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# Tag Info

1

It seems that $A$ dictates the evolution of the system over time. My best guess is that the relevant piece of information here is $A + A^T = 0 \implies$ $A$ has imaginary eigenvalues $\implies$ energy is conserved On the other hand, $A + A^T \preceq 0 \implies$ $A$ has eigenvalues with a non-positive real part $\implies$ energy decays with a steady state ...

0

You are on the right track, but the integrals involving $v(x)$ and $v_x(x)$ have to be expressed explicitly to go further :

2

No for $p=1$, yes for $1<p<\infty$. If $\phi\in \Delta C^\infty_c$ then $\int\phi=0$; this shows that $\Delta C^\infty_c$ is not dense in $L^1$. One might think at first that this shows the same thing for other $p$, but it doesn't, because the integral is not a bounded linear functional. Suppose from now on that $1<p<\infty$. Suppose that $K\... 1 The trace theorem you mentioned is an "if and only if" characterization, hence the question you asked is equivalent to: is it true that$f \in W^{1-\frac 1p,p}(D')$extends to a function$\bar{f} \in W^{1-\frac 1p,p}(D)$(here$D' \subseteq D = \partial \Omega$). As in the case of extension theorems for standard Sobolev spaces, the answer depends on$D$. 1 Assume that$v_1^2+v_2^2\neq 0$. Write$u(x,y)=U(z,w)$with $$z:=v_1x+v_2y\text{ and }w:=v_2x-v_1y\,.$$ Hence, $$x=\frac{v_1z+v_2w}{v_1^2+v_2^2}\text{ and }y=\frac{v_2z-v_1w}{v_1^2+v_2^2}\,.$$ Observe that$z$and$w$are independent (by checking that$\frac{\partial z}{\partial w}=0$and$\frac{\partial w}{\partial z}=0$). Then, you can see that $$\frac{\... 2 To be clear, you want to find P_n(t) for a birth-death process with constant birth rate q(i,i+1)=\lambda and constant death rate q(i,i-1)=\mu? This particular birth and death process is exactly the M/M/1 queue. As you can see, under "Transient solution", there is a solution for the probability mass function dependent on time for a particular state. 2 Since \|f - f_n \|_p \to 0, we can extract a subsequence f_{n_j} so that \| f - f_{n_j} \|_p \le \frac{1}{2^j}. Put$$g = \lvert f \rvert + \sum^\infty_{j=1} \lvert f - f_{n_j} \rvert.$$Then g \in L^p and by the triangle inequality we have$$\lvert f_{n_j} \rvert \le g \,\,\, (\text{almost everywhere}).$$From this subsequence, you can extract a ... 1 Differentiate both sides of the equation f(\lambda x) = \lambda^{a}f(x) in \lambda to obtain: $$\sum_{i = 1}^{n}x_{i}(\partial_{i}f)(\lambda x) = a\lambda^{a-1}f(x)$$ Setting \lambda = 1 yields \sum_{i = 1}^{n}x_{i}(\partial_{i}f)(x) = af(x), known as Euler's identity. Differentiating the equation displayed above in \... 1 Use the Co-area formula with g=|\nabla u|^{-1} \chi_{\Omega_t} to get$$ \int_{\Omega_t} dA=\int_t^\infty \left(\int_{\partial \Omega_x} \frac{ds}{|\nabla u|}\right)dx.$$Differentiating this with respect to t gives the formula you're looking for. 1 You can find the general statement and proof in Chapter 6 of the book "Sobolev Spaces" by Robert A. Adams and John. J. F. Fournier. 0 The claim$$\int_{\partial B_r(x)} \frac \partial {\partial r}u(y)ds_y=\frac \partial {\partial r}\int_{\partial B_r(x)} u(y)ds_y$$is not true since the domain of integration depends on r. You also later claim that |\partial B_r(x)| is constant, when it is in fact proportional to r^{n-1}. 0$$(1-2f_{x}^{2})f_{t}f_{xx} + 2f_{x}f_{tx}(1+f_{x}^{2})=0$$Try Separation of variables:$$f_{t}f_{xx} - 2f_{t}f_{xx}f_{x}^{2} + 2f_{x}f_{tx} + 2 f_{tx} f_{x}^{3}=0$$write f = X(x)T(t) Then you have:$$X\dot{T}\ddot{X} T - 2X\dot{T}\ddot{X} \dot{X}^2 T^3 + 2\dot{T} \dot{X}^2 T + 2\dot{T} \dot{X}^4 T^3=0(X\ddot{X} + 2 \dot{X}^2)\dot{T}T = (2X\ddot{X}... 1 Let$L$be a$k^{th}$-order linear differential operator, i.e. one which satisfies$L(\alpha u + \beta v) = \alpha L u + \beta Lv$for all$u,v\in C^k$and constants$\alpha,\beta$(of course this notion can be weakened past$C^k$, but this will do for here). We say that the equation$L u =f$, for some given function$f, is linear. A semilinear equation ... 1 If we set $$x(\xi, t) = f(\xi + \beta t, t) = f(z, t)$$ we find (by the chain rule) \begin{align} x_{t} &= f_{z} \cdot z_{t} + f_{t} \\ &= \beta f_{z} + f_{t} \\ x_{\xi} &= f_{z} \cdot z_{\xi} \\ &= f_{z} \\ x_{\xi \xi} &= f_{zz} \cdot z_{\xi} \\ &= f_{zz} \end{align} Substituting into our original PDE, we get \begin{align} x_{t}... 1 Another particular solution you missed can be found as:\frac{1}{(D+D')^2}x=\frac{1}{D'^2}$$[1+\frac{D}{D'}]^{-2}x=\frac{1}{D'^2}(1-2D/D'+....]x=\frac{1}{D'^2}(x-2y)=\frac{xy^2}{2}-\frac{y^3}{3} 0 Here some general issues that can come up when solving this numerically and how one can resolve them. Consider the equation u_{rr} + \frac{2}{r}u_r = f(u,r). Discretized on a grid with n points in the radial direction from [R_{\rm min},R_{\rm max}] this becomes (an example; there are other possible discretizations)$$\frac{u_{i+1}+u_{i-1}-2u_i}{h^2} +...

1

The only way to have a spatially smooth solution is for $\partial u/\partial r$ go to zero as $r$ goes to zero. This means you can use L'Hopitals's rule in evaluating that second term: $\displaystyle \frac{1}{r}\frac{\partial u}{\partial r} \rightarrow \frac{\partial_r(\partial u/\partial r)}{\partial_r(r)} = \frac{\partial^2 u}{\partial r^2}$

1

You can transform to Sturm-Liouville form by dividing by $x$: $$(xM')'+\frac{\lambda}{x}M = 0.$$ Whenever you have such a form, you can get rid of the multiplier in the derivative. There is a standard trick when you change the independent variable. In this case, let $M(x) = P(\int\frac{1}{x}dx)$. Then $$xM'(x) = ... 0 I tried to prove the Part 1 as follows. Please check if it is valid. We prove Part 1 by contradiction. Suppose the conclusion is not true, then \max_{\overline{U}} u\neq \max_{\partial U} u. But since \partial U\subset \overline{U}, we have \max_{\overline{U}} u>\max_{\partial U} u. Because u\in C(\overline{U}), with U bounded, it follows ... 0 y''=Ay'+B\sin(2Cy) E.g. y(0):=a and y'(0):=b. => y''(0)=Ab+B\sin(2Ca), y'''(0)=Ay''(0)+2CB\cos(2Cy(0))y'(0)=A^2 b+AB\sin(2Ca)+2CBb\cos(2Ca) and so on. You can construct a Taylor series for y(t), if you know y^{(n)}(0). 0 Hints Notice the general identity$$\psi \nabla ^2 \phi=\nabla \cdot (\psi \nabla \phi)-\nabla \psi \cdot \nabla \phi \tag{1}$$Integrate over the domain Use the divergence theorem Put \psi=\phi 1$$x^2\frac{d^2M}{dx^2}+x\frac{dM}{dx}+\lambda M(x)=0$$The change of variable \quad z=e^x\quad\to\quad x=\ln|z[\quad\to\quad \frac{dz}{dx}=z \quad isn't a good idea because it leads to a more complicated form of ODE : \frac{dM}{dx}=\frac{dM}{dz}\frac{dz}{dx}=z\frac{dM}{dz} \frac{d^2M}{dx^2}=\left(\frac{d}{dz}\frac{dM}{dx}\right)\frac{dz}{dx}=\frac{d}{... 2 The "Maximum Principle" is a very useful tool to answer to such a question (as John Barber rightly did). I will not come back with this principle to repeat what was already said and which is more general for all times from t=0 to t\to\infty. In the present case, since the wording of the question concerns only t\to\infty and where it is not question of ... 3 There is a property of certain PDEs known as the Maximum Principle. (See these notes (pdf warning) for its application to the heat equation.) The maximum principle states that any solution u(x,t) of the heat equation in a region of (x,t) must have its extrema (minimum and maximum values) on the boundary of that region. In this case, the region is (x,t)... 3 Using the integrating factor method borrowed from first order linear ODEs, your equation becomes \frac{\partial}{\partial t}(e^{\lambda^2 t}u)=k \frac{\partial^2}{\partial x^2} \left ( e^{\lambda^2 t} u \right ). Thus e^{\lambda^2 t} u solves the ordinary heat equation. 0 If \lambda=0 is an eigenvalue, then X(x) will be of the form X(x) = mx + c. Plug this into the boundary conditions, solve for m and c, and voilà! The converse for part (a) is just as easy - if the condition is satisfied, then show there exist m and c so that X(x) satisfies the boundary conditions. 0 This is nothing more than integration by parts. Multiply your equation by u, integrate and take the derivative w.r.t. to t out (I let the formal verification to you):$$ \frac{1}{2}\frac{d}{dt}\int_\Omega |u(x,t)|^2 dx = \int_\Omega ( u\Delta u )(x,t)dx + \int_\Omega f(x,t) dx. $$Using integration by parts on the form u\Delta u and integrating in time ... 1 Why not L^1: In studying elliptic equation, it is most convenient to consider L^p space for 1<p<\infty. The reason is that one does not have nice L^p-estimates$$\|u\|_{W^{2,p}(\Omega)}\le C (\|f\|_{L^p(\Omega)} + \|u\|_{L^p(\Omega)}$$for p=1 (Here we assume that p(x) is nice). Note that the above estimates is crucial in establishing ... 0 You want an operator that is symmetric. Suppose$$ Lf = a\frac{d^2}{dx^2}f+b\frac{d}{dx}f+cf $$You want a and b to be such that$$ \int_{a}^{b}(Lf)gdx = \int_{a}^{b}f(Lg)dx+\mbox{evaluation terms}. $$Symmetry of operators plays a critical role in Sturm-Liouville problems. It turns out that L needs to have the form$$...

0

Let $S$ be the graph of the function $x^2-y^2$ in the neigbourhood of $(0,0)$ positive normals in the $(x,z)$ plane ($y=0$) are not disjoint, and neagtive normal in the $(z,y)$ plane are not disjoint.

