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This question is intended to kind of rekindle this old question, which was apparently very hard didn't receive a satisfactory answer. I'm aware that the hope of a definite answer is quite slim, but I'm still very curious to know:
Suppose $f:U\to \Bbb R$ is a function, $U$ is an open subset of $\Bbb R^n$, and $x_0\in U$, and the partial derivatives $\partial_i|_{x_0}f,\,i=1,\cdots,n$ exist. Then what regularity conditions on $\partial_i|_{x_0}f,\,i=1,\cdots,n$ are a sufficient and necessary condition for $f$ to be differentiable at $x_0$?
The mere existence of $\partial_i|_{x_0}f$ is far from enough. The continuity of all of them is over-sufficient. The weakest sufficient condition AFAIK is this one, i.e. the continuity of all but one of them, which is nevertheless still over-sufficient.
Given that we can easily construct a function differentiable at $x_0$ yet has all its partial derivatives discontinuous at $x_0$, like $$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\dfrac{1}{\sqrt{x^2+y^2}}\right) & \text{ if $(x,y) \ne (0,0)$}\\0 & \text{ if $(x,y) = (0,0)$}.\end{cases}$$ this problem I believe is intrinsically hard.
Has any research been done that can shed some light on this complicated problem? |
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We were doing $\ce{O2}$ estimation by Winkler's method
The procedure and reactions involved in this experiment are: Carefully fill a 300-mL glass stoppered bottle brim-full with sample water. Immediately add 2mL of manganese sulfate to the collection bottle. Add 2 mL of alkali-iodide reagent (NaOH + KI) in the same manner. Stopper the bottle with care to be sure no air is introduced. Mix the sample by inverting several times. Add 2 mL of concentrated sulfuric acid.Then mix well. In a glass flask, titrate 100 mL of the sample with sodium thiosulfate to a pale straw color. Titrate by slowly dropping titrant solution from a calibrated pipette into the flask and continually stirring or swirling the sample water. Add 2 mL of starch solution so a blue color forms. Continue slowly titrating until the sample turns clear.
$$\begin{align}\ce{MnSO4 + 2NaOH &-> \underset{{white ppt.}}{Mn(OH)2} + Na2SO4}\\[0.6em] % \ce{2Mn(OH)2 + O2 &-> \underset{{orange-brown cloudy ppt.}}{2MnO(OH)2}}\\[0.6em] % \ce{2Na2SO4 + 2H2SO4 &-> 4 NaHSO4}\\[0.6em] % \ce{MnO(OH)2 + NaHSO4 + 2KI &-> \underset{{golden-brown colour}}{I2} + MnSO4 + K2SO4 \\&\quad\quad + 2Na2SO4 + 3H2O}\\[0.6em] % \ce{I2 + 2Na2S2O3 &-> 2NaI + Na2S4O6}\end{align}$$
And at the end of the experiment we needed to calculate the $\ce{O2}$ concentration in mg/l by the following equation:
$$c(\ce{O2})\ \mathrm{(mg/L)} = \frac{(V_1 \times N \times E \times 1000) }{[V_4 {(V_2-V_3)/V_2}]}$$
Here,
$V_1=$ Volume of titrant $\ce{Na2S2O3}$ (ml)
$V_2=$ Total volume of water sample (ml)
$V_3=$ Volume of ($\ce{MnSO4 + KI}$ solution) added (ml)
$V_4=$ Volume of analyte taken in titration
$E=$ Equivalent weight of $\ce{O2}$
$N=$ Normality of $\ce{Na2S2O3}$ solution
So how can I derive this equation?
Reference followed: Ghosh, K. C. & Manna, B. (2005) Practical Zoology
A different book though had a different formula :
Conc. of O2 (mg/l)= 0.2 x (Volume of sample water in ml) x (Strength of $\ce{Na2S2O3}$) |
I am trying to solve the equation
$\frac{d}{dt}V(t)=r(t)V(t)+\pi-\mu(x+t)(b_d-V(t))$
numerically using the R function 'ode'. This is a Thiele differential equation for a life insurance reserve with premium rate $\pi$, mortality intensity $\mu$ (for an $x+t$ year old), death benefit $b_d$ and interest rate process $r$.
I want to do this in a so called unit-linked setting, where the returns on the policy are generated by investments in stocks, hence I have assumed a Black Scholes model for simplicity.
I generate a Geometric Brownian Motion. As seen the time interval is 40 years and the step size of the simulation is $40/100000$. For each simulated point I calculate the returns as
$r[i]=\frac{S[i+1]-S[i]}{S[i]}$
Such that I have $100000$ return values. Plotting these two for two very different scenarios yields
My problem is, at I would expect the reserve process to vary a lot more, much like the simulated Geometric Brownian Motion. In essence, it is too smooth, i think. Also if I generate GBM trajectories which end in, say, the value 500 and 10 respectively, the difference in the final value of the reserve varies very little.
Does anyone know why this is, am I doing something wrong? The R-code is attached.
maturity <- 40simulation.length <- 100001dt <- maturity/(simulation.length-1)timeline <- seq(0,maturity, dt)BM <- GBM <- EV <- rep(0, times=simulation.length)EV[1] <- GBM[1] <- S0for(i in 2:simulation.length){ BM[i] <- BM[i-1]+sqrt(dt)*rnorm(1) GBM[i] <- GBM[1]*exp((mu-(sigma^2)/2)*(i-1)*dt+sigma*BM[i]) EV[i] <- EV[1]*exp(mu*(i-1)*dt)}return <- rep(0,length(GBM))returns[1] <- 0for (i in 2:length(GBM)-1){ returns[i] <- (GBM[i+1] - GBM[i]) / GBM[i]}dV <- function(t,V, parms){ list(returns[t/0.0004+1] * V + premiumRate - mortalityIntensity(t+25) * (deathBenefit - V))}out <- ode(y = 0, times = timeline, func = dV, parms = NULL)interpolatedReserve <- approxfun(times,out[,2], method="linear")
The problem is not that the
approxfun function is not good enough, because I used it on the GBM to plot the green trajectory.
I use 60000 as the premium rate (around 1000 dollars per month paid to pension) and a death benefit of 1.000.000 (realistic numbers in Danish Kroner). Is it because the returns $r\cdot V$ are too small to be noticed compared to the yearly premium rate? I would just suspect the reserve plot to follow the movements of the GBM roughly? |
First of all, rates of convergence are usually given in the form$$ \|u-u_h\| \leq C N^\alpha,$$rather than equality. Furthermore, rates are
asymptotic, i.e., only have to hold for $N\to \infty$. This means that you're unlikely to find a single $C$ and $\alpha$ such that your equation holds.
Another reason why your approach doesn't work is that what you're trying to fit is
not a polynomial, since $\alpha$ is a) unknown and b) not an integer (for one thing, it must be negative since the error goes down as $N$ goes up).
What people usually do is look at a
double-logarithmic plot: If you take the logarithm of your equation (or my inequality), you get$$ log(\|u-u_h\|) \leq \log(CN^\alpha) = \log(C) + \alpha\log(N).$$This is a linear polynomial $ax+b$ in $\log (N)$ with coefficients $a=\alpha$ and $b=\log(C)$ (from which you can find $C=\exp(b)$).
So if you apply
polyfit to $\log(N)$, $\log(\|u-u_h\|)$, you should get an array $[a,b]$ with $\alpha=a$ and $C=\exp(b)$. |
Category: Ring theory Problem 624
Let $R$ and $R’$ be commutative rings and let $f:R\to R’$ be a ring homomorphism.
Let $I$ and $I’$ be ideals of $R$ and $R’$, respectively. (a) Prove that $f(\sqrt{I}\,) \subset \sqrt{f(I)}$. (b) Prove that $\sqrt{f^{-1}(I’)}=f^{-1}(\sqrt{I’})$
Add to solve later
(c) Suppose that $f$ is surjective and $\ker(f)\subset I$. Then prove that $f(\sqrt{I}\,) =\sqrt{f(I)}$ Problem 618
Let $R$ be a commutative ring with $1$ such that every element $x$ in $R$ is idempotent, that is, $x^2=x$. (Such a ring is called a
Boolean ring.) (a) Prove that $x^n=x$ for any positive integer $n$.
Add to solve later
(b) Prove that $R$ does not have a nonzero nilpotent element. Problem 543
Let $R$ be a ring with $1$.
Suppose that $a, b$ are elements in $R$ such that \[ab=1 \text{ and } ba\neq 1.\] (a) Prove that $1-ba$ is idempotent. (b) Prove that $b^n(1-ba)$ is nilpotent for each positive integer $n$.
Add to solve later
(c) Prove that the ring $R$ has infinitely many nilpotent elements. |
I felt I needed to write an additional answer to try to clear my mind about the question. Here is the try, step by step. Caveat: for simplicity, I used the same notation $C$ of a function of reals $u$ and $v$, for its rewriting in $x=u+iv$ and $\bar x$, and on complex $x$ alone. I hope it is not confusing for the reader.
Let $C(x)$ be a function (we don't assume it real for now) of a complex vector $x$. If $x$ were real, it could come from the left (negative) or the right (positive) direction, and the notion of derivative is simpler. For instance, take $C(x)=x^2$; a derivative at $x=x_0$ is given by the limit (if it exists) of $$\frac{C(x_0+\Delta x) -C(x_0) }{\Delta x}=\frac{x_0^2+2x_0\Delta x+\Delta x^2 - x_0^2}{\Delta x}=2x_0+\Delta x\,.$$
So, the limit is well-defined here.
When dealing with optimization in real variables, this can be satisfactory if one performs least-square optimization, as detailed by Matt L.. With reals, $x$ can approach $x_0$ only from the left (negative) or the right (positive), which makes differentiation relatively easy. With complex numbers, derivation seems more involved, as $x$ can approach $x_0$ from all angles/directions. And $x\to x^2$ is no longer the graph of a 2D nice positive parabola. So what happens when one want to optimize a simple nice convex function such as:
$$C(x)=|x|^2 = x\cdot \overline{x}\,?$$
$$\frac{C(x_0+\Delta x) -C(x_0) }{\Delta x}=\frac{x_0\overline{x_0}+x_0\overline{\Delta x}+\overline{x_0}\Delta x+|\Delta x|^2 - x_0\overline{x_0}}{\Delta x}=x_0\frac{\overline{\Delta x}}{{\Delta x}}+\overline{x_0}+\overline{\Delta x}\,.$$
In the real case, $\frac{\overline{\Delta x}}{{\Delta x}}=1$ and $\overline{x_0}={x_0}$, and we go back to normal with a $2x_0$ derivative. Sadly, in the complex case, the term with $\frac{\overline{\Delta x}}{{\Delta x}}$ is unresolved, as it has not clear limit (it has unit modulus), unless with $x_0=0$.
Yet, we would like to have a broader notion of derivation (for extremum extraction), allowing this function to be differentiable. So we can reformulate it with $x=u+iv$ (or "get back to real") and look for stationnary points where:
$$ \frac{\partial C(u,v)}{du} = \frac{\partial C(u,v)}{dv} =0\,.$$
This may work, but require to convert an all $x$ (complex) function in real and imaginary parts, which can be cumbersome. Since $u=\frac{1}{2}(x+\bar x)$, $v=\frac{1}{2}(x-\bar x)$, one can rewrite $C(u,v)$ as ${C}(x,\overline{x})$.With the chain rule:$$\frac{\partial}{\partial u}=\frac{\partial x}{\partial u}\frac{\partial}{\partial x}+\frac{\partial\bar x}{\partial u}\frac{\partial}{\partial\bar x}=\frac{\partial}{\partial x}+\frac{\partial}{\partial\bar x}\,.$$
Similarly, one gets:
$$\frac{\partial}{\partial v}=i\left(\frac{\partial}{\partial x}-\frac{\partial}{\partial\bar x}\right)\,.$$
Then, resolving the above (invertible) system in both $x$ and $\bar x$, that can be treated as another pair of "independent" variables, and one get the Wirtinger derivatives:
$$\frac{\partial}{\partial x}=\frac{1}{2}\left(\frac{\partial}{\partial u}-i\frac{\partial}{\partial v}\right)$$and $$\frac{\partial}{\partial \bar x}=\frac{1}{2}\left(\frac{\partial}{\partial u}+i\frac{\partial}{\partial v}\right)\,.$$
Wirtinger calculus has its merits: since $\frac{\partial}{\partial x} \bar x = \frac{\partial}{\partial \bar x} x =0 $, $\bar x $ can be regarded as a constant when differentiating with respect to $ x $ and vice-versa (hence the notion of "independence" used above. For the squared error ${C}(x,\overline{x}) = x.\bar x$ undergoes the Wirtinger derivation as:
$$\frac{\partial}{\partial \bar x}{C}(x,\overline{x}) = x$$and $$\frac{\partial}{\partial x}{C}(x,\overline{x}) = \bar x\,.$$
In the general case, Cauchy–Riemann equations do appear: if $C$ is Fréchet-differentiable (T. Tao, Notes 1: complex differentiation, exercise 23), $C$ is holomorphic if and only if $\frac{\partial}{\partial \bar x}{C}(x,\overline{x})=0$ everywhere on some open domain of $\mathbb{C}$. In other words, ${C}$ need to be $\overline{x}$-free to be holomorphic. Roughly, a real-function over $\mathbb{C}$ is rarely holomorphic, unless it is constant. This is somewhat intuitive: to be real, a $x$ in the expression of ${C}$ should be counterbalanced by some $\overline{x}$ that un-complexes it. So, Wirtinger derivatives elegantly re-express CS conditions in a compact way.To find stationnary points, you now can look where $\partial_u$ and $\partial_v$ vanish, or equivalently where $\partial_{x}$ and $\partial_\overline{x}$ do. You have not gain too much, apart from delaing with the complex variables directly, instead of converting everything to real and imaginary parts.
Now, what happen if $C$ is real? Then, there is some redundancy in the equations, that can be exploited. For instance, both partial derivatives $\frac{\partial}{\partial u}$ and $\frac{\partial}{\partial v}$ ought to be real. In other words, they are equal to their conjugate. And it turns out that this amounts to the derivative of ${C}(x,\overline{x})$ with respect to the first variable being zero.
So now,
only one equation suffices to find the stationnary points of a real function of a complex variable:
$$\frac{\partial C(x)}{\partial x}=0\,,$$and the steepest ascent points (under some conditions) to the direction of:
$$\frac{\partial C(x)}{\partial \bar x}\,.$$
Resultantly, a natural extention of gradient descent is therefore written as:
$$x_{k+1}=x_{k}-\mu \frac{\partial C(x)}{\partial \bar x}x_{k}\,,$$which can find good approximations of the extremas, instead of trying to solve the vanishing equation in $\partial x$.Hoping I did not made mistakes in the notations, more information can be found in:
Wirtinger’s Calculus in general Hilbert Spaces, Pantelis Bouboulis, 2010 An Introduction to Complex Differentials and Complex Differentiability, Raphael Hunger, 2007, Technical Report, TUM-LNS-TR-07-06 Complex differentiation is quite complex, Stephen Tu, 2016 Notes 1: complex differentiation, Terence Tao, 2016 dz and dz bar: How to derive the Wirtinger derivatives What is the intuition behind the Wirtinger derivatives? |
Consider the system of equations formulated in a previous question (optical fibers continuity of tangent field components across the core-cladding interface):
$$ \left\{ \begin{array}{c} e_z^{(1)}(r,\phi)|_{r = a} = e_z^{(2)}(r,\phi)|_{r = a}\\ e_{\phi}^{(1)}(r,\phi)|_{r = a} = e_{\phi}^{(2)}(r,\phi)|_{r = a}\\ h_z^{(1)}(r,\phi)|_{r = a} = h_z^{(2)}(r,\phi)|_{r = a}\\ h_{\phi}^{(1)}(r,\phi)|_{r = a} = h_{\phi}^{(2)}(r,\phi)|_{r = a} \end{array} \right. $$
Its determinant should be $0$, to obtain a non-trivial solution for the unknown amplitudes $C_1$, $C_2$, $D_1$, $D_2$. According to Maxwell's equations, $e_{\phi}$ and $h_{\phi}$ depend on the derivative of $e_z$ and $h_z$ with respect to $r$ and $\phi$. Books use some tricks to avoid having a function of $\phi$ in the computation of these derivatives. See for example John A. Buck,
Fundamentals of Optical Fibers, which assumes (Table 3.2):
$$A(\phi) = \cos (\nu \phi), \ \mathrm{for} \ e_{\phi}\\ A(\phi) = \sin (\nu \phi), \ \mathrm{for} \ h_{\phi}$$
The resulting eigenvalue equation is:
$$\left[ \frac{J'_{\nu}(k_{c_1} a)}{k_{c_1}a J_{\nu}(k_{c_1} a)} + \frac{K'_{\nu}(|k_{c_2}| a)}{|k_{c_2}|a K_{\nu}(|k_{c_2}| a)} \right] \left[ \frac{n_1^2}{n_2^2}\frac{J'_{\nu}(k_{c_1} a)}{k_{c_1}a J_{\nu}(k_{c_1} a)} + \frac{K'_{\nu}(|k_{c_2}| a)}{|k_{c_2}|a K_{\nu}(|k_{c_2}| a)} \right] = \nu^2 \left( \frac{1}{k_{c_1}^2 a^2} + \frac{1}{|k_{c_2}|^2 a^2} \right)\left( \frac{n_1^2}{n_2^2} \frac{1}{k_{c_1}^2 a^2} + \frac{1}{|k_{c_2}|^2 a^2}\right)$$
Consider instead (respectively for the Electric and Magnetic fields)
$$A(\phi) = C_3 \sin(\theta) + C_4 \cos(\theta) = C' \cos (\nu \phi + \alpha)\\ A(\phi) = D_3 \sin(\theta) + D_4 \cos(\theta) = D' \cos (\nu \phi + \beta)$$
(in fact, according to this answer, $A(\phi) = A \sin(\theta) + B \cos(\theta)$ can be re-written as $A'\sin(\theta+\rho)$ or $A'\cos(\theta+\alpha)$). In the most general case, $\alpha \neq \beta$.
1) How would the eigenvalue equation be in this more general case?
2) Is there any textbook dealing with this computation? |
Category: Group Theory
Group Theory Problems and Solutions.
Popular posts in Group Theory are:
Problem 625
Let $G$ be a group and let $H_1, H_2$ be subgroups of $G$ such that $H_1 \not \subset H_2$ and $H_2 \not \subset H_1$.
(a) Prove that the union $H_1 \cup H_2$ is never a subgroup in $G$.
Add to solve later
(b) Prove that a group cannot be written as the union of two proper subgroups. Problem 616
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$.
Add to solve later
(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. Problem 613
Let $m$ and $n$ be positive integers such that $m \mid n$.
(a) Prove that the map $\phi:\Zmod{n} \to \Zmod{m}$ sending $a+n\Z$ to $a+m\Z$ for any $a\in \Z$ is well-defined. (b) Prove that $\phi$ is a group homomorphism. (c) Prove that $\phi$ is surjective.
Add to solve later
(d) Determine the group structure of the kernel of $\phi$. If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 497
Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively. Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.
Also determine whether the statement is true if $G$ is a non-abelian group.Add to solve later |
As a quick note, the equations of motion that come from that Lagrangian are
$$\frac{d}{dt}\left(\dot q_1 + \dot q_2\right) = -2kq_1^3$$$$\frac{d}{dt}\left(\dot q_2 + \dot q_1\right) = -2kq_2^3$$
The Dirac procedure for singular Lagrangians goes as follows:
Step 1: Calculate the generalized momenta as usual$$p_1 \equiv \frac{\partial L}{\partial \dot q_1} = \dot q_1+\dot q_2$$$$p_2 \equiv \frac{\partial L}{\partial \dot q_2} = \dot q_1+\dot q_2 = p_1$$
Clearly this is not invertible. We have one "good" equation (defining $p_1$ in terms of the generalized velocities) and one "bad" equation ($G\equiv p_2-p_1 = 0$, an algebraic relation between the momenta in which the velocities do not appear).
$G=0$ is called a
primary constraint of the Dirac procedure - an algebraic relation between momenta and (possibly) coordinates, in which the generalized velocities are absent.
Step 2: Compute the naive Hamiltonian
If we compute the Hamiltonian as usual, we find
$$H_0 = p_1\dot q_1 + p_2 \dot q_2 - L = p_1(p_1-\dot q_2) + p_2 \dot q_2 - \frac{1}{2}p_1^2 + \frac{k}{2}(q_1^4+q_2^4)$$$$ = \frac{p_1^2}{2} + (p_2-p_1)\dot q_2 + \frac{k}{2}(q_1^4+q_2^4)$$$$ = \frac{p_1^2}{2} + \frac{k}{2}(q_1^4+q_2^4)$$
If you calculate the Hamilton equations, you will see that they don't match with the Lagrangian equations:
$$\dot p_1 = -\frac{\partial H_0}{\partial q_1} = -2kq_1^3$$$$\dot p_2 = -\frac{\partial H_0}{\partial q_2} = -2kq_2^3$$$$\dot q_1 = \frac{\partial H_0}{\partial p_1} = p_1$$$$\dot q_2 = \frac{\partial H_0}{\partial p_2} = 0$$
Step 3: Extend the phase space and construct the complete Hamiltonian
We now extend the phase space by introducing a new variable $v$, and defining it to Poisson-commute with the regular phase space variables, i.e.$$\{v,q_i\} = \{v,p_i\} = 0$$
The
complete Hamiltonian is obtained by multiplying $v$ by our primary constraint $G$ and adding it to $H_0$:
$$H = \frac{p_1^2}{2} + \frac{k}{2}(q_1^4+q_2^4) + v(p_2-p_1)$$The new Hamiltonian equations are
$$\dot p_1 = -\frac{\partial H}{\partial q_1} = -2kq_1^3$$$$\dot p_2 = -\frac{\partial H}{\partial q_2} = -2kq_2^3$$$$\dot q_1 = \frac{\partial H}{\partial p_1} = p_1-v$$$$\dot q_2 = \frac{\partial H}{\partial p_2} = v$$
Step 4: Obtain additional algebraic relations
Because $G$ is identically zero, it must be that$$\dot G = \dot p_2 - \dot p_1 = 0$$$$\implies T\equiv q_2^3-q_1^3 = 0$$
We call $T$ a
secondary constraint of the Dirac procedure - a constraint obtained through differentiation of a primary constraint, and then simplified by using the Hamilton equations obtained from the complete Hamiltonian (though in this case, the naive Hamiltonian would have done just as well).
Step 5: Determine $v$ and eliminate it from the complete Hamiltonian
Differentiating the secondary constraint allows us to determine $v$:
$$\dot T = 3(q_1^2 \dot q_1 - q_2^2 \dot q_2) = 3(q_1^2[p_1-v] - q_2^2[v])$$$$= 3(q_1^2 p_1 - (q_1^2+q_2^2)v) = 0$$$$\implies v = \frac{q_1^2 p_1}{q_1^2 + q_2^2}$$
and so
$$H = \frac{p_1^2}{2} + \frac{k}{2}(q_1^4+q_2^4) + \frac{q_1^2 p_1}{q_1^2+q_2^2}(p_2-p_1)$$
And that's it, we're done.
You can confirm that this, along with the primary and secondary constraints, reproduces the proper equations of motion:
$$\dot p_1 = -\frac{\partial H}{\partial q_1} = -2kq_1^3 - \frac{2q_1(1-q_1^2)(p_1p_2-p_1^2)}{q_1^2+q_2^2}$$$$\dot p_2 = -2kq_2^3 +\frac{2q_2q_1^2(p_1p_2-p_1^2)}{q_1^2+q_2^2}$$$$\dot q_1 = p_1 + \frac{q_1^2}{q_1^2+q_2^2}(p_2-2p_1)$$$$\dot q_2 = \frac{q_1^2}{q_1^2+q_2^2} p_1$$$$G \equiv p_2-p_1 = 0$$$$T\equiv x_2^3-x_1^3 = 0$$
which simplifies to
$$\dot p_1 = \frac{d}{dt}(\dot q_1 + \dot q_2) = -2kq_1^3 = -2kq_2^3$$
In summary, singular Lagrangian systems have several common features
Defining equations for the generalized momenta yield (some) algebraic equations between phase space variables which do not include generalized velocities, and the system is therefore non-invertible. These equations are called primary constraints, and their derivatives yield secondary constraints The procedure for obtaining the complete Hamiltonian extends the phase space and uses the new variables a bit like Lagrange multipliers to add the primary constraints to the naive Hamiltonian At least some of the "Lagrange multipliers" can be eliminated from the new Hamilton equations by using the primary and secondary constraints, and the resulting system of equations (Hamilton equations + constraints) reproduces the original dynamics This was not included in this example, but any multipliers which remain undetermined at the end of this procedure enter the solutions as arbitrary functions, which wouldn't have been determined by the Lagrangian equations of motion either. |
I am intrigued by this question because it gets at the heart of so many grey areas of the financial system in which it becomes almost impossible to know how many assets derive their values from some unseen or ill-prescribed, but presumed extant, underlying process.
Calibration can be interpreted as means of deriving an expectation which is the probabilistic point estimate, subject to certain parameters, $p_1,\,p_2,\,...p_n$, i.e.,:
$\mathbb{E}[X_T] = f(X_t,\,p_1,\,p_2,\,...p_n)$
This is, in essence,
of asset pricing. However, the law of arbitrage supersedes the strong law when it is possible to show that: The Strong Law
$f(\mathbb{E}[(H_T^i)_{i \in I}]) \ne (C_{i,T})_{i \in I}$
only if/when it is possible to partake in both $H^i$ and $C_i$, such as in the assumption regarding complete markets.
However, in the absence of a complete market, or when faced with complicated pay-off scenarios, any expectation of a $\mathbb{Q}$ martingale may indeed be parametric at best (i.e., the expectation must taken through calibration).
My skepticism is perhaps best demonstrated by the following passage out of Baxter's and Rennie's
Financial Calculus (kudos to @DaneelOlivaw for making me aware of this):
Almost everything appeared safe to price via expectation and the
strong law, and only forwards and close relations seemed to have an
arbitrage price. Since 1973, however, and the infamous Black-Scholes
paper, just how wrong this is has slowly come out. Nowhere in this
book will we use the strong law again. […] All derivatives can be
built from the underlying −− arbitrage lurks everywhere.
Perhaps... but my personal,
fallible experience tells me otherwise. While the no-arbitrage range of possible values for an equity option may be known presuming that the price of the equity is known, what is the fair value of an equity? I.e., how can we construct a replicating payoff for this equity in a way that is not a tautology (i.e., a thing which defines itself but nothing more)? To my knowledge, there exist no market for accounting values of assets and liabilities. More explicitly, how can we show the value of a thing, $C_t$, as follows:
$C_{i,t} = \int_t^T f(\mathbb{E}[H_{i,t}]P_t) \, dt$; $P_t := e^{-rt}$
when $C_{i,t}$ is a function of human perception regarding the unknown future values of $T$ and $H_{i,t}$, even if we take the risk-neutral expectation and short-rate as gospel?
Given that no perfect model for human behavior exists (otherwise that model would equal reality and its creator, a god), an imperfect (practical) answer to mitigating
is to derive an expectation which make use of the fewest possible parameters. Fewer parameters means less calibration, which means decreased odds of over-fitting. GIGO
A thing which is descriptive of the past, present, and/or future, and which is also not highly calibrated has a better likelihood of being prescriptive than a thing which is more highly descriptive but also more highly calibrated.
Is there a model for that? And don't say degrees of freedom...
Does this imply that highly-specified models (e.g., Heston) which calibrate expectations to observations are less robust? Not necessarily if the fit is not
garbage (i.e., not spurious; i.e., it states something which is true regarding the nature of uncertainty), but in aggregate, I believe so.
I take the broad corpus of economic literature's failure to predict anything but the past as evidence. |
...
The Giant Radio Array for Neutrino Detection (GRAND) is a planned large-scale observatory of ultra-high-energy (UHE) cosmic particles, with energies exceeding 10^8 GeV. Its goal is to solve the long-standing mystery of the origin of UHE cosmic rays. To do this, GRAND will detect an unprecedented number of UHE cosmic rays and search for the undiscov... ...
A novel method to analyse the spatial distribution of neutrino candidates recorded with the ANTARES neutrino telescope is introduced, searching for an excess of neutrinos in a region of arbitrary size and shape from any direction in the sky. Techniques originating from the domains of machine learning, pattern recognition and image processing are us...
Over the last years, radio detection has matured to become a competitive method for the detection of air showers. Arrays of thousands of antennas are now envisioned for the detection of cosmic rays of ultra high energy or neutrinos of astrophysical origin. The data exploitation of such detectors requires to run massive air-shower simulations to eva... ...
The Astrophysical Multimessenger Observatory Network (AMON) has been built with the purpose of enabling near real-time coincidence searches between different observatories and telescopes. Its mission is to evoke, exploit and explore multimessenger sources. AMON aims to promote the advancement of multimessenger astrophysics by allowing its participa...
In October 2017, a multi-layer small multi-wire (MW) detector for cosmic ray (CR) muon measurements was installed at King Abdulaziz City for Science and Technology (KACST), Riyadh, Saudi Arabia (Rc = 14.4 GV). This detector is considered to be the first of its kind used for long term CR measurements. In this study, CR data collected from KACST-MW d...
We study the statistics of weak lensing convergence peaks, such as their abundance and two-point correlation function (2PCF), for a wide range of cosmological parameters $\Omega_m$ and $\sigma_8$ within the standard $\Lambda$CDM paradigm, focusing on intermediate-height peaks with signal-to-noise ratio (SNR) of $1.5$ to $3.5$. We find that the cosm...
We study the impact of baryonic processes and massive neutrinos on weak lensing peak statistics that can be used to constrain cosmological parameters. We use the BAHAMAS suite of cosmological simulations, which self-consistently include baryonic processes and the effect of massive neutrino free-streaming on the evolution of structure formation. We ...
We present a model of inflation in which the inflaton field is charged under a triplet of $U(1)$ gauge fields. The model enjoys an internal $O(3)$ symmetry supporting the isotropic FRW solution. With an appropriate coupling between the gauge fields and the inflaton field, the system reaches an attractor regime in which the gauge fields furnish a sm...
The recent detection of gravitational waves (GWs) from the neutron star binary inspiral GW170817 has opened a unique avenue to probe matter and fundamental interactions in previously unexplored regimes. Extracting information on neutron star matter from the observed GWs requires robust and computationally efficient theoretical waveform models. We d...
Published in EPJ Web of Conferences
Observation of supernovae (SN) through their neutrino emission is a fundamental point to understand both SN dynamics and neutrino physical properties. JUNO is a 20kton liquid scintillator detector, under construction in Jiangmen, China. The main aim of the experiment is to determine neutrino mass hierarchy by precisely measuring the energy spectrum... |
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Change to browse by: References & Citations Bookmark(what is this?) Physics > Physics and Society Title: Epidemic SIR model on a face-to-face interaction network: new mobility induced phase transitions
(Submitted on 21 Mar 2018 (v1), last revised 24 May 2018 (this version, v3))
Abstract: In this work, we study the epidemic SIR model on a system which takes into consideration face-to-face interaction networks. This approach has been used as prototype to describe people interactions in different kinds of social organizations and, here, it is considered by means of three features of human interactions: the mobility, the duration of the interaction among people, and the dependence of the number of interactions of each person on the time evolution of the system. For this purpose, the initial configuration of the system is set as a regular square lattice where the nodes are the individuals which, in turn, are able to move in a random walk along the network. So, the connectivity among the individuals evolve with time and is defined by the positions of the individuals at each iteration. In a time unit, each individual is able move up to a distance $v$ creating different networks along the time evolution of the system. In addition, the individuals are interacting with each other only if they are within the interaction distance $\delta$ and, in this case, they are considered as neighbors. If a given individual is interacting with other ones, he performs the random walk with a diffusion probability $\omega$. Otherwise, the diffusion occurs with probability 1. The study was carried out through non-equilibrium Monte Carlo Simulations and we take into account the asynchronous updating scheme. The results show that, for a given $v>0$, there exist a critical line in the $(c, \delta)$ space, where $c$ is the immunization rate. We also obtain the dynamic critical exponent $\theta$ for some points belonging to this line and show that this model does not belong to the directed percolation universality class. Submission historyFrom: Paulo Freitas Gomes [view email] [v1]Wed, 21 Mar 2018 12:47:30 GMT (1454kb,D) [v2]Tue, 10 Apr 2018 18:08:44 GMT (1454kb,D) [v3]Thu, 24 May 2018 17:56:21 GMT (1616kb,D) |
(a) If $AB=B$, then $B$ is the identity matrix. (b) If the coefficient matrix $A$ of the system $A\mathbf{x}=\mathbf{b}$ is invertible, then the system has infinitely many solutions. (c) If $A$ is invertible, then $ABA^{-1}=B$. (d) If $A$ is an idempotent nonsingular matrix, then $A$ must be the identity matrix. (e) If $x_1=0, x_2=0, x_3=1$ is a solution to a homogeneous system of linear equation, then the system has infinitely many solutions.
Let $A$ and $B$ be $3\times 3$ matrices and let $C=A-2B$.If\[A\begin{bmatrix}1 \\3 \\5\end{bmatrix}=B\begin{bmatrix}2 \\6 \\10\end{bmatrix},\]then is the matrix $C$ nonsingular? If so, prove it. Otherwise, explain why not.
(a) Suppose that a $3\times 3$ system of linear equations is inconsistent. Is the coefficient matrix of the system nonsingular?
(b) Suppose that a $3\times 3$ homogeneous system of linear equations has a solution $x_1=0, x_2=-3, x_3=5$. Is the coefficient matrix of the system nonsingular?
(c) Let $A$ be a $4\times 4$ matrix and let\[\mathbf{v}=\begin{bmatrix}1 \\2 \\3 \\4\end{bmatrix} \text{ and } \mathbf{w}=\begin{bmatrix}4 \\3 \\2 \\1\end{bmatrix}.\]Suppose that we have $A\mathbf{v}=A\mathbf{w}$. Is the matrix $A$ nonsingular?
Suppose that $B=\{\mathbf{v}_1, \mathbf{v}_2\}$ is a basis for $\R^2$. Let $S:=[\mathbf{v}_1, \mathbf{v}_2]$.Note that as the column vectors of $S$ are linearly independent, the matrix $S$ is invertible.
Prove that for each vector $\mathbf{v} \in V$, the vector $S^{-1}\mathbf{v}$ is the coordinate vector of $\mathbf{v}$ with respect to the basis $B$.
Prove that the matrix\[A=\begin{bmatrix}0 & 1\\-1& 0\end{bmatrix}\]is diagonalizable.Prove, however, that $A$ cannot be diagonalized by a real nonsingular matrix.That is, there is no real nonsingular matrix $S$ such that $S^{-1}AS$ is a diagonal matrix.
