increase of SMR band under the stimulating electrode.  MAR – ALL OF THE ABOVE 3.2.1. Is there are connection between C3 and P4 in resting state brain? Why is alpha decreased significantly in P4 (cortex parietal) in the contralateral hemisphere.  CONECTIONS BETWEEN MOTOR CORTEX AND SENSORY MOTOR CORTEX?? – JANA 3.3. Anodal tDCS makes C3 to be more related to the rest of the head, by increasing the connectivity of the SMR under the stimulating electrode. INTERPRETANTION OF THIS POINT TO GIULIO/LAURA, ANYTHING PUBLISHED BEFORE?Here you have a reference using EEG coherence: Rafael Polanıia, Michael A. Nitsche, and Walter Paulus. "Modulating Functional Connectivity Patterns and Topological Functional Organization of the Human Brain with Transcranial Direct Current Stimulation". Human Brain Mapping (2010). 3.4. This is an excitability independent change (no TMS change associated). Can EEG pick up changes in the brain than TMS cannot?                                                       LABAR? Conclusion These findings show that t-DCS is capable of inducing modulation of ongoing oscillatory brain rhythms captured by EEG, in spinal cord injury patients. The combined use of EEG and t-DCS sets the stage for optimizing t-DCS protocols targeting motor cortex and may have application in treatment of motor dysfunction and chronic painFigures 

Undistinguished particles and the Gibbs factor (a) \Omega = {N!} S = k_B \space \log(\Omega) = k_B(N\log(V) - B\log(N) + N) = Nk_B\left(1+\log\left({N}\right) \right) The average volume of a region for single particle is v = V/N with the entropy of s = kBlog(v). Therefore, the configuration entropy is related to N times the entropy of a particle in its region with average volume. (b) S = Nk_B\left({2}-\log(\rho\lambda^3)\right) = Nk_B\left({2}+\log\left({N}\lambda^3\right)\right) The configuration entropy for an ideal gas particle is also related to the small average volume v. S_c = Nk_B\log(v) But the ideal gas particles also occupy the momentum space and each particle has the average length of λ in each dimension. So S_p = 3Nk_B\log(\lambda) = Nk_B\log(\lambda^3) Combine the configuration and momentum entropy for one particle, S = S_c+S_p = Nk_B\log(v*\lambda^3) = Nk_B\log\left({N}*\lambda^3\right) The total entropy is related to the N times the space occupied both in configuration and momentum space.

Ergodicity The system of classical particles bouncing back and forth is not ergodic since each particles can only occupy certain close volume in phase space, with fixed x, y position, Px, Py momenta and ±pz. Therefore, the given volume in phase space of the ensemble will not evolve with time. Heat engine (a) True. (b) True, since the work in done on the gas. (c) False, it’s a refrigerator. (d) True. W = (P_0)(3V_0) + ^{V_0} {V} dV = 3P_0V_0 - 4P_0V_0\log4 (e) True, a part does no work so ΔU = Q. \Delta U = {2} nRT_c - {2} nRT_h = -{2} (3P_0V_0) = -{2}P_0V_0 where n is the number of particles in mole and R = 8.317 is the ideal gas constant. (f) True. The whole cycle ΔU = 0, the heat input and the net work done _onto_ the gas must equal the heat output. Q_h + W = Q_c