0

This is not possible. Take $r =1$ and let's work on the real line. Suppose there were such a function $f.$ Then $f(x) = 1, 0\le x \le 1,$ $f(x) = 0, x\ge 2,$ and $|f'(x)|\le 1$  everywhere. Note that $f'(1) = 0.$ Let $g(x)$ be the line connecting $(1,1)$ with $(2,0).$ Then the slope of $g$ is $-1.$ Because $f'(1) = 0,$ $f$ is above $g$ on $(1,1+\epsilon]$ ...

1

$\newcommand{\dd}{\partial}$If $u$ is a (differentiable) function of two real variables, then loosely, "$du$" isn't a number, but an ordered pair of numbers $$\frac{\dd u}{\dd t}(t, s) = \lim_{h \to 0} \frac{u(t + h, s) - u(t, s)}{h},\qquad \frac{\dd u}{\dd s}(t, s) = \lim_{k \to 0} \frac{u(t, s + k) - u(t, s)}{k}.$$ Technically, at each point $(t, s)$, ...

1

Yes. The necessary and sufficient condition is that $A$ is bounded. Indeed, the domain of $AE_I$ is the whole Hilbert space, and the domain of $E_IA$ is the domain $\mathcal D(A)$ of $A$.

1

Your answer doesn't make much sense - complex analysis does not apply in any sense when the dimension is odd. I'm guessing $\partial/\partial n$ denotes the normal derivative along the boundary, in which case the divergence theorem provides the answer: since $u$ is harmonic, $\nabla u$ is divergence-free, and thus $$\iint_{\partial \Omega} \frac{\partial u}{\... 0 Hint: Use multipliers 1,-1,1 and then equate to the third term. Edit: Using above you will get,$$\frac{dx-dy+du}{2(x-y+u)}=\frac{du}{u}$$Now solve, and you will get \frac{x-y+u}{u^2}=c_2. Then combine with the one you have already found and use the initial condition. 1 In your first equation you can divide with \cos (\sqrt{\lambda} L) and get$$ \tan (\sqrt{\lambda} L)= - \beta \sqrt{\lambda}, $$Define f(x) = \tan (\sqrt{x} L)+ \beta \sqrt{x} and notice,$$\lim _{x\rightarrow^{+} \frac{(\pi +4n\pi)^2}{4L^2} }f(x)=-\infty$$and that$$\lim _{x\rightarrow^{-} \frac{(\pi +4(n+1)\pi)^2}{4L^2} }f(x)=\infty$$... 0 Yes, the tensor product is dense. To approximate a function f in C_c^{\infty}((a_1,b_1) \times (a_2,b_2)) you can use functions of the form p(x,y)\chi_1(x)\chi_2(y) where p is any polynomial and \chi_1,\chi_2 are cut-off functions that are equal to 1 on most of the interval, so that \chi_1(x)\chi_2(y)\equiv 1 on the support of f. Then the ... 2 First of all, the wrong thing that you think is wrong, is totally wrong! :) So, yes, you are right! If you take the divergence and curl of the following equation$$ \textbf{F}=-\nabla \phi+\nabla \times \textbf{A} $$then you will get, respectively$$\boxed{ \begin{align} \nabla \cdot \textbf{F} &= - \nabla^2 \phi \\ \nabla \times \textbf{F} &= - ...

2

If ${\bf F} = - \nabla \phi + \nabla \times {\bf A}$, we get $\nabla \cdot {\bf F} = - \Delta \phi$. So we want to get $\phi$ by solving a Laplace equation. Then we get $\bf A$ by solving $\nabla \times {\bf A} = {\bf F} + \nabla \phi$. In your first example, ${\bf F}_1$ is indeed divergence-free, so you can take any $\phi$ whose Laplacian is $0$. You ...

-1

Take a look at the The Laplace Transform, Joel L. Schiff 1991 Springer Verlag, ISBN 0-387-98698-7, Examples on p182 onwards.

0

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# Horseshoe vignette

#### July 18th, 2019

In this vignette, the main functions of the package horseshoe are explained. This vignette is split up into three sections:

1. Brief introduction to the horseshoe
2. The normal means problem
3. Linear regression

First, install the package by typing

install.packages("horseshoe")

and then load the package. In this vignette we also use ggplot2 to make plots.

library(horseshoe)

# 1. Brief introduction to the horseshoe

The horseshoe (Carvalho et al 2010) is a Bayesian method for ‘needle-in-a-haystack’ type problems where there is some sparsity, meaning that there are some signals amid mostly noise.

We introduce the horseshoe in the context of the normal means model, which is given by $Y_i = \beta_i + \varepsilon_i, \quad i = 1, \ldots, n,$ with $$\varepsilon_i$$ i.i.d. $$\mathcal{N}(0, \sigma^2)$$. The horseshoe prior is given by \begin{align*} \beta_i &\sim \mathcal{N}(0, \sigma^2 \tau^2 \lambda_i^2)\\ \lambda_i &\sim C^+(0, 1), \end{align*} where $$C^+$$ denotes the half-Cauchy distribution. Optionally, hyperpriors on $$\tau$$ and $$\sigma$$ may be specified, as is described further in the next two sections.

To illustrate the shrinkage behaviour of the horseshoe, let’s plot the posterior mean for $$\beta_i$$ as a function of $$y_i$$ for three different values of $$\tau$$.

tau.values <- c(0.005, 0.05, 0.5)
y.values <- seq(-5, 5, length = 100)
df <- data.frame(tau = rep(tau.values, each = length(y.values)),
y = rep(y.values, 3),
post.mean = c(HS.post.mean(y.values, tau = tau.values[1], Sigma2=1),
HS.post.mean(y.values, tau = tau.values[2], Sigma2=1),
HS.post.mean(y.values, tau = tau.values[3], Sigma2=1)) )

ggplot(data = df, aes(x = y, y = post.mean, group = tau, color = factor(tau))) +
geom_line(size = 1.5) +
scale_color_brewer(palette="Dark2") +
geom_abline(lty = 2) + geom_hline(yintercept = 0, colour = "grey") +
theme_classic() + ylab("") + labs(color = "Tau") +
ggtitle("Horseshoe posterior mean for three values of tau")

Smaller values of $$\tau$$ lead to stronger shrinkage behaviour of the horseshoe. Observations that are in absolute value at most equal to $$\sqrt{2\sigma^2\log(1/\tau)}$$ are shrunk to values close to zero (Van der Pas et al (2014)). For larger observed values, the horseshoe posterior mean will tend to the identity (that is, barely any shrinkage, the estimate will be very close to the observed value). The optimal value of $$\tau$$ is the proportion of true signals. This value is typically not known in practice but can be estimated, as described further in the next sections.

# 2. The normal means problem

The normal means model is: $Y_i = \beta_i + \varepsilon_i, \quad i = 1, \ldots, n,$ with $$\varepsilon_i$$ i.i.d. $$\mathcal{N}(0, \sigma^2)$$.

### Computing the posterior mean only, with known variance $$\sigma^2$$

The function HS.post.mean computes the posterior mean of $$(\beta_1, \ldots, \beta_n)$$. It does not require MCMC and is suitable when only an estimate of the vector $$(\beta_1, \ldots, \beta_n)$$ is desired. In case uncertainty quantification or variable selection is also of interest, or no good value for $$\sigma^2$$ is available, please see below for the function HS.normal.means.

The function HS.post.mean requires the observed outcomes, a value for $$\tau$$ and a value for $$\sigma$$. Ideally, $$\tau$$ should be equal to the proportion of nonzero $$\beta_i$$’s. Typically, this proportion is unknown, in which case it is recommended to use the function HS.MMLE to find the marginal maximum likelihood estimator for $$\tau$$.

As an example, we generate 50 data points, the first 10 of which are coming from true signals. The first 10 $$\beta_i$$’s are equal to five and the remaining $$\beta_i$$’s are equal to zero. Let’s first plot the true parameters (black) and observations (blue).

df <- data.frame(index = 1:50,
truth <- c(rep(5, 10), rep(0, 40)),
y <- truth + rnorm(50) #observations
)

ggplot(data = df, aes(x = index, y = truth)) +
geom_point(size = 2) +
geom_point(aes(x = index, y = y), size = 2, col = "blue") +
theme_classic() + ylab("") +
ggtitle("Black = truth, Blue = observations")

We estimate $$\tau$$ using the MMLE, using the known variance.

(tau.est <- HS.MMLE(df$y, Sigma2 = 1)) #> [1] 0.9231211 We then use this estimate of $$\tau$$ to find the posterior mean, and add it to the plot in red. post.mean <- HS.post.mean(df$y, tau.est, 1)
df$post.mean <- post.mean ggplot(data = df, aes(x = index, y = truth)) + geom_point(size = 2) + geom_point(aes(x = index, y = y), size = 2, col = "blue") + theme_classic() + ylab("") + geom_point(aes(x = index, y = post.mean), size = 2, col = "red") + ggtitle("Black = truth, Blue = observations, Red = estimates") If the posterior variance is of interest, the function HS.post.var can be used. It takes the same arguments as HS.post.mean. ### Posterior mean, credible intervals and variable selection, possibly unknown $$\sigma^2$$ The function HS.normal.means is the main function to use for the normal means problem. It uses MCMC and results in an object that contains all MCMC samples as well as the posterior mean for all parameters ($$\beta_i$$’s, $$\tau$$, $$\sigma$$), the posterior median for the $$\beta_i$$’s, and credible intervals for the $$\beta_i$$’s. The key choices to make are: • How to handle $$\tau$$. The recommended option is “truncatedCauchy” (a half-Cauchy prior truncated to $$[1/n, 1]$$). See the manual for other options. • How to handle $$\sigma$$. The recommended option is “Jeffreys” (Jeffrey’s prior). See the manual for other options. Other options that can be set by the user are the level of the credible intervals (default is 95%), and the number of MCMC samples (default is 1000 burn-in samples and then 5000 more). Let’s continue the example from the previous section. We first create a ‘horseshoe object’. hs.object <- HS.normal.means(df$y, method.tau = "truncatedCauchy", method.sigma = "Jeffreys")

We extract the posterior mean of the $$\beta_i$$’s and plot them in red.

df$post.mean.full <- hs.object$BetaHat

ggplot(data = df, aes(x = index, y = truth)) +
geom_point(size = 2) +
geom_point(aes(x = index, y = y), size = 2, col = "blue") +
theme_classic() + ylab("") +
geom_point(aes(x = index, y = post.mean.full), size = 2, col = "red") +
ggtitle("Black = truth, Blue = observations, Red = estimates")

We plot the marginal credible intervals (and remove the observations from the plot for clarity).

df$lower.CI <- hs.object$LeftCI
df$upper.CI <- hs.object$RightCI

ggplot(data = df, aes(x = index, y = truth)) +
geom_point(size = 2) +
theme_classic() + ylab("") +
geom_point(aes(x = index, y = post.mean.full), size = 2, col = "red") +
geom_errorbar(aes(ymin = lower.CI, ymax = upper.CI), width = .1, col = "red") +
ggtitle("Black = truth, Red = estimates with 95% credible intervals")

Finally, we perform variable selection using HS.var.select. In the normal means problem, we can use two decision rules. We will illustrate them both. The first method checks whether zero is contained in the credible interval, as studied by Van der Pas et al (2017).

df$selected.CI <- HS.var.select(hs.object, df$y, method = "intervals")

The result is a vector of zeroes and ones, with the ones indicating that the observations is suspected to correspond to an actual signal. We now plot the results, coloring the estimates/intervals blue if a signal is detected and red otherwise.