The following problems are Midterm 1 problems of Linear Algebra (Math 2568) at the Ohio State University in Autumn 2017.There were 9 problems that covered Chapter 1 of our textbook (Johnson, Riess, Arnold).The time limit was 55 minutes.
This post is Part 3 and contains Problem 7, 8, and 9.Check out Part 1 and Part 2 for the rest of the exam problems.
Problem 7. Let $A=\begin{bmatrix}-3 & -4\\8& 9\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}-1 \\2\end{bmatrix}$.
(a) Calculate $A\mathbf{v}$ and find the number $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$.
(b) Without forming $A^3$, calculate the vector $A^3\mathbf{v}$.
Problem 8. Prove that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular.
Problem 9.Determine whether each of the following sentences is true or false.
(a) There is a $3\times 3$ homogeneous system that has exactly three solutions.
(b) If $A$ and $B$ are $n\times n$ symmetric matrices, then the sum $A+B$ is also symmetric.
(c) If $n$-dimensional vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4$ is also linearly dependent for any $n$-dimensional vector $\mathbf{v}_4$.
(d) If the coefficient matrix of a system of linear equations is singular, then the system is inconsistent.
An $n\times n$ matrix $A$ is called nonsingular if the only vector $\mathbf{x}\in \R^n$ satisfying the equation $A\mathbf{x}=\mathbf{0}$ is $\mathbf{x}=\mathbf{0}$.Using the definition of a nonsingular matrix, prove the following statements.
(a) If $A$ and $B$ are $n\times n$ nonsingular matrix, then the product $AB$ is also nonsingular.
(b) Let $A$ and $B$ be $n\times n$ matrices and suppose that the product $AB$ is nonsingular. Then:
The matrix $B$ is nonsingular.
The matrix $A$ is nonsingular. (You may use the fact that a nonsingular matrix is invertible.)
Let $A$ be an $n\times (n-1)$ matrix and let $\mathbf{b}$ be an $(n-1)$-dimensional vector.Then the product $A\mathbf{b}$ is an $n$-dimensional vector.Set the $n\times n$ matrix $B=[A_1, A_2, \dots, A_{n-1}, A\mathbf{b}]$, where $A_i$ is the $i$-th column vector of $A$.
Prove that $B$ is a singular matrix for any choice of $\mathbf{b}$.
For each of the following $3\times 3$ matrices $A$, determine whether $A$ is invertible and find the inverse $A^{-1}$ if exists by computing the augmented matrix $[A|I]$, where $I$ is the $3\times 3$ identity matrix. |
Various stringed instruments of Chinese make on display in a shop.
String instruments or stringed instruments are chordophones. Some common instruments in the string family are violin, guitar, sitar, electric bass, viola, cello, harp, double bass, rabab, banjo, mandolin, ukulele, and bouzouki. History
Early string instruments have been excavated in ancient Mesopotamia, like the lyres of Ur, which date to 2500 BC.
[1] Types of instruments Construction
String instruments can be divided in three groups.
Lutes - instruments in which the strings are supported by a neck and a bout ("gourd"), for instance a guitar, a violin, a saz. Harps - instruments in which the strings are contained within a frame. Zithers - instruments with the strings mounted on a body, such as a guqin, a cimbalom, an autoharp, or a piano.
It is also possible to divide the instruments in groups focused on how the instrument is played.
Types of playing techniques For a full list, see List of string instruments.
All string instruments produce sound from one or more vibrating strings, transferred to the air by the body of the instrument (or by a pickup in the case of electronically amplified instruments). They are usually categorized by the technique used to make the strings vibrate (or by the primary technique, in the case of instruments where more than one may apply.) The three most common techniques are plucking, bowing and striking.
Plucking
Plucking is a method of playing on instruments such as the Veena, banjo, ukulele, guitar, harp, lute, mandolin, oud, and sitar, using either a finger, thumb, or quills (now plastic plectra) to pluck the strings.
Instruments normally played by bowing (see below) may also be plucked, a technique referred to by the Italian term
pizzicato. Bowing
Bowing (
Italian: Arco) is a method used in some string instruments, including the violin, viola, cello, and the double bass (of the violin family) and the old viol family. The bow consists of a stick with many hairs stretched between its ends. Bowing the instrument's string causes a stick-slip phenomenon to occur, which makes the string vibrate.
Ravanahatha has been recognized as one of the oldest string instruments in the history, Ancestors of the modern bowed string instruments are the rebab of the Islamic Empires, the Persian kamanche and the Byzantine lira. Other bowed instruments are the rebec, hardingfele, nyckelharpa, kokyū, erhu, igil, sarangi and K'ni. The hurdy gurdy is bowed by a wheel.
Rarely, the guitar has been played with a bow (rather than plucked) for unique effects.
Striking
The third common method of sound production in stringed instruments is to strike the string. The piano uses this method of sound production.
Violin family string instrument players are occasionally instructed to strike the string with the stick of the bow, a technique called
col legno. This yields a percussive sound along with the pitch of the note. A well-known use of col legno for orchestral strings is the Gustav Holst's "Mars" movement from The Planets suite. Other methods
The aeolian harp employs a very unusual method of sound production: the strings are excited by the movement of the air.
Some instruments that have strings have attached keyboards that the player uses instead of directly manipulating the strings. These include the piano, the clavichord, and the harpsichord.
With these keyboard instruments too, strings are occasionally plucked or bowed by hand. Composers such as Henry Cowell wrote music that requires that the player reach inside the piano and pluck the strings directly, "bow" them with bow hair wrapped around the strings, or play them by rolling the bell of a brass instrument such as a trombone on the array of strings.
Other keyed string instruments, small enough for a strolling musician to play, include the plucked autoharp, the bowed nyckelharpa, and the hurdy gurdy, which is played by cranking a rosined wheel.
Steel-stringed instruments (such as the guitar, bass, violin, etc.) can be played using a magnetic field. An E-Bow is small hand-held battery-powered device that magnetically excites the strings of an electric string instrument to provide a sustained, singing tone.
3rd bridge is a plucking method where the player frets a string and strikes the side opposite the bridge. The technique is mainly used on electric instruments, because these have a pickup that amplifies only the local string vibration. It's possible on acoustic instruments as well, but less effective. For instance, a player might press on the 7th fret on a guitar and pluck it at the head side to make a tone resonate at the opposed side. At electric instruments this technique generates multitone sounds reminiscent of a clock or bell.
Changing the pitch of a vibrating string
There are three ways to change the pitch of a vibrating string. String instruments are tuned by varying the strings' tension because adjusting length or mass per unit length is impractical. Instruments with a fingerboard are then played by adjusting the length of the vibrating portion of the strings. The following observations all apply to a string that is infinitely flexible strung between two fixed supports. Real strings have finite curvature at the bridge and nut, and the bridge, because of its motion, are not exactly nodes of vibration. Hence the following statements about proportionality are (usually rather good) approximations.
Length
Pitch can be adjusted by varying the length of the string. A longer string results in a lower pitch, while a shorter string results in a higher pitch. The frequency is inversely proportional to the length:
f \propto \frac{1}{l}
A string twice as long produces a tone of half the frequency (one octave lower).
Tension
Pitch can be adjusted by varying the tension of the string. A string with less tension (looser) results in a lower pitch, while a string with greater tension (tighter) results in a higher pitch. The frequency is proportional to the square root of the tension:
f \propto \sqrt{T} Linear density
The pitch of a string can also be varied by changing the linear density (mass per unit length) of the string. The frequency is inversely proportional to the square root of the linear density:
f \propto {1 \over \sqrt{\mu}}
Given two strings of equal length and tension, the string with higher mass per unit length produces the lower pitch.
String length or scale length
The length of the string from nut to bridge on bowed or plucked instruments ultimately determines the distance between different notes on the instrument. For example, a double bass with its low range needs a scale length of around 42 inches (110 cm), whilst a violin scale is only about 13 inches (33 cm). On the shorter scale of the violin, the left hand may easily reach a range of slightly more than two octaves without shifting position, while on the bass' longer scale, a single octave or a ninth is reachable in lower positions.
Contact points along the string
In bowed instruments, the bow is normally placed perpendicularly to the string, at a point half way between the end of the fingerboard and the bridge. However, different bow placements can be selected to change timbre. Application of the bow close to the bridge (known as
sul ponticello) produces an intense, sometimes harsh sound, which acoustically emphasizes the upper harmonics. Bowing above the fingerboard ( sul tasto) produces a purer tone with less overtone strength, emphasizing the fundamental, also known as flautando, since it sounds less reedy and more flute-like.
Similar timbral distinctions are also possible with plucked string instruments by selecting an appropriate plucking point, although the difference is perhaps more subtle.
In keyboard instruments, the contact point along the string (whether this be hammer, tangent, or plectrum) is a choice made by the instrument designer. Builders use a combination of experience and acoustic theory to establish the right set of contact points.
In harpsichords, often there are two sets of strings of equal length. These "choirs" usually differ in their plucking points. One choir has a "normal" plucking point, producing a canonical harpsichord sound; the other has a plucking point close to the bridge, producing a reedier "nasal" sound rich in upper harmonics.
Production of multiple notes
A string at a certain tension and length only produces one note, so to obtain multiple notes, string instruments employ one of two methods. One is to add enough strings to cover the range of notes desired; the other is to allow the strings to be
stopped. The piano and harp are examples of the former method, where each note on the instrument has its own string or course of multiple strings. (Many notes on a piano are strung with a "choir" of three strings tuned alike.)
Some zithers combine stoppable (melody) strings with a greater number of "open" harmony or chord strings. On instruments with stoppable strings, such as the violin or guitar, the player can shorten the vibrating length of the string, using their fingers directly (or more rarely through some mechanical device, as in the nyckelharpa or the hurdy gurdy). Such instruments usually have a
fingerboard attached to the neck of the instrument, that provides a hard flat surface athe player can stop the strings against. On some string instruments, the fingerboard has frets, raised ridges perpendicular to the strings that stop the string at precise intervals, in which case the fingerboard is also called a fretboard.
Moving frets during performance is usually impractical. The bridges of a koto, on the other hand, may be moved by the player, occasionally in the course of a single piece of music. Many modern Western harps include levers, either directly moved by fingers (on Celtic harps) or controlled by foot pedals (on orchestral harps), to raise the pitch of individual strings by a fixed amount. The middle Eastern zither, the qanun, is equipped with small levers called
mandal that allow each course of multiple strings to be incrementally retuned "on the fly" while the instrument is being played. These levers raise or lower the pitch of the string course by a microtone, less than a half step. Sympathetic strings
Some instruments are employed with sympathetic strings, additional strings not meant to be plucked. These strings resonate along with the played notes. This system is for instance present on a sarangi.
Sound production Acoustic instruments
A vibrating string on its own makes only a very quiet sound, so string instruments are usually constructed in such a way that this sound is coupled to a hollow resonating chamber, a soundboard, or both. On the violin, for example the tout strings pass over a bridge resting on a hollow box. The strings vibrations are distributed via the bridge and soundpost to all surfaces of the instrument, and are thus made louder. The correct technical explanation is that they allow a better match to the acoustic impedance of the air.
It is sometimes said that the sounding board or soundbox "amplifies" the sound of the strings. Technically speaking, no amplification occurs, because all of the energy to produce sound comes from the vibrating string. What really happens is that the sounding board of the instrument provides a larger surface area to create sound waves than that of the string. A larger vibrating surface moves more air, hence produces a louder sound.
All lute type instruments traditionally have a bridge, which holds the string at the proper action height from the fret/finger board. On acoustic instruments, the bridge performs an equally important function of transmitting string energy into the "sound box" of the instrument, thereby increasing the sound volume. The specific design, and materials the used in the construction of the bridge of an instrument, have a dramatic impact upon both the sound and responsiveness of the instrument.
Achieving a tonal characteristic that is effective and pleasing to the player's and listener's ear is something of an art, and the makers of string instruments often seek very high quality woods to this end, particularly spruce (chosen for its lightness, strength and flexibility) and maple (a very hard wood). Spruce is used for the sounding boards of instruments from the violin to the piano. Instruments such as the banjo use a drum, covered in natural or synthetic skin as their soundboard.
Acoustic instruments can also be made out of artificial materials, such as carbon fiber and fiberglass (particularly the larger instruments, such as cellos and basses).
In the early 20th century, the Stroh violin used a diaphragm-type resonator and a metal horn to project the string sound, much like early mechanical gramophones. Its use declined beginning about 1920, as electronic amplification came into use.
Electronic amplification
Most string instruments can be fitted with piezoelectric or magnetic pickups to convert the string's vibrations into an electrical signal that is amplified and then converted back into sound by loudspeakers. Some players attach a pickup to their traditional string instrument to "electrify" it. Another option is to use a solid-bodied instrument, which reduces unwanted feedback howls or squeals.
Amplified string instruments can be much louder than their acoustic counterparts, which allows them to be used in relatively loud rock, blues, and jazz ensembles. Amplified instruments can also have their amplified tone modified by using electronic effects such as distortion, reverb, or wah-wah.
Bass-register string instruments such as the double bass and the electric bass are amplified with bass instrument amplifiers that are designed to reproduce low-frequency sounds. To modify the tone of amplified bass instruments, a range of electronic bass effects are available, such as distortion and chorus.
Symphonic Strings
The string instruments usually used in the orchestra, and often called the "symphonic strings" are:
[2]
When the instrumentation of an orchestral work is said to include "strings," is it very often this combination of string parts that is indicated. Orchestral works rarely omit any of these string parts, but fairly often will include additional string instruments, especially harp and piano. In the Baroque orchestra, harpsichord is almost always used, and often theorbo.
See also References ^ Michael Chanan (1994). Musica Practica: The Social Practice of Western Music from Gregorian Chant to Postmodernism. Verso. p. 170. ^ The Concise Oxford Dictionary of Music. Oxford University Press. 1964. p. 412. External links
The physics of the bowed string Instruments in Depth: The Viola, an online feature presented by Bloomingdale School of Music (2010)
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The rate of turn is inversely proportional to the (True) airspeed. For an aircraft in a level, coordinated turn, the rate of turn is given by
$\mathrm{Rate\ of\ turn} = \frac{1091 \tan\theta}{V}$
where
Rate of turn is in degrees per second, $\theta$ is the bank angle in degrees, and $V$ is the TAS in knots.
So, as the airspeed decreases, the rate of turn increases and vice versa—as long as the angle of bank is kept constant.
Image from cfinotebook.net
For an aircraft in a coordinated turn (i.e., no skidding or slipping), the vertical component of lift is equal to the weight, while the horizontal component is equal to the centrifugal force.
Forces in a turn, image from Instrument Flying Handbook
As the aircraft turns, if the airspeed increases with the bank angle held constant, the radius of turn increases with the square of the speed ($r = \frac{V^{2}}{11.26 \tan\theta}\ \mathrm{ft}$). Hence, the distance traveled during the turn increases as the square of the speed. Even though the aircraft is flying faster, the distance to be flown increases faster than the speed. As a result, the time taken to complete the turn is increased and the turn rate decreases. |
For what values of $x$ is the number $121$ a square in base $x$?
This is a very easy question, but I still think it's pretty fun. Perhaps it's something to pass on to younger people who are taking algebra.
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For what values of $x$ is the number $121$ a square in base $x$?
This is a very easy question, but I still think it's pretty fun. Perhaps it's something to pass on to younger people who are taking algebra.
All of them (except base $2$ which can't have $121$.)
Added pedantry:
If you're ok with negative bases, the answer is all positive and negative bases except $-2$, $-1$, and $2$. Base $0$ and $1$ make no sense at all to me (and $-1$ very little) so I would also claim it doesn't hold for these, mostly because "holding" isn't really provably true or false, but also because 121 is not a valid number in those bases, as also it is not valid in base 2 and -2.
Also, I am considering integer bases only.
Why is it a square for these bases?
For whatever base $n$ you're working in, simply expanding the number $121$, it means $n^2 + 2n + 1$. And that is $(n+1)^2$ as a matter of algebra. And that is a square.
The base-$x$ number $121$ is a square
whenever $x\geq 3$.
For $x\geq 3$, we can multiply in base $x$: \begin{array}{ccc}&1&1\\\times&1&1\\\hline&1&1\\1&1&\\\hline1&2&1\end{array}We never needed to "carry", so this computation looks the same for every $x$. |
For a given set of filter specifications, we generally obtain the filter system function, $$H(z)$$, assuming that the filter coefficients can be represented with infinite precision. However, when implementing the filter in the real world, we have to use a finite number of bits to represent each coefficient of $$H(z)$$. This coefficient quantization can somehow change the location of the filter poles and zeros.
As a result, after implementing a filter, we may observe that the frequency response of the filter is quite different from that of the original design. The error in the pole and zero locations depends on several factors. This article will discuss some of these factors and show how we can design filters which exhibit smaller sensitivity to the coefficient quantization.
Before continuing our discussion, let’s review an example of the coefficient quantization.
Example 1
The transfer function of an Infinite Impulse Response (IIR) filter is given by:
$$H(z)=\frac{\sum_{k=0}^{M-1}b_{k}z^{-k}}{\sum_{k=0}^{N-1}a_{k}z^{-k}}$$
We can use the MATLAB
ellip function to design an elliptic filter. For example, [b, a]=ellip(7,0.5, 50, 0.3) gives a seventh-order elliptic lowpass filter with 0.5 dB ripples in the passband and 50 dB attenuation in the stopband. The passband edge of the filter will be at the normalized frequency of $$0.3 \pi$$. The coefficients of this filter are given in the following table. We will consider these coefficients as the unquantized ones.
k b k(unquantized) a k(unquantized) 0 0.012218357882143 1.000000000000000 1 -0.009700754662078 -4.288900601525732 2 0.024350450826845 9.216957436091198 3 0.002532504848041 -12.195350561406707 4 0.002532504848041 10.633166152311462 5 0.024350450826845 -6.062798190498858 6 -0.009700754662078 2.098067018562072 7 0.012218357882143 -0.342340135743532
The magnitude of this filter’s frequency response is shown in Figure 1.
Figure 1. The magnitude of the frequency response of the unquantized filter. Figure 1.The magnitude of the frequency response of the unquantized filter.
Figure 2 shows the poles (blue crosses) and zeros (red dots) of the transfer function. Since the poles are inside the unit circle, the filter is stable.
Figure 2. The poles and zeros of the unquantized system function. Figure 2.The poles and zeros of the unquantized system function.
We will quantize the coefficients using one bit for the sign and nine bits for the magnitude of the coefficients. Since all the $$b_k$$ coefficients are smaller than $$2^{-5}=0.03125$$, we can consider a scaling factor of $$2^{(number \; of \; bits + 5)}=2^{(9+5)}=16384$$ for these coefficients and achieve a more accurate representation. The quantized $$b_k$$ coefficients are listed in Table 2. For example, to calculate the quantized value of $$b_1$$, we will first apply the scaling factor and obtain:
$$b_{1} \times 2^{(9+5)}=0.012218357882143 \times 2^{(9+5)}=200.1856$$
Now, we can round the result to $$200$$. The binary representation of $$200$$, which is $$011001000$$, will be used to implement the coefficients. However, we should keep in mind that we need to interpret the results with a rescaling factor of $$2^{-(9+5)}$$. Then, the decimal equivalent of the quantized coefficient can be obtained by multiplying $$200$$ by $$2^{-(9+5)}$$ which gives $$0.0122$$.
In this particular example, with a scaling factor smaller than $$2^{(9+5)}$$, several bits of the binary representation of the $$b_k$$ coefficients would be zero for all the coefficients and we would lose the accuracy. For example, suppose that we allocate nine bits to represent the fractional value of the $$b_k$$ coefficients. Hence, we should apply a scaling factor of $$2^{9}$$ which gives:
$$b_{1} \times 2^9=0.012218357882143 \times 2^9=6.2558$$
Obviously, in this case, a rescaling factor of $$2^{-9}$$ must be considered when interpreting the result of the calculations. Rounding $$6.2558$$ and converting it to the binary representation, we obtain $$00000110$$. We observe that, although we are using nine bits to represent this number, most of them are zero. The reader can verify that, with this scaling factor, several bits will be zero even for the largest $$b_k$$, i.e. $$0.024350450826845$$.
To quantize the $$a_k$$ coefficients, we note that the magnitude of the integer part of these coefficients is smaller than $$16$$. Hence, we can allocate four bits for the integer part and five bits for the fractional part. As a result, the scaling factor of the $$a_k$$ coefficients will be $$2^5$$. For example, with $$a_2=9.216957436091198$$, we have:
$$a_{2} \times 2^5=9.216957436091198 \times 2^5=294.9426 \approx 295$$
Hence, the quantized decimal value for the coefficient will be $$9.2188$$. Similarly, we can find the quantized values of other $$a_k$$ coefficients as given in Table 2 below.
k b k(quantized) a k(quantized) 0 0.0122 1.0000 1 -0.0097 -4.2813 2 0.0244 9.2188 3 0.0025 -12.1875 4 0.0025 10.6250 5 0.0244 -6.0625 6 -0.0097 2.0938 7 0.0122 -0.3438
Figure 3 compares the frequency response of the quantized filter (the curve in red) with that of the unquantized system (the curve in blue). We observe that quantizing the coefficients has adversely affected the frequency response.
Figure 3. The frequency response of the quantized filter (in red) versus that of the unquantized system (in blue). Figure 3.The frequency response of the quantized filter (in red) versus that of the unquantized system (in blue).
The poles (blue crosses) and zeros (the red dots) of the quantized filter are shown in Figure 4. As shown in this figure, two of the poles are moved out of the unit circle and the quantized filter is unstable. This example shows that after designing a filter, we need to examine the effect of the coefficient quantization. If the quantized filter does not meet the target specifications, we need to redesign the filter. In the rest of the article, we will see that implementing a high-order filter as a cascade of second-order sections can significantly reduce the sensitivity to the coefficient quantization.
Figure 4. The poles and zeros of the quantized filter. Figure 4.The poles and zeros of the quantized filter.
Analysis of Sensitivity to Coefficient Quantization
To examine the sensitivity of the poles and zeros of a filter to the coefficient quantization, let’s consider a polynomial, $$D(z)$$, with $$N$$ roots:
$$D(z)=1+\sum_{k=0}^{N}a_{k}z^{-k}$$
Equation 1 Equation 1
Equation 1 can represent the system function of a Finite Impulse Response (FIR) filter or either the numerator or the denominator of an IIR filter. Analyzing the sensitivity of roots of $$D(z)$$ to the coefficient quantization allows us to have a better insight into how the roots and poles of a digital filter will move under finite precision conditions.
We can write $$D(z)$$ in terms of its factors as:
$$D(z)=1+\sum_{k=0}^{N}a_{k}z^{-k}=\prod_{k=1}^{N}(1-p_{k}z^{-1})$$
Where $$p_k$$ denotes the roots of the polynomial. With a finite number of bits to represent each coefficient, we expect that $$a_k$$ will change to $$a_k+ \Delta a_k$$ where $$\Delta a_k$$ is the error resulted from using a finite precision representation. Consequently, we expect the roots of the polynomial to change from $$p_k$$ to $$p_k+ \Delta p_k$$. The error in the root location, $$\Delta p_i$$, can be found as:
$$\Delta p_{i}=-\sum_{k=1}^{N}\frac{p_{i}^{N-k}}{\prod_{l=1, l \neq i}^{N}(p_{i}-p_{l})}\Delta a_{k}$$
Equation 2 Equation 2
To see the proof of Equation 2 refer to section 9.5 of this book. Equation 2 has two important implications that will be discussed next.
Avoid Clusters of Poles and Zeros
Firstly, the error in the $$i$$th root, $$\Delta p_i$$, is equal to the error in the $$k$$th coefficient multiplied by the following factor:
$$F_{k}=\frac{p_{i}^{N-k}}{\prod_{l=1, l \neq i}^{N}(p_{i}-p_{l})}$$
Equation 3 Equation 3
This factor can be very large when the polynomial has other roots close to the $$i$$th pole, i.e. $$p_{i}-p_{l}$$ is small. In other words, when we have a cluster of roots, the error in the root locations will be much higher for a given $$\Delta a_k$$. Since a narrow band filter has generally tightly clustered roots, we expect that the frequency response of these filters will be highly sensitive to the coefficient quantization.
Avoid High-Order Filter Sections
Rewriting Equation 2 as Equation 4 below, we observe that each and every coefficient of the polynomial contributes some error to the location of a particular pole:
$$\Delta p_{i}=-\sum_{k=1}^{N}F_{k}\Delta a_{k}$$
Equation 4 Equation 4
This means that as the number of roots of a polynomial increases, the sensitivity to the quantization error will increase. This is due to the fact that each root of $$D(z)$$ in Equation 1 depends on the value of all the coefficients $$a_{k}$$. For a polynomial of degree $$N$$, there are $$N$$ coefficients that need to be quantized. And, obviously, each of these quantized coefficients will contribute a particular amount of error to the overall error.
To summarize, we should avoid clusters of poles and zeros and use low-order filter sections. These two goals can be achieved by using single-pole sections to implement a high-order filter. However, a filter has generally complex poles and zeros and use of single-pole sections mandates complex arithmetic which increases the computational complexity. The next best alternative is to use second-order sections. In this case, we can pair complex-conjugate roots and avoid the complex arithmetic. Since finding the cascade form of a high-order filter involves tedious mathematics, we can use the MATLAB function
tf2sos, which stands for transfer function to second-order section, to obtain the cascade form of a given transfer function.
Example 2
We will use the
tf2sos function to convert the transfer function of Example 1 into the cascade form. Then we will quantize the coefficients of these second-order sections and compare the frequency response of the obtained structure with that of the unquantized system.
The following lines of code define the transfer function of Example 1 and convert that to second-order sections:
N=[0.012218357882143 -0.009700754662078 0.024350450826845 0.002532504848041 0.002532504848041 0.024350450826845 -0.009700754662078 0.012218357882143]; % This line defines the numerator of H(z) D=[1.000000000000000 -4.288900601525731 9.216957436091192 -12.195350561406695 10.633166152311450 -6.062798190498850 2.098067018562069 -0.342340135743531]; % This defines the denominator of H(z) [sos, G]=tf2sos(N, D) % converts the transfer function defined by N and D into a cascade of second-order sections
The result will be:
sos =
1.0000000000 1.0000000000 0.0000000000 1.0000000000 -0.6790830001 0.0000000000 1.0000000000 0.0102799961 1.0000000000 1.0000000000 -1.2818759037 0.6209275764 1.0000000000 -0.8106030432 1.0000000000 1.0000000000 -1.1804902667 0.8437961219 1.0000000000 -0.9936260871 1.0000000000 1.0000000000 -1.1474514311 0.9621803579
and:
$$G=$$0.0122183579.
Each row of sos above gives the transfer function of one of the second-order sections. The first three numbers of each row represent the numerator of the corresponding second-order section and the second three numbers give its denominator. For example, the second-order section obtained from the second row is:
$$H_{2}(z)=\frac{1.0000000000+0.0102799961z^{-1}+1.0000000000z^{-2}}{1.0000000000-1.2818759037z^{-1}+0.6209275764z^{-2}}$$
We will quantize the coefficients using one bit for the sign and nine bits for the magnitude of the coefficients. We need to choose an appropriate scaling factor. Since the coefficients of the second-order sections are less than 2, we will use one bit for the integer part and eight bit for the fractional part, i.e. the scaling factor will be $$2^8$$. Hence, we obtain the transfer function of the first section as:
$$H_{1}(z)=\frac{1.00000000+1.00000000z^{-1}}{1.00000000-0.67968750z^{-1}}$$
The quantized transfer function of the other second-order sections will be:
$$H_{2}(z)=\frac{1.00000000+ 0.01171875z^{-1}+1.00000000z^{-2}}{1.00000000-1.28125000z^{-1}+ 0.62109375z^{-2}}$$
$$H_{3}(z)=\frac{1.00000000-0.81250000z^{-1}+1.00000000z^{-2}}{1.00000000-1.17968750z^{-1}+0.84375000z^{-2}}$$
$$H_{4}(z)=\frac{1.00000000-0.99218750z^{-1}+1.00000000z^{-2}}{1.00000000-1.14843750z^{-1}+0.96093750z^{-2}}$$
Figure 5 compares the frequency response of $$H(z)=GH_{1}(z)H_{2}(z)H_{3}(z)H_{4}(z)$$ (which is shown in red) with that of the unquantized filter (shown in blue). As you can see the two graphs are barely distinguishable from each other.
Figure 5. The frequency response of the cascade structure (in red) versus that of the unquantized system (in blue). Figure 5.The frequency response of the cascade structure (in red) versus that of the unquantized system (in blue).
The reader can easily use MATLAB
roots() function to verify that the poles of the quantized cascade structure are inside the unit circle and the system is stable. This example shows that implementing a high-order filter as a cascade of second-order sections can significantly reduce the sensitivity to the coefficient quantization.
Summary When implementing a digital filter in the real world, we have to use a finite number of bits to represent each coefficient of $$H(z)$$. The frequency response of the quantized filter might be quite different from that of the original design. Implementing a high-order filter as a cascade of second-order sections can significantly reduce the sensitivity to the coefficient quantization.
Supporting Information
FIR Filter Design by Windowing: Concepts and the Rectangular Window Undesired Effects of a Window Function in FIR Filter Design The Bartlett Versus the Rectangular Window From Filter Specs to Window Parameters in FIR Filter Design Design of FIR Filters Using the Frequency Sampling Method Structures for Implementing Finite Impulse Response Filters |
Recommended Level
Beginner
Nodal Analysis
Nodal analysis is a form of analysis that uses Kirchhoff’s Current Law (KCL) and node equations to solve for circuit voltage values where the schematic diagram does not have any conductor paths crossing. A term typically used for this purpose is said to represent a planar circuit.
This is used to determine the voltage at each node( or connection point of two or more components) with respect to a reference node. The reference node is frequently called ground where the voltage at ground is equal to zero Volts.
In viewing schematic diagrams with either voltage sources or current sources, a reference node would usually be assigned to the negative terminal for a voltage source and to the opposite end when an arrow is shown for a current source. Another method of selecting the reference node would be to select the middle node when viewing all nodes.
There are two kinds of electrical sources: 1) independent and 2) dependent.
An independent source provides a fixed value of voltage or current to the connected circuit. Independent sources are power supplies and batteries. Power supplies provide a constant fixed value whereas batteries will not provide a constant fixed value over time without recharging them.
A dependent source is either a voltage source or current source whose value depends upon a voltage or current value somewhere else in the circuit. Dependent sources are useful tools in analyzing amplifiers. Two characteristics of amplifiers are voltage gain (A
V) and current gain (Ai). There are four basic linear dependent sources:
1. Voltage-controlled voltage source where the output is V, and A
V is the constant of proportionality (voltage gain), and V CD is the parameter being sensed. The following equation is associated with a voltage-controlled voltage source:
$$V = A_{V}V_{CD}$$
2. Current-controlled voltage source where the output is V, and R
M is the constant of proportionality (resistance), and I C is the parameter being sensed. The following equation is associated with a current-controlled voltage source:
$$V = R_{M}I_{C}$$
3. Current-controlled current source where the output is I, and Ai is the constant of proportionality (current gain), and I
C is the parameter being sensed. The following equation is associated with a current-controlled current source:
$$I = A_{I}I_{C}$$
4. Voltage-controlled current source where the output is I, and GM is the constant of proportionality (conductance), and VCD is the parameter being sensed. The following equation is associated with a voltage-controlled current source:
$$I = (G_{M})(V_{CD})$$
A nodal analysis with a dependent source occurs when there are two DC voltage sources and a DC current source as shown in Figure 1. Note the value for E1 is expressed in terms of an unknown value. E1 = 2Vx. Note that the voltage across resistor R1 is expressed as Vx. Note that the voltage across resistor R3 is expressed as V0. This information is to be used later in calculating node voltages.
Fig. 1
If the node or test point voltage were positive, it would read a positive value on the voltmeter. If the node or test point voltage were negative, it would read a negative value on the voltmeter.
An example problem for the circuit shown in Figure 1 would be to find the following:
A. The voltage across resistor R3 (V0).
B. The current through resistor R1 (IR1), R2 (IR2), R3 (IR3) and R4 (IR4).
C. Let I1 = 2 mAmps, I2 = 2 mAmps, E1 = 2Vx, E2 = 4 Volts, R1 = 1 Kilo Ohms, R2 = 2 kilo Ohms, R3 = 3 kilo Ohms, and R4 = 4 kilo Ohms.
D. Let the voltage across R1 (ER1) = Vx and E1 = V1 – V2.
The 1st step would be to identify the reference node or ground and then all the nodes of the circuit in Figure 1. Usually any independent source would be connected to ground and is shown in Figure 1.
In this circuit, there are nodes at the top of DC voltage source E1 labeled V1, below E1 labeled V2, below resistor R3 labeled V3, and above DC source E2 labeled V4. These nodes are shown in Figure 2.
Fig. 2
The 2nd step would be to identify the super node, which has the dependent source with an unknown value of E1 that is equal to 2Vx. Draw a red line around V1 and V2 that has E1 within as shown in Figure 3.
Fig. 3
The 3rd step would be to determine the independent source, which are the voltage source E2, and current sources I1 and I2. Draw a blue line around V4 and E2, I1 and I2 as shown in Figure 4.
Fig. 4
The 4th step is to identify currents at the super node with a blue line and arrow labeled a, b, c, d and e as shown in Figure 5.
Fig. 5
The 5th step is to identify currents at the V3 node with a red line and arrow labeled f, g and h as shown in Figure 6.
Fig. 6
The next step will utilize Figure 5 that identifies the super node so Kirchhoff’s Current Law (KCL) equations can be identified. Remember, Kirchhoff’s Current Law (KCL) states the algebraic sum of all currents entering and exiting a node must equal zero.