ggplot(data = df, aes(x = index, y = truth)) +
geom_point(size = 2) +
theme_classic() + ylab("") +
geom_point(aes(x = index, y = post.mean.full, col = factor(selected.CI)),
size = 2) +
geom_errorbar(aes(ymin = lower.CI, ymax = upper.CI, col = factor(selected.CI)),
width = .1) +
theme(legend.position="none") +
ggtitle("Black = truth, Blue = selected as signal, Red = selected as noise")

The other variable selection method is the thresholding method of Carvalho et al (2010). The posterior mean can be written as $$c_iy_i$$ where $$y_i$$ is the observation and $$c_i$$ some number between 0 and 1. A variable is selected if $$c_i \geq c$$ for some user-selected threshold $$c$$ (default is $$c = 0.5$$). In the example:

df$selected.thres <- HS.var.select(hs.object, df$y, method = "threshold")

ggplot(data = df, aes(x = index, y = truth)) +
geom_point(size = 2) +
theme_classic() + ylab("") +
geom_point(aes(x = index, y = post.mean.full, col = factor(selected.thres)),
size = 2) +
geom_errorbar(aes(ymin = lower.CI, ymax = upper.CI, col = factor(selected.thres)),
width = .1) +
theme(legend.position="none") +
ggtitle("Black = truth, Blue = selected as signal, Red = selected as noise")

# 3. Linear regression

The linear regression model is given by $Y = X\beta + \varepsilon,$

where $$Y$$ and $$\varepsilon$$ are vectors of length $$n$$, $$\beta$$ is a vector of length $$p$$ and $$X$$ is an $$n \times p$$-matrix. We assume $$\varepsilon \sim \mathcal{N}(0, I_n)$$. The main function for the horseshoe for the linear regression model is horseshoe and it implements the algorithm of Bhattacharya et al (2016).

The options of horseshoe are the same as for HS.normal.means (discussed above, although in case of linear regression it is less clear which prior to use for $$\tau$$). We illustrate the use of horseshoe via an example.

We create a 50 by 100 design matrix $$X$$ filled with realizations of independent normal random variables. The first 10 entries of the vector $$\beta$$ are set equal to six (the signals) and the remaining 90 entries are set equal to zero (the noise).

X <- matrix(rnorm(50*100), 50)
beta <- c(rep(6, 10), rep(0, 90))
y <- X %*% beta + rnorm(50)

We use the horseshoe and plot the posterior mean and marginal 95% credible interval per parameter in red. The true parameter values are shown in black.

hs.object <- horseshoe(y, X, method.tau = "truncatedCauchy", method.sigma ="Jeffreys")
#> [1] 1000
#> [1] 2000
#> [1] 3000
#> [1] 4000
#> [1] 5000
#> [1] 6000

df <- data.frame(index = 1:100,
truth = beta,
post.mean = hs.object$BetaHat, lower.CI <- hs.object$LeftCI,
upper.CI <- hs.object$RightCI ) ggplot(data = df, aes(x = index, y = truth)) + geom_point(size = 2) + theme_classic() + ylab("") + geom_point(aes(x = index, y = post.mean), size = 2, col = "red") + geom_errorbar(aes(ymin = lower.CI, ymax = upper.CI), width = .1, col = "red") + ggtitle("Black = truth, Red = estimates with 95% credible intervals") We again perform variable selection. The function, HS.var.select, is the same as described above for the normal means problem. Here we show how it works when variables are selected by checking whether 0 is in the credible interval. For the thresholding procedure, please refer to the normal means example above. We perform variable selection: df$selected.CI <- HS.var.select(hs.object, df\$y, method = "intervals")

The result is a vector of zeroes and ones, with the ones indicating that the observations is suspected to correspond to an actual signal. We now plot the results, coloring the estimates/intervals blue if a signal is detected and red otherwise.

ggplot(data = df, aes(x = index, y = truth)) +
geom_point(size = 2) +
theme_classic() + ylab("") +
geom_point(aes(x = index, y = post.mean, col = factor(selected.CI)),
size = 2) +
geom_errorbar(aes(ymin = lower.CI, ymax = upper.CI, col = factor(selected.CI)),
width = .1) +
theme(legend.position="none") +
ggtitle("Black = truth, Blue = selected as signal, Red = selected as noise")

# References

Bhattacharya A., Chakraborty A., and Mallick B.K (2016), Fast sampling with Gaussian scale-mixture priors in high-dimensional regression. Biometrika 103(4), 985–991.

Carvalho, C. M., Polson, N. G., and Scott, J. G. (2010), The Horseshoe Estimator for Sparse Signals. Biometrika 97(2), 465–480.

van der Pas, S. L., Kleijn, B. J. K., and van der Vaart, A. W. (2014), The horseshoe estimator: Posterior concentration around nearly black vectors. Electronic Journal of Statistics 8(2), 2585–2618.

van der Pas, S.L., Szabo, B., and van der Vaart, A. (2017), Uncertainty quantification for the horseshoe (with discussion). Bayesian Analysis 12(4), 1221-1274.

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# Introduction

## Representative case: KOI 1843.03

• Small planet candidate in the Kepler sample
• 4.2-hour orbital period
• Discovery paper: Rappaport et al. (2013)
• Used interpolated polytrope models
• Assumed a fluid-like planet filling its Roche lobe
• Measured $0.61 R_\oplus$ radius and $0.44 M_\oplus$ mass
• Inferred an iron-rich composition — exo-Mercury!

## The problem with rocky materials

• Polytrope models $P \propto \rho^\gamma$ do not accurately represent rocky material, which has nonzero density at zero pressure
• We can do better than Rappaport et al. (2013) by using a simple modified polytrope model, $\rho = c P^n + \rho_0$ (Seager et al. 2007)

## A circular problem

• We are interested in the full, three-dimensional shape of planets in extreme environments, because the shape influences the transit
• Relevant forces:
• Extreme gravitational force from the star
• Gravitational force of the planet on itself
• Centrifugal force from rotational motion
• There is no analytic way to compute the shape of the planet!

# Methods

## Hachisu method for multiple stars

• Hachisu (1986a,b) presents an elegant method for computing self-consistent structures of stars in multiple systems
• Cycles through enthalpy and density calculations until convergence is reached
• Requires computing the gravitational potential at all points (slow)
• Also requires an analytic or numerical approximation to the enthalpy as a function of density

## Applying Hachisu method

• Important update to the Hachisu method: Adding a point-mass star at fixed distance from the planet
• We have found that the most stable numerical procedure is as follows:
• Assume values for the core-mantle boundary pressure, central pressure, and distance from the star
• Solve for constants, including stellar mass, that meet boundary conditions
• Update density, compute potential, and loop

## Applying Hachisu method: Caveat

This method is “backwards” — we find the physical, measured parameters (radius, mass, etc.) only at the end of the simulation


text:
# 5.7 Least and greatest function values  (Page 3/3)

 Page 3 / 3

Problem 5: A function is defined as :

$f\left(x\right)=\frac{{e}^{x}}{1+\left[x\right]}$

If domain of the function is [0,∞), find the range of the function.

Solution :

Statement of the problem : The function contains greatest integer function in its denominator, whereas numerator of functon is an exponential function. Greatest integer function, GIF, is defined in discontinuous intervals of 1. It returns integral value equal to the starting value of discontinuous interval. The function, however, is continuous in sub-intervals of 1 i.e. in a finite interval. It is clear that function values and function nature will change in accordance with the values of GIF in sub-intervals of the domain. We need to determine nature of function in few of the initial sub-intervals and extrapolate the results to determine the range of function as required.

In order to determine nature of function, we write function and its derivative in the initial sub-intervals of the given domain. Note that domain of function starts with 0.

$0\le x<1,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}f\left(x\right)={e}^{x},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}f\prime \left(x\right)={e}^{x}$

$1\le x<2,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}f\left(x\right)=\frac{{e}^{x}}{2},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}f\prime \left(x\right)=\frac{{e}^{x}}{2}$

$2\le x<3,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}f\left(x\right)=\frac{{e}^{x}}{3},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}f\prime \left(x\right)=\frac{{e}^{x}}{3}$

$\text{-------------------------------------}$

The derivatives in each interval are positive for values of “x” in the domain. It means that function is strictly increasing in each of the intervals. There is no minimum and maximum as monotonic nature of function does not change. The least and greatest values, therefore, correspond to end function values in each finite sub-intervals. Using these least and greatest values, we determine range of initial sub-intervals as given here :

$0\le x<1\phantom{\rule{1em}{0ex}}⇒{e}^{0}\le y<{e}^{1}\phantom{\rule{1em}{0ex}}⇒1\le y

$1\le x<2\phantom{\rule{1em}{0ex}}⇒\frac{e}{2}\le y<\frac{{e}^{2}}{2}$

$2\le x<3\phantom{\rule{1em}{0ex}}⇒\frac{{e}^{2}}{3}\le y<\frac{{e}^{3}}{3}$

$\text{-------------------------------------}$

The plot of the function is drawn in the figure. Note that function values in adjacent intervals overlap each other, but exceeds the greatest value in the preceding intervals. Clearly, function value begins from “1” and continues towards infinity. Hence, range of the function is :

$\left[1,\infty \right)$

## Function with finite intervals

A continuous function, in finite interval, has finite numbers of minimum and maximum values. Using this fact and the monotonic nature function in the finite interval, we can determine least and greatest values of function. If “A” and “B” be the least and greatest values then range of the function is given by :

$\text{Range}=\left[\mathrm{A,B}\right]$

We need to compare minimum values, maximum values and end values to determine which are the least and greatest values. This aspect will be clear as we work out with the example here.

There are certain cases in which functions have exactly a pair of intervals characterized by opposite strict monotonic nature i.e. strictly increasing and strictly decreasing. In such cases, the point at which function changes its monotonic nature should be either minimum and maximum. This in turn means that function has exactly one minimum or maximum. Clearly, this minimum or maximum corresponds to least or greatest function value of the function in its domain.

Problem : Find range of continuous function :

$f\left(x\right)=\sqrt{\left(x-1\right)}+\sqrt{\left(5-x\right)}$

Solution : Given function is continuous. Also, it involves square root of polynomial. It means that function is defined in sub interval of real number set R. We, therefore, need to find the domain of the function first. Then we investigate the nature of the function in its domain and determine least (a) and greatest (b) function values.

For individual square roots to be real,

$x-1\ge 0\phantom{\rule{1em}{0ex}}⇒x\ge =1$

$5-x\ge 0\phantom{\rule{1em}{0ex}}⇒x\le 5$

Hence, domain of the function is intersection of individual domains and is a finite interval [1,5]. Now, we differentiate the function with respect to independent variable to determine the nature of function :

$⇒f\prime \left(x\right)=\frac{1}{2\sqrt{\left(x-1\right)}}-\frac{1}{2\sqrt{\left(5-x\right)}}$

In order to draw sign scheme, we find the roots of the equation by putting f’(x) = 0 and solving for “x” as :

$⇒2\sqrt{\left(x-1\right)}=2\sqrt{\left(5-x\right)}$

$⇒x-1=5-x\phantom{\rule{1em}{0ex}}⇒2x=6\phantom{\rule{1em}{0ex}}⇒x=3$

For test value, x = 4, f’(x) is negative. Hence, corresponding sign scheme of f’(x) is as given here :

Since function is increasing to the left and decreasing to the right of root value, the function value at root value is maximum. Clearly, there is only one maximum. Thus, it is greatest value also. Further, either of two function values at end points is least value of the function in the domain . Now, function values at end points and root point are :

$⇒f\left(1\right)=0+2=2$

$⇒f\left(3\right)=\sqrt{2}+\sqrt{2}=2\sqrt{2}$

$⇒f\left(5\right)=2+0=2$

Clearly, function at x = 3 is greatest value and either values at x = 1 and 5 is least. Hence, range of the function is :

$\left[2,2\sqrt{2}\right]$

## Composite function

We shall use concept of least value, greatest value and monotonic nature to evaluate range of composite function. As pointed out in the beginning of the module, we shall proceed evaluation inside out i.e. proceeding from innermost to outermost.

Problem : A function is defined as :

$f\left(x\right)=\left[{\mathrm{log}}_{e}\left\{{\mathrm{sin}}^{-1}\sqrt{\left({x}^{2}+x+1\right)}\right\}\right]$

The outermost square bracket, here, represents greatest integer function. Find range of the function.

Solution : The argument of arcsine function is a square root. But we know that domain of arcsine function is [-1,1]. If we know the least or greatest value of quadratic expression within sqaure root, then we shall be able to futher narrow down the value of argument of arcsine function. The derivative of expression within square root is :

Now, the argument of arcsine function is a square root. The derivative of expression within square root is :

$f\prime \left(x\right)=2x+1$

The root of this expression when equated to zero is x = -1/2. The function value at test poin, x = 1 is 3, which is a positive number. Thus, corresponding sign scheme of the derivative expression is shown in the figure.

The polynomial function is strictly decreasing to the left and strictly increasing to the right of root value. This means that function value at root point is minimum. Also, as there is only one minimum, this value is the least function value in the domain. Thus, least value of square root is :

$⇒\sqrt{{x}^{2}+x+1}=\sqrt{\left\{{\left(-\frac{1}{2}\right)}^{2}+\left(-\frac{1}{2}\right)+1\right\}}=\frac{\sqrt{3}}{2}$

But, we know that the maximum value of the argument of arcsine function is “1”. Also, square root is a positive number. It means that :

$\sqrt{{x}^{2}+x+1}\le 1$

Thus, we conclude that the value of square root expression should lie in the interval given by :

$⇒\frac{\sqrt{3}}{2}\le \sqrt{{x}^{2}+x+1}\le 1$

We see that arcsine is an increasing function within the interval specified by domain of the function. Hence, we can take sine inverse of each term without any change in the direction of inequality :

$⇒{\mathrm{sin}}^{-1}\frac{\sqrt{3}}{2}\le {\mathrm{sin}}^{-1}\sqrt{{x}^{2}+x+1}\le {\mathrm{sin}}^{-1}1$

$⇒\frac{\pi }{3}\le {\mathrm{sin}}^{-1}\sqrt{{x}^{2}+x+1}\le \frac{\pi }{2}$

Logarithmic function is also an increasing function. Hence, we can take logarithm of each term without any change in the direction of inequality :

$\mathrm{log}{}_{e}\frac{\pi }{3}\le \mathrm{log}{}_{e}\left({\mathrm{sin}}^{-1}\sqrt{{x}^{2}+x+1}\right)\le \mathrm{log}{}_{e}\pi /2$

Since ${\mathrm{log}}_{e}\pi /2<1$ and ${\mathrm{log}}_{e}\pi /3>0$ , we can state the inequality as :

$0<{\mathrm{log}}_{e}\left({\mathrm{sin}}^{-1}\sqrt{{x}^{2}+x+1}\right)<1$

We know that greatest integer function returns "0" in the interval [0,1). Hence,

$f\left(x\right)=\left[{\mathrm{log}}_{e}\left({\mathrm{sin}}^{-1}\sqrt{{x}_{2}+x+1}\right)\right]=0$

Clearly, function returns zero for all values of “x” in the domain of the function. Therefore, range of the function is :

$\text{Range}=\left\{0\right\}$

Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
What is power set
Period of sin^6 3x+ cos^6 3x
Period of sin^6 3x+ cos^6 3x


text:
## Tuesday, December 02, 2008

### Heat expansion of the spectral action and loop diagrams

Ali Chamseddine and Alain Connes have a new paper,
The uncanny precision of the spectral action,
which argues that under some assumptions they find worth considering, two terms in an expansion of an action give the correct initial 6820 decimal places of an action.

It's a typical paper where I feel that there is something extremely cool and exciting going on, but once I spend ten minutes with each such thing, it seems to evaporate. Recall that they want to write a theory of everything as a simple system in noncommutative geometry. The action is
S = Tr [ f(D/Lambda) ]
It's the so-called spectral action. The trace should probably be taken over a first-quantized (one-particle) Hilbert space only, generating the classical spacetime action of a field theory in some way. Here, "D" is the Dirac operator - including all dimensions they consider.

The letter "Lambda" stands for a cosmological constant. You would think that it is extremely small but they actually study "Lambda goes to infinity" limit. That's bizarre and I am afraid that you won't find any justification of such an approach in the paper but some other things are fun so you shouldn't stop reading yet.

The letter "f" is a function of a real variable "x" that drops to zero if "x" goes to plus minus infinity, something like "exp(-x^2)" except that the authors want to prefer a slightly different function with the same asymptotics but with derivatives of all orders vanishing at x=0. Again, not sure what justifies such a strange and arbitrary function.

Clearly, for some contrived choices of functions and parameters, certain approximations to some quantities may be very accurate, but another question is whether such an observation under certain conditions is interesting, non-trivial, or relevant for anything in science or just an artificially created curiosity from the realm of recreational mathematics. If the accurate values of both sides of the (approximate) equation are physically irrelevant, so is the (approximate) identity!

But what's funny and possibly non-trivial is a "heat expansion" of the spectral action, "Tr [f(D/Lambda)]", for large values of Lambda. You might think that it is just a Taylor expansion except that it is apparently not. I feel that in their non-commutative geometrical context, they have an alternative calculation scheme for loop diagrams in field theory but all the links seem very obscure so far.

For example, I think that the numerator "11" in the equation below (27), obtained from zeta functions, could be equivalent to the "11" in the one-loop beta functions of gauge theories. Analogously, the calculation of the curvature-cubed terms around page 27 could be a reorganization of the calculation of two-loop diagrams in general relativity except that they don't get the desired (or at least expected) non-zero result.

Note that the Goroff-Sagnotti two-loop result, proving non-renormalizability of general relativity, is a cubic function of the Weyl tensor only, and the rational coefficient is 209/2880. The number 209 could be related to 208 in some of their formulae.

If a reader knows something about the relationship between the heat expansions and loop diagrams, or why it's not there if it's not there, I would like to know! Thanks. Another topic worth research is the conceptual origin of the cancellations of the "a4" and "a6" terms in their expansion and their possible applications in the solution to various hierarchy problems.

#### 1 comment:

1. Triggered by your post I had a look at the paper. Since answering your questions would be too much for a comment I have started writing my own post. Yesterday, I wrote a first installment on some non-commutate geometry background

http://atdotde.blogspot.com/2008/12/spectral-actions-imprecisely.html

I hope to have time tonight to write to write a part II actually covering the spectral action. As a spoiler: I don't think there is much surprising going on: The increadible precission expresses that you make only a small error when you replace the discrete spectrum of the Dirac (or any other differential operator) on S^3xS^1 of the size of the universe with a continuous momentum space.

The cancelation is about ordinary Sealy-deWit heat-kernel coefficients on S3 and has nothing to do with NCG. I have no explanation but I also cannot see a significance.


text:
# Seemingly random left margin after page break

My document seems to be switching left margins at random and I do not know what is causing this. I am running MacTex and compiling it like this

latex thesis.tex
bibtex A
bibtex B
latex thesis.tex
dvips -t letter -o thesis.ps thesis.dvi
ps2pdf14 -dPDFSETTINGS=/prepress -dEmbedAllFonts=true thesis.ps thesis.pdf

\documentclass[twoside,11pt,openright]{report}

\usepackage[latin1]{inputenc}
\usepackage[american]{babel}
\usepackage{a4}
\usepackage{latexsym}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{epsfig}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[labeled]{multibib}
\usepackage{color}
\usepackage{datetime}

\usepackage{etoolbox}
\makeatletter
\patchcmd{\chapter}{\if@openright\cleardoublepage\else\clearpage\fi}{}{}{}
\makeatother

\renewcommand*\ttdefault{txtt}

\newcommand{\todo}[1]{{\color[rgb]{.5,0,0}\textbf{\$\blacktriangleright\$#1\$\blacktriangleleft\$}}}

\newcites{A,B}{Primary Bibliography,Secondary Bibliography}

Below is a link to a screenshot of what looks like, I inserted a ruler to highlight the issue. There is no command in the paragraph except the \todo you can see in red.

The \patchcmd is simply to avoid pagebreaks for empty chapters, the problem persists if I remove it.

-
It is because you are using twoside option. Try change to oneside. The spacing you refer to will only be the same if you have the same margin sizes on the left and right side of the text. Try printing it in duplex and you will see why it does this. –  zeroth May 16 '12 at 8:18
You're doing two contradictory actions: \usepackage{a4} (which is obsolete, by the way) and dvips -t letter ... –  egreg May 16 '12 at 8:21
Indeed. \usepackage[a4paper]{geometry} please - and read l2tabu-en. –  Martin Schröder May 16 '12 at 8:24
Whoa! Thanks for all the help guys. Seems like my template is a bit outdated and not very optimized. Having trouble with eps to pdf conversion now, but I'm sure others have solved that problem and wrote about it somewhere. btw. I was quite sure it wasn't random, that's why i wrote "seemingly" :-) –  andsens May 17 '12 at 10:46

It's not random. You're using the twoside option, so the margins are different on even (left hand) and odd (right hand) pages.

If you need "symmetric" setup on the pages, remove the options.

There's another source of problems: you're using the obsolete package a4 that should set the text block for being positioned on ISO A4 paper, but then you convert the DVI file by passing to dvips the option -t letter, that is, printing on American letter paper. Be consistent: if you plan to output on letter paper, then say

\documentclass{report}

If you instead plan to output on ISO A4 paper, use

\documentclass[a4paper]{report}

The openright option is useless if you don't call the twoside option.

Consider also to use the modern package graphicx for the inclusion of graphic objects, rather than the obsolete epsfig (which is included in the distributions only for back compatibility).

If you have a fairly recent TeX distribution (MacTeX 2010 or later), there is no need to do the "LaTeX-dvips-ps2pdf" route: run directly pdflatex. If you use graphicx instead of epsfig, the EPS files will be automatically converted to PDF for inclusion.