The following KCL current equations can be written for the blue lines and arrows at the super node for a, b, c, d and e:
$$ -I_{1} + I_{b} + I_{c} + I_{d} + I_{e} = 0 $$
Note that:
$$ I_{b} = \frac{V_{2}}{R_{1}}, I_{c} = \frac{V_{2}-V_{3}}{R_{2}}, I_{d} = \frac{V_{1}-V_{3}}{R_{3}}, I_{e} = \frac{V_{1}-E_{2}}{R_{4}} $$
The equation now becomes:
$$ -I_{1} + \frac{V_{2}}{R_{1}} + \frac{V_{2}-V_{3}}{R_{2}} + \frac{V_{1}-V_{3}}{R_{3}} + \frac{V_{1}-E_{2}}{R_{4}} = 0 $$
Substitute circuit values:
$$ -(\text{2 m}) + \frac{V_{2}}{\text{1 k}} + \frac{V_{2}-V_{3}}{\text{2 k}} + \frac{V_{1}-V_{3}}{\text{3 k}} + \frac{V_{1}-E_{2}}{\text{4 k}} = 0 $$
Multiply both sides by 12 k (Least Common Denominator):
$$ [-(\text{2 m}) + \frac{V_{2}}{\text{1 k}} + \frac{V_{2}-V_{3}}{\text{2 k}} + \frac{V_{1}-V_{3}}{\text{3 k}} + \frac{V_{1}-E_{2}}{\text{4 k}} = 0] (\text{12 k}) $$
Expand:
$$-24 + 12V_{2} + 6(V_{2} – V_{3}) + 4(V_{1} – V_{3}) + 3(V_{1} – 4) = 0$$
$$-24 + 12V_{2} + 6V_{2} – 6V_{3} + 4V_{1} – 4 V_{3} + 3V_{1} – 12 = 0$$
Combine terms:
$$7V_{1} + 18V_{2} – 10V_{3} = 36$$ [Equation 1]
The next step will utilize Figure 6 that shows the currents at the V3 node so the equations can be written for the red lines and arrows for f, g and h. Please note the directions of currents through resistor R2 and R3 are opposites when viewing the blue and red lines and arrows for c and g, and for d and h. This will be important later on in checking results.
$$ -I_{2} + \frac{V_{3} - V_{2}}{R_{2}} + \frac{V_{3}-V_{1}}{R_{3}} = 0 $$
Substitute circuit values:
$$ -(\text{2 m}) + \frac{V_{3} - V_{2}}{\text{2 k}} + \frac{V_{3}-V_{1}}{\text{3 k}} = 0 $$
Multiply both sides by 6 k (Least Common Denominator):
$$ [-(\text{2 m}) + \frac{V_{3} - V_{2}}{\text{2 k}} + \frac{V_{3}-V_{1}}{\text{3 k}} = 0 ] (\text{6 k}) $$
Expand:
$$-12 + 3(V_{3} – V_{2}) + 2(V_{3} – V_{1}) = 0$$
$$-12 + 3V_{3} – 3V_{2} + 2V_{3} – 2V_{1} = 0$$
Combine terms:
$$-2V_{1} – 3V_{2} + 5V_{3} = 12$$ [Equation 2]
There are two equations and three unknowns. Another equation is needed. When reviewing Figure 5 the given information concerning the DC voltage source E1 and nodes V2 and the voltage across resistor R1 that is Vx, another equation can be obtained.
Fig. 5
The following information is known:
$$E_{1} = V_{1} – V_{2}$$
$$E_{1} = 2V_{X}$$
$$V_{1} – V_{2} = 2V_{X}$$ [Equation A]
$$V_{2} = V_{X}$$ [Equation B]
Using Equation B with V2 = Vx in Equation A, another equation for V1 can be obtained:
$$V_{1} – V_{2} = 2V_{X}$$
$$V_{1} – V_{X} = 2V_{X}$$
Solve for V1:
$$V_{1} = 2V_{X} + V_{X}$$
$$V_{1} = 3V_{X}$$ [Equation 3]
Now Equation 1 and Equation 2 need to have the V1 and V2 terms in terms of Vx and V3 by using Equation A and Equation B:
Recall Equation 1: $$7V_{1} + 18V_{2} – 10V_{3} = 36$$
$$7(3V_{X}) + 18(V_{X}) – 10V_{3} = 36$$
$$21V_{X} + 18V_{X} – 10V_{3} = 36$$
$$39V_{X} – 10V_{3} = 36$$ [Equation C]
Recall Equation 2: $$-2V_{1} – 3V_{2} + 5V_{3} = 12$$
$$-2(3V_{X}) – 3(V_{X}) + 5V_{3} = 12$$
$$-6V_{X} – 3V_{X} + 5V_{3} = 12$$
$$-9V_{X} + 5V_{3} = 12$$ [Equation D]
Now there are two equations and two unknowns that can be solved.
$$39V_{X} – 10V_{3} = 36$$ [Equation C]
$$-9V_{X} + 5V_{3} = 12$$ [Equation D]
When 2 is multiplied on both sides of Equation D, the two equations could be added together causing the V3 terms to cancel leaving one equation with one unknown.
Multiply both sides of Equation D by 2:
$$(-9V_{X} + 5V_{3} = 12) (2)$$
Expand:
$$-18V_{X} + 10V_{3} = 24$$ [Equation D]
Add the new Equation D to Equation C:
$$39V_{X} – 10V_{3} = 36$$ [Equation C]
$$-18V_{X} + 10V_{3} = 24$$ [Equation D]
$$21V_{X} = 60$$
Solve for Vx:
$$\underline{V_{X} = 2.86 \text{ v}}$$
Recall: $$V_{2} = V_{X}$$, Substitute $$V_{X} = 2.86 \text{ volts}$$
$$\underline{V_{2} = 2.86 \text{ v}}$$
Recall: $$V_{1} = 3V_{X}$$, Substitute $$V_{X} = 2.86 \text{ volts}$$
$$V_{1} = 3(2.86 \text{ v})$$
$$\underline{V_{1} = 8.58 \text{ v}}$$
When Equation 1 and the calculated values for V1 and for V2 are used, V3 can be calculated:
Recall: $$7V_{1} + 18V_{2} – 10V_{3} = 36$$
Substitute values for V1 and for V2:
$$7(8.58) + 18(2.86) – 10V_{3} = 36$$
Expand:
$$60.06 + 51.48 – 10V_{3} = 36$$
Combine terms:
$$111.54 – 10V_{3} = 36$$
Solve for V3:
$$-10V_{3} = -111.54 + 36$$
Divide both sides by –10:
$$\underline{V_{3} = 7.55 \text{ v}}$$
The value of V0 can be calculated by using Figure 5:
Recall: $$V_{0} = V_{1} – V_{3}$$
Substitute values for V1 and for V3:
$$V_{0} = 8.58 \text{ v} – 7.55 \text{ v}$$
$$\underline{V_{0} = 1.03 \text{ v}}$$
Now that all node voltages are known, currents can be calculated for resistor R1 (IR1), R2 (IR2), R3 (IR3) and R4 (IR4).
Recall: $$I_{R_{1}} = \frac{V_{2}}{R_{1}}$$
$$I_{R_{1}} = \frac{2.86 \text{ v}}{\text {1 k }\Omega}$$
$$\underline{I_{R_{1}} = 2.86 \text{ mA}}$$
Recall: $$I_{R_{2}} = \frac{V_{2} – V_{3}}{R_{2}}$$
$$I_{R_{2}} = \frac{2.86 \text{ v} – 7.55 \text{ v}}{\text {2 k }\Omega}$$
$$\underline{I_{R_{2}} = -4.69 \text{ v}}$$
$$\underline{I_{R_{2}} = -2.35 \text{ mA}}$$
Recall: $$I_{R_{3}} = \frac{V_{1} – V_{3}}{R_{3}}$$
$$I_{R_{3}} = \frac{8.58 \text{ v} – 7.55 \text{ v}}{\text {3 k}\Omega}$$
$$I_{R_{3}} = \frac{1.03 \text{ v}}{\text {3 k}\Omega}$$
$$\underline{I_{R_{3}} = 0.34 \text{ mA}}$$
Recall: $$I_{R_{4}} = \frac{V_{1} – V_{4}}{R_{4}}$$
$$I_{R_{4}} = \frac{8.58 \text{ v} – 4 \text{ v}}{\text {4 k}\Omega}$$
$$I_{R_{4}} = \frac{4.58 \text{ v}}{\text {4 k}\Omega}$$
$$\underline{I_{R_{4}} = 1.15 \text{ mA}}$$
To confirm KCL current calculations, consider those associated at the super node:
$$- I_{1} + I_{R_{1}} + I_{R_{2}} + I_{R_{3}} + I_{R_{4}} = 0$$
Substitute circuit values:
$$(-2 \text{ mA}) + 2.86 \text{ mA} - 2.35 \text{ mA} + 0.34 \text{ mA} + 1.15 \text{ mA} = 0$$
Combine terms:
$$4.35 \text{ mA} – 4.35 \text{ mA} = 0$$
To confirm KCL current calculations at the V3 node:
$$-I_{2} + I_{R_{2}} + I_{R_{3}} = 0$$
Note IR2 and IR3 are opposites signs from the super node calculations
$$(-2 \text{ mA}) + 2.35 \text{ mA} – 0.34 \text{ mA} = 0$$
Combine terms:
$$(-2.34 \text{ mA}) + 2.35 \text{ mA} \approx 0$$
A voltage controlled current source is where the output current (IS) is a linear function of a connected component that has a referenced voltage (VX) in the following relationship:
IS = (A) (IX) Where A is a multiplier and IX needs to be determined.
The following diagram of a VCCS is illustrated in Figure 1.
Figure 1. Voltage Controlled Current Source
Consider the following circuit that consists of a dependent voltage source I2 having a value of (-2mA) (VR1), a connection link to a resistor R1 that has a value 1K Ohms with a voltage drop of VR1, an independent voltage source V1 having a value of 4 Volts and an independent current source I1 having a value of 1 mAmp as illustrated in Figure 2.
Figure 2.
Before using Kirchhoff’s Current Law (KCL) at node B, the voltage at node A can be determined using the value of the independent voltage source V1 of 4 Volts:
$$V_{A} = V_{1} = 4 \text{ volts}$$
To determine the KCL at node A, currents need to be identified. Current IA is positive because it enters the node whereas current IB and IC are negative because they leave the node as illustrated in Figure 3.
Figure 3.
The algebraic sum of all currents at node VA are equal to zero:
$$I_{A} - I_{B} - I_{C} = 0$$
Note that:
$$I_{B} = \frac{V_{B} - V_{A}}{R_{1}}$$
$$I_{C} = I_{1}$$
So that:
$$I_{A} – \frac{V_{B} – V_{A}}{R_{1}} – I_{1} = 0$$
Substitute circuit values:
$$I_{A} – \frac{V_{B} – 4}{1 \text{ k}\Omega} – 1 \text{ mA} = 0$$
Multiply both sides by 1k:
$$[I_{A} – \frac{(V_{B} – 4)}{1 \text{ k}\Omega} – 1 \text{ mA} = 0] (1 \text{ k})$$
Expand:
$$(1 \text{ k})I_{A} – (V_{B} – 4) – 1 = 0$$
$$(1 \text{ k})I_{A} – V_{B} + 4 – 1 = 0$$
Combine terms:
$$(1 \text{ k})I_{A} – V_{B} + 3 = 0$$
Unknowns on left, knowns on right
$$(1 \text{ k})I_{A} – V_{B} = -3$$ [Equation 1]
To determine the KCL at node B, currents need to be identified. Current IA, IB, and IC are positive because they enter the node whereas current IE is negative because it leaves the node as illustrated in Figure 4.
Figure 4.
The algebraic sum of all currents at node VB are equal to zero:
$$I_{B} + I_{C} + I_{D} - I_{E} = 0$$
Note that:
$$I_{B} = \frac{V_{A}-V_{B}}{R_{1}}$$
$$I_{C} = I_{1}$$
$$I_{D} = I_{2}$$
$$I_{E} = \frac{V_{B}}{R_{2}}
So that:
$$\frac{V_{A} - V_{B}}{R_{1}} + I_{1} + I_{2} - \frac{V_{B}}{R_{2}} = 0$$
Substitute circuit values:
$$\frac{4 - V_{B}}{1 \text{ k}\Omega} + 1 \text{ mA} - (2 \text{ m})V_{R_{1}} - \frac{V_{B}}{2 \text{ k}\Omega} = 0$$
Multiply both sides by 2 k:
$$[\frac{4 - V_{B}}{1 \text{ k}\Omega} + 1 \text{ mA} - (2 \text{ m})V_{R_{1}} - \frac{V_{B}}{2 \text{ k}\Omega} = 0] (2 \text{ k})$$
Expand:
$$2(4 - V_{B}) + 2 - 4V_{R_{1}} - V_{B} = 0$$
$$8 - 2V_{B} + 2 - 4V_{R_{1}} - V_{B} = 0$$
Combine terms:
$$-3V_{B} - 4V_{R_{1}} + 10 = 0$$
Unknowns on left, knowns on right
$$-3V_{B} - 4V_{R_{1}} = -10$$ [Equation 2]
Use Ohm's Law to find ER4:
$$E_{R_{4}} = I_{1}R_{4}$$
Substitute circuit values:
$$E_{R_{4}} = (1 \text{ mA}) (3 \text{ k}\Omega)$$
$$\underline{E_{R_{4}} = 3 \text{ v}}$$
Note that R4 is parallel to R1. That makes ER4 equal to VR1.
$$\underline{V_{R_{1}} = 3 \text{ v}}$$
Use Ohms Law to find IB:
\[ I_{B} = \frac{V_{R1}}{R_{1}} \]
Substitute circuit values:
$$I_{B} = \frac{3 \text{ v}}{1 \text{ k}\Omega}$$
$$\underline{I_{B} = 3 \text{ mA}}$$
Recall currents at node A:
$$I_{A} - I_{B} - I_{C} = 0$$
Solve for IA:
$$I_{A} = I_{B} + I_{C} $$
Substitute circuit values:
$$I_{A} = (3 \text{ mA}) + (1 \text{ mA})$$
$$\underline{I_{A} = 4 \text{ mA}}$$
Note that current ID is current I2:
$$I_{D} = (-2 \text{ m}) (V_{R_{1}})$$
Substitute circuit values:
$$I_{D} = (-2 \text{ m}) (3 \text{ v})$$
$$\underline{I_{D} = - 6 \text{ mA}}$$
Recall currents at node B:
$$I_{B} + I_{C} + I_{D} - I_{E} = 0$$
Solve for IE:
$$I_{E} = I_{B} + I_{C} + I_{D}$$
Substitute circuit values:
$$I_{E} = (3 \text{ mA}) + (1 \text{ mA}) + (-6 \text{ mA})$$
$$\underline{I_{E} = - 2 \text{ mA}}$$
Note that VR1 can be determined from node A and node B:
$$V_{R_{1}} = V_{A} -V_{B}$$
Solve for VB:
$$V_{B} = V_{A} - V_{R_{1}}$$
Substitute circuit values:
$$V_{B} = (4 \text{ v}) - (3 \text{ v})$$
$$\underline{V_{B} = 1 {\text{ v}}}$$
Use the equation associated with the dependent current source to find current I2:
$$I_{2} = (-2 \text{ m}) (V_{R_{1}})$$
Substitute circuit values:
$$I_{2} = (-2 \text{ m}) (3 \text{ v})$$
$$\underline{I_{2} = - 6 \text{ mA}}$$
Use Ohms Law to find the voltage across resistor R3:
$$E_{R_{3}} = (I_{2}) (R_{3})$$
Substitute circuit values:
$$E_{R_{3}} = (-6 \text{ mA}) (4 \text{ k}\Omega)$$
$$\underline{E_{R_{3}} = - 24 \text{ v}}$$
To confirm KVL loop involving the independent voltage source V1 and resistors R1 and R2 the following equation is obtained:
$$V_{1} = E_{R_{1}} + E_{R_{2}}$$
Substitute circuit values and note ER2 equals VB
$$(4 \text{ v}) = (3 \text{ v}) + (1 \text{ v}) $$
Nodal analysis with dependent sources utilized Kirchhoff’s Current Law with Algebra and Ohm’s Law to substitute an unknown voltage for a node and to find other circuit values. By taking the time to carefully label the nodes, by identifying the proper node voltages and polarities, problem solving is made easier and can avoid mistakes. |
You need to evaluate $H(z)$ on the unit circle $z=e^{iw}$ in order to get the frequency response (assuming that the system is stable, i.e. the region of convergence contains the unit circle). But your $\mathcal{Z}$-transform looks a bit unusual because for causal signals (or filter impulse responses) you should get negative powers of $z$:
$$H(z)=\sum_{n=0}^{\infty}x(n)z^{-n}$$
In your case $x(n)=c^n$, $n\ge 0$, so you get
$$H(z)=\frac{1}{1-cz^{-1}}=\frac{z}{z-c}\tag{1}$$
From (1) you see that the pole is at $z_{\infty}=c$. Note that you need to determine the poles of $H(z)$ for general $z$, not on the unit circle $z=e^{iw}$ because stable systems cannot have poles on the unit circle (this is also why you got a complex frequency, which does not make sense). The frequency response is
$$H(e^{iw})=\frac{1}{1-ce^{-iw}}$$
It is true that for real filter coefficients you get complex conjugate pairs of poles and zeros. But this is not the whole truth. You can also get real-valued poles and zeros, and for a first-order system like yours, this must be the case. You need at least a second-order system to get a complex pole pair. In your case you simply got one real pole (assuming that $c$ is a real-valued constant). |
A complete combinatorial proof using Allen's comment:
Let $(W,S)$ be a Coxeter system, and let $Dem(T) \in W$ be the
Demazure product or greedy product of a word $T$ in $S$.
Claim: $Dem(T)$ is the unique Bruhat maximal element in the set $\big\{ \prod Q : Q \subseteq T\big\}$ (where $Q\subseteq T$ means that $Q$ is a subsequence of $T$, and where $\prod Q$ means the product of the entries of $Q$ in the order in which they appear in $Q$).
The idea for the proof is to start with a subword $Q$ of $T$ and compare it with the subword $D$ of $T$ picked by the greedy product (you find the formal proof below). You scan through $Q$ from left to right and if you see a letter that is picked in $D$ but not in $Q$, you insert it into $Q$. If this goes up in Bruhat order, we are fine in doing so, and if you go down in Bruhat order, you find by the exchange condition a letter to its right that you can remove in exchange for the inserted letter.By this procedure, you only go up in Bruhat order and we are done.
Corollary: The Demazure product is well-defined in Artin groups. This is, let $T$ be a word of $S$ and let $T'$ be obtained from $T$ by a braid move. Then $Dem(T) = Dem(T')$.
Proof of corollary: $T'$ is obtained from $T$ by replacing a consecutive substring $x = stst\ldots$ of length $m(s,t)$ by $y = tsts\ldots$. For any subword $Q$ of $T$, one can now choose the same subword in $T′$ as long as $Q$ does not contain all of $x$. But if this is the case, one can choose the subword $Q'$ of $T'$ where $Q'$ is obtained from $Q$ by using $y$ instead of $x$. $\square$
Proof of Claim:This is a consequence of the following lifting property in Bruhat order as described in Proposition 2.2.7 of Björner-Brenti's Combinatorics of Coxeter groups
Lemma 1 (lifting property) : Let $u < w$ in Bruhat order, and let $s$ be a right descent of $w$ but not of $u$. (Here, a right descent of an element $v \in W$ means a $t \in S$ satisfying $vt < v$.) Then $us \leq w$.
(Proof in Björner-Brenti, at least for the analogous statement about left descents; apply it to $u^{-1}$ and $w^{-1}$.)
Lemma 2: Let $u \leq w$ in Bruhat order, and let $s$ be a right ascent of $w$. (Here, a right ascent of an element $v \in W$ means a $t \in S$ satisfying $vt > v$.) Then $us \leq ws$.
Proof: Since $u \leq w$, we have that a reduced expression $a$ for $u$ which is a subword of a reduced expression $b$ for $w$. But since now $bs$ is a reduced expression for $ws$, it contains the expression $as$ (which might or might not be reduced) and we are done. $\square$
Final induction to prove the Claim: Let $T = t_1\cdots t_m$.The case $m \in \{0,1\}$ is trivial, so assume $m>1$, let $T' = t_1\cdots t_{m-1}$ and we know that $Dem(T')$ is the unique Bruhat maximal element in $\{ \prod Q : Q \subseteq T'\}$.
Let $Q$ be a subword of $T$. If $Q$ is a subword of $T'$ we are done since by assumption $\prod Q \leq Dem(T') \leq Dem(T)$, so we only treat the case that $Q$ uses the last letter $t_m$.
We have $Q \setminus t_m$ is a subword of $T'$ so $\prod \left(Q\setminus t_m\right) \leq Dem(T')$ by induction.
If $Dem(T) > Dem(T')$, we are in the situation of Lemma 2 and conclude $$\prod Q \leq Dem(T') \cdot t_m = Dem(T).$$
If $Dem(T) = Dem(T')$, we are in the situation of Lemma 1 and conclude $$\prod Q \leq Dem(T') = Dem(T). \quad \square$$
(As usual with MO proofs, please let me know if something is unclear or plainly wrong.) |
stat946w18/Implicit Causal Models for Genome-wide Association Studies Contents 1 Introduction and Motivation 2 Implicit Causal Models 3 Implicit Causal Models with Latent Confounders 4 Likelihood-free Variational Inference 5 Empirical Study 6 Conclusion 7 Critique 8 References 9 Implicit causal model in Edward Introduction and Motivation
There is currently much progress in probabilistic models which could lead to the development of rich generative models. The models have been applied with neural networks, implicit densities, and with scalable algorithms to very large data for their Bayesian inference. However, most of the models are focused on capturing statistical relationships rather than causal relationships. Causal relationships are relationships where one event is a result of another event, i.e. a cause and effect. Causal models give us a sense of how manipulating the generative process could change the final results.
Genome-wide association studies (GWAS) are examples of causal relationships. Genome is basically the sum of all DNAs in an organism and contain information about the organism's attributes. Specifically, GWAS is about figuring out how genetic factors cause disease among humans. Here the genetic factors we are referring to are single nucleotide polymorphisms (SNPs), and getting a particular disease is treated as a trait, i.e., the outcome. In order to know about the reason of developing a disease and to cure it, the causation between SNPs and diseases is investigated: first, predict which one or more SNPs cause the disease; second, target the selected SNPs to cure the disease.
The figure below depicts an example Manhattan plot for a GWAS. Each dot represents an SNP. The x-axis is the chromosome location, and the y-axis is the negative log of the association p-value between the SNP and the disease, so points with the largest values represent strongly associated risk loci.
This paper focuses on two challenges to combining modern probabilistic models and causality. The first one is how to build rich causal models with specific needs by GWAS. In general, probabilistic causal models involve a function [math]f[/math] and a noise [math]n[/math]. For working simplicity, we usually assume [math]f[/math] as a linear model with Gaussian noise. However problems like GWAS require models with nonlinear, learnable interactions among the inputs and the noise.
The second challenge is how to address latent population-based confounders. Latent confounders are issues when we apply the causal models since we cannot observe them nor know the underlying structure. For example, in GWAS, both latent population structure, i.e., subgroups in the population with ancestry differences, and relatedness among sample individuals produce spurious correlations among SNPs to the trait of interest. The existing methods cannot easily accommodate the complex latent structure.
For the first challenge, the authors develop implicit causal models, a class of causal models that leverages neural architectures with an implicit density. With GWAS, implicit causal models generalize previous methods to capture important nonlinearities, such as gene-gene and gene-population interaction. Building on this, for the second challenge, they describe an implicit causal model that adjusts for population-confounders by sharing strength across examples (genes).
There has been an increasing number of works on causal models which focus on causal discovery and typically have strong assumptions such as Gaussian processes on noise variable or nonlinearities for the main function.
Implicit Causal Models
Implicit causal models are an extension of probabilistic causal models. Probabilistic causal models will be introduced first.
Probabilistic Causal Models
Probabilistic causal models have two parts: deterministic functions of noise and other variables. Consider background noise [math]\epsilon[/math], representing unknown background quantities which are jointly independent and global variable [math]\beta[/math], some function of this noise, where
Each [math]\beta[/math] and [math]x[/math] is a function of noise; [math]y[/math] is a function of noise and [math]x[/math]，
The target is the causal mechanism [math]f_y[/math] so that the causal effect [math]p(y|do(X=x),\beta)[/math] can be calculated. [math]do(X=x)[/math] means that we specify a value of [math]X[/math] under the fixed structure [math]\beta[/math]. By other paper’s work, it is assumed that [math]p(y|do(x),\beta) = p(y|x, \beta)[/math].
An example of probabilistic causal models is additive noise model.
[math]f(.)[/math] is usually a linear function or spline functions for nonlinearities. [math]\epsilon[/math] is assumed to be standard normal, as well as [math]y[/math]. Thus the posterior [math]p(\theta | x, y, \beta)[/math] can be represented as
where [math]p(\theta)[/math] is the prior which is known. Then, variational inference or MCMC can be applied to calculate the posterior distribution.
Implicit Causal Models
The difference between implicit causal models and probabilistic causal models is the noise variable. Instead of using an additive noise term, implicit causal models directly take noise [math]\epsilon[/math] as input and outputs [math]x[/math] given parameter [math]\theta[/math].
[math] x=g(\epsilon | \theta), \epsilon \tilde s(\cdot) [/math]
The causal diagram has changed to:
They used fully connected neural network with a fair amount of hidden units to approximate each causal mechanism. Below is the formal description: Implicit Causal Models with Latent Confounders
Previously, they assumed the global structure is observed. Next, the unobserved scenario is being considered.
Causal Inference with a Latent Confounder
Similar to before, the interest is the causal effect [math]p(y|do(x_m), x_{-m})[/math]. Here, the SNPs other than [math]x_m[/math] is also under consideration. However, it is confounded by the unobserved confounder [math]z_n[/math]. As a result, the standard inference method cannot be used in this case.
The paper proposed a new method which include the latent confounders. For each subject [math]n=1,…,N[/math] and each SNP [math]m=1,…,M[/math],
The mechanism for latent confounder [math]z_n[/math] is assumed to be known. SNPs depend on the confounders and the trait depends on all the SNPs and the confounders as well.
The posterior of [math]\theta[/math] is needed to be calculate in order to estimate the mechanism [math]g_y[/math] as well as the causal effect [math]p(y|do(x_m), x_{-m})[/math], so that it can be explained how changes to each SNP [math]X_m[/math] cause changes to the trait [math]Y[/math].
Note that the latent structure [math]p(z|x, y)[/math] is assumed known.
In general, causal inference with latent confounders can be dangerous: it uses the data twice, and thus it may bias the estimates of each arrow [math]X_m → Y[/math]. Why is this justified? This is answered below:
Proposition 1. Assume the causal graph of Figure 2 (left) is correct and that the true distribution resides in some configuration of the parameters of the causal model (Figure 2 (right)). Then the posterior [math]p(θ | x, y)[/math] provides a consistent estimator of the causal mechanism [math]f_y[/math].
Proposition 1 rigorizes previous methods in the framework of probabilistic causal models. The intuition is that as more SNPs arrive (“M → ∞, N fixed”), the posterior concentrates at the true confounders [math]z_n[/math], and thus we can estimate the causal mechanism given each data point’s confounder [math]z_n[/math]. As more data points arrive (“N → ∞, M fixed”), we can estimate the causal mechanism given any confounder [math]z_n[/math] as there is an infinity of them.
Implicit Causal Model with a Latent Confounder
This section is the algorithm and functions to implementing an implicit causal model for GWAS.
Generative Process of Confounders [math]z_n[/math].
The distribution of confounders is set as standard normal. [math]z_n \in R^K[/math] , where [math]K[/math] is the dimension of [math]z_n[/math] and [math]K[/math] should make the latent space as close as possible to the true population structural.
Generative Process of SNPs [math]x_{nm}[/math].
Given SNP is coded for,
The authors defined a [math]Binomial(2,\pi_{nm})[/math] distribution on [math]x_{nm}[/math]. And used logistic factor analysis to design the SNP matrix.
A SNP matrix looks like this:
Since logistic factor analysis makes strong assumptions, this paper suggests using a neural network to relax these assumptions,
This renders the outputs to be a full [math]N*M[/math] matrix due the the variables [math]w_m[/math], which act as principal component in PCA. Here, [math]\phi[/math] has a standard normal prior distribution. The weights [math]w[/math] and biases [math]\phi[/math] are shared over the [math]m[/math] SNPs and [math]n[/math] individuals, which makes it possible to learn nonlinear interactions between [math]z_n[/math] and [math]w_m[/math].
Generative Process of Traits [math]y_n[/math].
Previously, each trait is modeled by a linear regression,
This also has very strong assumptions on SNPs, interactions, and additive noise. It can also be replaced by a neural network which only outputs a scalar,
Likelihood-free Variational Inference
Calculating the posterior of [math]\theta[/math] is the key of applying the implicit causal model with latent confounders.
could be reduces to
However, with implicit models, integrating over a nonlinear function could be suffered. The authors applied likelihood-free variational inference (LFVI). LFVI proposes a family of distribution over the latent variables. Here the variables [math]w_m[/math] and [math]z_n[/math] are all assumed to be Normal,
For LFVI applied to GWAS, the algorithm which similar to the EM algorithm has been used:
Empirical Study
The authors performed simulation on 100,000 SNPs, 940 to 5,000 individuals, and across 100 replications of 11 settings. Four methods were compared:
implicit causal model (ICM); PCA with linear regression (PCA); a linear mixed model (LMM); logistic factor analysis with inverse regression (GCAT).
The feedforward neural networks for traits and SNPs are fully connected with two hidden layers using ReLU activation function, and batch normalization.
Simulation Study
Based on real genomic data, a true model is applied to generate the SNPs and traits for each configuration. There are four datasets used in this simulation study:
HapMap [Balding-Nichols model] 1000 Genomes Project (TGP) [PCA] Human Genome Diversity project (HGDP) [PCA] HGDP [Pritchard-Stephens-Donelly model] A latent spatial position of individuals for population structure [spatial] The table shows the prediction accuracy. The accuracy is calculated by the rate of the number of true positives divide the number of true positives plus false positives. True positives measure the proportion of positives that are correctly identified as such (e.g. the percentage of SNPs which are correctly identified as having the causal relation with the trait). In contrast, false positives state the SNPs has the causal relation with the trait when they don’t. The closer the rate to 1, the better the model is since false positives are considered as the wrong prediction.
The result represented above shows that the implicit causal model has the best performance among these four models in every situation. Especially, other models tend to do poorly on PSD and Spatial when [math]a[/math] is small, but the ICM achieved a significantly high rate. The only comparable method to ICM is GCAT, when applying to simpler configurations.
Real-data Analysis
They also applied ICM to GWAS of Northern Finland Birth Cohorts, which measure 10 metabolic traits and also contain 324,160 SNPs and 5,027 individuals. The data came from the database of Genotypes and Phenotypes (dbGaP) and used the same preprocessing as Song et al. Ten implicit causal models were fitted, one for each trait to be modeled. For each of the 10 implicit causal models the dimension of the counfounders was set to be six, same as what was used in the paper by Song. The SNP network used 512 hidden units in both layers and the trait network used 32 and 256. et al. for comparable models in Table 2.
The numbers in the above table are the number of significant loci for each of the 10 traits. The number for other methods, such as GCAT, LMM, PCA, and "uncorrected" (association tests without accounting for hidden relatedness of study samples) are obtained from other papers. By comparison, the ICM reached the level of the best previous model for each trait.
Conclusion
This paper introduced implicit causal models in order to account for nonlinear complex causal relationships, and applied the method to GWAS. It can not only capture important interactions between genes within an individual and among population level, but also can adjust for latent confounders by taking account of the latent variables into the model.
By the simulation study, the authors proved that the implicit causal model could beat other methods by 15-45.3% on a variety of datasets with variations on parameters.
The authors also believed this GWAS application is only the start of the usage of implicit causal models. The authors suggest that it might also be successfully used in the design of dynamic theories in high-energy physics or for modeling discrete choices in economics.
Critique
This paper is an interesting and novel work. The main contribution of this paper is to connect the statistical genetics and the machine learning methodology. The method is technically sound and does indeed generalize techniques currently used in statistical genetics.
The neural network used in this paper is a very simple feed-forward 2 hidden-layer neural network, but the idea of where to use the neural network is crucial and might be significant in GWAS.
It has limitations as well. The empirical example in this paper is too easy, and far away from the realistic situation. Despite the simulation study showing some competing results, the Northern Finland Birth Cohort Data application did not demonstrate the advantage of using implicit causal model over the previous methods, such as GCAT or LMM.
Another limitation is about linkage disequilibrium as the authors stated as well. SNPs are not completely independent of each other; usually, they have correlations when the alleles at close locus. They did not consider this complex case, rather they only considered the simplest case where they assumed all the SNPs are independent.
Furthermore, one SNP maybe does not have enough power to explain the causal relationship. Recent papers indicate that causation to a trait may involve multiple SNPs. This could be a future work as well.
References
Tran D, Blei D M. Implicit Causal Models for Genome-wide Association Studies[J]. arXiv preprint arXiv:1710.10742, 2017.
Patrik O Hoyer, Dominik Janzing, Joris M Mooij, Jonas Peters, and Prof Bernhard Schölkopf. Non- linear causal discovery with additive noise models. In Neural Information Processing Systems, 2009.
Alkes L Price, Nick J Patterson, Robert M Plenge, Michael E Weinblatt, Nancy A Shadick, and David Reich. Principal components analysis corrects for stratification in genome-wide association studies. Nature Genetics, 38(8):904–909, 2006.
Minsun Song, Wei Hao, and John D Storey. Testing for genetic associations in arbitrarily structured populations. Nature, 47(5):550–554, 2015.
Dustin Tran, Rajesh Ranganath, and David M Blei. Hierarchical implicit models and likelihood-free variational inference. In Neural Information Processing Systems, 2017. |
For a harmonic oscillator in one dimension, there is an uncertainty relation between the number of quanta $n$ and the phase of the oscillation $\phi$. There are all kinds of technical complications arising from the fact that $\phi$ can't be made into a single-valued and continuous operator (Carruthers 1968), but roughly speaking, you can write an uncertainty relation like this:
$\Delta n \Delta \phi \ge 1$
The fact that the right-hand side is 1 rather than $\hbar$ is not just from using natural units such that $\hbar=1$; we can see this because $n$ and $\phi$ are both unitless. This suggests that this uncertainty relation is a classical one, like the time-frequency uncertainty relation that is the reason for referring to data transmission speeds in terms of bandwidth.
However, the only physical interpretation I know of seems purely quantum-mechanical (Peierls 1979):
[...] any device capable of measuring the field, including its phase, must be capable of altering the number of quanta by an indeterminate amount
If this uncertainty relation is classical, what is its classical interpretation? If it's not classical, then why doesn't the restriction vanish in the limit $\hbar\rightarrow0$?
Carruthers and Nieto, "Phase and Angle Variables in Quantum Mechanics," Rev Mod Phys 40 (1968) 411 -- can be found online by googling
Peierls, Surprises in theoretical physics, 1979 |
stat946w18/Implicit Causal Models for Genome-wide Association Studies Contents 1 Introduction and Motivation 2 Implicit Causal Models 3 Implicit Causal Models with Latent Confounders 4 Likelihood-free Variational Inference 5 Empirical Study 6 Conclusion 7 Critique 8 References 9 Implicit causal model in Edward Introduction and Motivation
There is currently much progress in probabilistic models which could lead to the development of rich generative models. The models have been applied with neural networks, implicit densities, and with scalable algorithms to very large data for their Bayesian inference. However, most of the models are focused on capturing statistical relationships rather than causal relationships. Causal relationships are relationships where one event is a result of another event, i.e. a cause and effect. Causal models give us a sense of how manipulating the generative process could change the final results.