-
@AndreasWallner Thanks for fixing the silly typo which I'm sorry of. –  egreg May 16 '12 at 10:02


text:
SeaHawk/HawkEye ▸ On-Orbit Calibration

## HawkEye On-Orbit Calibration

The HawyEye instrument consists of four CCD arrays with two bandpass filters per array to provide the 8 spectral bands for the sensor. Each band consists of an 1800 illuminated elements

Each scene includes the collection of 64 dark scans at the end of the imaging sequence. These dark scans are used to provide the electronic baseline

Calibration Equation $$L = K1[band] * DN * K2[band,pixel,time] * K3[band,temperature] * K4[band,pixel] * (1+K5[band,pixel]*DN) * K7[band,pixel]$$ where:
• DN = Digital Number from the
• K1 = Gain coefficients (at nadir pixel)
• K2 = Temporal correction coefficients
• K3 = Temperature correction coefficients
• K4 = Response vs Scan correction coefficients
• K5 = Non-linearity correction coefficients
• K7 = CCD smile correction coefficients (not currently used)

#### Vicarious Calibration

The vicarious calibration of HawkEye was accomplished using a process similar to that used for OCTS. A clear-water radiance model based on a climatological chlorophyll time-series of SeaWiFS data over the MOBY location was used to derive an estimate of the target water-leaving radiance. MOBY data were not availble for the few dates that HawkEye imaged the MOBY location.

### Geometric Registration

Since there are four separate CCDs, each supportig two spectral bands, the data from each band are offset in both the along track and cross track directions and must be registered to a common reference band. Additionally, the focal length varies slightly from band to band. Corrections for the physical offsets and the focal length differences are necessary to ensure that the data from each pixel for each band match the same geolocated area in the scene.


text:
equinv - Maple Help

DEtools

 equinv
 look for the most general ODE invariant under a given symmetry

 Calling Sequence equinv([xi, eta], y(x)) equinv([[xi1, eta1], [xi2, eta2],...], y(x)) equinv([xi, eta], y(x), n) equinv([[xi1, eta1], [xi2, eta2],...], y(x), n)

Parameters

 [xi, eta] - list of the coefficients of the symmetry generator (infinitesimals) y(x) - indeterminate function to be used in the result n - positive integer, indicating that an nth order ODE is to be returned

Description

 • equinv takes as first argument either a list of two algebraic expressions, to be seen as the infinitesimals of a one-parameter Lie group, or a list of such pairs representing many different symmetry groups. It returns, within the possibilities of the system, the most general ODE simultaneously invariant under those symmetry groups. The second argument is the indeterminate function of that ODE.
 • If the number n is given as extra argument, equinv looks for an invariant ODE of order n; otherwise, it will look for a first order ODE.
 • As a rule, if we represent the infinitesimal generator as

$\mathrm{\xi }\left(x,y\right)\left(\frac{\partial }{\partial x}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\right)+\mathrm{\eta }\left(x,y\right)\left(\frac{\partial }{\partial y}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}\right),$

 where x is the independent variable and y is the dependent one; equinv assumes that the order in which the infinitesimals xi and eta appear in the received list is $\left[\mathrm{\xi },\mathrm{\eta }\right]$.
 • In principle, there are no restrictions on the number of pairs of infinitesimals you can give at one time to equinv. However, the complexity of the problem increases with the number of pairs of infinitesimals given, and the problem may not have a solution. Also, when an nth order ODE is required (extra argument n) the given infinitesimals may contain _y1, _y2, ... which, as a convention, represent the derivative of the "indeterminate function" of the problem. In short, equinv also accepts dynamical symmetries as arguments.
 • equinv and buildsym may be useful in connection with the odeadvisor command and the $\mathrm{HINT}=\left[\mathrm{e1},\mathrm{e2}\right]$ option of dsolve, in order to study the relationship between symmetry patterns and ODE patterns (see dsolve/Lie).
 • This function is part of the DEtools package, and so it can be used in the form equinv(..) only after executing the command with(DEtools). However, it can always be accessed through the long form of the command by using DEtools[equinv](..).

Examples

Three pairs of infinitesimals not containing a 2-D subalgebra

 > $\mathrm{with}\left(\mathrm{DEtools}\right):$
 > $\mathrm{sym}≔\left[\left[1,1\right],\left[x,y\right],\left[{x}^{2},{y}^{2}\right]\right]$
 ${\mathrm{sym}}{≔}\left[\left[{1}{,}{1}\right]{,}\left[{x}{,}{y}\right]{,}\left[{{x}}^{{2}}{,}{{y}}^{{2}}\right]\right]$ (1)

The most general second order ODE simultaneously invariant under these three point symmetries

 > $\mathrm{ODE2}≔\mathrm{equinv}\left(\mathrm{sym},y\left(x\right),2\right)$
 ${\mathrm{ODE2}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}^{{3}}{{2}}}{}{\mathrm{_C1}}{+}{2}{}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}^{{2}}{+}{2}{}\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)}{{-}{x}{+}{y}{}\left({x}\right)}$ (2)

The fact that ODE2 is invariant under the three pairs of infinitesimals above can be tested using the symtest command (see symtest):

 > $\mathrm{map}\left(\mathrm{symtest},\mathrm{sym},\mathrm{ODE2}\right)$
 $\left[{0}{,}{0}{,}{0}\right]$ (3)

Two pairs of dynamical symmetries

 > $\mathrm{dyn_sym}≔\left[\left[\mathrm{_y1},0\right],\left[0,\frac{1}{\mathrm{_y1}}\right]\right]$
 ${\mathrm{dyn_sym}}{≔}\left[\left[{\mathrm{_y1}}{,}{0}\right]{,}\left[{0}{,}\frac{{1}}{{\mathrm{_y1}}}\right]\right]$ (4)

The most general second order ODE simultaneously invariant under these two dynamical symmetries

 > $\mathrm{ODE2}≔\mathrm{equinv}\left(\mathrm{dyn_sym},y\left(x\right),2\right)$
 ${\mathrm{ODE2}}{≔}\frac{{{ⅆ}}^{{2}}}{{ⅆ}{{x}}^{{2}}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right){=}\frac{{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}^{{2}}}{{\mathrm{_F1}}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right)}^{{2}}{+}{2}{}\left(\frac{{ⅆ}}{{ⅆ}{x}}\phantom{\rule[-0.0ex]{0.4em}{0.0ex}}{y}{}\left({x}\right)\right){}{x}{-}{2}{}{y}{}\left({x}\right)}$ (5)
 > $\mathrm{map}\left(\mathrm{symtest},\mathrm{dyn_sym},\mathrm{ODE2}\right)$
 $\left[{0}{,}{0}\right]$ (6)


text:
# In the Gaussian Integers $$\alpha^4 + \beta^4 \ne \gamma^4$$

by James T. Cross (University of the South)

Mathematics Magazine
April, 1993

Subject classification(s): Algebra and Number Theory | Number Theory
Applicable Course(s): 4.3 Number Theory

A simple proof for Last Fermat Theorem when n=4 in Gaussian Integers

A pdf copy of the article can be viewed by clicking below. Since the copy is a faithful reproduction of the actual journal pages, the article may not begin at the top of the first page.


text:
# Solving y’ + a(x) y = b(x), Part 3

Yet another way to find the solution of

$y' + a(x) y = b(x)\quad\quad\quad(1)$

is to seek a function $f(x) > 0 \; \forall x$ such that the result of multiply (1) by $f(x)$, namely

$f(x)y' + f(x) a(x) y =f(x) b(x)\quad\quad\quad(2)$

can be written as

$(f(x) y)' = b(x) f(x)$,

i.e.,

$f(x) y =\int b(x) f(x)\; dx + c_{3}$

where $c_{3}$ is a constant.

Or,

$y = {1 \over f(x)} (\int b(x) f(x)\; dx + c_{3})\quad\quad\quad(3)$

since $f(x) > 0$.

Let us proceed to find such $f(x)$.

From (2) we see that if

$f'(x)=a(x) f(x)$

then the left side of (2) is certainly $(f(x) y)'$ and consequently  for $f(x) > 0 \;\forall x$,

${1 \over f(x)} f'(x) = a(x)\quad\quad\quad(4)$

or,

$log(f(x)) = \int a(x) dx + c_{1}$

where $c_{1}$ is a constant.

Therefore, a solution to (4) is

$f(x) = c_{2}e^{\int a(x)\;dx}\quad\quad\quad(5)$

where $c_{2} = exp(c_{1})$. This is a positive function $\forall x$ indeed.

With (5), (3) becomes

$y = e^{-\int a(x)\;dx} (\int b(x) e^{ \int b(x) dx} dx + c)\quad\quad\quad(6)$

where $c = {c_{3} \over c_{2}}$.

In fact, for any constant c,

$e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c)$

is a solution of (1):

$(e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx+ c))' +$

$a(x)\dot(e^{-\int a(x)\;dx} (\int b(x) e^{\int a(x) dx} dx + c))$

$= -a(x) (e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c))$

$+ e^{-\int a(x) dx} b(x) e^{\int a(x) dx}$

$+ a(x)(e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c))$

$=b(x)$

Therefore, the solution of (1) is

$y = e^{-\int a(x) dx} (\int b(x) e^{\int a(x) dx} dx + c)$

where $c$ is any constant.

Exercise: prove that $e^x$  is a positive function.


text:
# “DIM” terminology

I've been bothered recently by a MATLAB terminology question: How should we describe the dim argument in functions such as sum? The documentation for sum says:

  B = sum(A,dim) sums along the dimension of A specified by
scalar dim. The dim input is an integer value from 1 to N,
where N is the number of dimensions in A. Set dim to 1 to
compute the sum of each column, 2 to sum rows, etc.

For example:

a = [1 2 3; 4 5 6; 7 8 9]
a =

1     2     3
4     5     6
7     8     9

sum(a, 1)
ans =

12    15    18

sum(a, 2)
ans =

6
15
24

For some people this terminology seems to be quite confusing. I started thinking about this when I saw a user question about it recently. In that question, the mental model leading to confusion seems to be this: dimension 1 is the row dimension, so sum(a,1) should sum each row, but MATLAB sums each column.

This a hard mental model to shake, apparently, and it makes me wonder if the phrase "along the dimension" isn't clear enough.

When you look at equations instead of ambiguous prose, it's easier to see the logic of the dim argument. Consider, for example, an M-by-N matrix A. sum(A,1) is doing this:

The variable of summation, i, is the first subscript of A.

sum(A,2) is doing this:

The variable of summation, j, is the second subscript of A.

This pattern extends nicely to higher dimensions.

So, dear readers, can you suggest a better way to document the meaning of the dim argument? Or should we just leave it be?

Published with MATLAB® 7.10

|


text:
# “Undefined” error using LyX

I always get these errors when I try to show a file;

LaTeX Error: \eqdeclaration undefined.
LaTeX Error: \pagedeclaration undefined.
LaTeX Error: \nomname undefined.

I can't find a fix for that. I didn't find anything about it in the tutorial or in the manuals.

Some one had the same errors years ago. Found it in the LyX mailing list. But I can't find an answer. Can somebody help me?

I am using Manjaro Linux!

• Please post a minimal example. Does this happen with any document? If so, make a very simple document and paste it in here (a .lyx file is just plain text). For more information see wiki.lyx.org/FAQ/MinimalExample – scottkosty Oct 27 '15 at 23:53

You must define somewhere these control sequences. This can be done in the body of the document in a ERT box, and/or in the LaTeX preamble, directly ( writing yourself in plain TeX \def\eqdeclaration#1{#1 ...} or LaTeX code as as \newcommand{\eqdeclaration}[1]{#1...}, etc.) or including a .tex file with these macros (\input{mymacros}) or a package (.sty file) or selecting a document class (.cls file) that have these definitions or load these definitions of another file.