Genome-wide association studies (GWAS) are examples of causal relationships. Genome is basically the sum of all DNAs in an organism and contain information about the organism's attributes. Specifically, GWAS is about figuring out how genetic factors cause disease among humans. Here the genetic factors we are referring to are single nucleotide polymorphisms (SNPs), and getting a particular disease is treated as a trait, i.e., the outcome. In order to know about the reason of developing a disease and to cure it, the causation between SNPs and diseases is investigated: first, predict which one or more SNPs cause the disease; second, target the selected SNPs to cure the disease.
The figure below depicts an example Manhattan plot for a GWAS. Each dot represents an SNP. The x-axis is the chromosome location, and the y-axis is the negative log of the association p-value between the SNP and the disease, so points with the largest values represent strongly associated risk loci.
This paper focuses on two challenges to combining modern probabilistic models and causality. The first one is how to build rich causal models with specific needs by GWAS. In general, probabilistic causal models involve a function [math]f[/math] and a noise [math]n[/math]. For working simplicity, we usually assume [math]f[/math] as a linear model with Gaussian noise. However problems like GWAS require models with nonlinear, learnable interactions among the inputs and the noise.
The second challenge is how to address latent population-based confounders. Latent confounders are issues when we apply the causal models since we cannot observe them nor know the underlying structure. For example, in GWAS, both latent population structure, i.e., subgroups in the population with ancestry differences, and relatedness among sample individuals produce spurious correlations among SNPs to the trait of interest. The existing methods cannot easily accommodate the complex latent structure.
For the first challenge, the authors develop implicit causal models, a class of causal models that leverages neural architectures with an implicit density. With GWAS, implicit causal models generalize previous methods to capture important nonlinearities, such as gene-gene and gene-population interaction. Building on this, for the second challenge, they describe an implicit causal model that adjusts for population-confounders by sharing strength across examples (genes).
There has been an increasing number of works on causal models which focus on causal discovery and typically have strong assumptions such as Gaussian processes on noise variable or nonlinearities for the main function.
Implicit Causal Models
Implicit causal models are an extension of probabilistic causal models. Probabilistic causal models will be introduced first.
Probabilistic Causal Models
Probabilistic causal models have two parts: deterministic functions of noise and other variables. Consider background noise [math]\epsilon[/math], representing unknown background quantities which are jointly independent and global variable [math]\beta[/math], some function of this noise, where
Each [math]\beta[/math] and [math]x[/math] is a function of noise; [math]y[/math] is a function of noise and [math]x[/math]，
The target is the causal mechanism [math]f_y[/math] so that the causal effect [math]p(y|do(X=x),\beta)[/math] can be calculated. [math]do(X=x)[/math] means that we specify a value of [math]X[/math] under the fixed structure [math]\beta[/math]. By other paper’s work, it is assumed that [math]p(y|do(x),\beta) = p(y|x, \beta)[/math].
An example of probabilistic causal models is additive noise model.
[math]f(.)[/math] is usually a linear function or spline functions for nonlinearities. [math]\epsilon[/math] is assumed to be standard normal, as well as [math]y[/math]. Thus the posterior [math]p(\theta | x, y, \beta)[/math] can be represented as
where [math]p(\theta)[/math] is the prior which is known. Then, variational inference or MCMC can be applied to calculate the posterior distribution.
Implicit Causal Models
The difference between implicit causal models and probabilistic causal models is the noise variable. Instead of using an additive noise term, implicit causal models directly take noise [math]\epsilon[/math] as input and outputs [math]x[/math] given parameter [math]\theta[/math].
[math] x=g(\epsilon | \theta), \epsilon \tilde s(\cdot) [/math]
The causal diagram has changed to:
They used fully connected neural network with a fair amount of hidden units to approximate each causal mechanism. Below is the formal description: Implicit Causal Models with Latent Confounders
Previously, they assumed the global structure is observed. Next, the unobserved scenario is being considered.
Causal Inference with a Latent Confounder
Similar to before, the interest is the causal effect [math]p(y|do(x_m), x_{-m})[/math]. Here, the SNPs other than [math]x_m[/math] is also under consideration. However, it is confounded by the unobserved confounder [math]z_n[/math]. As a result, the standard inference method cannot be used in this case.
The paper proposed a new method which include the latent confounders. For each subject [math]n=1,…,N[/math] and each SNP [math]m=1,…,M[/math],
The mechanism for latent confounder [math]z_n[/math] is assumed to be known. SNPs depend on the confounders and the trait depends on all the SNPs and the confounders as well.
The posterior of [math]\theta[/math] is needed to be calculate in order to estimate the mechanism [math]g_y[/math] as well as the causal effect [math]p(y|do(x_m), x_{-m})[/math], so that it can be explained how changes to each SNP [math]X_m[/math] cause changes to the trait [math]Y[/math].
Note that the latent structure [math]p(z|x, y)[/math] is assumed known.
In general, causal inference with latent confounders can be dangerous: it uses the data twice, and thus it may bias the estimates of each arrow [math]X_m → Y[/math]. Why is this justified? This is answered below:
Proposition 1. Assume the causal graph of Figure 2 (left) is correct and that the true distribution resides in some configuration of the parameters of the causal model (Figure 2 (right)). Then the posterior [math]p(θ | x, y)[/math] provides a consistent estimator of the causal mechanism [math]f_y[/math].
Proposition 1 rigorizes previous methods in the framework of probabilistic causal models. The intuition is that as more SNPs arrive (“M → ∞, N fixed”), the posterior concentrates at the true confounders [math]z_n[/math], and thus we can estimate the causal mechanism given each data point’s confounder [math]z_n[/math]. As more data points arrive (“N → ∞, M fixed”), we can estimate the causal mechanism given any confounder [math]z_n[/math] as there is an infinity of them.
Implicit Causal Model with a Latent Confounder
This section is the algorithm and functions to implementing an implicit causal model for GWAS.
Generative Process of Confounders [math]z_n[/math].
The distribution of confounders is set as standard normal. [math]z_n \in R^K[/math] , where [math]K[/math] is the dimension of [math]z_n[/math] and [math]K[/math] should make the latent space as close as possible to the true population structural.
Generative Process of SNPs [math]x_{nm}[/math].
Given SNP is coded for,
The authors defined a [math]Binomial(2,\pi_{nm})[/math] distribution on [math]x_{nm}[/math]. And used logistic factor analysis to design the SNP matrix.
A SNP matrix looks like this:
Since logistic factor analysis makes strong assumptions, this paper suggests using a neural network to relax these assumptions,
This renders the outputs to be a full [math]N*M[/math] matrix due the the variables [math]w_m[/math], which act as principal component in PCA. Here, [math]\phi[/math] has a standard normal prior distribution. The weights [math]w[/math] and biases [math]\phi[/math] are shared over the [math]m[/math] SNPs and [math]n[/math] individuals, which makes it possible to learn nonlinear interactions between [math]z_n[/math] and [math]w_m[/math].
Generative Process of Traits [math]y_n[/math].
Previously, each trait is modeled by a linear regression,
This also has very strong assumptions on SNPs, interactions, and additive noise. It can also be replaced by a neural network which only outputs a scalar,
Likelihood-free Variational Inference
Calculating the posterior of [math]\theta[/math] is the key of applying the implicit causal model with latent confounders.
could be reduces to
However, with implicit models, integrating over a nonlinear function could be suffered. The authors applied likelihood-free variational inference (LFVI). LFVI proposes a family of distribution over the latent variables. Here the variables [math]w_m[/math] and [math]z_n[/math] are all assumed to be Normal,
For LFVI applied to GWAS, the algorithm which similar to the EM algorithm has been used:
Empirical Study
The authors performed simulation on 100,000 SNPs, 940 to 5,000 individuals, and across 100 replications of 11 settings. Four methods were compared:
implicit causal model (ICM); PCA with linear regression (PCA); a linear mixed model (LMM); logistic factor analysis with inverse regression (GCAT).
The feedforward neural networks for traits and SNPs are fully connected with two hidden layers using ReLU activation function, and batch normalization.
Simulation Study
Based on real genomic data, a true model is applied to generate the SNPs and traits for each configuration. There are four datasets used in this simulation study:
HapMap [Balding-Nichols model] 1000 Genomes Project (TGP) [PCA] Human Genome Diversity project (HGDP) [PCA] HGDP [Pritchard-Stephens-Donelly model] A latent spatial position of individuals for population structure [spatial] The table shows the prediction accuracy. The accuracy is calculated by the rate of the number of true positives divide the number of true positives plus false positives. True positives measure the proportion of positives that are correctly identified as such (e.g. the percentage of SNPs which are correctly identified as having the causal relation with the trait). In contrast, false positives state the SNPs has the causal relation with the trait when they don’t. The closer the rate to 1, the better the model is since false positives are considered as the wrong prediction.
The result represented above shows that the implicit causal model has the best performance among these four models in every situation. Especially, other models tend to do poorly on PSD and Spatial when [math]a[/math] is small, but the ICM achieved a significantly high rate. The only comparable method to ICM is GCAT, when applying to simpler configurations.
Real-data Analysis
They also applied ICM to GWAS of Northern Finland Birth Cohorts, which measure 10 metabolic traits and also contain 324,160 SNPs and 5,027 individuals. The data came from the database of Genotypes and Phenotypes (dbGaP) and used the same preprocessing as Song et al. Ten implicit causal models were fitted, one for each trait to be modeled. For each of the 10 implicit causal models the dimension of the counfounders was set to be six, same as what was used in the paper by Song. The SNP network used 512 hidden units in both layers and the trait network used 32 and 256. et al. for comparable models in Table 2.
The numbers in the above table are the number of significant loci for each of the 10 traits. The number for other methods, such as GCAT, LMM, PCA, and "uncorrected" (association tests without accounting for hidden relatedness of study samples) are obtained from other papers. By comparison, the ICM reached the level of the best previous model for each trait.
Conclusion
This paper introduced implicit causal models in order to account for nonlinear complex causal relationships, and applied the method to GWAS. It can not only capture important interactions between genes within an individual and among population level, but also can adjust for latent confounders by taking account of the latent variables into the model.
By the simulation study, the authors proved that the implicit causal model could beat other methods by 15-45.3% on a variety of datasets with variations on parameters.
The authors also believed this GWAS application is only the start of the usage of implicit causal models. The authors suggest that it might also be successfully used in the design of dynamic theories in high-energy physics or for modeling discrete choices in economics.
Critique
This paper is an interesting and novel work. The main contribution of this paper is to connect the statistical genetics and the machine learning methodology. The method is technically sound and does indeed generalize techniques currently used in statistical genetics.
The neural network used in this paper is a very simple feed-forward 2 hidden-layer neural network, but the idea of where to use the neural network is crucial and might be significant in GWAS.
It has limitations as well. The empirical example in this paper is too easy, and far away from the realistic situation. Despite the simulation study showing some competing results, the Northern Finland Birth Cohort Data application did not demonstrate the advantage of using implicit causal model over the previous methods, such as GCAT or LMM.
Another limitation is about linkage disequilibrium as the authors stated as well. SNPs are not completely independent of each other; usually, they have correlations when the alleles at close locus. They did not consider this complex case, rather they only considered the simplest case where they assumed all the SNPs are independent.
Furthermore, one SNP maybe does not have enough power to explain the causal relationship. Recent papers indicate that causation to a trait may involve multiple SNPs. This could be a future work as well.
References
Tran D, Blei D M. Implicit Causal Models for Genome-wide Association Studies[J]. arXiv preprint arXiv:1710.10742, 2017.
Patrik O Hoyer, Dominik Janzing, Joris M Mooij, Jonas Peters, and Prof Bernhard Schölkopf. Non- linear causal discovery with additive noise models. In Neural Information Processing Systems, 2009.
Alkes L Price, Nick J Patterson, Robert M Plenge, Michael E Weinblatt, Nancy A Shadick, and David Reich. Principal components analysis corrects for stratification in genome-wide association studies. Nature Genetics, 38(8):904–909, 2006.
Minsun Song, Wei Hao, and John D Storey. Testing for genetic associations in arbitrarily structured populations. Nature, 47(5):550–554, 2015.
Dustin Tran, Rajesh Ranganath, and David M Blei. Hierarchical implicit models and likelihood-free variational inference. In Neural Information Processing Systems, 2017. |
14 0 Homework Statement A yo-yo is placed on a conveyor belt accelerating ##a_C = 1 m/s^2## to the left. The end of the rope of the yo-yo is fixed to a wall on the right. The moment of inertia is ##I = 200 kg \cdot m^2##. Its mass is ##m = 100kg##. The radius of the outer circle is ##R = 2m## and the radius of the inner circle is ##r = 1m##. The coefficient of static friction is ##0.4## and the coefficient of kinetic friction is ##0.3##. Find the initial tension in the rope and the angular acceleration of the yo-yo. Homework Equations ##T - f = ma## ##\tau_P = -fr## ##\tau_G = Tr## ##I_P = I + mr^2## ##I_G = I + mR^2## ##a = \alpha R##
First off, I was wondering if the acceleration of the conveyor belt can be considered a force. And I'm not exactly sure how to use Newton's second law if the object of the forces is itself on an accelerating surface.
Also, I don't know whether it rolls with or without slipping. I thought I could use ##a_C = \alpha R## for the angular acceleration, but the acceleration of the conveyor belt is not the only source of acceleration, since the friction and the tension also play a role. I can't find a way to combine these equations to get the
Also, I don't know whether it rolls with or without slipping.
I thought I could use ##a_C = \alpha R## for the angular acceleration, but the acceleration of the conveyor belt is not the only source of acceleration, since the friction and the tension also play a role.
I can't find a way to combine these equations to get the |
Tagged: linear algebra Problem 26
In this problem, we will show that the concept of non-singularity of a matrix is equivalent to the concept of invertibility.
That is, we will prove that: (a)Show that if $A$ is invertible, then $A$ is nonsingular. (b)Let $A, B, C$ be $n\times n$ matrices such that $AB=C$.
Prove that if either $A$ or $B$ is singular, then so is $C$.
(c)Show that if $A$ is nonsingular, then $A$ is invertible. Add to solve later Problem 25
An $n \times n$ matrix $A$ is called
if the only solution of the equation $A \mathbf{x}=\mathbf{0}$ is the zero vector $\mathbf{x}=\mathbf{0}$. nonsingular
Otherwise $A$ is called
. singular (a) Show that if $A$ and $B$ are $n\times n$ nonsingular matrices, then the product $AB$ is also nonsingular. (b) Show that if $A$ is nonsingular, then the column vectors of $A$ are linearly independent. (c) Show that an $n \times n$ matrix $A$ is nonsingular if and only if the equation $A\mathbf{x}=\mathbf{b}$ has a unique solution for any vector $\mathbf{b}\in \R^n$.
Add to solve later
Restriction Do not use the fact that a matrix is nonsingular if and only if the matrix is invertible. Problem 19
Let $A=(a_{i j})$ and $B=(b_{i j})$ be $n\times n$ real matrices for some $n \in \N$. Then answer the following questions about the trace of a matrix.
(a) Express $\tr(AB^{\trans})$ in terms of the entries of the matrices $A$ and $B$. Here $B^{\trans}$ is the transpose matrix of $B$. (b) Show that $\tr(AA^{\trans})$ is the sum of the square of the entries of $A$.
Add to solve later
(c) Show that if $A$ is nonzero symmetric matrix, then $\tr(A^2)>0$. Problem 15
Let $p_1(x), p_2(x), p_3(x), p_4(x)$ be (real) polynomials of degree at most $3$. Which (if any) of the following two conditions is sufficient for the conclusion that these polynomials are linearly dependent?
(a) At $1$ each of the polynomials has the value $0$. Namely $p_i(1)=0$ for $i=1,2,3,4$. (b) At $0$ each of the polynomials has the value $1$. Namely $p_i(0)=1$ for $i=1,2,3,4$.
(
University of California, Berkeley) Problem 12
Let $A$ be an $n \times n$ real matrix. Prove the followings.
(a) The matrix $AA^{\trans}$ is a symmetric matrix. (b) The set of eigenvalues of $A$ and the set of eigenvalues of $A^{\trans}$ are equal. (c) The matrix $AA^{\trans}$ is non-negative definite.
(An $n\times n$ matrix $B$ is called
non-negative definite if for any $n$ dimensional vector $\mathbf{x}$, we have $\mathbf{x}^{\trans}B \mathbf{x} \geq 0$.)
Add to solve later
(d) All the eigenvalues of $AA^{\trans}$ is non-negative. Problem 11
An $n\times n$ matrix $A$ is called
nilpotent if $A^k=O$, where $O$ is the $n\times n$ zero matrix. Prove the followings. (a) The matrix $A$ is nilpotent if and only if all the eigenvalues of $A$ is zero.
Add to solve later
(b) The matrix $A$ is nilpotent if and only if $A^n=O$. Read solution Problem 9
Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues.
Show that (1) $$\det(A)=\prod_{i=1}^n \lambda_i$$ (2) $$\tr(A)=\sum_{i=1}^n \lambda_i$$
Here $\det(A)$ is the determinant of the matrix $A$ and $\tr(A)$ is the trace of the matrix $A$.
Namely, prove that (1) the determinant of $A$ is the product of its eigenvalues, and (2) the trace of $A$ is the sum of the eigenvalues.
Read solution Problem 5
Let $T : \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation.
Let $\mathbf{0}_n$ and $\mathbf{0}_m$ be zero vectors of $\mathbb{R}^n$ and $\mathbb{R}^m$, respectively. Show that $T(\mathbf{0}_n)=\mathbf{0}_m$.
(
The Ohio State University Linear Algebra Exam)
Add to solve later |
Hi everyone!
I have a problem at hand that needs to calculate a series of covariant derivatives like
\nabla_{\alpha} \nabla_{\beta} \nabla_{\gamma} \nabla_{\lambda}\partial_{\nu}\phi
Where \nabla is the usual covariant derivative in GR and \partial is the usual partial derivative.
In fact, this expression is a term of an infinite series, thematically
g^{\mu \nu} \partial_{\mu}\phi (\sum\limits_n \nabla^{2n}) \partial_{\nu}\phi
where I am only interested in the some first terms of the series as follow
(\nabla)^2 = g^{\alpha \beta} \nabla_\alpha \nabla_\beta
(\nabla)^4 = g^{\alpha \beta} g^{\mu \nu} \nabla_\alpha \nabla_\beta \nabla_\mu \nabla_\nu
(\nabla)^6= ...
Here \nabla is the usual covariant derivative for a covector.
\nabla_\alpha \partial_{\beta}\phi = \partial_{\alpha \beta}\phi - \Gamma^\lambda_{\alpha \beta} \partial_{\lambda}\phi
where \Gamma represents (symmetric) Christoffel symbols.
I have read the the excellent paper written by Leo Brewin but the way that he implements the covariant derivative is not clear for me and I can't generalize it to my case. Can anyone, please, give me a hint how to implant this expression in Cadabra? |
The setting is basically earth. Our planet is unfortunately on a collision course with a large asteroid. However, humans have discovered and decoded a message from an ancient, advanced alien race (our parents). These aliens have left behind technology and knowledge that could save the planet, but this technology is encased in the earth's core (so that we may only access it when we are "ready"). If the whole world worked together for the next 5-10 years, would it be possible to detonate our way down to the outer core using our nuclear arsenal? I know that lateral pressure is a problem and if you dig a cylindrical hole it will collapse, but could we dig a cone-shaped hole?
Instead of telling you it's impossible, I'll make a list of the problems you need to solve:
Pressure: Pressure at Earth's center is $3.65 \times 10^{11} \ \mbox{Pa}$. Whatever enclosure you build is subject to that. If you made a solid block of diamond (one of the least compressible materials, with a bulk modulus of $4.43 \times 10^{11} \ \mbox{Pa}$), you'll find that it shrinks to $82.3\ \%$ of its size. If you make it out of "steel" (say, $\sim1.50 \times 10^{11}\ \mbox{Pa}$), it becomes $33.9\ \%$ its size. That's bad news, especially since your vehicle needs to be hollow. Most humans are not happy being compacted to $34\ \%$ their volume. You can't solve this by using unobtainium because whatever atoms unobtainium is made of need to actually exist. Bond dissociation energies are the physical limit of strength. Density and Viscosity: Earth's inner core is $12.8\ \mbox{g/cm}^3$. Something like lead has $11.34\ \mbox{g/cm}^3$. Your ship is going to float, and will have to actively propel itself downward. When it reaches the inner core, it will need to move through something solid. To fix this, you need propulsion and drilling. But, both are subject to the same crushing pressures mentioned above. Temperature: Temperature at the Earth's core is at least $5\,000\ ^\circ\mbox{C}$. Because of thermodynamics, the Earth's core will try to make your vehicle the same temperature. Most humans not so much "happy" at $5\,000\ ^\circ\mbox{C}$ as they are "charred-lumps-of-their-constituent-elements".
It is worth noting that if humans have difficulty solving these challenges, your aliens will have difficulty solving them too. If your aliens
can solve them, this raises some serious unintended consequences.
Here's a possible solution. It is in the exterior realms of possibility and undoubtedly has
Problems, but perhaps another worldbuilding question could help fix them:
Make a large, very long steel rod, and hollow out many small interior regions. Suspend from the foremost region your vehicle in a vacuum. Similarly, put nuclear warheads in the rear regions. The outer hull can compress, leaving inner components unharmed. After sinking through the mantle normally, the rear regions of the device successively detonate, pushing the device deeper (this is an inverted Orion nuclear pulse drive, with all attendant problems). Count on sacrificial outer hull to absorb heat and pressure for long enough to get to the center.
Speculative/imaginary/magic tech that would make easier solutions (use with caution):
Force fields Neutronium Teleportation of matter Teleportation of energy (heat especially) Reactionless drive Arbitrary adjustment of magnetism of nearby materials Universe editation Asking your bloody aliens to come up with a nicer plan and stop being ruddy showoffs already.
No.
On this scale, the Earth is not solid and rigid. It's more like extremely hot jello, with a thin and weak crust, a layer of hot floppy jello, the "mantle", a liquid outer core (actually molten iron) that's about 1,400 miles thick, and an inner core of iron about 750 miles in radius.
Films and TV programmes that show journeys to the centre of the Earth are
exceptionally scientifically inaccurate, even by Hollywood standards.
A "cone-shaped hole" isn't possible, the Earth will just flow to fill it in once you get down a hundred miles or so. No, there isn't anything strong enough to brace the hole with. The only way to retrieve something from the Earth's core is to dismantle the planet, which will do more damage than any asteroid hit.
I would agree with the NO answer already given.
For comparison: The deepest humans have ever dug is only a little over 12 kilometers.
And these are drilling shafts much less than a meter in diameter.
Also consider that blowing a hole in the earth with all of our nuclear weapons to reach the core would most likely make the earth just as lifeless as the possible asteroid impact.
David J Stevenson has proposed a method to reach the Earth's core. It requires a nuclear device of only a few megatons to crack open the crust. The planetary mission vehicle descends using a large mass of about one million tons of molten iron to sink down to the core. This journey should take roughly one week.
The real technical problems your inner-earthonauts need to solve are how to survive the temperatures and pressures imposed on their vehicle during the descent for, at least, one week. We can safely assume there will be a human habitable base where the ancient alien technology is stored. So once they get there it is plain sailing.
But Stevenson has solved the technical problems and has the numbers to prove it too, of reaching the Earth's core. So this problem is already been solved. His probe isn't manned. Getting humans down there safely still remains to be solved. Possibly an extremely strong and highly refrigerated capsule needs to be built. Hopefully someone else on Worldbuilding SE has the answer.
I'll approach this problem from a different angle than I've seen in the current answers.
The radius of the Earth is about 3959 miles.
A cone shaped hole (assuming 1/10 ratio of base to height) will have a hole of almost 400 miles across at the surface.
Even if the composition of the earth was "only" dirt and rock, you would have to move 163,000,000 cubic miles of material during the excavation. Using nukes can break that material up for you, but you are still going to have to move that material out of the hole.
The material excavated by this project could be put into 271,666 piles, each larger than Mount Everest.
Changing the ratio of the cone to 1/100 would result in a much smaller number, but you would still be talking about moving many multiples of Mount Everest. Please note that pit mines generally use much more gradual slopes (actually wider than they are deep), which would result in a continent wide hole at the surface.
As another comparison, the amount of material moved is about half the volume of the entirety of the world's Oceans.
TL;DR: Yes, but not the way you thought: Aliens left their message in the form a of punch card that can be read with a neutrino beam.
Neutrinos are elementary particles that only interact weekly with other particles of matter and can therefore travel through the earth. Because it is so advanced, the alien civilisation could prepare a material stopping neutrinos and embedded a “punch card” made of this material at the centre of the earth. It is already possible to product “neutrino beams” and to detect them, so we can imagine reading the “punch card” by emitting an intense neutrino beam towards the center from one side of the earth and reading it from the other side. Because we generate a very intense beam, it is easily distinguished from “universe's noise”.
Considering that about 1,000 above-ground nuclear tests were conducted between 1946 and 1964 by the superpowers, huge numbers of underground tests followed, and nations not subscribing to treaty conducted many more, the fact that the earth is still here and not noticeably different should dispel any notion about the power of even fusion devices for excavation purposes of such magnitude. One large volcanic explosion subsumes the power of many fusion devices (see article on Krakatoa in Wikipedia, e.g.). Add to this that once (or if) you get through the crust you hit magma. Underneath this impenetrable barrier, what you might find is purely theoretical.
Perhaps more disheartening is the fact that nuclear weapons shot at an asteroid would have little to no chance of affecting it. The reaction is quite momentary, and in the vacuum of space it does nothing other than get very bright and very hot for an instant. There's no surrounding matter to create a blast effect. There's little chance of intercepting something at aggregate velocity of perhaps 60,000mph with any chance of timing the reaction properly. Perhaps if, as in movies, you could bore a (very) deep hole in the thing and detonate the device there, the thermal shock would either fracture it or at least eject enough matter to alter its course a bit. But the odds of landing on a 40,000mph object with almost no gravity and then conducting a drilling operation difficult even on earth are, to put it mildly, not encouraging.
So, supposing we have alien directions that fix the location of the artifact (it's not moving relative to a location on the crust), perhaps the solution would be to drill with the intent of causing an eruption, using the pressure of the core to push out through a weakened mantle, saving us from drilling all the distance to the artifact (perhaps), thereby ejecting the nearly indestructible artifact, and recovering the item from the ejecta.
Let's hope the artifact isn't buried under New York City....
Obviously, the environmental consequences would be catastrophic, if we can't engineer for them. I'm not proposing solutions to such engineering problems here - I don't have any. This is just a post to suggest a new line of reasoning if someone would like to follow up on it.
Maybe the aliens put it under a mid oceanic rift?
If you knew, where it was, maybe this would be better.
Disassemble all those nuclear weapons and build power plants. Make a giant, focus magnet so that the magnetic forces were aimed at the alien device.
Using all the electricity these power plants could muster, maybe they could pull it up. Throw is some graphene for good measure. At least with my idea you wouldn't have to deal with crazy amounts of heat and pressure. |
Notch Filters
Typical target reflectance is as high as possible in the high reflectance range and as small as possible in the rest of the specified spectral range.
Design approaches: rugate type coatings and conventional multilayer stacks. The advantage of the rugate and quasi-rugate coatings is that they suppress ripples in the transmission spectral ranges.
The disadvantages of conventional multilayer solutions is that they might contain layers of four, five, and more materials.
Notch filters with narrow and extremely narrow high-reflection zones can be obtained using efficient numerical design algorithms. Two-component solutions typically contain thin layers which may be a problem in the course of the practical realization of the obtained designs. However, it has been demonstrated by several companies that production of multilayer notch filters containing thin layers is realistic. See for example, these references [1], [2], [3], and [4].
There are two periodic multilayer structures that can be used as good starting designs for notch filters. Using these structures you may achieve the target width of the high reflection zone and target values of reflectance in this zone. If we denote layer materials as A and B and their refractive indices as \(n_1\) and
, then two structures \(n_s\) and S1 provide good starting design for notch filters:
S2
Analytical estimations of the reference wavelength reflectance \(R(\lambda_0)\) and the width of the high-reflection zone \(\Delta\lambda_0\) are obtained allow you to find the number of layer pair \(n\) and fraction
Reference wavelength reflectance are calculated as follows:
\( R(\lambda_0)=\frac bc, \)
where \( n_a, n_s\) are refractive indices of incident medium and substrate
\( b=a_{11}^2 (n_1-n_s)^2+a^2 (n_a n_s\alpha - 1)^2 a_{12}^2\)
\( c=a_{11}^2 (n_1+n_s)^2+a^2 (n_a n_s\alpha + 1)^2 a_{12}^2\)
\( a=\pi p (n_1=n_2)/2, \; \alpha=-1/(n_1 n_2)\)
\(a_{11}=0.5\left[ (1+\sqrt{-\alpha} |a|)^{-n}+(1-\sqrt{-\alpha} |a|)^{-n}\right] \)
\(a_{12}=0.5\left[ (1-\sqrt{-\alpha} |a|)^{-n}-(1+\sqrt{-\alpha} |a|)^{-n}\right] /(\sqrt{-\alpha} |a|)\)
\( n_1=2.35, n_2=1.45, n_s=1.52, n_a=1\), number of layer pairs \(n=64, \lambda_0=532\) nm, \(p=0.12\).
\( n_1=2.0, n_2=1.5, n_s=1.52, n_a=1\), number of layer pairs \(n=32, \lambda_0=532\) nm, \(p=0.2\).
The width of the high reflection zone can be estimated as:
Structure
\(\Delta\lambda_0=\displaystyle\frac {4\lambda_0}{\pi} \frac{\sqrt{C^2-1}\;\sin(\pi p/2)}{\left[C+1-(1-p)^2(C-1)\cos(\pi p)\right]}\)
ghgh
Structure
\(\Delta\lambda_0=\displaystyle\frac {2\lambda_0}{\pi} \frac{\sqrt{C^2-1}\;\sin(\pi p/2)}{\left[C+1-(1-2p)^2(C-1)\cos(\pi p)\right]}\)
where \( C=\displaystyle \frac 12\left(\frac{n_1}{n_2}+\frac{n_2}{n_1}\right)\)
If you have a specified R(lam0) value and width of high reflection zone then using
Equation 2 --> from n Equation 1
See the details on the next page. |
Problem 15
Let $p_1(x), p_2(x), p_3(x), p_4(x)$ be (real) polynomials of degree at most $3$. Which (if any) of the following two conditions is sufficient for the conclusion that these polynomials are linearly dependent?
(a) At $1$ each of the polynomials has the value $0$. Namely $p_i(1)=0$ for $i=1,2,3,4$. (b) At $0$ each of the polynomials has the value $1$. Namely $p_i(0)=1$ for $i=1,2,3,4$.
(
University of California, Berkeley) Problem 12
Let $A$ be an $n \times n$ real matrix. Prove the followings.
(a) The matrix $AA^{\trans}$ is a symmetric matrix. (b) The set of eigenvalues of $A$ and the set of eigenvalues of $A^{\trans}$ are equal. (c) The matrix $AA^{\trans}$ is non-negative definite.
(An $n\times n$ matrix $B$ is called
non-negative definite if for any $n$ dimensional vector $\mathbf{x}$, we have $\mathbf{x}^{\trans}B \mathbf{x} \geq 0$.)
Add to solve later
(d) All the eigenvalues of $AA^{\trans}$ is non-negative. Problem 11
An $n\times n$ matrix $A$ is called
nilpotent if $A^k=O$, where $O$ is the $n\times n$ zero matrix. Prove the followings. (a) The matrix $A$ is nilpotent if and only if all the eigenvalues of $A$ is zero.
Add to solve later
(b) The matrix $A$ is nilpotent if and only if $A^n=O$. Read solution Problem 9
Let $A$ be an $n\times n$ matrix and let $\lambda_1, \dots, \lambda_n$ be its eigenvalues.
Show that (1) $$\det(A)=\prod_{i=1}^n \lambda_i$$ (2) $$\tr(A)=\sum_{i=1}^n \lambda_i$$
Here $\det(A)$ is the determinant of the matrix $A$ and $\tr(A)$ is the trace of the matrix $A$.
Namely, prove that (1) the determinant of $A$ is the product of its eigenvalues, and (2) the trace of $A$ is the sum of the eigenvalues.
Read solution Problem 5
Let $T : \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation.
Let $\mathbf{0}_n$ and $\mathbf{0}_m$ be zero vectors of $\mathbb{R}^n$ and $\mathbb{R}^m$, respectively. Show that $T(\mathbf{0}_n)=\mathbf{0}_m$.
(
The Ohio State University Linear Algebra Exam)
Add to solve later
Problem 3
Let $H$ be a normal subgroup of a group $G$.
Then show that $N:=[H, G]$ is a subgroup of $H$ and $N \triangleleft G$.
Here $[H, G]$ is a subgroup of $G$ generated by commutators $[h,k]:=hkh^{-1}k^{-1}$.
In particular, the commutator subgroup $[G, G]$ is a normal subgroup of $G$Add to solve later |
I want to get a better grasp of what a rigorous formal proof is. So I was hoping to find proofs of interesting results using natural deduction or Hilbert system or similar. The "interesting result" could be anything from infinitude of primes to Lagrange's theorem. "Interesting result" is not $x \wedge y \implies y \wedge x$ or similar. In particular I'm interested in how mathematical objects (like primes or groups in the examples) fit into the proofs.
Nobody uses
Hilbert systems to actually write down fully formal proofs of anything interesting. The explosion in size that results from repeated use of the Deduction Theorem makes them completely unfeasible for practical work.
Using natural deduction instead avoids this problem, but still one needs to have some sort of facility for using abbreviations for defined notions; otherwise the properties you want to prove will
themselves become completely impenetrable spaghetti balls of primitive notions.
This means that systems of the kind usually presented by logic texts are not really well suited to doing actual work, but only for theoretical investigations into the
limits of what can be proved. For actual work you need a system with native support for defined notions, something like formalized metatheorems, and so forth.
What you'll want is probably to download an actual
proof assistant software with an existing library of basic concepts, and look at how the proofs in that library look -- for example Isabelle/HOL (which was all the rage a decade or two ago when I had a connection with the area, but may be obsoleted by something else these days, for all I know). Be aware that there's something of a learning curve if you start out with just knowledge of abstract textbook-style proof systems, but the basic concepts ought to be recognizable.
I will present proofs of two interesting formulas (both proofs are hopefuly interesting enough). I will not use any specific calculi, but the reasoning I will show is precisely how you would prove it in Hilbert style calculus or in natural deduction.
An interesting proof can be the one of Peirce law $(A\to B)\to A)\to A$:
Assume (1) $ (A\to B)\to A$ and further assume (2) $\neg A$. From (2) you obtain $A\to B$, thus from (1) you get $A$. It means that (1) and $\neg A$ implies A, contradiction. You can conclude $(1)$ implies $A$. Which is the Peirce law.
Another interesting proof, this time in first order classical logic, is for the drinking guru formula $\exists x (D(x)\to \forall y D(y))$. Meaning D is a unary predicate symbol. Lets say that D(x) means $x$ is drinking. Then the formula stands for: There is someone who whenever he is drinking, everyone else is drinking. Proof:
We will use the following instace of the law of excluded middle: $\forall x D(x)\vee\neg\forall x D(x)$. It is now enough to show that the desired formula is consequence of both of these - this is a proof by cases argument - which is formalised by the Hilbert style axiom $(A\to C)\to ((B\to C)\to (A\vee B\to C))$. Assume $\forall x D(x)$, then easily one get the following chain: $\forall y D(y)$, $D(x)\to \forall y D(y)$ (using weakening), $\exists x (D(x)\to \forall y D(y)) $ (exist. quantification). Now assume $\neg \forall x (D(x))$, from this we prove $\exists x (\neg D(x))$ (interderivability of quantifiers). Next from $\neg D(x)$ you can prove $D(x)\to \forall y D(y) $ (this is is true because from a false premis everything follows). Therefore from $\exists x (\neg D(x)) $ you conclude the desired $\exists x (D(x)\to \forall D(y))$. |
In statistical mechanics, for a system of $N$ particles $x_1, \ldots, x_N$ in three dimensions, the Gibbs free energy is defined in terms of the Hamiltonian $H$ as $$ G = -k_\mathrm B T \log \int_{\mathbb{R}^{3N}} \mathrm e^{-H(x_1, \dots, x_N)}\,\mathrm dx_1 \dots \mathrm dx_N.$$ People talk about the "free energy difference between conformations". For instance, for some region $A \subset \mathbb{R}^N$, the particles could be in a certain conformation, and for some other region $B \subset \mathbb{R}^N$, the particles could be in another conformation.
Question
In this paper, equation (3), what is the (implied) definition of $F_A$ or $F_B$ in terms of equation (2)?
Equivalent Question
Is the free energy difference between the two conformations defined as $$G_B - G_A = -k_\mathrm B T\log \frac{\int_B \mathrm e^{-H(x)}\,\mathrm dx}{\int_A \mathrm e^{-H(x)}\,\mathrm dx}?$$
Auxiliary details
The above definition made sense to me, until I was told that "it does not make sense to talk about the free energy of a conformation". An alternative I see is to think of two Hamiltonians $H_A$ and $H_B$, and instead define the free energy difference as
$$G_B - G_A = -k_\mathrm B T \log \frac{\int_{\mathbb{R}^{3N}} \mathrm e^{-H_B(x)}\,\mathrm dx}{\int_{\mathbb{R}^{3N}}\mathrm e^{-H_A(x)}\,\mathrm dx}. $$
Are either/both/neither of these definitions correct?
I apologize if the question is exceedingly obvious. I am unable to find a straightforward answer by Googling. Any references are welcome. |
Here is an example (the group $H_K^\mathbf{Q}$ defined below, which is actually a split extension abelian-by-countable).
Let $(Q,\le)$ be a total ordering. Let $K$ be a countable (possibly finite) field. Let $V^Q_K=K^{(Q)}$ be the free $K$-module with basis $(e_q)_{q\in Q}$. Let $G^Q_K$ be the group of $K$-module automorphisms $g$ of $V_K$ such that $g(e_q)-e_q\in\mathrm{Vect}_Q(\{e_r:r<q\})$ for every $q$ (it can be thought as a upper unipotent group of $Q\times Q$-matrices over $K$).
Define a
cut as a subset $I$ of $Q$ such that both $I$ and its complement $I^c$ are nonempty, and such that $i<j$ for all $i\in I$ and $j\in I^c$. For every cut $I$ with complement $J=I^c$ and $g\in G^Q_K$, let $p_I(g):V^I_K\to V^I_K$ be its restriction to the stable subspace $V^I_K$, and let $r_I(g):V^J_K\to V^J_K$ be the induced automorphism modulo the stable subspace $V^I_K$. (Visually, $I$ defines a $2\times 2$ block decomposition, $p_I(g)$ is the northwest block and $r_I(g)$ is the southeast block of $g$.)
Obviously, $p_I$ and $r_I$ are homomorphisms, and if $P_I$ and $R_I$ are their kernels, then $N_I=P_I\cap R_I$ is is a normal abelian subgroup in $G^Q_K$.
Now we specify to $Q=\mathbf{Q}$, the rationals with the usual ordering. Still, $N_I$ is not of countable index in $G^\mathbf{Q}_K$. So we define $H^Q_K$ as the normal subgroup of $G^Q_K$ of elements $g$ such that for every cut $I$, both matrices $p_I(g)$ and $r_I(g)$ are finitely supported (i.e. finitely many non-diagonal elements are nonzero). Then $N_I\cap H^\mathbf{Q}_K$ is normal abelian of countable index in $H_K^\mathbf{Q}$.
It remains to see that $H_K^\mathbf{Q}$ has no abelian characteristic subgroup of countable index. This is rather easy: the first observation (an easy play with commutators(*)) is that every nontrivial normal subgroup $N$ of $H_K^\mathbf{Q}$ contains an elementary matrix $e_{ij}(s)$ for some $i<j$ and $s\neq 0$. All such matrices are conjugate under the full automorphism group of $H_K^\mathbf{Q}$ (to vary $s$, use diagonal matrices, and to vary $(i,j)$, use that the group of order-preserving permutations of $\mathbf{Q}$ is 2-transitive). Hence if $N$ is characteristic, it contains all elementary matrices, and thus the (non-abelian) subgroup they generate.
Hence the only abelian characteristic subgroup of $H_K^\mathbf{Q}$ is the trivial subgroup $\{1\}$: we have to check that it does not have countable index, i.e. $H_K^\mathbf{Q}$ is uncountable: indeed for every subset $L$ of $\mathbf{N}\smallsetminus\{0\}$, consider the matrix $k_L$ with same entries as the identity except $(k_L)_{-i,i}=1$ for all $i\in L$. Then $L\mapsto k_L$ is injective, $k_L$ belongs to $H_K^\mathbf{Q}$ for every $L$, hence $H_K^\mathbf{Q}$ has continuum cardinal.
Edits:
(1) (*) Starting from a non-identity element $g$ in $G_K^\mathbf{Q}$, its commutator with a suitable elementary matrix yields a non-identity element $h$ such that $h-1$ has finite rank; in particular if $g\in H_K^\mathbf{Q}$ then so does $h$, and it follows that $h$ is finitely supported. Then $h$ lies in the nilpotent subgroup $G_K^I$ for some finite subset $I$ of $\mathbf{Q}$. So the normal subgroup of $H_K^\mathbf{Q}$ generated by $g$ contains a central element of $G_K^I$, showing that any nontrivial normal subgroup of $H_K^\mathbf{Q}$ contains a non-identity elementary matrix.
(2) Actually, a similar argument shows that the only normal abelian subgroup of $H_K^\mathbf{Z}$ stable under the shift automorphism $\alpha$ (conjugation by the automorphism $(e_n\mapsto e_{n+1})_{n\in\mathbf{Z}}$) is the trivial subgroup $\{1\}$. Note that $H_K^\mathbf{Z}$ is residually nilpotent, while $H_K^\mathbf{Q}$ is a perfect group.
(3) Incidentally, with the notation of (2), the semidirect product $\langle\alpha\rangle\ltimes H_K^\mathbf{Z}$ has a countable index abelian subgroup but no normal one. I don't know whether the Podoski-Szegedy example is similar (I don't have access to their article right now). |
Note: I first posted question on math.stackexchange and I got one reply, which was a bit helpful (I'm still trying to understand it fully), but did not explore the two solution cases that I mentioned. Besides I would appreciate additional insight on this question, which I hope its not too trivial for mathoverflow. Thanks.
I'd like to find the $n$-dimensional inverse Fourier transform of $\frac{1}{\| \mathbf{\omega} \|^{2\alpha}}$ i.e. $$ \int_{\mathbb{R}^n} \frac{1}{ \| \mathbf{\omega} \|^{2\alpha}} e^{2 \pi i \mathbf{\omega}\cdot \mathbf{x} } d \mathbf{\omega} $$ where $\mathbf{x} = ( x_0 , x_1 , \cdots , x_n )$ is a spatial parameter in $\mathbb{R}^n$, $\mathbf{\omega} = ( \omega_0 , \omega_1 , \cdots , \omega_n )$, and $$ \| \omega\| = \omega_0^2 + \omega_1^2 + \cdots + \omega_n^2 $$ All I've been able to come up with in the one-dimensional case is that the integral $$ \int_{-\infty}^{+\infty} \frac{1}{ \| \omega \|^{2\alpha}} e^{2 \pi i \omega x } d \mathbf{\omega} $$ diverges because the lower power terms $\omega^p$ terms, for which $p < 2\alpha$, in expansion of the exponential $$ e^{2 \pi i \omega x } = \sum_{p = 0}^{\infty} \frac{(2 \pi i \omega x)^p}{p!} $$ do not prevent $\frac{1}{\| \omega \|^{2\alpha}}$ from blowing up at the origin.
I know that one possible way of regularizing this integral is to include a test function and consider the limit of the resulting integral, but I don't quite know how to do so. I've tried reading Gelfand and Shilov's Gneralized Functions vol 1 and while I understand bits of it on the whole its a bit heavy for me.
Based on the papers that I've read I know that there are two cases (the latter of which appears to me more general) and two solutions in each.
Case 1: 2$\alpha$ is an odd/even integer Case 2: 2$\alpha$ is integer or otherwise
I'd appreciate help, if possible, coming up with both solutions. |
Tagged: normal subgroup If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 470
Let $G$ be a finite group of order $p^n$, where $p$ is a prime number and $n$ is a positive integer.
Suppose that $H$ is a subgroup of $G$ with index $[G:P]=p$. Then prove that $H$ is a normal subgroup of $G$.
(
Michigan State University, Abstract Algebra Qualifying Exam) Problem 332
Let $G=\GL(n, \R)$ be the
general linear group of degree $n$, that is, the group of all $n\times n$ invertible matrices. Consider the subset of $G$ defined by \[\SL(n, \R)=\{X\in \GL(n,\R) \mid \det(X)=1\}.\] Prove that $\SL(n, \R)$ is a subgroup of $G$. Furthermore, prove that $\SL(n,\R)$ is a normal subgroup of $G$. The subgroup $\SL(n,\R)$ is called special linear group |
Article - Physically Based Rendering
Introduction
The pursuit of realism is pushing rendering technology towards a detailed simulation of how light works and interacts with objects.
Physically based rendering is a catch all term for any technique that tries to achieve photorealism via physical simulation of light.
Currently the best model to simulate light is captured by an equation known as
the rendering equation. The rendering equation tries to describe how a "unit" of light is obtained given all the incoming light that interacts with a specific point of a given scene. We will see the details and introduce the correct terminology in a moment. It's important to notice that we won't try to solve the full rendering equation, instead we will use the following simplified version:
\[L_o(p,\omega_o) = \int\limits_{\Omega} f_r(p,\omega_i,\omega_o) L_i(p,\omega_i) \mathbf n \cdot \omega_i d\omega_i\]
To understand this equation we first need to understand how the light works, and then, we will need to agree on some common terms. To give you a rough idea of what the formula means, in simple terms, we could say that the formula describes
the colour of a pixel given all the incoming 'coloured light' and a function that tells us how to mix them.
Physics terms
If we want to properly understand the rendering equation we need to capture the meaning of some physical quantities; the most important of these quantities is called
radiance (represented with \(L\) in the formula).
Radiance is a tricky thing to understand, as it is a combination of other physics quantities, therefore, before formally define it, we will introduce a few other quantities.
Radiant Flux: The radiant flux is the measure of the total amount of energy, emitted by a light source, expressed in Watts. We will represent the flux with the Greek letter \(\Phi\).
Any light source emits energy, and the amount of emitted energy is function of the wavelength.
Figure 1: Daylight spectral distribution
In figure 1 we can see the spectral distribution for day light; the radiant flux is the area of the function (to be exact, the area is the
luminous flux, as the graph is limiting the wavelength to the human visible spectrum). For our purposes we will simplify the radiant flux with an RGB colour, even if this means losing a lot of information.
Solid angle: It's a way to measure how large an object appears to an observer looking from a point. To do this we project the silhouette of the object onto the surface of a unit sphere centred in the point we are observing from. The area of the shape we have obtained is the solid angle. In Figure 2 you can see the solid angle \(\omega\) as a projection of the light blue polygon on the unit sphere.
Figure 2: Solid angle
Radiant Intensity: is the amount of flux per solid angle. If you have a light source that emits in all directions, how much of that light (flux) is actually going towards a specific direction? Intensity is the way to answer to that, it's the amount of flux that is going in one direction passing through a defined solid angle. The formula that describes it is \( I =\frac{d\Phi}{ d\omega} \), where \(\Phi\) is the radiant flux and \(\omega\) is the solid angle.
Figure 3: Light intensity
Radiance: finally, we get to radiance. Radiance formula is:
\[L=\frac{d\phi^2}{ dA d\omega cos\theta} \]
where \(\Phi\) is the radiant flux, \(A\) is the area affected by the light, \(\omega\) is the solid angle where the light is passing through and \(cos\theta\) is a scaling factor that "fades" the light with the angle.
Figure 4: Radiance components
We like this formula because it contains all the physical components we are interested in, and we can use it to describe a single
"ray" of light. In fact we can use radiance to describe the amount of flux, passing through an infinitely small solid angle, hitting an infinitely small area, and that describes the behaviour of a light ray. So when we talk about radiance we talk about some amount of light going in some direction to some area.
When we shade a point we are interested in all the incoming light into that point, that is the sum of all the radiance that hit a hemisphere centred on the point itself; the name for this entity is irradiance. Irradiance and radiance are our main physical quantities, and we will work on both of them to achieve our physically based rendering.
The rendering equation
We can now go back on the rendering equation and try to fully understand it.
\[L_o(p,\omega_o) = \int\limits_{\Omega} f_r(p,\omega_i,\omega_o) L_i(p,\omega_i) \mathbf n \cdot \omega_i d\omega_i\]
We know understand that \(L\) is radiance, and it's function of some point in the world and some direction plus the solid angle (we will always use infinitely small solid angles from now on, so think of it simply as a direction vector). The equation describes the outgoing radiance from a point \(L_o(p,\omega_o)\), which is all we need to colour a pixel on screen.
To calculate it we need the normal of the surface where our pixel lies on (\(\mathbf n\)), and the irradiance of the scene, which is given by \(L_i(p,\omega_i) \forall \omega_i\). To obtain the irradiance we sum them all the incoming radiance, hence the integral sign in front of the equation. Note that the domain of the integral \(\Omega\) is a semi-sphere centered at the point we are calculating and oriented so that the top of the hemisphere itself can be found by moving away from the point along the normal direction.
The dot product \(\mathbf n \cdot \omega_i\) is there to take into account the angle of incidence angle of the light ray. If the ray is perpendicular to the surface it will be more localized on the lit area, while if the angle is shallow it will be spread across a bigger area, eventually spreading across too much to actually being visible.
Now we can see that the equation is simply representing the outgoing radiance given the incoming radiance weighted by the cosine of the angle between every incoming ray and the normal to the surface. The only bit we still need to introduce is \(f_r(p,\omega_i,\omega_o)\), that is the
BRDF. This function takes as input position, incoming and outgoing ray, and outputs a weight of how much the incoming ray is contributing to the final outgoing radiance. For a perfectly specular reflection, like a mirror, the BRDF function is 0 for every incoming ray apart for the one that has the same angle of the outgoing ray, in which case the function returns 1 (the angle is measured between the rays and the surface normal). It's important to notice that a physically based BRDF has to respect the law of conservation of energy, that is \( \forall \omega_i \int\limits_{\Omega} f_r(p,\omega_i,\omega_o) (\mathbf n \cdot \omega_i) d\omega_o \leq 1\), which means that she sum of reflected light must not exceed the amount of incoming light.
Translate to code
So, now that we have all this useful knowledge, how do we apply it to actually write something that renders to the screen? We have two main problems here.
First of all, how can we represent all these radiance functions in the scene?
And secondly, how do we solve the integral fast enough to be able to use this in a real-time engine?
The answer to the first question is simple,
environment maps. For our purposes we'll use environment maps (cubemaps, although spherical maps would be more suited) to encode the incoming radiance from a specific direction towards a given point.
If we imagine that every pixel of the cubemap is a small emitter whose flux is the RGB colour, we can approximate
\(L(p,\omega)\), with p being the exact center of the cubemap, to a texture read from the cubemap itself, so \(L(p,\omega) \approx texCUBE(cubemap, \omega) \). Obviously it would be too much memory consuming to have a cubemap for every point in the scene(!), therefore we trade off some quality by creating a certain number of cubemaps in the scene and every point picks the closest one. To reduce the error we can correct the sampling vector with the world position of the cubemap to be more accurate. This gives us a way to evaluate radiance, which is:
\[L(p,\omega) = texCUBE(cubemap, \omega_p) \]
where \(\omega_p\) is the sampling vector corrected by the point position and cubemap position in the world.
The answer for our second problem, how to solve the integral, is a bit more tricky, because in some cases, we won't be able to solve it quickly enough. But if the BRDF happens to depend only on the incoming radiance, or even better, on nothing (if it's constant), then we can do some nice optimization. So let's see how this happens if we plug in
Lambert's BRDF, which is a constant factor (all the incoming radiance contributes to the outgoing ray after being scaled by a constant).
Lambert
Lambert's BRDF sets \(f_r(p,\omega_i,\omega_o) = \frac{c}{ \pi } \) where \(c\) is the surface colour. If we plug this into the rendering equation we get:
\[L_o(p,\omega_o) = \int\limits_{\Omega} \frac{c}{ \pi } L_i(p,\omega_i) (\mathbf n \cdot \omega_i) d\omega_i = \]
\[L_o(p,\omega_o) = \frac{c}{ \pi } \int\limits_{\Omega} L_i(p,\omega_i) (\mathbf n \cdot \omega_i) d\omega_i \]
Now, the integral depends on \(\omega_i\) and nothing else, which means we can precalculate it (solving it with a Monte Carlo integration for example) and store the result into another cubemap. The value will be stored in \(\omega_o\) direction, which means that knowing what output direction we have we can sample the cubemap and obtain the reflected light in that very direction. This reduces the whole rendering equation to a single sample from a pre-calculated cubemap, specifically:
\[L(p,\omega_o) = texCUBE(lambertCubemap, \omega_{op}) \]
where \(\omega_{op}\) is the outgoing radiance corrected by the point position and cubemap position in the world.
So, now we have all the elements, and we can finally write a shader. I'll show that in a moment, but for now, let's see the results.
Quite good for a single texture read shader uh? Please note how the whole lighting changes with the change of the envinroment (the cubemap rendered is not the convolved one, which looks way more blurry as shown below).
Figure 5: Left the radiance map, right the irradiance map (Lambert rendering equation)
Now let's present the shader's code. Please note that for simplicity I'm not using the Monte Carlo integration but I've simply discretized the integral. Given infinite samples it wouldn't make any difference, but in a real case it will introduce more banding than Monte Carlo. In my tests it was good enough given that I've dropped the resolution of the cubemap to a 32x32 per face, but it's worth bearing this in mind if you want to experiment with it.
The first shader we need is the one that generates the blurry envmap (often referred to as the
convolved envmap, since it is the result of the convolution of the radiance envmap and the kernel function \( (\mathbf n \cdot \omega_i) \) ).
Since in the shader we will integrate in spherical coordinates we will change the formula to reflect that.
\[L_o(p,\theta_o,\phi_o) = \frac{c}{ \pi } \int\limits_{\Phi} \int\limits_{\Theta} L_i(p,\theta_i,\phi_i) cos(\theta_i) sin(\theta_i) d\theta_i d\phi_i \]
You may have noticed that there is an extra \(sin(\theta_i)\) in the formula; that is due to the fact that the integration is made of small uniform steps. When we are using the solid angle this is fine as the solid angles are evenly distributed on the integration area, but when we change to spherical coordinates we will get more samples where \(\theta\) is zero and less where it goes to \(\frac{\pi}{2}\). If you create a sphere in your favorite modeling tool and check it's wireframe you'll see what I mean. The \(sin(\theta_i)\) function is there to compensate the distribution as \(d\omega_i = \sin(\theta) d\theta d\phi\).
The double integral is solved by applying a Monte Carlo estimator on each one; this leads to the following discrete equation that we can finally transform into shader code:
\[L_o(p,\theta_o,\phi_o) = \frac{c}{ \pi }\frac{2 \pi}{ N_1 } \frac{\pi}{ 2 N_2 } \sum_{}^{N_1} \sum_{}^{N_2} L_i(p,\theta_i,\phi_i) cos(\theta_i) sin(\theta_i) \]
\[L_o(p,\theta_o,\phi_o) = \frac{\pi c}{ N_1 N_2 } \sum_{}^{N_1} \sum_{}^{N_2} L_i(p,\theta_i,\phi_i) cos(\theta_i) sin(\theta_i) \]
...
float4 PixelShaderFunction(VertexShaderOutput input) : COLOR
{
float3 normal = normalize( float3(input.InterpolatedPosition.xy, 1) );
if(cubeFace==2)
normal = normalize( float3(input.InterpolatedPosition.x, 1, -input.InterpolatedPosition.y) );
else if(cubeFace==3)
normal = normalize( float3(input.InterpolatedPosition.x, -1, input.InterpolatedPosition.y) );
else if(cubeFace==0)
normal = normalize( float3( 1, input.InterpolatedPosition.y,-input.InterpolatedPosition.x) );
else if(cubeFace==1)
normal = normalize( float3( -1, input.InterpolatedPosition.y, input.InterpolatedPosition.x) );
else if(cubeFace==5)
normal = normalize( float3(-input.InterpolatedPosition.x, input.InterpolatedPosition.y, -1) );
float3 up = float3(0,1,0);
float3 right = normalize(cross(up,normal));
up = cross(normal,right);
float3 sampledColour = float3(0,0,0);
float index = 0;
for(float phi = 0; phi < 6.283; phi += 0.025)
{
for(float theta = 0; theta < 1.57; theta += 0.1)
{
float3 temp = cos(phi) * right + sin(phi) * up;
float3 sampleVector = cos(theta) * normal + sin(theta) * temp;
sampledColour += texCUBE( diffuseCubemap_Sampler, sampleVector ).rgb *
cos(theta) * sin(theta);
index ++;
}
}
return float4( PI * sampledColour / index), 1 );
}
...
I've omitted the vertex shader and the variables definition and the source shader I've used is in HLSL. Running this for every face of the convolved cubemap using the normal cubemap as input gives us the irradiance map. We can now use the irradiance map as an input for the next shader, the model shader.
...
float4 PixelShaderFunction(VertexShaderOutput input) : COLOR
{
float3 irradiance= texCUBE(irradianceCubemap_Sampler, input.SampleDir).rgb;
float3 diffuse = materialColour * irradiance;
return float4( diffuse , 1);
}
...
Very short and super fast to evaluate.
This concludes the first part of the article on physically based rendering. I'm planning to write a second part on how to implement a more interesting BRDF like Cook-Torrance's BRDF. |
Given a Boolean function $f$ over the set of variables $X =\{ x_1,...,x_n \}$, the influence of $x_i$ is defined as the probability that changing only $x_i$ on random input changes $f$.
Given a function $f$ presented in $\text{CNF}$, and a coordinate, are there any known methods for determine if the coordinate has positive influence?
Given a Boolean function $f$ over the set of variables $X =\{ x_1,...,x_n \}$, the influence of $x_i$ is defined as the probability that changing only $x_i$ on random input changes $f$.
Given a CNF $\phi$ and a variable $x_i$, deciding whether $\phi$ depends on $x_i$ is NP-complete. It is clearly in NP: all we need to verify that $\phi$ depends on $x_i$ is two inputs different on $x_i$ on which $\phi$ evaluates to different values.
In the other direction, we reduce from SAT. Given a CNF $\psi = C_1 \land \cdots \land C_m$, let $x$ be a new variable and consider the CNF $\phi = \psi \land x$. If $\psi$ is satisfiable then $\phi$ depends on $x$, since substituting such a satisfying assignment reduces $\phi$ to $x$. If $\psi$ is not satisfiable then $\phi$ is always false, and in particular doesn't depend on $x$. |
Definition(1). Let $\mathscr{L}$ be a first-order language. An $\mathscr{L}$-theory $T$ is said to have quantifier-elimination whenever if for all $\mathscr{L}$-formula $\phi(\bar{x})$ there exists a quantifier-free $\mathscr{L}$-formula $\psi_\phi(\bar{x})$ such that $$T\models \forall \bar{x}\left(\phi(\bar{x})\leftrightarrow\psi_\phi(\bar{x}) \right) $$ Definition(2). Let $\mathcal{C}$ be a class of $\mathscr{L}$-structures, for a first-order language $\mathscr{L}$. We say $\mathcal{C}$ eliminates quantifiers if for every $\mathscr{L}$-formula $\phi (\bar{x}) $ there is a quantifier-free formula $\psi_\phi(\bar{x})$ such that $\phi(\bar{x})$ is equivalent to $\psi_\phi(\bar{x})$ in every structure in $\mathcal{C}$. Definition(3). Let $\mathscr{L}$ be a first-order language, and let $\mathcal{C}$ be a class of finite $\mathscr{L}$-structures. $\mathcal{C}$ is called a Fraisse class if it satisfies the heriditary property (HP), the joint embedding property (JEP), and the amalgamation property (AP).
By Fraisse Theorem, any Fraisse class has a generec model $\mathcal{M}_\mathcal{C}$ such that it is homogeneous and its theory is $\omega$-categorical (i.e. up to isomorphism there is exactly one model of size $\aleph_0$.)
Question(1). Let $\mathcal{C}$ be a Fraisse calss with generic model $\mathcal{M}$ and generic theory $T$ (i.e. $T=Th(\mathcal{M})$). Is it always true that $T$ eliminates quantifiers? If no, what is the counterexample? and under what conditions we can conclude $T$ has quantifier elimination? Question(2). In general, how can we prove a class of $\mathscr{L}$-structures has quantifier elimination? There are some techniques in Marker's book, but they usually are used to show a theory has quantifier-elimination not a class. In Hodges's book there is a theorm which says who a class eliminates quantifiers but the theorem is clearly equivalent to the definition and it does not give us any tool. Question(3). Is it true that there is not general ways to show a theory has quantifier elimination? I mean for each theory we need to find a specific way related to our theory? (If this is correct, that's so bad!)
Any refrences whould be helpful. |
I am a new LaTeX user. I already searched through the topic and I have not found an answer yet. I am using the
exercise package to build a large Multiple Choice math question set complete with an answer key and full explanations. This is the format of a typical Multiple Choice question:
\item \(\frac{19^2+19}{19}\)\begin{enumerate}[(A)] % (A), (B), (C), (D),(E)\item 396\item 361\item 38\item 21\item 20 \end{enumerate}\textbf{Explanation}\parWe may simplify the fraction by separating it into two fractions as follows:\par\(\frac{19^{2}+19}{19}\)=\(\frac{19^2}{19}+\frac{19}{19}=19+1=20\)\par\textbf{Answer:E}\par
The number of questions has grown beyond 50 and I have started getting the following error:
! LaTeX Error: Counter too large.See the LaTeX manual or LaTeX Companion for explanation.Type H <return> for immediate help. ... l.418 \item I f $x=\sqrt{5}{-37}$, then which of the following must be true??
I may be doing something wrong with the
alphalph package, so kindly help me resolve this problem. If I were to use the following command, what do I write instead of
\thesection and
section.
\renewcommand{\thesection}{\alphalph{\value{section}}}
Sorry if this question has already been answered before. |
Let's assume that an excess of 2 cm of water evaporates across the entire surface of the Earth, and settles at an altitude of 3000 m. It is possible to compute the increased moment of inertia of the Earth, and thus the theoretical slowdown.
Moment of inertia of the earth can be found from data on the NASA website http://nssdc.gsfc.nasa.gov/planetary/factsheet/earthfact.html
moment of inertia I/MR^2 = 0.3308
mass = 5.97E24 kg
radius = 6.371E6 m
from which we get $I = 8.03\cdot 10^{37} kg m^2$ (confirmed by Wolfram Alpha).
A shell of water radius $R$ and thickness $dr$, density $\rho$ has a moment of inertia $I=\frac23 m r^2 = \frac23 (4\pi R^2 \rho dr) R^2$
For the given dimensions, this becomes 2.8E29 kg m$^2$.
When it rises to an altitude of 3000 m, the moment of inertia increases to
$$I = \frac23 (4\pi R^2\rho dr)(R+h)^2$$
an increase of about $\frac{2h}{R}$ or roughly 0.1%
But that's 0.1% increase for the moment of inertia represented by the water, which itself is only 3.5E-9 of the moment of inertia of the earth.
This will therefore change the rate of rotation by 3.5E-12 , meaning that a day would get longer by 0.28 µs
By comparison, the big earthquake in Japan in 2011 increased the rate of rotation of the earth by 1.8 µs (see http://www.nasa.gov/topics/earth/features/japanquake/earth20110314.html), and the tidal drag of the Moon on the Earth is causing the day to lengthen by 2.3 ms per
century (much less on a per-day basis, but it keeps going day in and day out...)
Finally - the mass of water I estimated (2 cm covering the entire surface of the Earth including both sea and land) would be approximately 10$^{16}$ kg ; now the density of a cumulonimbus ('fluffy') cloud is about 0.3 gram / m$^3$ (source); if you need clouds that have a mass of 20 kg per square meter of Earth's surface, this implies clouds that would be 60 km high. Or a lot more dense. Either way, my estimate of the amount of water that might evaporate is pretty optimistic, meaning that the calculated (tiny) shift in the length of the day is an upper estimate.
On the other hand - if it's that muggy, the day might
seem a lot longer.... |
Hybrid Interaction Point Process Model
Creates an instance of a hybrid point process model which can then be fitted to point pattern data.
Usage
Hybrid(...)
Arguments …
Two or more interactions (objects of class
"interact") or objects which can be converted to interactions. See Details.
Details
A
hybrid (Baddeley, Turner, Mateu and Bevan, 2013) is a point process model created by combining two or more point process models, or an interpoint interaction created by combining two or more interpoint interactions.
The
hybrid of two point processes, with probability densities \(f(x)\) and \(g(x)\) respectively, is the point process with probability density $$h(x) = c \, f(x) \, g(x)$$ where \(c\) is a normalising constant.
Equivalently, the hybrid of two point processes with conditional intensities \(\lambda(u,x)\) and \(\kappa(u,x)\) is the point process with conditional intensity $$ \phi(u,x) = \lambda(u,x) \, \kappa(u,x). $$ The hybrid of \(m > 3\) point processes is defined in a similar way.
The function
ppm, which fits point process models to point pattern data, requires an argument of class
"interact" describing the interpoint interaction structure of the model to be fitted. The appropriate description of a hybrid interaction is yielded by the function
Hybrid().
The arguments
… will be interpreted as interpoint interactions (objects of class
"interact") and the result will be the hybrid of these interactions. Each argument must either be an interpoint interaction (object of class
"interact"), or a point process model (object of class
"ppm") from which the interpoint interaction will be extracted.
The arguments
… may also be given in the form
name=value. This is purely cosmetic: it can be used to attach simple mnemonic names to the component interactions, and makes the printed output from
print.ppm neater.
Value
An object of class
"interact" describing an interpoint interaction structure.
References
Baddeley, A., Turner, R., Mateu, J. and Bevan, A. (2013) Hybrids of Gibbs point process models and their implementation.
Journal of Statistical Software 55:11, 1--43. http://www.jstatsoft.org/v55/i11/ See Also Aliases Hybrid Examples
# NOT RUN { Hybrid(Strauss(0.1), Geyer(0.2, 3)) Hybrid(Ha=Hardcore(0.05), St=Strauss(0.1), Ge=Geyer(0.2, 3)) fit <- ppm(redwood, ~1, Hybrid(A=Strauss(0.02), B=Geyer(0.1, 2))) fit ctr <- rmhcontrol(nrep=5e4, expand=1) plot(simulate(fit, control=ctr)) # hybrid components can be models (including hybrid models) Hybrid(fit, S=Softcore(0.5)) # plot.fii only works if every component is a pairwise interaction data(swedishpines) fit2 <- ppm(swedishpines, ~1, Hybrid(DG=DiggleGratton(2,10), S=Strauss(5))) plot(fitin(fit2)) plot(fitin(fit2), separate=TRUE, mar.panel=rep(4,4))# }
Documentation reproduced from package spatstat, version 1.60-1, License: GPL (>= 2) |
Legacy code is a concern for any company with a reasonably size team pretty quickly. In this article Wayne Lobb from Foliage provides us with industry metrics with respect to code growth (Unfortunately need to register to download). The quick summary is that on average a developer can either create…
December 2011
The first assumption is that developer productivity declines with a constant rate $r$: (1) $\Large \frac{dP(t)}{dt}$ = $\large -r \cdot P(t)$ Divide by $P(t)$: (2) $\Large \frac{1}{P(t)}$ $\cdot$ $\Large \frac{dP(t)}{dt}$ = $-r$ (3) $\Large \frac{dP(t)}{P(t)}$ = $-r dt$ (4) $\Large \int \frac{dP(t)}{P(t)}$ = $-r \int t dt$ (5) $\large ln(P(t))=…
At $t = 0$ $P_0$ is defined to be developer productivity without legacy obligations (1) $\large P(0) = P_0$ At time $t$ the developer productivity is given by the following equation. $r$ denotes the rate in which productivity declines over time: (2) $\large P(t) = P_0 \cdot e^{-r t} dt$… |
amp-mathml
Displays a MathML formula.
Required Script
<script async custom-element="amp-mathml" src="https://cdn.ampproject.org/v0/amp-mathml-0.1.js"></script>
Supported Layouts container Examples amp-mathml.amp.html
This extension creates an iframe and renders a MathML formula.
<amp-mathml layout="container" data-formula="\[x = {-b \pm \sqrt{b^2-4ac} \over 2a}.\]"> </amp-mathml> <amp-mathml layout="container" data-formula="\[f(a) = \frac{1}{2\pi i} \oint\frac{f(z)}{z-a}dz\]"> </amp-mathml> <amp-mathml layout="container" data-formula="$$ \cos(θ+φ)=\cos(θ)\cos(φ)−\sin(θ)\sin(φ) $$"> </amp-mathml>
This is an example of a formula of
<amp-mathml layout="container" inline data-formula="`x`"></amp-mathml>,
<amp-mathml layout="container" inline data-formula="\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)"></amp-mathml> placed inline in the middle of a block of text.