Among all these possibilities, without see a MWE, I'll guess watching my crystal ball: You need \usepackage{nomencl} in the LaTeX Preamble (menu Document > Settings... ) but this must be done automatically when you Insert any Nomenclature Entry..., so at the end, we simply need more information to know the source of these errors. First, make a plain tex file without Lyx named test.tex with any plain text editor:

\documentclass{article}
\usepackage{nomencl}
\begin{document}
\nomname
\eqdeclaration
\pagedeclaration
\end{document}

Then compile the next MWE in the system prompt with pdflatex test.tex. It should produce a test.pdf file with a non non-sense text:

Nomemclature, see equation (, page )")

... but without errors. Otherwise the problem is not related with LyX, but with the LaTeX installation.

• @Max My MWE cannot just copied & pasted in LyX. I added some clarification. I strongly suggest you first understand the very basic structure of LaTeX documents (explained in thousands of places) ad how are compiled without LyX, then you can understand how LyX make most of this work for you, and what you can do in case of LaTeX errors. – Fran Oct 30 '15 at 18:22


text:
# Isol

2010 Mathematics Subject Classification: Primary: 03D50 [MSN][ZBL]

The recursive equivalence type of an isolated (that is, finite or immune, cf. Immune set) set of natural numbers. The set of all isols has the cardinality of the continuum and is a semi-ring under the operations of addition and multiplication defined for arbitrary recursive equivalence types. This semi-ring is called the arithmetic of isols. It has a number of properties of the arithmetic of natural numbers; in particular, all universal Horn formulas whose elementary subformulas represent equality of so-called combinatorial functions are true in it. An example of such a formula is the cancellation law: $X+Z=Y+Z\Rightarrow X=Y$. Isols can be regarded as recursive analogues of the cardinalities of Dedekind-finite sets, that is, sets not equivalent to any of their proper subsets.

A subset of the natural numbers $\mathbf N$ is an isolated set if it contains no infinite recursively-enumerable subset of $\mathbf N$.


text:
# Feature Normalization

Categories:

Updated:

Normalizing features mean putting each of features in roughly the same range.

# Why Normalize Features?

We can speed up gradient descent by normalizing features. This is because $\theta$ will descend quickly on small ranges and slowly on large ranges. Theta will oscillate inefficiently down to the optimum when the variables are very uneven.

# How?

Feature Normalization is composed of two steps.

## 1. Feature Scaling

Feature scaling means dividing the input values by the range (i.e. the maximum value minus the minimum value) of the input variables.

Commonly, we use ‘standard deviation’ as a ‘range’ parameter.

where $\overline {x_j}$ is the average of vector $x_j$.

## 2. Mean Normalization

Mean normalization involves subtracting the average value for an input variable from the values for that input variable, resulting in a new average value for the input variable of just zero.

## Resulting Formula

We combine these two steps and get following formula.

of just zero. To implement both of these techniques, adjust your input values as shown in this formula:

Categories:


text:
£199.99 £139.99 for one year of Unlimited learning. Offer ends on 28 February 2023 at 23:59 (UTC). T&Cs apply

Turing Machines Explained

A Turing machine consists of an infinitely long tape, which has been divided up into cells. Each cell can contain either a 1, a 0, or an empty space. Above one cell of the tape is a head, which can either move left or right, and can read the symbols written in the cells. The head is also capable of erasing symbols and writing new symbols into the cells.

While Babbage and Lovelace can be considered the parents of computers and programming, Alan Turing is often called the father of computer science. During the war effort, Turing worked on cracking encrypted messages used by the German armed forces, and helped design machines to make this possible.

It was when working on a mathematical problem concerning Decidability and The Halting Problem, though, that Turing realised he needed a mathematical description of a machine that could solve algorithms. These machines came to be known as Turing machines.

What is a Turing Machine?

A Turing machine consists of an infinitely long tape, which has been divided up into cells. Each cell can contain either a 1, a 0, or an empty space. Above one cell of the tape is a head, which can either move left or right, and can read the symbols written in the cells. The head is also capable of erasing symbols and writing new symbols into the cells.

The direction that the head moves, which values it erases, and which values it enters, are dependent on a set of instructions provided to the machine.

In the machine shown below, the head is in state 2, and the instructions say that if it reads a 0, then erase it, write in a 1, move one cell to the right and change your state to 5. (There’ll also be instructions in the case it reads a 1.)

State 5 might say, if you’re reading a 1, erase it, write a 0, and move one cell to the left. Then go to state 7.

It seems incredible, but Alan Turing proved that any problem that is computable can be computed by a Turing machine using this simple system.

Build a Turing Machine

Now you are going to build a Turing machine and see how that might work.

You have your tape, your head, some initial instructions, and you will place some 1s and 0s on your tape to begin with.

The first symbol seen by the head is a 0, and the instructions say that when reading a 0, it should write a 1 and move left.

Moving left, it then reads a 1, which according to the instructions means it should write a 0 and move left.

It then reads another 1, writes a 0 and moves left, and finally reads a blank at which point the program halts.

So this Turing machine is designed to flip bits. It changes 0s to 1s and 1s to 0s.

The instructions for this Turing machine only had one state, but more complex Turing machines can be built using multiple states. You are going to have a look at one of these now.

You start in state A, with the head positioned underneath the first symbol. Looking at the instructions, you will see that in state A, with a symbol of 0, there’s no write instruction, the head should simply move to the right, and the Turing machine should remain in state A.

Next, it reads another 0 and again moves to the right. Then, it reads a 1, and moves to the right again. Now, it reads a blank, which means it’s got to move to the left, but it also changes to state B.

Looking at the instructions for state B, if it’s reading a 1, it erases it and writes a 0, and then moves left, staying in state B. Next, it reads a 0, which means it writes a 1, moves left and switches to state C.

In state C, there are no write actions perform. The head just keeps moving right until it eventually reaches a blank, at which point it moves left and the program halts.

Now it’s your turn to have a go. Get yourself a piece of paper, a pencil, and an eraser, and copy out the tape, the head, and the numbers exactly as you see them on the screen. With the machine in this state, run through the instructions again, and then once more. See if you can figure out what this Turing machine does.


text:
April 23, 2012

## Text-to-Speech in Adobe Captivate 5.5 – Create and Reuse!

There are different ways of adding audio to an eLearning project in Adobe Captivate 5.5 and one of the ways is to convert Text to Audio with the powerful functionality of  Text-to-Speech in Adobe Captivate. As I was thinking about it, a thread in LinkedIn drew my attention, that talked about creating Text-to-Speech audio and then reusing that audio in a different project. Leslie Bivens, an Adobe Captivate Expert, has added her comments with a solution and I thought of creating a tutorial based on that… Thank you Leslie

Watch this video to understand more about how to create Text-to-Speech and reuse it in Adobe Captivate.

July 13, 2011

## Captivate Hands-on Training: Enhancing Adobe Captivate Content with Audio

Topic: Enhancing Adobe Captivate Content with Audio

Description: Join Vish and Dr. Pooja Jaisingh for a hands-on session on using audio in Adobe Captivate. They will take you through different options to add, edit, and record audio to either slides or to the entire project. They will also show you how to add Text to Speech voices.

What’s in it for you? Continue reading…

August 6, 2010

## Transforming the narration of text using Loquendo Tags.

Through my previous blog series on text-to-speech, you have learnt about:

In this blog post, let us learn how to tweak the speech generated using Loquendo voices.

Loquendo allows you to control  how the text will be read out by the voices, such as the language in which the text will be read, the voice to be used, speaking rate, loudness, the interpretation of numbers, the stress prominence of a word and its pronunciation.

• You can specify these aspects by:Setting parameters in the system configuration files; or by
• Inserting commands directly into the input text (in slide notes)

We will explore the second approach, which is to tweak the speech by inserting commands directly into slide notes. The commands are grouped as follows:

I have used these controls in an Adobe Captivate project (.cptx and SWF file) and attached it here for your reference. To see the usage of tags, open the .cptx file, and select Audio > Speech Management. To hear how the voices modify the actual text, open the SWF file, plug-in your headphones and listen to the narration.

Do try out these tags and let us know your experience.

Stay tuned for my next blog post on “How to change the pronunciations of words used to generate speech”

## Global Controls

The family of Global Controls includes commands that change the value of some of the Reading Parameters of Loquendo TTS , which affect the quality of the output speech:

• Voice and Language
• Prosodic aspects of the voice (speaking rate, volume, voice pitch and timbre)
• Sound effects
• Text interpretation.

Voice Control: forces a voice switch between voices.

\voice=<mnemonic>

Example:

\voice=Simon hello. \voice=Stefan hi.

(“hello” is read by the voice “Simon”, then “hi” is read by the voice “Stefan”).

Language Control: forces a language switch between languages. The mnemonic must be the name of an installed language.

\language=<mnemonic>

Example:

\language=English Paris \language=French Paris.

(the first occurrence of the word Paris will be pronounced: p”}rIs , and the second: paR”i).

## Prosodic Control

The following commands allow the quality of the output voice to be controlled by modifying its rhythm, intonation, volume and timbre. The output speech is modified from the word following the command, up until the end of the prompt.

Speed Control: Allows the speaking rate to be modified, expressed in an abstract scale 0-100.

\speed=<num>

Example:

\speed=60 (Scale 0-100)

\speed=60 This text is read at a faster speed.

Pitch Control: allows the fundamental frequency (tone or pitch) to be modified, expressed in an abstract scale 0-100

\pitch=<num>

Example:

\pitch=60 (Scale 0-100)

\pitch=60 This text is read at a 60Hz frequency rating.

Volume Control: allows the volume (loudness) to be modified, expressed in an abstract scale 0-100 or in decibels (dB).

\volume=<num>

Example:

\volume=60 (Scale 0-100)

\volume=60 This text will be read at a 60 decibel rating.

Timbre Control: allows the voice timbre to be modified by a shift in frequency, expressed in an abstract scale 0-100.

\timbre =<num>

Example:

\timbre=60 (Scale 0-100)

\timbre=60 This text is read at a timbre value of 60.

## Sound Effects

The following commands create certain sound effects by acting on acoustic parameters of the speech output signal. For example, Reverb gives the impression of a large hall or a church, while delay (or echo) repeats the audio signal at every diminishing volume.

Reverb Effect: Creates reverbations with an intensity of <gain> and a delay of <delay> milliseconds

\reverb=<gain>,<delay>

Example:

\reverb=80,500 (0<gain<100, 0<delay<2000)

\reverb=0,0 (removes the reverb effect)

Robot Control: Applies the ‘robotization’ effect to the voice currently active in the system. There are 9 robots available: Robby, Gort, Twiki, Torg, Tobor, Ash, Hector, Max and Lynjx.

\robot=<robotName>

Example:

\robot=Max

\robot (removes the robotization effect)

Whisper effect: Applies the whisper effect to the voice currently active in the system. The possible values are: on, off.

Example:

\whisper=on

(the effect is active)

\whisper=off

(the effect is not active)

## Text Interpretation

The following commands control certain general aspects of text interpretation. Here, I describe how to adjust them synchronously with the text by means of a User Control embedded in the text.

The general syntax for these User Controls is the following:

\@<key>=<value>

where <key> is the name of the Reading Parameter to be changed and <value> is its chosen value.

Example:

\@TextEncoding=utf8 ( will interpret all the text as UTF8 format text. Characters like Ä/ä, Ö/ö, and Ü/ü will be read properly)

Pause insertion
Inserts a pause (silence) in the absence of punctuation marks. The effect is not applied  if punctuations already present in the text.

 \pause inserts a medium-length pause (120 ms), preceded by a ‘comma intonation’ \pause, inserts a medium-length pause (120 ms), preceded by a ‘comma intonation’ \pause. inserts a long pause (500ms), preceded by a ‘conclusive intonation’ \pause? inserts a long pause (500ms), preceded by a ‘question intonation’

Example:

Here \pause is a comma pause. (inserts a 120ms ‘comma intonation’ pause between “Here” and “is”)

Here \pause, is a comma pause. (leaves unaltered the ‘comma pause’ between “Here” and “is”)

Here \pause. is a conclusive pause. (inserts a 500ms ‘conclusion’ pause between “Here” and “is”)

Here \pause? is a question pause. (inserts a 500ms ‘question’ pause between “Here” and “is”)

Pause duration
When followed by a punctuation mark, forces the duration of the corresponding pause to <num> milliseconds. In the absence of punctuation, inserts a ‘comma intonation’ pause of <num> milliseconds.

\pause=<num> sets to <num> milliseconds the duration of the following pause

Example:

This \pause=10 , is a comma pause. (reduces to 10ms the following ‘comma intonation’ pause)

This \pause=10, is a comma pause. (reduces to 10ms the following ‘comma intonation’ pause)

This \pause=10 is a comma pause. (inserts a 10ms ‘comma intonation’ pause)

No final pause \pause=0. (reduces the final silence to a minimum duration, while keeping the conclusive intonation)

## Special Events

These commands trigger particular actions at the moment when the synthesis output reaches the exact point in the text where they have been inserted.

Play sound
Plays one of the paralinguistic sounds recorded for the voice in use. For most voices, the following sounds at least are available: Cough, Cry, Eh, Kiss, Laugh, Mmm, Oh, Sniff, Swallow, Throat, Whistle, and Yawn.

\item=<sound name>

Example:

\item=Laugh

## Audio Mixer features

The audio mixer allows synthetic speech to be mixed with sound files.

". wav" files are only supported and played.

Example 1:
This is \audio(play=<audioPath>/music.wav) a test.

Result:
“This is” will be read, then the music.wav will be played, then “a test” will be read.

Example 2:
This is \audio(mix=music.wav) a test.

Result:
Speech and music.wav will be mixed and heard together

Example 3:
This is \audio(mix=music.wav) \audio(volume=50) a test.

Result:
The volume of the audio file is set to 50% (from the start).

Attachments :

Link2 : Published SWF file containing all the tags mentioned above

Posted by Ashwin Bharghav4:04 PM Permalink
July 29, 2010

## Suggestions for optimal use of the Text to speech engine

Captivate 4 introduced text to speech (TTS) technology in Rapid eLearning authoring. Given the adoption and feedback, in Captivate 5 we’ve introduced more voices. Tweaking these voices seem to be one of the most discussed topics on our forums. There were a couple of blog posts last year on tweaking the Captivate 4 voices (VTML tags, User dictionary). These continue to be applicable in Captivate 5 for the NeoSpeech voices. Our other partner, Loquendo, also offers the ability to insert commands in the input text to modify the way words are pronounced. In the next week, we will have a few posts detailing this. But prior to that, here are some best practices to follow when using text-to-speech:

The TTS process exploits only a subset of the complex knowledge base on which a human reader implicitly relies. While it can access grammatical and phonetic knowledge, the artificial system does not come to a true comprehension of the text, lacking the necessary semantic and pragmatic skills. This is why the system cannot deal with ambiguous or misspelled text, nor give different emotional colors to its voices according to text semantics. The system tries to pronounce exactly what is written, applying the standard orthographic conventions for interpreting characters, symbols, numbers, word sentence delimiters and punctuation. The cues to a proper intonation are mainly punctuation marks and syntactic relationships between words.

This means that the best synthesis results will be obtained with well-formed sentences, correct and standard orthography, unambiguous contexts and rich and appropriate punctuation. If you are able to prepare or select in advance the texts that will be fed into the TTS system, then the main rule to follow is: “Write texts according to the standard orthographic and grammatical rules of the language”

Loquendo suggests that you keep to the following simple guidelines:

• Spell words carefully (using the correct character set for the language)
• Use capital letters when grammatically appropriate and apply standard conventions for representing numbers and abbreviations
• Separate words according to the standard orthographic conventions (insert blanks between words and after punctuation marks, when appropriate)
• Avoid ambiguities
• Write short sentences with correct syntactic structure
• Insert punctuation marks frequently and carefully
Posted by Shameer Ayyappan5:22 PM Permalink
June 23, 2010

## Rocket speed intro to TTS in Cp5

Got 4 minutes to spare. I’ve uploaded a short and to the point video overview of Text-To-Speech using Adobe Captivate 5. It includes a demo of the new French, German & British English voices as well as a step by step walk through of the simple process of adding TTS elements to your Captivate projects.

One of the most exciting aspects of Captivate 5′s TTS handling is the support for non-included voices. We know that people are trying to localize in many, many languages. We also hear from folks who prefer different, more specialized voices for different reasons. In order to support the broader needs and interests of eLearning developers, Captivate 5 actually examines the voices already installed on your computer. If you have special voices, it will find them and offer them as choices during the recording session.

As i’ve looked at this area, TTS, over the past several years, it does seem (especially with some of the recent advancements in ‘emotional’ and ‘evocative’ voices) that TTS is growing more and more viable, at least as a low cost solution – and in some cases, its really just the right solution. Have a look at the video to see how it is handled in Adobe Captivate 5.

As usual, the video is available in higher res – click the 360p tab on the video to choose different resolutions.

Posted by Allen Partridge4:15 AM Permalink
July 31, 2009

## Text-to-Speech – User Dictionary Editor

Adobe Captivate 4 Text-to-Speech feature gives a Dictionary Editor tool to add or change the pronunciation for a word. This tool is available at <Adobe Captivate Installation Dir>\VT\<agent>\M16\bin\UserDicEng.exe. The default version of this tool is not unlocked. So when a word is added/changed and its pronunciation is tested, it announces that it is a trial license. This post explains how to fix this.

Follow the steps below to correct this problem in your installation –

1. Open the Windows Explorer and browser to the directory where Adobe Captivate is installed. By default it is ‘C:\Program Files\Adobe\Adobe Captivate 4′.

2. If the Text-to-Speech engine is installed, then there will be a ‘VT’ directory in this directory. This directory will have directories for ‘Kate’ and ‘Paul’ agents.

3. Go to directory ‘VT/Kate/M16/bin’ directory.

4. Rename the ‘UserDicEng.exe’ as ‘UserDicEng_backup.exe’.

5. Download the new UserDicEng.exe file from here and put it in this directory.

6. Repeat the same steps for ‘Paul’ agent.

Now, you should not hear the trial license message when you add/change the pronunciation of any word.

Quick Tip – We have seen that ‘Create’ is not pronounced properly by the speech engine. So, you can fix this by specifying the target as ‘cree ate’ for this source word ‘create’.

Posted by Ashish Garg9:42 PM Permalink


text:
# A sum involving Fibonacci numbers, $\sum_{k=1}^\infty F_k/k!$

Let $F_k$ be Fibonacci numbers. I am looking for a closed form of the sum $\sum_{k=1}^\infty F_k/k!$.

I tried to use Wolfram Alpha, but it is not doing the sum Fibonacci[k]/k! , k=1 to infinity.

Can someone tell what is the problem with WA and what this sum equals to?

A hint:

You can prove for yourself or find in a book a formula of the form $${\rm fib}(k)=a\lambda^k + b\mu^k$$ with certain constants $a$, $b$, $\lambda$, $\mu$. The requested sum can then be easily written as a sum of two exponentials.

Im getting the solution: $$\frac{e^\phi - e^\varphi}{\sqrt{5}}$$

Since it is known that $F_n = \frac{\phi^n - \varphi^n}{\sqrt{5}}$.

$$\sum_{k=0}^\infty \frac{F_k}{k!} = \sum_{k=0}^\infty \frac{\phi^k - \varphi^k}{k!\sqrt{5}} = \frac{1}{\sqrt{5}}\left[\sum_{k=0}^\infty \frac{\phi^k}{k!} - \sum_{k=0}^\infty\frac{\varphi^k}{k!}\right]$$

Youre welcome.

• Im sure the down voting was well-deserved. – CogitoErgoCogitoSum Dec 6 '14 at 22:09
• Why bump an old question? – Cyclohexanol. Dec 6 '14 at 22:22
• I dont check the dates on these questions. It was near the top of the list already when I clicked on it. Besides, how does a legitimate and valid answer deserve a down thumb, regardless of how old it is? – CogitoErgoCogitoSum Dec 6 '14 at 22:25
• I did not downvote, just stating a possible reason. – Cyclohexanol. Dec 6 '14 at 22:27

Using Mma, Limit[Sum[Fibonacci[k]/k!, {k, 1, p}], p -> Infinity] gives an answer and
Sum[Fibonacci[k]/k!, {k, 1, Infinity}] gives the same. Is your problem with Wolfram Alpha ?

• computation time out every time. Now please tell the numerical value now. – Waqar Ahmad Nov 14 '13 at 12:00
• The numerical value is 2.014322733458315736581346. But I think you should establish the analytical value using what lab bhattacharjee sent you. The formula is quite simple. – Claude Leibovici Nov 14 '13 at 12:07
• According to Maple, with: evalf(sum(fibonacci(k)/k!,k=1..infinity)); Giving: 2.014322733. But, does there exist an "exact" solution? Hmm, "the formula is quite simple". I'm quite curious .. – Han de Bruijn Nov 14 '13 at 12:09
• @HandeBruijn. Yes and it is (-1 + E**Sqrt(5))/(Sqrt(5)*E**(2/(1 + Sqrt(5)))) – Claude Leibovici Nov 14 '13 at 12:10
• @Claude_Leibovici: Numerically the same, OK, but how to prove it? – Han de Bruijn Nov 14 '13 at 12:16

Solve y'' = y' + y both via power series and then by elementary methods (substitution of exponentials) and setting boundary conditions y(0) = 0, and y'(0) = 1. The result is the following identity:

$$\sum_{n=1}^\infty \frac{F_n}{n!} x^n = \frac{2}{\sqrt{5}} e^\frac{x}{2} \sinh(\frac{\sqrt{5}}{2}x)$$

If you don't feel like solving the differential equation yourself you can easily verify by substituting each side of the above equation into the differential equation and boundary conditions and then remind yourself via the existence and uniqueness theorem that the two solutions must be the same. Meaning they are the same at the point x=1.

Setting x=1 gives

$$\sum_{n=1}^\infty \frac{F_n}{n!} = 2 \sqrt{\frac{e}{5}} \sinh(\frac{\sqrt{5}}{2}) = \frac{e^{\phi_+} - e^{\phi_-}}{\sqrt{5}}$$

where $$\phi_\pm = \frac{1 \pm \sqrt{5}}{2}$$ are the golden ratio and it's congugate. I.e. the two solutions to the quadratic equation $$y^2 = y + 1$$. (Note that this is the same as the original differential equation with the order of the derivative replaced by the power of y: $$y^{(n)} \rightarrow y^n$$. Is there a reason for this or is it just an interesting coincidence?)