<amp-mathml layout="container" inline data-formula="\( \cos(θ+φ) \)"></amp-mathml> This shows how the formula will fit inside a block of text and can be styled with CSS.
Specifies the formula to render.
If specified, the component renders inline (
inline-block in CSS).
See amp-mathml rules in the AMP validator specification.
You've read this document a dozen times but it doesn't really cover all of your questions? Maybe other people felt the same: reach out to them on Stack Overflow.Go to Stack Overflow Found a bug or missing a feature?
The AMP project strongly encourages your participation and contributions! We hope you'll become an ongoing participant in our open source community but we also welcome one-off contributions for the issues you're particularly passionate about.Go to GitHub |
States of Matter: Gases and Liquids Melting and Boiling Points, Molecular Speeds Most Probable Velocity C p (or) Vp:
\tt C_p=\sqrt{\frac{2RT}{M}}=\sqrt{\frac{2PV}{M}}=\sqrt{\frac{2P}{d}}
RMS Velocity (C or u): \tt u=\sqrt{\frac{u_1^2+u_2^2+u_3^2+...+u_n^2}{n}} \tt PV=\frac{1}{3}\ MC^2 \tt C=\sqrt{\frac{3PV}{M}}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3P}{d}}
\tt C=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.314\times 10^{-7}\times T}{M}}
\tt C=1.58\sqrt{\frac{T}{M}}\times 10^4 cm/sec Relation between Cp : \tt \overline{C} : C \tt =\sqrt{\frac{2RT}{M}}:\sqrt{\frac{8RT}{M}}:\sqrt{{\frac{3RT}{M}}} \tt =\sqrt{2}:\sqrt{\frac{8}{\pi}}:\sqrt{3} = 1.414 : 1.596 : 1.732
Cp = 0.816 × C
\tt \overline{C} = 0.9213 × C \tt \frac{C_1}{C_2}=\sqrt{\frac{T_1}{T_2}\cdot\frac{M_2}{M_1}} Part1: View the Topic in this Video from 0:17 to 5:35 Part2: View the Topic in this Video from 0:02 to 7:21
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1.
Root mean square velocity (u rms) : It is the square root of the mean of the squares of the velocity of a large number of moleucles of the same gas.
\tt u_{rms} = \sqrt{\frac{u_1^2 + u_2^2 + u_3^2 + .... u_n^2}{n}}
\tt u_{rms} = \sqrt{\frac{3PV}{\left(mN_{0}\right)}} = \sqrt{\frac{3RT}{\left(mN_{0}\right)}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3P}{d}}
Where, mN
0= M
Where, k = Boltzmann constant = \tt \frac{R}{N_{0}}
2. When temperature alone is given then, \tt u_{rms} = 1.58 \ \times \sqrt{\frac{T}{M}} \times 10^{4} cm / sec
3. If P and T both are given, use equation in terms of temperature, i.e., \tt u_{rms} = \sqrt{\frac{3RT}{M}} and \ not \sqrt{\frac{3pV}{M}}
4.
Average velocity ( vav) : It is the average of the various velocities possessed by the molecules.v_{av} = \frac{v_{1} + v_{2} + v_{3} + ...... v_{n}}{n}; v_{av} =\tt \sqrt{\frac{8RT}{\pi M}} = \sqrt{\frac{8kT}{\pi m}}
5.
Most probable velocity (∝ mp) : It is the velocity possessed by maximum number of molecules of a gas at a given temperature.
\tt \alpha_{mp} = \sqrt{\frac{2RT}{M}} = \sqrt{\frac{2PV}{M}} = \sqrt{\frac{2P}{d}} |
No, I don't think it's
necessary to use MathJax. In fact, one of our top answerers here doesn't use MathJax at all. This is mostly because answers, and questions, are based in physics and not in maths (though often the maths & physics are intertwined so it's kinda difficult to distinguish them). So if you can explain your question, no matter the subject, without maths, that's perfectly fine.
That said, it's
useful to write your equations in MathJax because it helps readability for more complex statements. Obviously writing $\psi$ versus ψ will result in anyone picking up, but if you've got an equation or an integral, it's going to be harder to follow what's being asked without it written up in MathJax. I've actually seen a case where someone wrote something along the lines of "integral of f(x) from 0 to infinity" in the question because they didn't know how to write it as,$$\int_0^\infty f(x)\,\mathrm dx.$$The point taken is the same in both, but it helps readers read more quickly. 1 Hence, you ought to write your mathy things with MathJax.
Note that, in many cases, the users here are friendly enough and help fix MathJax issues, so even if you don't get it quite right, someone will help you. And if don't know the notation fully, there's the ever useful Detexify which analyzes your input to find the matching LaTeX/MathJax command for it (I've not tried it on iOS, so I can't speak to how good it is there, but on Android & desktops & tablets, it's just fine).
1. Not the case I was thinking of when writing this, but it just happened today again: https://physics.stackexchange.com/posts/464590/revisions |
Electroweak theory has two coupling constants before and after Spontaneous Symmetry Breaking (SSB) each one for $SU(2)_L$ and $U(1)_Y$, though they are connected by Weinberg angle after SSB. My question is, how is unification complete with two independent couplings before SSB. The motive for unification is a single unified force with certain range and (coupling)strength.
The two parts of the electroweak gauge group do not separately describe the weak and the electromagnetic force.
The unified group here is not $$ SU(2)_\mathrm{weak} \times U(1)_\mathrm{em}$$ but rather $$ SU(2)_L \times U(1)_Y $$ and the electric charge arises as a linear combination of hypercharge and weak isospin.
Therefore, although the group is not simple, the weak and electromagnetic forces have been unified, giving rise to two
other forces.
Edit: To adress the matter of coupling constants:
Indeed, before SSB there are two independent coupling constants $g'$ (for the $U(1)$) and $g$ (for $SU(2)$). One way to relate them to parameters after SSB is to think of the couplings constant $g$ vanishing, but a new parameter arising: the
Weinberg angle. The Weinberg angle $\theta_W$ determines, what linear combination of of the neutral vector bosons $W_3$ from $SU(2)$ and $B$ from $U(1)$ turn into the massive $Z$ boson and what combination turns into the massless $\gamma$.
The Weinberg angle is determined through the gauge couplings as$$ \cos\theta_w = \frac{g}{\sqrt{g^2 + g'^2}}.$$In other words, in the breaking $SU(2)_L \times U(1)_Y$
both groups get broken, but there exists a linear combination of generators that remains unbroken. The $U(1)$ spanned by this generator does not relate 1:1 to either gauge group before SSB, though!
The coupling constant for the photon now relates to the couplings before SSB through the Weinberg angle $$ e = g \sin\theta_w = g' \cos\theta_w.$$
Unification in the EW theory means that weak interactions that are short range are not very difference at short distances than electromagnetism, both are described by spin-1 gauge boson exchange with couplings that are similar in size. They differ mainly at large distances where the mass of the mediators becomes important, killing the interactions at distances $r>1/m_{W}$. In this sense, EW unification is just meant for the conceptual understanding of these two seemingly different forces that are instead described by very similar equations and coupling constants. |
Disclaimer: This is not really an answer, but rather a "long comment". However, I hope it can provide some useful insights. Anyway, I think that this question can be basically considered a duplicate of Is there a way to obtain the classical partition function from the quantum partition function in the limit $h→0$?.
The first version of the partition function you report,
$$\tag{1}\label{1} Z_{qm}=\sum_{\text{states}} e^{-\beta E_i}$$
is intrinsically quantum mechanical, and has little meaning in classical mechanics, unless you are considering some model system with a discrete number of states (like the Ising model).
Note that the index $i$ of \ref{1} must be interpreted as running over the
states of the system, not over the energy levels. In quantum mechanics, the number of states of a system is in general discrete, and therefore an expression like \ref{1} is meaningful. You can transform it in a sum over the energy levels if you want, by writing
$$\tag{2}\label{2} Z_{qm} = \sum_{\text{energy levels}} g_i e^{-\beta E_i}$$
where $g_i$ is the degeneracy of the energy level $E_i$, i.e. the number of different quantum states corresponding to this energy level.
In classical mechanics, you have a continuum of states, which we call the phase space. Every point $P=(\mathbf x_1, \dots, \mathbf x_N,\mathbf q_1, \dots, \mathbf q_N)$ in phase space corresponds to a different physical state. Therefore an expression like \ref{1} has no meaning in classical physics. The corresponding classical expression is indeed
$$\tag{3}\label{3} Z_{cm} = \frac 1 {h^{3N} N!} \int e^{-\beta H (p,q)} d^N \mathbf p d^N \mathbf x$$
However, we know that classical mechanics is an approximation of quantum mechanics. Therefore, under some condition we must be able to approximate \ref{1} with \ref{2}:
$$Z_{qm} \approx Z_{cm}$$
To rigorously prove that we can do this approximation is quite cumbersome. In K. Huang,
Statistical Mechanics, second edition paragraph 9.2 you can find a rigorous proof of this result for the case of non-interacting particles (ideal gas), but the general case is quite cumbersome.
You can find another proof in this article (there is a paywall, though), which also mainly considers an ideal gas.
Another, more simple derivation in the case of a single particle in 1D can be found on these lecture notes (par. 2.1.1).
I have been thinking about a way to explain the general idea of such approximation is simple terms without relying too much on quantum mechanical concepts, but I admit that I found no explanation that would not dumb down the concept excessively. In other words, I cannot provide you any explanation that wouldn't be a "lie", and the best suggestion that I can give you is to actually learn some quantum mechanics and then take a look at the derivation from one of the sources I cited.
In particular, I will note that even though a non quantum mechanical derivation can be attempted, you will never be able to get from it:
The factor $h^{3N}$, which come from phase space quantization. In some sense, as also explained by knzhou in his answer, this come from the fact that a quantum state occupies approximately a volume $h$ in phase space. The factor $N!$, which comes from the indistinguishability of quantum particles. In purely classical mechanics, this factor must be put in Z by hand, to avoid double counting of stats which only differ by a permutation of identical particles. Notice that even if in classical mechanics particles are always distinguishable, they are still identical, i.e. the classical Hamiltonian remains unchanged if you exchange the label of two atoms. Because of this, you need the factor $N!$ in classical mechanics too. However, it comes form quantum indistinguishability. |
I'll summarize the comments from chi, and sketch the proof that
There can be no proof of $\mathrm{False}:=\forall X:*.X$ in the CoC in head normal form. Furthermore, this fact can be proven in a weak theory, say Peano Arithmetic (though the excluded middle is not required).
This fact implies that if the CoC is normalizing, then it is consistent, and furthermore this implication does not use classical logic.
for 1., you proceed by induction on the term structure of a hypothetical closed term $t$ of type $\mathrm{False}$. Actually terms in normal form must be in the form: $$ \lambda x_1:T_1\ldots\lambda x_n:T_n.y\ t_1\ldots t_m$$
where $n$ and $m$ may be $0$. If $n$ is zero, then $t=y\ t_1\ldots t_m$, which is not possible since $t$ is closed (typed in the empty context).
Otherwise, we may apply inversion and conclude that $T_1=*$ (and $x_1=X$), and we get$$ X:*\vdash\lambda x_2\ldots\lambda x_n.y\ t_1\ldots t_m\ :\ X$$to be derivable in the CoC. Now since $X$ is not a $\Pi$, we can apply inversion again to conclude that $n=1$ and in fact the above term is simply$$ X:*\vdash y\ t_1\ldots t_m\ :\ X$$Inversion yet again shows that$$X:*\vdash y\ :\ \Pi y_1:U_1\ldots\Pi y_m:U_m.X $$but $y$ must be $X$, since it is the only variable around! Therefore $m=0$, and $X=*$, which is impossible, contradiction.
Now all this reasoning is intuitionistic, as I've proven a
negative (there can be no proof of...) and proofs of negations are always constructive. I do rely heavily on inversion, and you'll just have to take it on faith that this also can be proven in arithmetic, without the excluded middle, which is non-trivial.
Now for 2. we
define consistency to mean "does not prove $\mathrm{False}$"! Again a negative statement. Now if the CoC is normalizing, one can take any normal form of a proof of $\mathrm{False}$ and use the above argument to get a contradiction. Again a constructive argument!
Finally to tie it all together. Now suppose you had enough arithmetic to carry out the above arguments in CoC. Note that this is
almost possible: you actually need to add the axiom $0\neq 1$ to get anything off the ground. You can then prove that normalization implies consistency within the CoC, and you also have enough arithmetic for the second incompleteness theorem to apply. Therefore you cannot (if CoC is consistent!) prove normalization, as then you would have a full proof of consistency within CoC. |
I understood the definition of
affine $k$-chain and that he defines $\int \limits_{\Gamma} \omega$ as $(82)$. But I can't understand the last two above examples. What does they mean?
Can anyone explain them detailed?
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When you have a binary operation $\newcommand{\bop}{\mathop{\scriptstyle\top}}\bop \colon S\times S \to S$, that induces a corresponding operation on the space of functions $D \to S$ by applying the operation pointwise,
$$(f \bop g)(x) := f(x) \bop g(x).$$
The pointwise sum or product of real-valued functions are very familiar examples.
The same construction applied to affine $k$-simplices gives the concept of affine $k$-chains. However, one can view affine $k$-simplices as functions in different ways, and it's the non-direct way that gives rise to $k$-chains.
To avoid any ambiguity, let me use different notations for the two ways to view $k$-simplices as functions that Rudin mentions.
First, by definition an affine $k$-simplex is a function (sufficiently regular) $\sigma \colon Q^k \to \mathbb{R}^n$. We have an addition on $\mathbb{R}^n$, and that induces an addition on the space of functions $Q^k \to \mathbb{R}^n$. Let's denote this addition by $\oplus$. Then $(\sigma_1 \oplus \sigma_2) \colon u \mapsto \sigma_1(u) + \sigma_2(u)$ is an affine $k$-simplex if $\sigma_1$ and $\sigma_2$ are affine $k$-simplices. In the context of integration of differential forms, this operation is however uninteresting and rarely - if ever - considered.
The interesting concept arises when one views an affine $k$-simplex (or more generally a $k$-surface) in $E$ as a map $\Omega^k(E) \to \mathbb{R}$ via integration, where $\Omega^k(E)$ denotes the space of (continuous) $k$-forms in $E \subset \mathbb{R}^n$. Formally, since that is not exactly the same thing as the $k$-surface $\Phi$, it should be denoted differently, but doing that would be cumbersome, so it is customary to abuse notation and denote this map also by $\Phi$. For this discussion, I will however use the notation $I_\Phi$ for the map $\omega \mapsto \int_{\Phi} \omega$ to disambiguate. Thus every $k$-surface $\Phi$ in $E$ defines a map $I_{\Phi} \colon \Omega^k(E) \to \mathbb{R}$, and we have the induced addition
$$(I_{\Phi} + I_{\Psi}) \colon \omega \mapsto I_{\Phi}(\omega) + I_{\Psi}(\omega) = \int_{\Phi} \omega + \int_{\Psi} \omega.$$
It is this induced addition that pertains to affine $k$-chains. An affine $k$-chain in $E$ "is" a map $\Omega^k(E) \to \mathbb{R}$ which we can write as the sum of finitely many $I_{\sigma_i}$, where each $\sigma_i$ is an affine $k$-simplex in $E$. But out of convenience, one drops the $I$s and writes $k$-chains as $\sigma_1 + \dotsc \sigma_r$ rather than $I_{\sigma_i} + \dotsc + I_{\sigma_r}$.
In the penultimate paragraph of the section, Rudin explains that one could be tempted to interpret the notation $\sigma_1 + \sigma_2$ as $\sigma_1 \oplus \sigma_2$, but that is not how $(83)$ is to be interpreted. In the last paragraph, he gives an example illustrating that these two additions are very different. If $\sigma_1$ and $\sigma_2$ are affine $k$-simplices with $\sigma_2 = -\sigma_1$ in the sense of $(80)$, then by theorem 10.27 we have $I_{\sigma_1} + I_{\sigma_2} = 0$, and that means $\sigma_1 + \sigma_2 = 0$ in the sense of $k$-chains, but generally we don't have $\sigma_1 \oplus \sigma_2 = 0$, where the last $0$ is the constant map $0 \colon u \mapsto 0 \in \mathbb{R}^n$. |
Let $M,N$ be oriented $d$-dimensional Riemannian manifolds, $M$
compact*, and let $f:M \to N$ be smooth.
Consider the Dirichlet energy functional: $E_{M,N}(f)=\int_M \|df\|^2 \operatorname{Vol}_g$. ($\operatorname{Vol}_g$ is the Riemannian volume form of $M$).
Fix another $d$-dimensional Riemannian manifold $W$.It's easy to see that for any isometric immersion $\phi:N \to W$,
$$ E_{M,N}(f)=E_{M,W}(\phi \circ f) \, \text{ for any } \, f:M \to N. \tag{1}$$
Let us call a map $\phi$ which sarisfies $(1)$ a
symmetry of the Dirichlet's integral.
In this terminology, every isometric immersion is a symmetry.
Question:Is every symmetry an isometric immersion?
Does anything change if restrict the symmetries to be invertible maps? (or diffeomorphisms)? Does this notion of symmetry has a name somewhere in the literature?
I guess the answer is positive.
*We can relax the compactness assumption, but then we need to restrict the discussion to maps which are constant outside a compact domain.
Comments:
$(1)$ The "pointwise analog" is quite trivial:
If $B \in M_d$, and for every $A \in M_d$, $ \|BA\|=\|A\|$ (where $\| \|$ is the Euclidean norm), then $B$ is an orthogonal matrix.
The challenge is that the notion of a Dirichlet's-symmetry is a
global one, while the concept of isometric immersion is local.
I think the rough idea should be to choose maps $f$ which are very "localized" (are constant outside small balls). However, this does not seem trivial, since the differential would have to pass from a given linear map to zero, so its norm would vary. (Think of a bump function which goes quickly from one to zero, you can't make the integral of the derivative small).
$(2)$ The answer could a-priori depend on the manifolds. Even the case where $M$ is an Euclidean ball in $\mathbb{R}^d$, $N=W=\mathbb{R}^d$, the answer does not seem to be trivial (see the previous comment). |
Let $f$ be measurable, finite a.e, and Lebesgue integrable on $\mathbb{R}$ . Show that for all $\epsilon > 0$ there exists a measurable set $E$ (i.e $\mu(E) < \infty$) such that:
$\int_{E} |f| > \int_{\mathbb{R}} |f| - \epsilon$
My attempt
Suppose that for all $\epsilon > 0$ there exists set of finite measure $E$ such that:
$\int_{E} |f| \leq \int_{\mathbb{R}} |f| - \epsilon$
Then:
$ \int_{\mathbb{R}\setminus{E}} |f| \geq \epsilon$
Since $\epsilon > 0$ was arbitrary then we have:
$ \int_{\mathbb{R}\setminus{E}} |f| = \infty$
But $\mu(\mathbb{R}\setminus{E}) \geq \mu(\mathbb{R}) - \mu(E) = \infty$
This contradicts the fact that $g$ is finite a.e.
I don’t think my answer is correct because it seems pretty hand-wavy. Any thoughts on a better way to do this? |
As you may expect from the title of this section, this will be the most difficult and complicated section of this chapter so far. Our aim will be to calculate the field and potential surrounding a simple dipole.
A simple dipole is a system consisting of two charges, \(+Q \text{ and }−Q\), separated by a distance \(2L\). The dipole moment of this system is just \(p = 2QL\). We’ll suppose that the dipole lies along the
x-axis, with the negative charge at \(x = −L\) and the positive charge at \(x = +L\) . See Figure \(III\).5. \(\text{FIGURE III.5}\)
Let us first calculate the electric field at a point P at a distance \(y\) along the \(y\)-axis. It will be agreed, I think, that it is directed towards the left and is equal to
\[E_1 \cos \theta +E_2 \cos \theta,\text{ where }E_1=E_2=\dfrac{Q}{4\pi\epsilon_0 (L^2+y^2 )}\text{ and } \cos \theta = \dfrac{L}{(L^2+y^2)^{1/2}}.\nonumber\]
Therefore
\[E=\dfrac{2QL}{4\pi\epsilon_0(L^2+y^2)^{3/2}}=\dfrac{p}{4\pi\epsilon_0(L^2+y^2)^{3/2}}.\label{3.7.1}\]
For large \(y\) this becomes
\[\label{3.7.2}E=\dfrac{p}{4\pi\epsilon_0y^3}.\]
That is, the field falls off as the cube of the distance.
To find the field on the \(x\)-axis, refer to Figure \(III\).6.
\(\text{FIGURE III.6}\)
It will be agreed, I think, that the field is directed towards the right and is equal to
\[\label{3.7.3}E=E_1-E_2=\dfrac{Q}{4\pi\epsilon_0} \left (\dfrac{1}{(x-L)^2}-\dfrac{1}{(x+L)^2}\right ).\]
This can be written \(\dfrac{Q}{4\pi\epsilon_0x^2} \left (\dfrac{1}{(1-L/x)^2}-\dfrac{1}{(1+L/x)^2}\right )\), and on expansion of this by the binomial theorem, neglecting terms of order \((L / x)^2\) and smaller, we see that at large
x the field is
\[\label{3.7.4}E=\dfrac{2p}{4\pi\epsilon_0x^3}.\]
Now for the field at a point P that is neither on the axis (\(x\)-axis) nor the equator (\(y\)-axis) of the dipole. See Figure \(III\).7.
\(\text{FIGURE III.7}\)
It will probably be agreed that it would not be particularly difficult to write down expressions for the contributions to the field at P from each of the two charges in turn. The difficult part then begins; the two contributions to the field are in different and awkward directions, and adding them vectorially is going to be a bit of a headache.
It is much easier to calculate the
potential at P, since the two contributions to the potential can be added as scalars. Then we can find the x- and y-components of the field by calculating \(∂V / ∂x\) and \(∂V/∂y\).
Thus
\[V=\dfrac{Q}{4\pi\epsilon_0}\left ( \dfrac{1}{\{(x-L)^2+y^2\}^{1/2}}-\dfrac{1}{\{(x+L)^2+y^2\}^{1/2}}\right ) . \label{3.7.5}\]
To start with I am going to investigate the potential and the field at a
large distance from the dipole – though I shall return later to the near vicinity of it.
At
large distances from a small dipole (see Figure \(III\).8), we can write, \(r^2=x^2+y^2\), \(\text{FIGURE III.8}\)
and, with \(L^2 << r^ 2\) , the expression 3.7.5 for the potential at P becomes
\[V=\dfrac{Q}{4\pi\epsilon_0}\left ( \dfrac{1}{(r^2-2Lx)^{1/2}}-\dfrac{1}{(r^2+2Lx)^{1/2}}\right )=\dfrac{Q}{4\pi\epsilon_0 r}\left ( (1-2Lx/r^2)^{-1/2}-(1+2Lx/r^2)^{-1/2}\right ).\nonumber\]
When this is expanded by the binomial theorem we find, to order
L/ r , that the potential can be written in any of the following equivalent ways:
\[\label{3.7.6}V=\dfrac{2QLx}{4\pi\epsilon_0 r^3}=\dfrac{px}{4\pi\epsilon_0 r^3}=\dfrac{p\cos \theta}{4\pi\epsilon_0 r^2}=\dfrac{\textbf{p}\cdot \textbf{r}}{4\pi\epsilon_0 r^3}.\]
Thus the equipotentials are of the form
\[\label{3.7.7}r^2=c\cos \theta ,\]
where
\[\label{3.7.8}c=\dfrac{p}{4\pi\epsilon_0 V}.\]
Now, bearing in mind that \(r^2+x^2+y^2\), we can differentiate \(V=\dfrac{px}{4\pi\epsilon_0 r^3}\) with respect to \(x\) and \(y\) to find the \(x\)- and \(y\)-components of the field. Thus we find that
\[\label{3.7.9}E_x =\dfrac{p}{4\pi\epsilon_0}\left ( \dfrac{3x^2-r^2}{r^5}\right ) \text{ and }E_y=\dfrac{pxy}{4\pi\epsilon r^5}.\]
We can also use polar coordinates to find the radial and transverse components from \(E_r=-\dfrac{∂V}{∂r}\text{ and }E_\theta = -\dfrac{1}{r}\dfrac{∂V}{∂\theta}\text{ together with }V=\dfrac{p\cos \theta}{4\pi\epsilon_0 r^2}\) to obtain
\[\label{3.7.10}E_r = \dfrac{2p\cos \theta}{4\pi\epsilon_0 r^3},\quad E_\theta = \dfrac{p\sin \theta }{4\pi\epsilon_0 r^3}\text{ and }E=\dfrac{p}{4\pi\epsilon_0 r^3}\sqrt{1+3 \cos^2 \theta}.\]
The angle that \(\textbf{E}\) makes with the axis of the dipole at the point \((r, θ)\) is \(θ + \tan^{-1}\dfrac{1}{2}\tan \theta\).
For those who enjoy vector calculus, we can also say \(\textbf{E}=-\dfrac{1}{4\pi\epsilon_0}\nabla \left ( \dfrac{\textbf{p}\cdot \textbf{r}}{r^3}\right ) \), from which, after a little algebra and quite a lot of vector calculus, we find
\[\label{3.7.11}\textbf{E}=\dfrac{1}{4\pi\epsilon_0}\left ( \dfrac{3(\textbf{p}\cdot \textbf{r})\textbf{r}}{r^5}-\dfrac{\textbf{p}}{r^3}\right ).\]
This equation contains all the information that we are likely to want, but I expect most readers will prefer the more explicit rectangular and polar forms of equations \ref{3.7.9} and \ref{3.7.10}.
Equation \ref{3.7.7} gives the equation to the equipotentials. The equation to the lines of force can be found as follows. Referring to Figure \(III\).9, we see that the differential equation to the lines of force is
\(\text{FIGURE III.9}\)
\[\label{3.7.12}r\dfrac{d\theta}{dr}=\dfrac{E_\theta}{E_r}=\dfrac{\sin \theta}{2\cos \theta}=\dfrac{1}{2}\tan \theta ,\]
which, upon integration, becomes
\[\label{3.7.13}r=a\sin^2 \theta .\]
Note that the equations \(r^2= c\cos θ\) (for the equipotentials) and \(r= a \sin^2 \theta\) (for the lines of force) are orthogonal trajectories, and either can be derived from the other. Thus, given that the differential equation to the lines of force is \(r\dfrac{d\theta}{dr}= \dfrac{1}{2}\tan \theta\) with solution \(r=a\sin^2 \theta\), the differential equation to the orthogonal trajectories (i.e. the equipotentials) is \(-\dfrac{1}{r}\dfrac{dr}{d\theta}=\dfrac{1}{2}\tan \theta\), with solution \(r^2=c\cos \theta\).
In Figure \(III\).10, there is supposed to be a tiny dipole situated at the origin. The unit of length is \(L\), half the length of the dipole. I have drawn eight electric field lines (continuous), corresponding to
a = 25, 50, 100, 200, 400, 800, 1600, 3200. If r is expressed in units of \(L\), and if \(V\) is expressed in units of \(\dfrac{Q}{4\pi\epsilon_0 L}\), the equations \ref{3.7.7} and \ref{3.7.8} for the equipotentials can be written , \(r=\sqrt{\dfrac{2\cos \theta}{V}}\), and I have drawn seven equipotentials (dashed) for \(V\) = 0.0001, 0.0002, 0.0004, 0.0008, 0.0016, 0.0032, 0.0064. It will be noticed from Equation \ref{3.7.9a}, and is also evident from Figure \(III\).10, that \(E_x\) is zero for \(\theta = 54^\circ \, 44'\) . \(\text{FIGURE III.10}\)
At the end of this chapter I append a (geophysical) exercise in the geometry of the field at a large distance from a small dipole.
Equipotentials near to the dipole
These, then, are the field lines and equipotentials at a
large distance from the dipole. We arrived at these equations and graphs by expanding equation \ref{3.7.5} binomially, and neglecting terms of higher order than \(L/r\). We now look near to the dipole, where we cannot make such an approximation. Refer to Figure \(III\).7.
We can write Equation \ref{3.7.5} as
\[\label{3.7.14}V(x,y)=\dfrac{Q}{4\pi\epsilon_0}\left ( \dfrac{1}{r_1}-\dfrac{1}{r_2}\right ),\]
where \(r_1^2=(x-L)^2+y^2 \text{ and }r_2^2=(x+L)^2+y^2\). If, as before, we express distances in terms of \(L\) and \(V\) in units of \(\dfrac{Q}{4\pi\epsilon_0 L}\), the expression for the potential becomes
\[\label{3.7.15}V(x,y)\dfrac{1}{r_1}-\dfrac{1}{r_2},\]
where \(r_1^2=(x+1)^2+y^2 \text{ and }r_2^2=(x-1)^2+y^2\).
One way to plot the equipotentials would be to calculate \(L\) for a whole grid of \((x , y)\) values and then use a contour plotting routine to draw the equipotentials. My computing skills are not up to this, so I’m going to see if we can find some way of plotting the equipotentials directly.
I present two methods. In the first method I use equation \ref{3.7.15} and endeavour to manipulate it so that I can calculate \(y\) as a function of \(x\) and \(L\). The second method was shown to me by J. Visvanathan of Chennai, India. We’ll do both, and then compare them.
First Method.
To anticipate, we are going to need the following:
\[\begin{align} &r_1^2r_2^2=(x^2+y^2+1)^2-4x^2=B^2-A,\label{3.7.16} \\ &r_1^2+r_2^2 = 2(x^2+y^2+1)=2B,\label{3.7.17}\\ \text{and} \quad &r_1^4+r_2^4=2[(x^2+y^2+1)^2+4x^2]=2(B^2+A),\label{3.7.18} \\ \text{where}\quad \,&A=4x^2 \label{3.7.19} \\ \text{and}\quad &B=x^2+y^2+1 \label{3.7.20} \\ \end{align}\]
Now equation \ref{3.7.15} is \(r_1r_2 V=r_2-r_1\). In order to extract \(y\) it is necessary to square this twice, so that \(r_1 \text{ and }r_2\) appear only as \(r_1^2 \text{ and }r_2^2\). After some algebra, we obtain
\[\label{3.7.21}r_1^2r_2^2 [2-V^4r_1^2r_2^2+2V^2(r_1^2 +r_2^2)]=r_1^4+r_2^4.\]
Upon substitution of equations \ref{3.7.16},17,18, for which we are well prepared, we find for the equation to the equipotentials an equation which, after some algebra, can be written as a quartic equation in
B:
\[\begin{align}&a_0+a_1B+a_2B^2+a_3B^3+a_4B^4=0 \label{3.7.22} \\ \text{where}\quad &a_0=A(4+V^4A), \label{3.7.23} \\ &a_1=4V^2A, \label{3.7.24} \\ &a_2 = -2V^2A,\label{3.7.25} \\ &a_3=-4V^2,\label{3.7.26} \\ \text{and} \quad &a_4 = V^4 .\label{3.7.27} \\ \end{align}\]
The algorithm will be as follows: For a given \(V\) and \(x\), calculate the quartic coefficients from equations \ref{3.7.23}-\ref{3.7.27}. Solve the quartic equation \ref{3.7.22} for
B. Calculate y from equation \ref{3.7.20}. My attempt to do this is shown in Figure \(III\).11. The dipole is supposed to have a negative charge at (−1 , 0) and a positive charge at (+1 , 0). The equipotentials are drawn for \(V\) = 0.05, 0.10, 0.20, 0.40, 0.80. \(\text{FIGURE III.11}\) Second method (J. Visvanathan).
In this method, we work in polar coordinates, but instead of using the coordinates \((r,θ)\) , in which the origin, or pole, of the polar coordinate system is at the center of the dipole (see Figure \(III\).7), we use the coordinates \((r_1, \phi)\) with origin at the positive charge.
From the triangle, we see that
\[r_2^2=r_1^2+4L^2 +4Lr_1\cos \phi .\label{3.7.28}\]
For future reference we note that
\[\dfrac{∂r_2}{∂r_1}=\dfrac{r_1 +2L\cos \phi }{r_2}.\label{3.7.29}\]
Provided that distances are expressed in units of \(L\), these equations become
\[\label{3.7.30}r_2^2=r_1^2 + 4r_1 \cos \phi + 4,\]
\[\dfrac{∂r_2}{∂r_1}=\dfrac{r_1+2\cos \phi}{r_2}.\label{3.7.31}\]
If, in addition, electrical potential is expressed in units of \(\dfrac{Q}{4\pi\epsilon_0 L}\), the potential at P is given, as before (equation \ref{3.7.15}), by
\[\label{3.7.32}V(r_1,\phi )=\dfrac{1}{r_1}-\dfrac{1}{r_2}.\]
bearing in mind that \(r_2\) is given by equation \ref{3.7.31}.
By differentiation with respect to \(r_1\), we have
\[\label{3.7.34}f'(r_1)=-\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}\dfrac{∂r_2}{∂r_1}=-\dfrac{1}{r_1^2}+\dfrac{r_1+2\cos \phi}{r_2^3},\]
and we are all set to begin a Newton-Raphson iteration: \(r_1=r_1-f/f'\). Having obained \(r_1\), we can then obtain the \((x,\, y)\) coordinates from \(x = 1 + r_1 \cos φ \text{ and }y = r_1 \sin φ\).
I tried this method and I got exactly the same result as by the first method and as shown in Figure \(III\).11.
So which method do we prefer? Well, anyone who has worked through in detail the derivations of equations \ref{3.7.16} -\ref{3.7.27}, and has then tried to program them for a computer, will agree that the first method is very laborious and cumbersome. By comparison Visvanathan’s method is much easier both to derive and to program. On the other hand, one small point in favor of the first method is that it involves no trigonometric functions, and so the numerical computation is potentially faster than the second method in which a trigonometric function is calculated at each iteration of the Newton-Raphson process. In truth, though, a modern computer will perform the calculation by either method apparently instantaneously, so that small advantage is hardly relevant.
So far, we have managed to draw the
equipotentials near to the dipole. The lines of force are orthogonal to the equipotentials. After I tried several methods with only partial success, I am grateful to Dr Visvanathan who pointed out to me what ought to have been the “obvious” method, namely to use equation \ref{3.7.12}, which, in our \((r_1,\phi)\) coordinate system based on the positive charge, is \(r_1\dfrac{d\phi}{dr_1}=\dfrac{E_\phi}{E_{r_1}}\), just as we did for the large distance, small dipole, approximation. In this case, the potential is given by equations \ref{3.7.30} and \ref{3.7.32}. (Recall that in these equations, distances are expressed in units of L and the potential in units of \(\dfrac{Q}{4\pi\epsilon_0 L}\) .) The radial and transverse components of the field are given by \(E_{r_1}=-\dfrac{∂V}{∂{r_1}}\text{ and } E_\phi=-\dfrac{1}{r_1}\dfrac{∂V}{∂\phi}\), which result in
\[E_{r_1}=\dfrac{1}{r_1^2}-\dfrac{r_1+2\cos \phi }{r_2^3}\label{3.7.35}\]
and
\[E_\phi = \dfrac{2\sin \phi}{r_2^3}.\label{3.7.36}\]
Here, the field is expressed in units of \(\dfrac{Q}{4\pi\epsilon_0L^2}\), although that hardly matters, since we are interested only in the ratio. On applying \(r_1\dfrac{d\phi}{dr}=\dfrac{E_\phi}{E_{r_1}}\) to these field components we obtain the following differential equation to the lines of force:
\[d\phi = \dfrac{2r_1\sin \phi}{(r_1^2+4+4r_1\cos \phi)^{3/2}-r_1^2(r_1+2\cos \phi)}dr_1.\label{3.7.37}\]
Thus one can start with some initial \(φ_0\) and small \(r_2\) and increase \(r_1\) successively by small increments, calculating a new φ each time. The results are shown in Figure \(III\).12, in which the equipotentials are drawn for the same values as in Figure \(III\).11, and the initial angles for the lines of force are 30º, 60º, 90º, 120º, 150º.
\(\text{FIGURE III.12}\)
Before we leave this section, here is yet another method of calculating the potential near to a dipole, for those who are familiar with Legendre polynomials.
The potential at P is given by
\[\nonumber \begin{align}4\pi\epsilon_0V&=Q \left [ \dfrac{1}{(a^2+r^2-2ar\cos \theta )^{1/2}}-\dfrac{1}{(a^2+r^2+2ar\cos \theta )^{1/2}}\right ] \\ \nonumber &=\dfrac{Q}{a} \left [ \dfrac{1}{(1-2\rho \cos \theta + \rho^2)^{1/2}}-\dfrac{1}{1+2\rho \cos \theta + \rho^2)^{1/2}}\right ] \\ \end{align}\]
where \(\rho = r/a\).
It is well known (to those who are familiar with Legendre polynomials!) that
\[\nonumber (1-2\rho\cos \theta +\rho^2)^{-1/2}=P_0(\cos \theta)+P_1(\cos \theta)\rho +P_2(\cos \theta )\rho^2+P_3(x)\rho^3+...\]
where the \(P_n\) are the Legendre polynomials. Thus the potential can be calculated as a series expansion. Those who are unfamiliar with the Legendre polynomials can find something about them in my notes on celestial mechanics www.astro.uvic.ca/~tatum/celmechs/celm1.pdf |
Radioactive decay is the process by which unstable particles with high mass; fall apart into more stable particles with lower mass. Although the process itself is quantum mechanical in nature, the dynamics of radioactive decay are described by special relativity and are essentially identical to those of an inelastic collision in reverse. Decay may occur spontaneously (as a random process), but can also be stimulated, by the absorption of a (typically small, e.g. a photon or electron) particle by the unstable one - a process used in nuclear reactors. Because the absorbed particle also carries energy, in stimulated decay the masses of the resulting particles can add up to something more than the rest mass of the original particle. An important question in nuclear physics is what the
threshold energy of a given reaction is, i.e., the minimum energy the incoming particle must have for the process to be possible. This is not simply the differences in the mass-energy of the original and the resulting particles, as in the collision process momentum must also be conserved. To illustrate how to approach such a problem, let’s again consider a concrete example: the threshold energy for the reaction in which a proton (\(m_{\mathrm{p}}=938 \mathrm{MeV} / c^{2}\)), initially at rest, absorbs a photon, and then emits a neutral pion (\(m_{\pi} = 135 \mathrm{MeV} / c^{2}\)), see Figure 14.3.1 below.
Figuring out what the minimum required energy is in the lab frame is not easy, as you have to account for the kinetic energy of the particles after the reaction. There is however a system in which the reaction products are standing still: the
center-of-momentum frame, the relativistic analog of the center-of-mass frame of classical mechanics\(^{1}\). The center-of-momentum frame is defined as the frame in which the total momentum of all particles adds up to zero. In our specific example, before the collision, only the photon carries a momentum, equal to its energy \(E_{\gamma}\) divided by the speed of light. In general, the total momentum in the system can be a three-vector, equal to \(\boldsymbol{p}_{\mathrm{T}}=\sum_{i} \boldsymbol{p}_{i}\), while the total energy is given by \(E_{\mathrm{T}}=\sum_{i} E_{i}\). If we choose our coordinates such that the \(x\)-direction coincides with that of \(\boldsymbol{p}_{\mathrm{T}}\), the energy-momentum four-vector of the entire system becomes \(\overline{\boldsymbol{p}}_{\mathrm{T}}=\left(E_{\mathrm{T}} / c, p_{\mathrm{T}}, 0,0\right)\), where \(p_{\mathrm{T}}=\left|\boldsymbol{p}_{\mathrm{T}}\right|\). If we go to any different inertial frame \(S^{\prime}\) moving with velocity \(v\) in the positive \(x\) direction, the components of the energy-momentum four-vector are given by the Lorentz transform of \(\boldsymbol{p}_{\mathrm{T}}\):
\[\overline{\boldsymbol{p}}_{\mathrm{T}}^{\prime}=\gamma(v)\left(\frac{E_{\mathrm{T}}}{c}-\frac{v}{c} p_{\mathrm{T}}, p_{\mathrm{T}}-\frac{v}{c} \frac{E_{\mathrm{T}}}{c}, 0,0\right) \label{14.3.1}\]
so we end up in a frame in which the total momentum is zero if we pick
\[ v_{\mathrm{COM}}=\frac{c^{2} p_{\mathrm{T}}}{E_{\mathrm{T}}} \label{14.3.2}\]
for our velocity. In particular, we see that we can always make this transformation, and that the center-of-momentum frame is an inertial frame.
Back to our example: why do we care? The answer is almost tautological: if the total momentum is zero before the collision, it is also zero afterward - and so in the COM frame, the particles can all be standing still (see Figure 14.3.1b). That certainly corresponds to the lowest possible kinetic energy of the system, so the energy of the incoming photon is all converted to mass - and that must thus be the threshold energy we’re looking for. Interestingly, to answer our original question, we don’t even need to calculate what the actual velocity of the COM frame is, just the fact that it exists is sufficient. In the COM frame, we have, by conservation of four-momentum:
\[\overline{\boldsymbol{p}}_{\gamma}^{\prime}+\overline{\boldsymbol{p}}_{\mathrm{p}, \mathrm{i}}^{\prime} = \overline{\boldsymbol{p}}_{p}, f^{\prime}+\overline{\boldsymbol{p}}_{\pi}^{\prime} \label{14.3.3}\]
and therefore also
\[ \left(\overline{\boldsymbol{p}}_{\gamma}^{\prime}+\overline{\boldsymbol{p}}_{\mathrm{p}, \mathrm{i}}^{\prime}\right)^{2}=\left(\overline{\boldsymbol{p}}_{p}, f^{\prime}+\overline{\boldsymbol{p}}_{\pi}^{\prime}\right)^{2}=\left(m_{\mathrm{p}} c+m_{\pi} c\right)^{2} \label{14.3.4}\]
where the last equality follows from the fact that the reactants are standing still. Now the left-hand-side
of equation (/ref{14.3.4}) is the length of a four-vector, and we’ve proven that these lengths are invariant under Lorentz transformations - so it’s value is equal to that of \(\left(\overline{\boldsymbol{p}}_{\gamma}+\overline{\boldsymbol{p}}_{\mathrm{p}, \mathfrak{i}}\right)^{2}\) in the lab frame. In that frame, we have\(\overline{\boldsymbol{p}}_{\gamma}=\left(E_{\gamma} / c\right)(1,1,0,0)\) and \(\overline{\boldsymbol{p}}_{\mathrm{p}, \mathrm{i}}=\left(m_{\mathrm{p}} c, 0,0,0\right)\), so we end up with an easy equation form \(E_{\gamma}\):
\[\left(m_{\mathrm{p}}+m_{\pi}\right)^{2} c^{2}=\overline{\boldsymbol{p}}_{\gamma}^{2}+\overline{\boldsymbol{p}}_{\mathrm{p}, \mathrm{i}}^{2}+2 \overline{\boldsymbol{p}}_{\gamma}^{2} \cdot \overline{\boldsymbol{p}}_{\mathrm{p}, \mathrm{i}}=0+m_{\mathrm{p}}^{2} c^{2}+2 E_{\gamma} m_{\mathrm{p}} \label{14.3.5}\]
or
\[ E_{\gamma}=\frac{m_{\pi}^{2}+2 m_{\pi} m_{\mathrm{p}}}{2 m_{\mathrm{p}}} c^{2}=145 \mathrm{MeV} \label{14.3.6}\]
In this example, we thus need at least 10MeV of energy more than the mass of the particle we’ve created.
Note that in finding the threshold energy in the example, we again heavily relied on the four-vector properties of \(\overline{\boldsymbol{p}}\) - not only it’s length (like in the third method of section 14.2), but also the invariance of that length under Lorentz transformations. Using these properties results in easy equations to solve, while if you’d ignore them, you’d probably get stuck trying to figure out what the kinetic energy of the reaction products is.
\(^{1}\) As our system includes a photon, a center-of-mass frame doesn’t make sense here, as the photon has no mass - but it has nonzero momentum, so we can make a transformation to a system in which the total momentum vanishes. |
The problem is NP-Hard and the reduction is from partition.
So, assume you are given an instance of partition $S=${$x_1,\ldots,x_n$}. Let $t = (\sum_{i=1}^n x_i) / 2$. The question is whether there is a subset of $S$ such that its sum is $t$.
To this end, let $v_0 = (0, 2t, t)$ be a special vector. For every number $x_i$, create the vector
$v_i = (x_i, -x_i, 0)$,
for $i=1,\ldots, n$. Now, the claim is that there is a subset of the vectors with sum $\geq (t, t, t)$ (in absolute value) if and only if there is a subset of $S$ that add up to $t$.
So, consider a subset $X \subseteq S$, with $\alpha = \sum_{x\in X}$. We have that the corresponding sum of vectors (together with the special vector $v_0$) is
$(\alpha, 2t -\alpha, t)$.
Clearly, this vector is maximized when $\alpha=t$, as required.
Note, that you must include the special vector $v_0$ - otherwise the third coordinate would be zero in the sum of vectors.
QED
Note, that the problem is solvable in polynomial time if the numbers are polynomially small, doing dynamic programming (like the one used for solving subset sum if the numbers are small).
2d
If you set the special vector to be $(8t, 10t)$ then the above reduction would work verbatim in two dimensions. (The constant $8$ here is somewhat arbitrary...) |
When we follow the standard textbooks, or tradition, most of us teach the following definition of big-Oh notation in the first few lectures of an algorithms class: $$ f = O(g) \mbox{ iff } (\exists c > 0)(\exists n_0 \geq 0)(\forall n \geq n_0)(f(n) \leq c \cdot g(n)). $$ Perhaps we even give the whole list with all its quantifiers:
$f = o(g) \mbox{ iff } (\forall c > 0)(\exists n_0 \geq 0)(\forall n \geq n_0)(f(n) \leq c \cdot g(n))$ $f = O(g) \mbox{ iff } (\exists c > 0)(\exists n_0 \geq 0)(\forall n \geq n_0)(f(n) \leq c \cdot g(n))$ $f = \Theta(g) \mbox{ iff } (\exists c > 0)(\exists d > 0)(\exists n_0 \geq 0)(\forall n \geq n_0)(d \cdot g(n) \leq f(n) \leq c \cdot g(n))$ $f = \Omega(g) \mbox{ iff } (\exists d > 0)(\exists n_0 \geq 0)(\forall n \geq n_0)(f(n) \geq d \cdot g(n))$ $f = \omega(g) \mbox{ iff } (\forall d > 0)(\exists n_0 \geq 0)(\forall n \geq n_0)(f(n) \geq d \cdot g(n))$.
However, since these definitions are not so easy to work with when it comes to proving even simple things such as $5 n \log^4 n + \sqrt{n\log n} = o(n^{10/9})$, most of us quickly move to introduce the "trick of the limit":
$f = o(g)$ if $\lim_{n \rightarrow \infty} f(n)/g(n)$ exists and is $0$, $f = O(g)$ if $\lim_{n \rightarrow \infty} f(n)/g(n)$ exists and is not $+\infty$, $f = \Theta(g)$ if $\lim_{n \rightarrow \infty} f(n)/g(n)$ exists and is neither $0$ nor $+\infty$, $f = \Omega(g)$ if $\lim_{n \rightarrow \infty} f(n)/g(n)$ exists and is not $0$, $f = \omega(g)$ if $\lim_{n \rightarrow \infty} f(n)/g(n)$ exists and is $+\infty$.
My question is:
Would it be a big loss for teaching an undergraduate algorithms class to take the limit conditions as
the definitions of $o$, $O$, $\Theta$, $\Omega$, and $\omega$? That's what we all end-up using anyway and it seems pretty clear to me that skipping the quantifier definitions makes everybody's life easier.
I would be interested to know if you have encountered some convincing natural case where the standard $c,n_0$-definitions are actually required, and if not, whether you have a convincing argument to keep the standard $c,n_0$-definitions upfront anyway. |
Someone asked a question about self-avoiding random walks, and it made me think of the following:
Consider a piece that starts at a corner of an ordinary $8 \times 8$ chessboard. At each turn, it moves one step, either up, down, left, or right, with equal probability, except that it must stay on the board, of course, and it must not return to a square previously visited.
Clarification. On any given step, if the piece has $n$ available moves (excluding those that would put the piece on a previously visited square), it chooses randomly and uniformly from those $n$ moves. Example: Starting from a corner, at first move, $n = 2$, and either move is chosen with probability $1/2$. Next move, $n = 2$ also, because it cannot return to the corner, and so either of the two other moves are chosen with probability $1/2$. On the third move, if it is on the edge, $n = 2$, while if it is off the edge, $n = 3$. And so on.
It is possible for the piece to be deadlocked at some point prior to completing a tour of the chessboard. For instance, if it starts at lower left, and moves up, right, right, down, left, it is now stuck.
What is the probability that it completes the tour? Is there a method that answers this question besides exhaustive enumeration? What about $n \times n$ chessboards for $n \geq 3$? (The problem is trivial for $n = 1$ or $2$.)
Analysis for $n = 3$, as another clarification:
Let the chessboard be labelled $(1, 1)$ through $(3, 3)$, and the piece starts at $(1, 1)$. Without loss of generality, the piece moves to $(2, 1)$. Then:
With probability $1/2$, the piece moves to the center square $(2, 2)$ on its second move. From there, only one move permits completion of the tour—the move to $(1, 2)$—and in that case, the tour is guaranteed to complete. This move is chosen with probability $1/3$.
With probability $1/2$, the piece moves to $(3, 1)$ on its second move. It is then forced to move to $(3, 2)$.
With probability $1/2$, it then moves to $(3, 3)$ on its fourth move and is guaranteed to complete the tour.
Otherwise, also with probability $1/2$, it moves to the center square $(2, 2)$ on its fourth move. From there, it moves to $(1, 2)$ with probability $1/2$ (and is then guaranteed to complete the tour), or to $(2, 3)$ also with probability $1/2$ (and is then unable to complete the tour).
Thus, the probability of completing the tour on a $3 \times 3$ board is
$$ p_3 = \frac{1}{2} \times \frac{1}{3} + \frac{1}{2} \times \left( \frac{1}{2} + \frac{1}{2} \times \frac{1}{2} \right) = \frac{1}{6} + \frac{3}{8} = \frac{13}{24} $$
Update. An exhaustive enumeration in floating point yielded the following:
For $n = 8, p_n = 0.000006751027716$ |
257 23 Homework Statement Consider object going through circular motion with radial acceleration = ##2m/s## and radius given by ##\frac{4}{2t+2}## Find arc length of object swept through the first two seconds. Homework Equations --
My working:
##s=\int v## ##v= \sqrt{\frac{a_{c}}{r}}=\sqrt{\frac{a_{c}}{\frac{4}{2t+2}}}## ##s= \int_{0}^{2} \sqrt{\frac{2}{\frac{4}{2t+2}}}## My final answer seems to be wrong. Any ideas? Cheers
##s=\int v##
##v= \sqrt{\frac{a_{c}}{r}}=\sqrt{\frac{a_{c}}{\frac{4}{2t+2}}}##
##s= \int_{0}^{2} \sqrt{\frac{2}{\frac{4}{2t+2}}}##
My final answer seems to be wrong. Any ideas? Cheers |
You might want to have a look at the overlap-add (http://en.wikipedia.org/wiki/Overlap%E2%80%93add_method) and overlap-save (http://en.wikipedia.org/wiki/Overlap%E2%80%93save_method) methods. But, if all you are trying to do is additive synthesis, you don't absolutely have to use the IFFT to generate your signal.
You can set up a bank of $M \in {[1 \ldots N_{Osc}]}$ elementary oscilators of the form $x_m(n)=\alpha_m \times sin(\frac{2 \times \pi \times f_m \times n}{Fs}+\phi_m) $ where $m \in M, N_{Osc} \in \mathbb{N}$, $\alpha_m, \phi_m, F_s, f_m \in \mathbb{R}$ and final output $y(n) = \sum_{m \in M}{x_m(n)}$.
In this case, $\alpha_m, \phi_m$ are amplitude and phase coefficients for each oscillator at frequency $f_m$ and to make them variable in time, all you have to do is turn them to $\alpha_m(n), \phi_m(n)$, i.e. to make them dependent on time sample $n$. This can be an absolute definition, by actually defining a varying waveform or by using some form of interpolation to provide initial and final values for these quantities and let the intermediate values be calculated automatically.
This would work both for generating one sample at a time or a frame of samples at a time (by iteration) and would not have any problems with continuity. |
It would probably be useful before reading this and the next section to review Chapters 11 and 12.
Figure XVII.12 shows the same system as figure XVII.2, except that, instead of being left to vibrate on its own, the second mass is subject to a periodic force \( F\ =\ \hat{F}\sin\omega t\). For the time being, we’ll suppose that there is no damping. Either way, it is not a conservative force, and Lagrange’s equation will be used in the form of Equation 13.4.12. As in Section 17.2, the kinetic energy is
\[ T =\frac{1}{2}m_{1}\dot{x}_{1}^{2}\ +\ \frac{1}{2}m_{2}\dot{x}_{2}^{2}\label{17.12.1}\]
Lagrange’s equations are
\[ \frac{d}{dt}\frac{\partial T}{\partial \dot{x}_{1}}\ -\ \frac{\partial T}{\partial x_{1}}\ =\ P_{1} \label{17.12.2}\]
and
\[ \frac{d}{dt}\frac{\partial T}{\partial \dot{x}_{2}}\ -\ \frac{\partial T}{\partial x_{2}}\ =\ P_{2}. \label{17.12.3}\]
We have to identify the generalized forces \( P_{1}\) and \( P_{2}\).
In the nonequilibrium position, the extension of the left hand spring is \( x_{1}\) and so the tension in that spring is \( f_{1}\ =\ k_{1}x_{1}\). The extension of the right hand spring is \( x_{2}\ -\ x_{2}\) and so the tension in that spring is \( f_{2}\ =\ k_{2}(x_{2}-x_{1})\). If \( x_{1}\) were to increase by \( \delta x_{1}\), the work done on \( m_{1}\) would be \( (f_{2}-f_{1})\delta x_{1}\) and therefore the generalized force associated with the coordinate \( x_{1}\) is \( P_{1}\ =\ k_{2}(x_{2}-x_{1})-k_{1}x_{1}\). If \( x_{2}\) were to increase by \( \delta x_{2}\), the work done on \( m_{2}\) would be \( (F-f_{2})\delta x_{2}\) and therefore the generalized force associated with the coordinate \( x_{2}\) is \( P_{2}=\hat{F}\sin\omega t-k_{2}(x_{2}-x_{1})\). The lagrangian equations of motion therefore become
\[ m_{1}\ddot{x}_{1}\ +\ (k_{1}+k_{2})x_{1}\ -k_{2}x_{2}\ =\ 0 \label{17.12.4}\]
and
\[ m_{2}\ddot{x}_{2}\ +\ k_{2}(x_{2}-x_{1})\ =\ \hat{F}\sin\omega t. \label{17.12.5}\]
Seek solutions of the form \( \ddot{x}_{1}=-\omega^{2}x_{1}\) and \( \ddot{x}_{2}=-\omega^{2}x_{2}\). The equations become
\[ (k_{1}+k_{2}-m_{1}\omega^{2})x_{1}\ -\ k_{2}x_{2}\ =\ 0 \label{17.12.6}\]
and
\[ -k_{2}x_{1}\ +\ (k_{2}\ -\ m_{2}\omega^{2})x_{2}\ =\ \hat{F}\sin\omega t. \label{17.12.7}\]
We do
not, of course, now equate the determinants of the coefficients to zero (why not?!), but we can solve these equations to obtain
\[ x_{1}\ =\ \frac{k_{2}\hat{F}\sin\omega t}{(k_{1}+k_{2}-m_{1}\omega^{2})(k_{2}-m_{2}\omega^{2})-k_{2}^{2}} \label{17.12.8}\]
and
\[ x_{2}\ =\ \frac{(k_{1}+k_{2}-m_{1}\omega^{2})\hat{F}\sin\omega t}{(k_{1}+k_{2}-m_{1}\omega^{2})(k_{2}-m_{2}\omega^{2})-k_{2}^{2}}. \label{17.2.9}\]
The amplitudes of these motions (and how they vary with the forcing frequency \( \omega\)) are
\[ \hat{x}_{1}\ =\ \frac{k_{2}\hat{F}}{m_{1}m_{2}\omega^{4}\ -\ (m_{1}k_{2}+m_{2}k_{1}+m_{2}k_{2})\omega^{2}+k_{1}k_{2}} \label{17.12.10}\]
and
\[ \hat{x}_{2}\ =\ \frac{(k_{1}+k_{2}-m_{1}\omega^{2})\hat{F}}{m_{1}m_{2}\omega^{4}\ -\ (m_{1}k_{2}+m_{2}k_{1}+m_{2}k_{2})\omega^{2}+k_{1}k_{2}} \label{17.12.11}\]
where I have re-written the denominators in the form of a quadratic expression in \( \omega^{2}\).
For illustration I draw, in figure XVII.13, the amplitudes of the motion of \( m_{1}\)(continuous curve, in black) and of \( m_{2}\)(dashed curve, in blue) for the following data:
\( \hat{F}=1,\ k_{1}=k_{2}=1,\ m_{1}=3,\ m_{2}=2,\)
when the equations become
\[ \hat{x}_{1}=\frac{1}{6\omega^{4}-7\omega^{2}+1}=\frac{1}{(6\omega^{2}-1)(\omega^{2}-1)} \label{17.12.12}\]
and
\[ \hat{x}_{1}=\frac{2-3\omega^{2}}{6\omega^{4}-7\omega^{2}+1}=\frac{2-3\omega^{2}}{(6\omega^{2}-1)(\omega^{2}-1)} \label{17.12.13}\]
Where the amplitude is negative, the oscillations are out of phase with the force \( F\). The amplitudes go to infinity (remember we are assuming here zero damping) at the two frequencies where the denominators of Equations \( \ref{17.12.10}\) and \( \ref{17.12.11}\) are zero. The amplitude of the motion of \( m_{2}\) is zero when the numerator of Equation \( \ref{17.12.11}\) is zero. This is at an angular frequency of \( \sqrt{\frac{(k_{1}+k_{2})}{m_{1}}}\), which is just the angular frequency of the motion of \( m_{1}\) held by the two springs between two fixed points. In our numerical example, this is \( \omega\ =\ \sqrt{\frac{2}{3}}\ =\ 0.8165\). This is an example of
antiresonance. |
Question
I have the feeling gas cannot have an equivalent of Ohm's law, tying pressure and throughput via some kind of fluid resistance constant depending on the geometry of the obstacle considered. Certainly because gas can be compressible.
However I need a very rough estimate (not a number from experience, a first order model/formula) of the air throughput out of an obstacle of arbitrary geometry of which I know the hollow cross sectional area.
I have done my research but all I can find is Poiseuille's law or pipe flow formulas which apply to very long cylinders (what about if I'm looking at the "resistance" of a complex obstacle?)... And the venturi equation: $$p_i-p_o=1/2(\rho_o v_o²-\rho_i v_i²)$$ With (conservation of mass flow) $$\dot{m}=\rho_i A_i v_i=\rho_o A_o v_o$$ Which gives $$\dot{m}=\sqrt{\frac{2(p_i-p_o)}{\frac{1}{\rho_o A_o^2}-\frac{1}{\rho_i A_i^2}}}$$ Knowing that $$\rho=\frac{m}{V}=\frac{\frac{PVM}{RT}}{V}=\frac{PM}{RT}$$ (M is the molar mass of the gas, R the perfect gas constant)
Is it correct? It's not linear like Ohm's law, but it is a relationship.
Application
I would like this question to be generic, but as an application/illustration, attached is a simplified 3D model of the orifice - the scale is 15mm. I know the area of the side triangles and the front rectangle out of the conduit (top), and I'm wondering what the mass flow is through it. |
I'm trying to figure out some kind of immunization using a factor model I developed for interest rates. Here is the basic problem. Let's say that we have a bond portfolio containing $N$ bonds with weights $x_i$ and one liability. Call the present value of the liability $P_L$. I'll suppose that the bond prices are given by $$P_i=\Sigma_{t=1}^T F_{it}e^{-r_tt}$$ where $F_{it}$ is the future payoff of bond $i$ and $r_t$ is the interest rate at time $t$. Now the factor model for the term structure of the interest rates can be written $$\Delta r_t=\Sigma_{j=1}^k \beta_{jt} \Delta f_j +\epsilon_t$$ for $k$ independent factors and standard normal error term $\epsilon_t$. I'm assuming I have small but not necessarily parallel-shifts in the term structure, I want a first order condition for factor immunization. Can someone help?
The present value is $$ P_i= \sum_{t=1}^T F_{i,t} \exp(-r_t t), $$ what happens if rates change to $r_t + \Delta r_t$ then the new price is $$ P_i^{new} = \sum_{t=1}^T F_{i,t} \exp(-(r_t+\Delta r_t) t). $$ by the exponential series $\exp(x)\approx 1 + x$ we can write $$ P_i^{new} - P_i =: \Delta P_i \approx -\sum_{t=1}^T F_{i,t} \Delta r_t t. $$ Observing the shifts in these rates we have some kind of key-rate duration setting.
If you can decompose your liablity $P_L$ in the is way too (denote the cash flows by $F_t$, then you can try to find quantities $Q_i$ such that $$ |\sum_{t=1}^T (Q_i F_{i,t} - F_t) \Delta r_t t| -> Min $$
Usually the moviments in the yield curve are decomposed by level, slope and curvature or parallel, steepness and bends in the term structure. These 3 factors can explain over 95% 98% of the total variance. Sometimes a second curvature point is used and the explanation reaches 99%. Instead of approximate it numerically why won't you fit the Svensson or Nelson Siegel model and use it as your factors? |
Defining a subgroup of elliptic curves with specific characteristics
Hey,
is there a way, to define a subgroup of an elliptic curve with two or more characteristics? I would like to take an elliptic curve over a finite field of order p and $p^4$, define the r-torsion subgroup (where $r$ is a prime, too) and reduce those to the set of points, which also lays in the Frobenius-eigenspace.
For example:
p= 13r=5R=GF(p)_.<x> = PolynomialRing(R)R4.<x> = R.extension(x^4 - 2, 'x')_.<y> = PolynomialRing(R)b= x^-1E = EllipticCurve(R, [1,0]) # y^2 = x^3+xE4 = EllipticCurve(R4, [b,0])
Well, it is easy to find a point on $Q\in E4$, such that $r*Q = (0:1:0)$, use
Q=ZZ(E4.order()/r *Q
, but checking, if $( x(Q)^p, y(Q)^p )=\pi(Q) = pQ$ is hard. I only need one point of that group at all, but my $p$ is even larger (~340 bits), so brute-forcing would be an option, if I could start it 6-12 month ago :)
Furthermore, if I concider to evaluate the secant or tangent on E and let me return a point on that curve, it will have projective coordinates, with $z(P)\neq1$. Shall I apply $\pi$ to all three coordinates? |
Quasi harmonic approximation¶
Using phonopy results of thermal properties, thermal expansion andheat capacity at constant pressure can be calculated under thequasi-harmonic approximation.
phonopy-qha is the script to runfitting and calculation to perform it. Mind that at leave 5 volumepoints are needed to run
phonopy-qha for fitting.
An example of the usage for
example/Si-QHA is asfollows.
To watch selected plots:
phonopy-qha -p e-v.dat thermal_properties.yaml-{-{5..1},{0..5}}
Without plots:
phonopy-qha e-v.dat thermal_properties.yaml-{-{5..1},{0..5}}
The first argument is the filename of volume-energy data (in the aboveexpample,
e-v.dat). The volumes and energies are given in\(\text{Angstrom}^3\) and eV, respectively. Theses energies areonly dependent on volume but not on temperature unless using
--efeoption. Therefore in the simplest case, these are taken as theelectronic total energies at 0K. An example of the volume-energy fileis:
# cell volume energy of cell other than phonon 140.030000 -42.132246 144.500000 -42.600974 149.060000 -42.949142 153.720000 -43.188162 158.470000 -43.326751 163.320000 -43.375124 168.270000 -43.339884 173.320000 -43.230619 178.470000 -43.054343 183.720000 -42.817825 189.070000 -42.527932
Lines starting with
# are ignored.
The following arguments of
phonopy-qha are the filenames of
thermal_properties.yaml’s calculated at the volumes given in thevolume-energy file. These filenames have to be ordered in the sameorder as the volumes written in the volume-energy file. Since thevolume v.s. free energy fitting is done at each temperature given in
thermal_properties.yaml, all
thermal_properties.yaml’s have tobe calculated in the same temperature ranges and with the sametemperature step.
phonopy-qha can calculate thermal properties atconstant pressure up to the temperature point that is one point lessthan that in
thermal_properties.yaml because of the numericaldifferentiation with respect to temperature points. Therefore
thermal_properties.yaml has to be calculated up to highertemperatures than that expected by
phonopy-qha.
Another example for Aluminum is found in the
example/Al-QHA directory.
If the condition under puressure is expected, \(PV\) terms may beincluded in the energies, or equivalent effect is applied using
--pressure option.
Experimentally, temperature dependent energies are supported by
--efe option. The usage is written athttps://github.com/atztogo/phonopy/blob/develop/example/Cu-QHA/README.
-h¶
Show help. The available options are shown. Without any option, the results are saved into text files in simple data format.
--tmax¶
The maximum temperature calculated is specified. This temperature hasto be lower than the maximum temperature calculated in
thermal_properties.yaml to let at least one temperature pointsfewer. The default value is
--tmax=1000.
--pressure¶
Pressure is specified in GPa. This corresponds to the \(pV\) term described in the following section Thermal properties in (T, p) space calculated under QHA. Note that bulk modulus obtained with this option than 0 GPa is incorrect.
-b¶
Fitting volume-energy data to an EOS, and show bulk modulus (without considering phonons). This is made by:
% phonopy-qha -b e-v.dat
--eos¶
EOS is chosen among
vinet,
birch_murnaghan, and
murnaghan. The default EOS is
vinet.
% phonopy-qha --eos='birch_murnaghan' -b e-v.dat
-p¶
The fitting results, volume-temperature relation, and thermal expansion coefficient are plotted on the display.
-s¶
The calculated values are written into files.
--sparse¶
This is used with
-s or
-p to thin out the number of plots ofthe fitting results at temperatures. For example with
--sparse=10,1 in 10 temperature curves is only plotted.
--efe¶
Experimental
Temperature dependent energies other than phonon free energy are included with this option. This is used such as:
% phonopy-qha -p --tmax=1300 --efe fe-v.dat e-v.dat thermal_properties.yaml-{00..10}
The temperature dependent energies are stored in
fe-v.dat. Thefile format is:
# volume: 43.08047896 43.97798894 44.87549882 45.77300889 46.67051887 47.56802885 48.46553883 49.36304881 50.26055878 51.15806876 52.05557874# T(K) Free energies 0.0000 -17.27885993 -17.32227490 -17.34336569 -17.34479760 -17.32843604 -17.29673896 -17.25081954 -17.19263337 -17.12356816 -17.04467997 -16.95752155 10.0000 -17.27886659 -17.32228126 -17.34337279 -17.34481060 -17.32844885 -17.29675204 -17.25083261 -17.19264615 -17.12358094 -17.04469309 -16.95753464 20.0000 -17.27887453 -17.32228804 -17.34338499 -17.34482383 -17.32846353 -17.29676491 -17.25084547 -17.19265900 -17.12359399 -17.04470709 -16.95754774...
This file doesn’t contain the information about cell volumes that areobtained from
e-v.dat file though the energy data in
e-v.datare not used. In
fe-v.dat, the lines starting with
# areignored. Rows and columns are the temperature and volume axes. Thefirst column gives temperatures. The following columns give thetemperature dependent energies.The temperature points are expected tobe the same as those in
thermal_properties.yaml at least up to themaximum temperature specified for
phonopy-qha.
An example is given in
example/Cu-QHA. The
fe-v.dat containselectronic free energy calculated following, e.g., Eqs. (11) and (12)in the paper by Wolverton and Zunger, Phys. Rev. B,
52, 8813(1994) (of course this paper is not the first one that showed theseequations):
with
and
where \(g\) is 1 or 2 for collinear spin polarized and non-spinpolarized systems, respectively. For VASP, a script to create
fe-v.dat and
e-v.dat by these equations is prepared as
phonopy-vasp-efe, which is used as:
% phonopy-vasp-efe --tmax=1500 vasprun.xml-{00..10}
where
vasprun.xml-{00..10} have to be computed for the same unitcells as those used for
thermal_properties.yaml. When
phonopywas run with
PRIMITIVE_AXES or
--pa option, the unit cells forcomputing electronic eigenvalues have to be carefully chosen to agreewith those after applying
PRIMITIVE_AXES, or energies are scaled aposteriori.
The physical units of V and T are \(\text{Angstrom}^3\) and K, respectively. The unit of eV for Helmholtz and Gibbs energies, J/K/mol for \(C_V\) and entropy, GPa for for bulk modulus and pressure are used.