The numerical value to 1234 decimal places is:

2.0143227334583157365813462554697591356591114695811241821088403766742128397097006637111011319457016312404491456095258793427009006861303485673596801638909160410193717669927023995171048184677900815088560097082872535199871890392172995661754732113194156588018029328948474213043876788216048899918772417664402571617571139330772358975323915258959445759063347851074795879633549994051398333466312388233898626203203269110709326570399986964305691732913351786602533868094239267790638117143928091599301761319009602970110828764189441445889515565022310301247296449293897072832251096862983584326320448058898294430874162784325966040344781269902152316671151601869355616381547811631999794833349897492383827197553543970970099927748899115338963618621987298237221140308693191390458816782367508072038660426129042190625209844154130995652380767395098326152648319830337117521840226032419661899959469674966107384253745331849405956049149602323282511659984227035172708526550586977824786544373977008821755560614931363394267293413160342932047952358802051132151029521708819143029550619292413657149885241166291353201988752249793933504301075491208793989947801376838994380077603990476851019102426649383266999848219352544560978465856984179911553255421975337456890201316269

A fun exercise is to write a C program to generate this (not as easy as you might assume due to the limits of native C data types).

• Simply stating a solution is not helpful in order to understand the problem in more detail. Please add you whole evaluation. – mrtaurho Nov 17 '18 at 17:33
• I have added further detail which should satisfy your complaint but I won't give more than that. It is a fun problem which should be solvable by an undergraduate differential equations student and I'm not about to ruin their fun. I'm sure their professor expects them to solve it themselves and not simply copy a solution off the internet. If it were my class I would also require them to write a computer program to calculate the sum of the series to 15 decimal places (i.e. size of double precision floating point). A student getting 200 decimal places from a program written in C would impress me. – 1o_o7 Nov 19 '18 at 10:31


text:
1. [SOLVED] Complex Trig proof

*This is exactly how the question reads from the book

Using the fact that a complex equation is really two real equations, find the double angle formulas (for Sin2 $\theta\$, Cos2 $\theta\$ ) by using this equation: (e^i $\theta\$)^n = (Cos $\theta\$+ iSin $\theta\$)^n = Cos n $\theta\$ + iSin n $\theta\$

I know if you put in 2 for n and just exapand it out I can get it but I figured that there might be some other way to prove it. Just seems a bit simple but any help or insight would be greatly appriciated.

2. Hello,
Originally Posted by gidget
*This is exactly how the question reads from the book

Using the fact that a complex equation is really two real equations, find the double angle formulas (for Sin2 $\theta\$, Cos2 $\theta\$ ) by using this equation: (e^i $\theta\$)^n = (Cos $\theta\$+ iSin $\theta\$)^n = Cos n $\theta\$ + iSin n $\theta\$

I know if you put in 2 for n and just exapand it out I can get it but I figured that there might be some other way to prove it. Just seems a bit simple but any help or insight would be greatly appriciated.
Nope, you'll just have to expan $(\cos \theta + i \sin \theta)^2$

Note that $\cos(2 \theta)$ is the real part of it, and $\sin(2 \theta)$ the imaginary part.

3. Hello, gidget!

Using the fact that a complex equation is really two real equations,
find the double-angle formulas for $\sin2\theta,\;\cos2\theta$
by using this equation: $(e^{i\theta})^n \:=\: (\cos\theta + i\sin\theta)^n \:=\: \cos(n]\theta) + i\sin(n\theta)$

I know if you put in $n=2$ and just expand it out. I can get it.
But I figured that there might be some other way to prove it.
. .
I don't know of a simpler way.
I'm truly grateful for this method of cranking out multiple-angle identities.

If we want a formula for $\cos3\theta$, we'd normally proceed like this:

$\cos3\theta \;=\;\cos(\theta + 2\theta)$

. . . . $= \;\cos\theta\cos2\theta - \sin\theta\sin2\theta$

. . . . $= \;\cos\theta (2\cos^2\!\theta - 1) - \sin\theta(2\sin\theta\cos\theta)$

. . . . $= \;2\cos^3\!\theta - \cos\theta - 2\sin^2\!\theta\cos\theta$

. . . . $= \;2\cos^3\!\theta - \cos\theta - 2(1-\cos^2\!\theta)\cos\theta$

. . . . $= \;2\cos^3\!\theta - \cos\theta - 2\cos\theta + 2\cos^3\!\theta$

Therefore: . $\boxed{\cos3\theta \;=\;4\cos^3\!\theta - 3\cos\theta}$

Then we must repeat the process for $\sin3\theta.$

Using Demoivre's Formula, we have:

$\cos(3\theta) + i\sin(3\theta) \;=\;(\cos\theta + i\sin\theta)^3$

. . . . . . . . . . . . $=\;\cos^3\!\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\!\theta - i\sin^3\!\theta$

. . . . . . . . . . . . $= \;\left(\cos^3\!\theta - 3\cos\theta\sin^2\!\theta\right) + i\left(3\cos^2\!\theta\sin\theta - \sin^3\!\theta\right)$

. . . . . . . . . . . . $=\;\left(\cos^3\!\theta - 3\cos\theta[1-\cos^2\!\theta]\right) + i\left(3[1-\sin^2\!\theta]\sin\theta - \sin^3\!\theta\right)$

. . . . . . . . . . . . $= \;\left(4\cos^3\!\theta - 3\cos\theta\right) + i\left(3\sin\theta - 4\sin^3\!\theta\right)$

Equating real and imaginary components, we have:

. . . . $\begin{array}{ccc} \boxed{\cos3\theta \;=\; 4\cos^3\!\theta - 3\cos\theta} \\ \\[-3mm] \boxed{\sin3\theta \;=\; 3\sin\theta - 4\sin^3\theta} \end{array}$

And we have both formulas . . . with much less work!


text:
# Meters (m) to Kilometers (km) Conversion

Round:
Enter Meter
Enter Kilometer

## How many meters in a kilometer?

There are 1000 meters in 1 kilometer (km). To convert kilometers to meters, multiply the kilometer value by 1000.

For example, to find out how many meters in a kilometer and a half, multiply 1.5 by 1000, that makes 1500 meters in a kilometer and a half.

1 Kilometer = 1000 Meters

How many meters in 1/2 kilometer?

There are 500 meters in 1/2 km.

How many meters in 1/3 kilometer?

There are 333.334 meters in 1/3 km.

How many meters in 1/4 kilometer?

There are 250 meters in 1/4 km.

How many meters in 1/8 kilometer?

There are 125 meters in 1/8 km.

## How many kilometers in a meter?

1 Meter is equal to 0.001 kilometer. To convert meters to km, multiply the meter value by 0.001 or divide by 1000.

1 Meter = 0.001 Kilometer

You may also use this length and distance units conversion calculator tool to convert between kilometers, meters and all other length and distance units.

What is Kilometer?

Kilometer (kilometre) is a metric length unit. 1 km = 1000 m. The symbol is "km".

What is Meter?

Meter (metre) is a metric system base length unit. 1 m = 0.001 km. The symbol is "m".

Create Conversion Table
Click "Create Table". Enter a "Start" value (5, 100 etc). Select an "Increment" value (0.01, 5 etc) and select "Accuracy" to round the result.


text:
# Integrate cos(x^5)

1. Oct 19, 2007

### Artaxerxes

Can you solve this $\int \cos x^5 dx$ ?

2. Oct 19, 2007

### arildno

No, I can't

3. Oct 19, 2007

### rocomath

nope ...

4. Oct 19, 2007

### Zurtex

It probably can't be done in terms of elementary functions. Mathematica has an answer of 2 different forms:

Take E(v,z) here to be this function http://functions.wolfram.com/GammaBetaErf/ExpIntegralE/ we have:

$$\int \cos x^5 dx = - \frac{1}{10} \left( E\left( \frac{4}{5} , -i x^5 \right) + E\left( \frac{4}{5} , i x^5 \right) \right)$$

Or in terms of the gamma function http://functions.wolfram.com/GammaBetaErf/Gamma/ :

$$\int \cos x^5 dx = - \frac{\left( x^{10} \right)^{\frac{4}{5}}}{10x^9}\left( \left(i x^5 \right)^{\frac{1}{5}} \Gamma \left( \frac{1}{5} , -i x^5 \right) + \left(-i x^5 \right)^{\frac{1}{5}} \Gamma \left( \frac{1}{5} , i x^5 \right) \right)$$

I don't really understand why mathematica didn't simplify it further, so I've tried to keep it to what mathematica outputted.

Last edited: Oct 19, 2007
5. Oct 19, 2007

### Artaxerxes

Thank you!

6. Oct 19, 2007

### DavidSmith

thats why series solutions are so conviennt

7. Oct 20, 2007

### Gib Z

The Taylor series for cos converges for all real x, so you should have just let x^5 = u in the taylor expansion of cos u, integrated term by term and you are left with an even nicer result than what mathematicia gives out.


text:
# Can you make that wonderful observation?

Geometry Level 1

A triangle has sides of magnitude $1$, $\sin x$, and $\cos x$.

where $0 < x < \frac { \pi }{ 2 } .$

Find the largest angle of the triangle in degrees.

Assume that the triangle is non-degenerate.

×


text:
## The norm of a derivation.(English)Zbl 0197.10501

Summary: In this paper, we determine the norm of the inner derivation $$\mathfrak Q_T\colon A \to TA - AT$$ acting on the Banach algebra $$\mathfrak B(H)$$ of all bounded linear operators on Hilbert space. More precisely, we show that $$\Vert\mathfrak Q_T\Vert = \inf\{2\,\Vert T - \lambad I\Vert : \lambda\] complex \(\}$$. If $$T$$ is normal, then $$\Vert\mathfrak Q_T\Vert can be specified in terms of the geometry of the spectrum of \(T$$.

### MSC:

 47-XX Operator theory 46-XX Functional analysis
Full Text:


text:
# Proof of a focus point on parabola and tangent line equal angles

## Homework Statement

Prove that a vertical line and a line going from a point on a parabola to the focus of the parabola form equal angles with the tangent line of the point on the parabola.

## Homework Equations

Focus = 1/4a (maybe relevant)

## The Attempt at a Solution

I know how to prove that the triangle from the vertical line, midpoint of Focus point to an arbitrary line and the point on the parabola is equal to a triangle that goes from focus point to point on parabola to midpoint.

However, I have no clue how to show that these two angles are the same. I can find the slope of each line, obviously, but where to go from here?

Related Precalculus Mathematics Homework Help News on Phys.org
Mentallic
Homework Helper
Take any arbitrary point $P(2ap,ap^2)$ on the parabola $x^2=4ay$ where $(0,a)$ is the focus. Now, find the gradient of the tangent to the parabola which touches at P, also take the gradient of the line connecting the focus and the point P. Now find the angle between these 2 lines with the equation:
$$tan\theta=\left |\frac{m_1-m_2}{1+m_1m_2} \right |$$
Now take the gradient of a vertical line which is $1/0$ (don't worry that it is undefined, with the tan function that just means $\theta=\pi/2$) and now show the angle between that tangent line and the vertical line is the same.