Bulk modulus \(B_T\) (GPa) vs \(T\) (
bulk_modulus-temperature.*)
Gibbs free energy \(G\) (eV) vs \(T\) (
gibbs-temperature.*)
Heat capacity at constant pressure \(C_p\) (J/K/mol) vs \(T\) computed by \(-T\frac{\partial^2 G}{\partial T^2}\) from three \(G(T)\) points (
Cp-temperature.*)
Heat capacity at constant puressure \(C_p\) (J/K/mol) vs \(T\) computed by polynomial fittings of \(C_V(V)\) (
Cv-volume.dat) and \(S(V)\) (
entropy-volume.dat) for \(\partial S/\partial V\) (
dsdv-temperature.dat) and numerical differentiation of \(\partial V/\partial T\), e.g., see Eq.(5) of PRB
81, 174301 by Togo et al.(
Cp-temperature_polyfit.*). This may give smoother \(C_p\) than that from \(-T\frac{\partial^2 G}{\partial T^2}\).
Volumetric thermal expansion coefficient \(\beta\) vs \(T\) computed by numerical differentiation (
thermal_expansion.*)
Volume vs \(T\) (
volume-temperature.*)
Thermodynamics Grüneisen parameter \(\gamma = V\beta B_T/C_V\) (no unit) vs \(T\) (
gruneisen-temperature.dat)
Helmholtz free energy (eV) vs volume (
helmholtz-volume.*). When
--pressureoption is specified, energy offset of \(pV\) is added. See also the following section (Thermal properties in (T, p) space calculated under QHA).
Here the word ‘quasi-harmonic approximation’ is used for an approximation that introduces volume dependence of phonon frequencies as a part of anharmonic effect.
A part of temperature effect can be included into total energy ofelectronic structure through phonon (Helmholtz) free energy atconstant volume. But what we want to know is thermal properties atconstant pressure. We need some transformation from function of
V tofunction of p. Gibbs free energy is defined at a constant pressure bythe transformation:
where
means to find unique minimum value in the brackets by changing volume. Since volume dependencies of energies in electronic and phonon structures are different, volume giving the minimum value of the energy function in the square brackets shifts from the value calculated only from electronic structure even at 0 K. By increasing temperature, the volume dependence of phonon free energy changes, then the equilibrium volume at temperatures changes. This is considered as thermal expansion under this approximation.
phonopy-qha collects the values at volumes and transforms into thethermal properties at constant pressure. |
Is it possible to create a Normal Map from a Texture, not a mesh? If so, how?
As I'm working on an Eye, and have already made the textures, I was wondering if it was possible to create Normal Maps out of those Textures.
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While normal maps from textures can also be created within blender, there is another FLOSS software called AwesomeBump. It's not quite as user friendly as CrazyBump but it gets the job done.
If you are on an AMD GPU system you will have to use the stable version 3.14 for now.
First thing normal map generators do is to converts the input image into a grayscale image, even if it is colored, that is because color information is practically useless in the generation process, the algorithm used to convert the RGB colors into single values varies from a generator to another, so it is recommended that you convert your input image into grayscale before using the normal map generator, the reason will become clear in the next section.
Imagine the image as a grid of vertices, each vertex represents a pixel, the z location (height) of each vertex is equal to its value (Which is just a scalar not an RGB vector; because it is a grayscale image, remember). So we realize that the image is just a 2D function that associates a value to each point in space, except it is not a continuous function, since it is only defined on some discrete points in space. This surface we just described has
normals at each vertex (just like your mesh has normals), and those normals is what a normal map stores, so the generator's job is to simply compute the normals of the surface and encode it in an image. It can be proven that the normals' components relates to partial derivatives of the surface using the equation:
\begin{equation} \vec{N_x} = \frac{-\frac{\partial F}{\partial x}}{\sqrt{{\frac{\partial F}{\partial x}}^2 + {\frac{\partial F}{\partial y}}^2 +1 }} \\ \vec{N_y} = \frac{-\frac{\partial F}{\partial y}}{\sqrt{{\frac{\partial F}{\partial x}}^2 + {\frac{\partial F}{\partial y}}^2 +1 }}\\ \vec{N_z} = \frac{1}{\sqrt{{\frac{\partial F}{\partial x}}^2 + {\frac{\partial F}{\partial y}}^2 +1 } } \end{equation}
Where $F$ is the function that represents the surface and $N$ is the unit normal vector. Partial derivatives describes the steepness and direction of the surface and they can be computed using what is know as
Symmetric Derivatives which I explained in more details in my answerhere.
Assuming we have the partial derivatives, we can compute the normal vector components using the previous equation. A normal map stores the $x$, $y$ and $z$ components of the normal vector into the $R$, $G$ and $B$ channels of the image respectively. If we attempted to do that, we will encounter a problem, the image can only store positive values that are less than one (Assuming a standard image format is used), the normal vector components will always be smaller than one because it is a unit vector, however the vector components can be negative denoting directions opposite to the fundamental vectors of the space, so we have to find a way to encode these data which belongs to the interval $[-1, 1]$ to the interval $[0, 1]$ which is the possible values for a pixel. This is a simple remapping task that can be performed using the equation:
$$ \begin{equation} R = \frac{(\vec{N}_x+1)}{2}\\G = \frac{(\vec{N}_y+1)}{2}\\\label{1} \end{equation} $$
Where $R$ and $G$ are the red and green channels respectively, notice that we didn't perform the same remapping to the $z$ component because it is always positive, think about the surface that represents the image again, its normals will always be facing the positive direction of the $z$ axis. By using the previous equation, we get the RGB values for the normal map, so we are ready to do the implementation.
We can make a node group that computes both the $\frac{\partial F}{\partial x}$ and $\frac{\partial F}{\partial y}$ partial derivatives given an input image. The implementation is as simple as a translation and a subtraction:
Refer to the article for more information, the rest of the implementation is as easy as applying the two equations above and creating an image out of the individual channels.
An example result
Especially if you are already using Photoshop in your workflow, the Nvidia Texture Tools are simple to use and pack some good features. |
This chapter of our tutorial is about legends. Legends are the classical stories from ancient Greece or other places which are usually devoured by adolescents. They have to read it and this is where the original meaning comes from. The word
legend stems from Latin and it means in Latin "to be read". So we can say legends are the things in a graph or plot which have to be read to understand the plot. It gives us valuable information about the visualized data.
Before legends have been used in mathematical graphs, they have been used in maps. Legends - as they are found in maps - describe the pictorial language or symbology of the map. Legends are used in line graphs to explain the function or the values underlying the different lines of the graph.
We will demonstrate in the following simple example how we can place a legend on a graph. A legend contains one or more entries. Every entry consists of a key and a label.
The pyplot function
legend(*args, **kwargs)
places a legend on the axes.
All we have to do to create a legend for lines, which already exist on the axes, is to simply call the function "legend" with an iterable of strings, one for each legend item.
# the module polynomials can be downloaded from # https://www.python-course.eu/examples/polynomials.py from polynomials import Polynomial import numpy as np import matplotlib.pyplot as plt p = Polynomial(-0.8, 2.3, 0.5, 1, 0.2) p_der = p.derivative() fig, ax = plt.subplots() X = np.linspace(-2, 3, 50, endpoint=True) F = p(X) F_derivative = p_der(X) ax.plot(X, F) ax.plot(X, F_derivative) ax.legend(['p', 'derivation of p'])This gets us the following output: <matplotlib.legend.Legend at 0x7fd2efcad908>
If we add a label to the plot function, the value will be used as the label in the legend command. There is anolther argument that we can add to the legend function: We can define the location of the legend inside of the axes plot with the parameter
loc:
If we add a label to the plot function, the values will be used in the legend command:
from polynomials import Polynomial import numpy as np import matplotlib.pyplot as plt p = Polynomial(-0.8, 2.3, 0.5, 1, 0.2) p_der = p.derivative() fig, ax = plt.subplots() X = np.linspace(-2, 3, 50, endpoint=True) F = p(X) F_derivative = p_der(X) ax.plot(X, F, label="p") ax.plot(X, F_derivative, label="derivation of p") ax.legend(loc='upper left')The previous Python code returned the following output: <matplotlib.legend.Legend at 0x7fd2efc15cf8>
It might be even more interesting to see the actual function in mathematical notation in our legend. Our polynomial class is capable of printing the function in LaTeX notation.
print(p) print(p_der) -0.8x^4 + 2.3x^3 + 0.5x^2 + 1x^1 + 0.2 -3.2x^3 + 6.8999999999999995x^2 + 1.0x^1 + 1
We can also use LaTeX in our labels, if we include it in '$' signs.
from polynomials import Polynomial import numpy as np import matplotlib.pyplot as plt p = Polynomial(2, 3, -4, 6) p_der = p.derivative() fig, ax = plt.subplots() X = np.linspace(-2, 3, 50, endpoint=True) F = p(X) F_derivative = p_der(X) ax.plot(X, F, label="$" + str(p) + "$") ax.plot(X, F_derivative, label="$" + str(p_der) + "$") ax.legend(loc='upper left')After having executed the Python code above we received the following result: <matplotlib.legend.Legend at 0x7fd2efb99a58>
In many cases we don't know what the result may look like before you plot it. It could be for example, that the legend will overshadow an important part of the lines. If you don't know what the data may look like, it may be best to use 'best' as the argument for loc. Matplotlib will automatically try to find the best possible location for the legend:
from polynomials import Polynomial import numpy as np import matplotlib.pyplot as plt p = Polynomial(2, -1, -5, -6) p_der = p.derivative() print(p_der) fig, ax = plt.subplots() X = np.linspace(-2, 3, 50, endpoint=True) F = p(X) F_derivative = p_der(X) ax.plot(X, F, label="$" + str(p) + "$") ax.plot(X, F_derivative, label="$" + str(p_der) + "$") ax.legend(loc='best') 6x^2 - 2x^1 - 5After having executed the Python code above we received the following output: <matplotlib.legend.Legend at 0x7fd2ef370f60>
We will go back to trigonometric functions in the following examples. These examples show that
loc='best'can work pretty well: import numpy as np import matplotlib.pyplot as plt X = np.linspace(-2 * np.pi, 2 * np.pi, 70, endpoint=True) F1 = np.sin(0.5*X) F2 = -3 * np.cos(0.8*X) plt.xticks( [-6.28, -3.14, 3.14, 6.28], [r'$-2\pi$', r'$-\pi$', r'$+\pi$', r'$+2\pi$']) plt.yticks([-3, -1, 0, +1, 3]) plt.plot(X, F1, label="$sin(0.5x)$") plt.plot(X, F2, label="$-3 cos(0.8x)$") plt.legend(loc='best') plt.show() import numpy as np import matplotlib.pyplot as plt X = np.linspace(-2 * np.pi, 2 * np.pi, 70, endpoint=True) F1 = np.sin(0.5*X) F2 = 3 * np.cos(0.8*X) plt.xticks( [-6.28, -3.14, 3.14, 6.28], [r'$-2\pi$', r'$-\pi$', r'$+\pi$', r'$+2\pi$']) plt.yticks([-3, -1, 0, +1, 3]) plt.plot(X, F1, label="$sin(0.5x)$") plt.plot(X, F2, label="$3 cos(0.8x)$") plt.legend(loc='best') plt.show()
The visualizations of function plots often makes annotations necessary. This means we draw the readers attentions to important points and areas of the plot. To this purpose we use texts, labels and arrows. We have already used axis labels and titles for this purpose, but these are 'annotations' for the whole plot. We can easily annotate points inside the axis or on the graph with the
annotate method of an axes object. In an annotation, there are two points to consider: the location being annotated represented by the argument xy and the location of the text xytext. Both of these arguments are (x,y) tuples.
We demonstrate how easy it is in matplotlib to to annotate plots in matplotlib with the annotate method. We will annotate the local maximum and the local minimum of a function. In its simplest form
annotate method needs two arguments
annotate(s, xy), where
s is the text string for the annotation and
xx is the position of the point to be annotated:
from polynomials import Polynomial import numpy as np import matplotlib.pyplot as plt p = Polynomial(1, 0, -12, 0) p_der = p.derivative() fig, ax = plt.subplots() X = np.arange(-5, 5, 0.1) F = p(X) F_derivative = p_der(X) ax.grid() maximum = (-2, p(-2)) minimum = (2, p(2)) ax.annotate("local maximum", maximum) ax.annotate("local minimum", minimum) ax.plot(X, F, label="p") ax.plot(X, F_derivative, label="derivation of p") ax.legend(loc='best') plt.show()
If you are not satisfied with the automatic positioning of the text, you can assign a tuple with a position for the text to the keyword parameter xytext:
from polynomials import Polynomial import numpy as np import matplotlib.pyplot as plt p = Polynomial(1, 0, -12, 0) p_der = p.derivative() fig, ax = plt.subplots() X = np.arange(-5, 5, 0.1) F = p(X) F_derivative = p_der(X) ax.grid() ax.annotate("local maximum", xy=(-2, p(-2)), xytext=(-1, p(-2)+35), arrowprops=dict(facecolor='orange')) ax.annotate("local minimum", xy=(2, p(2)), xytext=(-2, p(2)-40), arrowprops=dict(facecolor='orange', shrink=0.05)) ax.annotate("inflection point", xy=(0, p(0)), xytext=(-3, -30), arrowprops=dict(facecolor='orange', shrink=0.05)) ax.plot(X, F, label="p") ax.plot(X, F_derivative, label="derivation of p") ax.legend(loc='best') plt.show()
We have to provide some informations to the parameters of annotate, we have used in our previous example.
Parameter Meaning xy coordinates of the arrow tip xytext coordinates of the text location
The xy and the xytext locations of our example are in data coordinates. There are other coordinate systems available we can choose. The coordinate system of xy and xytext can be specified string values assigned to xycoords and textcoords. The default value is 'data':
String Value Coordinate System figure points points from the lower left corner of the figure figure pixels pixels from the lower left corner of the figure figure fraction 0,0 is lower left of figure and 1,1 is upper right axes points points from lower left corner of axes axes pixels pixels from lower left corner of axes axes fraction 0,0 is lower left of axes and 1,1 is upper right data use the axes data coordinate system
Additionally, we can also specify the properties of the arrow. To do so, we have to provide a dictionary of arrow properties to the parameter arrowprops:
arrowprops key description width the width of the arrow in points headlength The length of the arrow head in points headwidth the width of the base of the arrow head in points shrink move the tip and base some percent away from the annotated point and text **kwargs any key for matplotlib.patches.Polygon, e.g., facecolor
Of course, the sinus function has "boring" and interesting values. Let's assume that you are especially interested in the value of $3 * sin(3 * pi/4)$.
import numpy as np print(3 * np.sin(3 * np.pi/4)) 2.121320343559643
The numerical result doesn't look special, but if we do a symbolic calculation for the above expression we get $\frac{3}{\sqrt{2}}$. Now we want to label this point on the graph. We can do this with the annotate function. We want to annotate our graph with this point.
import numpy as np import matplotlib.pyplot as plt X = np.linspace(-2 * np.pi, 2 * np.pi, 100) F1 = np.sin(X) F2 = 3 * np.sin(X) fig, ax = plt.subplots() ax.xaxis.set_major_locator(plt.MultipleLocator(np.pi / 2)) #plt.xticks(np.arange(-2 * np.pi, 2.5 * np.pi, np.pi / 2)) ax.set_xticklabels([r'$-2\pi$', r'$-\frac{3\pi}{2}$', r'$-\pi$', r'$-\frac{\pi}{2}$', 0, r'$\frac{\pi}{2}$', r'$+\pi$', r'$\frac{3\pi}{2}$', r'$+2\pi$']) ax.plot(X, F1, label="$sin(x)$") ax.plot(X, F2, label="$3 sin(x)$") ax.legend(loc='lower left') x = 3 * np.pi / 4 # Plot vertical line: ax.plot([x, x],[-3, 3 * np.sin(x)], color ='blue', linewidth=2.5, linestyle="--") # Print the blue dot: ax.scatter([x,],[3 * np.sin(x),], 50, color ='blue') text_x, text_y = (3.5, 2.2) ax.annotate(r'$3\sin(\frac{3\pi}{4})=\frac{3}{\sqrt{2}}$', xy=(x, 3 * np.sin(x)), xytext=(text_x, text_y), arrowprops=dict(facecolor='orange', shrink=0.05), fontsize=12) plt.show()
There is anothe example, in which we play around with the arrows and annoate the extrema.
import numpy as np import matplotlib.pyplot as plt X = np.linspace(-4.1, 3.1, 150, endpoint=True) F = X**5 + 3*X**4 - 11*X**3 - 27*X**2 + 10*X + 24 fig, ax = plt.subplots() ax.plot(X, F) minimum1 = -1.5264814, -7.051996717492152 minimum2 = 2.3123415793720303, -81.36889464201387 ax.annotate("minima", xy=minimum1, xytext=(-1.5, -50), arrowprops=dict(arrowstyle="->", connectionstyle="angle3,angleA=0,angleB=-90")) ax.annotate(" ", xy=minimum2, xytext=(-0.7, -50), arrowprops=dict(arrowstyle="->", connectionstyle="angle3,angleA=0,angleB=-90")) maximum1 = -3.35475845886632, 56.963107876630595 maximum2 = .16889828232847673, 24.868343482875485 ax.annotate("maxima", xy=maximum1, xytext=(-1.5, 30), arrowprops=dict(arrowstyle="->", connectionstyle="angle3,angleA=0,angleB=-90")) ax.annotate(" ", xy=maximum2, xytext=(-0.6, 30), arrowprops=dict(arrowstyle="->", connectionstyle="angle3,angleA=0,angleB=-90")) zeroes = -4, -2, -1, 1, 3 for zero in zeroes: zero = zero, 0 ax.annotate("Zeroes", xy=zero, color="orange", bbox=dict(boxstyle="round", fc="none", ec="green"), xytext=(1, 40), arrowprops=dict(arrowstyle="->", color="orange", connectionstyle="angle3,angleA=0,angleB=-90")) plt.show()
The following program visualizes the various arrowstyles:
import matplotlib.pyplot as plt def demo_con_style(ax, connectionstyle): x1, y1 = 0.3, 0.2 x2, y2 = 0.8, 0.6 ax.plot([x1, x2], [y1, y2], ".") ax.annotate("", xy=(x1, y1), xycoords='data', xytext=(x2, y2), textcoords='data', arrowprops=dict(arrowstyle="->", color="0.5", shrinkA=5, shrinkB=5, patchA=None, patchB=None, connectionstyle=connectionstyle)) ax.text(.05, .95, connectionstyle.replace(",", ",\n"), transform=ax.transAxes, ha="left", va="top") fig, axs = plt.subplots(3, 5, figsize=(8, 4.8)) demo_con_style(axs[0, 0], "angle3,angleA=90,angleB=0") demo_con_style(axs[1, 0], "angle3,angleA=0,angleB=90") demo_con_style(axs[0, 1], "arc3,rad=0.") demo_con_style(axs[1, 1], "arc3,rad=0.3") demo_con_style(axs[2, 1], "arc3,rad=-0.3") demo_con_style(axs[0, 2], "angle,angleA=-90,angleB=180,rad=0") demo_con_style(axs[1, 2], "angle,angleA=-90,angleB=180,rad=5") demo_con_style(axs[2, 2], "angle,angleA=-90,angleB=10,rad=5") demo_con_style(axs[0, 3], "arc,angleA=-90,angleB=0,armA=30,armB=30,rad=0") demo_con_style(axs[1, 3], "arc,angleA=-90,angleB=0,armA=30,armB=30,rad=5") demo_con_style(axs[2, 3], "arc,angleA=-90,angleB=0,armA=0,armB=40,rad=0") demo_con_style(axs[0, 4], "bar,fraction=0.3") demo_con_style(axs[1, 4], "bar,fraction=-0.3") demo_con_style(axs[2, 4], "bar,angle=180,fraction=-0.2") for ax in axs.flat: ax.set(xlim=(0, 1), ylim=(0, 1), xticks=[], yticks=[], aspect=1) fig.tight_layout(pad=0.2) plt.show() |
Sides of a parallelogram: \(a\), \(b\)
Diagonals of a parallelogram: \({d_1},\) \({d_2}\) Consecutive angles: \(\alpha\), \(\beta\) Angle between the diagonals: \(\varphi\)
Diagonals of a parallelogram: \({d_1},\) \({d_2}\)
Consecutive angles: \(\alpha\), \(\beta\)
Angle between the diagonals: \(\varphi\)
Perimeter: \(P\)
Area: \(S\) Altitude: \(h\)
Area: \(S\)
Altitude: \(h\)
A parallelogram is a quadrilateral with two pairs of parallel sides. In a paralellogram, the opposite sides are equal and the opposite angles are equal. The sum of the consecutive angles of a parallelogram is equal to \(180^\circ:\) \(\alpha + \beta = 180^\circ\) A parallelogram in which all four angles are right angles is called a rectangle. A parallelogram in which all four sides are equal is called a rhombus. Relationship between the sides and the diagonals of a parallelogram \(d_1^2 + d_2^2 =\) \( 2\left( {{a^2} + {b^2}} \right)\) Altitude of a parallelogram \(h = b\sin \alpha =\) \( b\sin \beta\) Perimeter of a parallelogram \(P = 2\left({a + b} \right)\) Area of a parallelogram \(S = ah =\) \(ab\sin \alpha =\) \({\large\frac{1}{2}\normalsize} {d_1}{d_2}\sin \varphi \) |
Calculation of a triple integral in Cartesian coordinates can be reduced to the consequent calculation of three integrals of one variable.
Consider the case when a three dimensional region \(U\) is a type I region, i.e. any straight line parallel to the \(z\)-axis intersects the boundary of the region \(U\) in no more than \(2\) points. Let the region \(U\) be bounded below by the surface \(z = {z_1}\left( {x,y} \right),\) and above by the surface \(z = {z_2}\left( {x,y} \right)\) (Figure \(1\)).
The projection of the solid \(U\) onto the \(xy\)-plane is the region \(D\) (Figure \(2\)).
We suppose that the functions \({z_1}\left( {x,y} \right)\) and \({z_2}\left( {x,y} \right)\) are continuous in the region \(D.\) Then for any function \(f\left( {x,y,z} \right)\) continuous in the region \(U\) we can write the relationship:
\[
{\iiint\limits_U {f\left( {x,y,z} \right)dV} } = {\iint\limits_D {\left[ {\int\limits_{{z_1}\left( {x,y} \right)}^{{z_2}\left( {x,y} \right)} {f\left( {x,y,z} \right)dz} } \right]dA}} \]
Thus, calculating a triple integral is reduced to calculating a double integral, where the integrand is an one-dimensional integral. In the given case, we need first to calculate the inner integral with respect to the variable \(z,\) and then the double integral with respect to the variables \(x\) and \(y.\)
If \(D\left( {x,y} \right)\) is a type \(I\) region in the \(xy\)-plane (see Iterated Integrals), which is bounded by the lines
\[
{x = a,\;\;x = b,\;\;}\kern-0.3pt {y = {f_1}\left( x \right),\;\;}\kern-0.3pt {y = {f_2}\left( x \right),} \]
where \({f_1}\left( x \right),\) \({f_2}\left( x \right)\) are continuous functions on the intervals \(\left[ {a,b} \right]\) and \({f_1}\left( x \right) \le {f_2}\left( x \right),\) then writing the double integral as an iterated integral, we get
\[
{\iiint\limits_U {f\left( {x,y,z} \right)dV} } = {{\int\limits_a^b {dx} \int\limits_{{f_1}\left( x \right)}^{{f_2}\left( x \right)} {dy} }\kern0pt{ \int\limits_{{z_1}\left( {x,y} \right)}^{{z_2}\left( {x,y} \right)} {f\left( {x,y,z} \right)dz}. }} \]
If \(D\left( {x,y} \right)\) is the region of type \(II\) and bounded by the lines
\[
{y = c,\;\;y = d,\;\;}\kern-0.3pt {x = {\varphi _1}\left( y \right),\;\;}\kern-0.3pt {x = {\varphi _2}\left( y \right),} \]
where the functions \({\varphi _1}\left( y \right),\) \({\varphi _2}\left( y \right)\) are continuous on the interval \(\left[ {c,d} \right]\) so that \({\varphi _1}\left( y \right) \le {\varphi _2}\left( y \right),\) we can rewrite the formula for the triple integral as
\[ {\iiint\limits_U {f\left( {x,y,z} \right)dV} } = {{\int\limits_c^d {dy} \int\limits_{{\varphi _1}\left( y \right)}^{{\varphi _2}\left( y \right)} {dx} }\kern0pt{ \int\limits_{{z_1}\left( {x,y} \right)}^{{z_2}\left( {x,y} \right)} {f\left( {x,y,z} \right)dz} .}} \]
The last two formulas are an application of
Fubini’s theorem to triple integral. They enable us to reduce the computation of triple integrals to iterated integrals.
In the particular case when the region of integration \(U\) is the rectangular box \(\left[ {a,b} \right] \) \(\times \left[ {c,d} \right] \) \(\times \left[ {p,q} \right],\) the triple integral is given by
\[
{\iiint\limits_U {f\left( {x,y,z} \right)dxdydz} } = {\int\limits_a^b {dx} \int\limits_c^d {dy} \int\limits_p^q {f\left( {x,y,z} \right)dz} .} \]
Similar formulas for triple integrals exist for the solid regions \(U\) of type \(II\) or \(III.\) The projection of a solid region of type \(II\) is the region \(D\left( {y,z} \right)\) in the \(yz\)-plane, and, respectively, the projection of a solid region of type \(III\) is the region \(D\left( {x,z} \right)\) in the \(xz\)-plane.
Finally, if a solid region of integration \(U\) is more complex than considered above, we can break the region \(U\) into two or more smaller regions, each of which can be integrated separately.
Solved Problems
Click a problem to see the solution.
Example 1Evaluate the integral Example 2Evaluate the integral Example 3Calculate the triple integral Example 4Express the triple integral \(\iiint\limits_U {dxdydz} \) in terms of iterated integrals in six different ways. The region \(U\) lies in the first octant and is bounded by the cylinder \({x^2} + {z^2} = 4\) and the plane \(y = 3\) (Figure \(7\)). Find the value of the integral. Example 1.Evaluate the integral
Solution.
Applying the Fubini’s theorem, we can calculate the iterated integral starting from the inner one:
\[ {I }={ \int\limits_0^2 {\int\limits_0^z {\int\limits_0^y {xyzdxdydz} } } } = {\int\limits_0^2 {dz} \int\limits_0^z {dy} \int\limits_0^y {xyzdz} } = {\int\limits_0^2 {dz} \int\limits_0^z {dy} \left[ {\left. {\left( {\frac{{{x^2}yz}}{2}} \right)} \right|_{x = 0}^{x = y}} \right] } = {\int\limits_0^2 {dz} \int\limits_0^z {\frac{{{y^3}z}}{2}dy} } = {\frac{1}{2}\int\limits_0^2 {dz} \int\limits_0^z {{y^3}zdy} } = {\frac{1}{2}\int\limits_0^2 {dz} \left[ {\left. {\left( {\frac{{{y^4}z}}{4}} \right)} \right|_{y = 0}^{y = z}} \right] } = {\frac{1}{2}\int\limits_0^2 {\frac{{{z^5}}}{4}dz} } = {\frac{1}{8}\int\limits_0^2 {{z^5}dz} } = {\frac{1}{8}\left. {\left( {\frac{{{z^6}}}{6}} \right)} \right|_0^2 } = {\frac{{64}}{{48}} = \frac{4}{3}.} \] |
For maximum sensitivity preamble detection with time referencing, consider using a barker code, or for more choices with longer lengths a PRN (pseudo-random noise) sequence, or for even more choices when more distinction is needed (CDMA) Gold codes are a possibility. (Also not further detailed but have other advantages are Kasami codes and Zadoff-Chu codes).
All these codes offer the characteristic of having a strong autocorrelation at time offset = 0 and importantly a much smaller autocorrelation for any other time offsets. Such as that demonstrated in the graphic below for a PRN sequence using an LFSR (Linear Feedback Shift Register):
The codes provide some noise immunity through processing gain; under condition of white noise, the processing gain which is the improvement of SNR is given as $10Log_{10}(N)$ where N is the length of the sequence. This comes out from addition of correlated signal components results in the standard deviation of the signal component growing at rate N while the noise (if white, meaning uncorrelated from sample to sample) will have a standard deviation that grows at $\sqrt{N}$. We detect the presence of the sequences using correlation, which is a multiply and sum:
$$Corr(\tau) = \Sigma r(t-\tau)x(t)$$
With the formula above representing a cross-correlation function of the received signal r(t) with the seequence of interest x(t), at various time offsets $\tau$.
There are several ways this is done in practice and depends on the detailed implementation approach but for shorter sequences this can be done with an FIR filter, with the coefficients of the filter as the code sequence upsampled to match the sample rate of the received signal (best to do this at the lowest rate possible of course). For longer sequences it is often more practical to create a code generator with a programmable offset (to shift the start position in time) and simply multiply the received sequence by the code and accumulate the output. A threshold detection can indicate the successful presence of the code.
Below shows the result of a sliding correlation on a GPS signal, where the code repeats every 1 ms. (I would never use an FIR filter to correlate a GPS sequence which is 1023 chips long, but this shows what a sliding correlation would look like).
Other approaches involve doing the correlation in the frequency domain using this relationship:
$$Corr_{xy} = IFFT(FFT(x)FFT^*(y))$$
Which states that taking the inverse Fourier Transform of the complex conjugate product of the Fourier Transforms of x and y results in the
circular correlation of x an y.
Further, a very important take-away is the sensitivity of any of these correlations to frequency offset, specifically that the magnitude goes down as a Sinc function with the first null at 1/T.
Please see this post directly where I detail that: GPS CA Signal Acquisition
This informs the frequency capture range based on the length of code used. If you are well inside this main lobe in frequency (I typically use 1/2 the null-null frequency range, but if the SNR is poor you may need to be more constrained). Typical acquisition approaches will either slowly ramp the frequency offset or step the frequency offset during correlation (The frequency is offset by using an IQ rotator on the received signal prior to correlation). Stepping the offset involves setting the frequency at an initial guess, correlating and if the acquisition threshold is not exceeded, stepping the frequency by 1/T (if using 1/2 the null-null frequency range) and repeating. If ramping, the ramp rate must be much less than 1/T without inducing an additional frequency offset loss.
The frequency offset is accurately estimated by comparing two successive correlations, since frequency is the change in phase versus the change in time ($d\phi/dt)$ by simply comparing the phase and time between two successive correlations, we can determine the frequency offset assuming we are in the frequency range of acquisition as detailed above (importantly these correlations are complex correlations with two real outputs I and Q such that carrier phase is determined).
Note that the frequency computation detailed at this link Obtaining phase and magnitude of 2 spectral components after ADCusing the computation
$$K = I[n]Q[n-1] - I[n-1]Q[n]$$ where K is propotional to frequency offset equally applies to the successive I and Q correlation outputs, referred to as I[n], Q[n] and I[n-1]Q[n-1].
Much more practical with modern processing and shorter preamble sequences would be a one shot frequency offset and delay estimation that I have detailed here:
GPS signal acquisition
That said, a robust preamble solution could involve sending multiple copies of one code to aid initial carrier and clock acquisition (and channel estimation!) followed by a final negative of the code sequence to mark the beginning of the data. (Just invert the code and the correlation will be shifted 180°). The trade space is over-head of the preamble versus acquisition sensitivity. It is typical to require more SNR during acquisition, especially if a fast acquisition is needed, hence the significance of correlation and processing gain. |
15.1 Two spaceships are connected by a strong cable. Both ships are initially at rest, and ignite their engines at the same time. Their accelerations are identical at every point in time. Does the cable stay intact?
Hint: use a spacetime diagram to analyze what happens.
15.2 In classical mechanics, the Doppler shift in the wavelength of an object moving towards the observer at
speed u is given by equation (9.7.1) repeated again below:
\[\lambda_{\mathrm{obs}}=\frac{v-u}{v} \lambda\ \label{15.E.1}\]
where \(v\) is the speed of the wave (usually sound). To get this result, we compared what happens with the source and the wave in a fixed time interval \(\Delta t\). As you now know, this result cannot hold at speeds close to that of light, because in that case there will be a significant effect due to time dilation. In this problem, we’ll, therefore, redo the calculation to account for relativistic effects.
We consider a distant source of light that moves with velocity \(\boldsymbol{u}\). At time \(t = 0\) (for both the source and the stationary observer), the source emits a signal (this could be a wave crest, but the argument holds for any signal). A time \(\Delta t^{\prime}\)) later, as measured on the clock moving with the source, the source emits a second signal, see Figure 15.E.1.
Determine the time interval \(\Delta t\) between the emission of the first and second signal as measured on the clock of the stationary observer. Determine the change in distance \(\Delta x\) between the (stationary) observer and the (moving) source in the time interval between the two signals, as measured by the stationary observer. Now determine the time interval \(\Delta t_{obs}\) between the arrival of the first and second signal at the location of the observer. From your answers at (a-c), show that the observed frequency \(\nu_{obs}\) is related to the source’s frequency \(\nu_{s}\) through \[v_{\mathrm{obs}}=\frac{\sqrt{1-u^{2} / c^{2}}}{1+(u / c) \cos \theta} v_{\mathrm{s}}\label{15.E.2}\] Note that equation (\ref{15.E.2}) reduces to (\ref{15.E.1}) in the case that the source is moving radially away from the observer. From equation (\ref{15.E.2}), find the expression for the (relativistic) transverse Doppler shift for the case that the source is moving in a direction perpendicular to the observer’s line of sight (i.e., \(\theta = 90^{o}\)). How can you tell that in this case, the Doppler shift is exclusively due to time dilation? |
I'm confused about the impact that a mean reverting stock price process has on the value of an option on it.
Several sources say that there is indeed an impact on the price of an option:
Lo and Wang (1995)
Yet, another source seems to say that mean reversion has no impact on the price of an option:
"The drift term of the process has no impact on the price of a call option, since we know that under the correct pricing measure we need the discounted stock price to have zero drift. This is achieved by changing the drift of the original process, rendering any initial drift term irrelevant"
Mark Joshi, Quant Job Interview Questions and Answers.
So I guess my ultimate question is, if the stock price follows the following process:
$$ dS_t=\alpha(\mu-S_t)dt+\sigma S_tdZ$$
Is the price of an option on the stock just equal to the BSM price where $\sigma_{BSM} = \sigma$?
It would make sense to me that there is no effect, because the replicating portfolio argument still works, and we end up with the same PDE and boundary conditions, which would give the same price. |
This question already has an answer here:
Can anybody show me how to do the boxes in the image attached in latex.
Thanks ahead
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This question already has an answer here:
Can anybody show me how to do the boxes in the image attached in latex.
Thanks ahead
\documentclass[a4paper,10pt]{article}\usepackage{tikz}\usetikzlibrary{positioning}\begin{document}\begin{tikzpicture}[mybox/.style={minimum width=4cm,draw,thick,align=center,minimum height=1.8cm}]\node[mybox,label=above:$\alpha$] (alpha) {1,2,3};\node[right=1cm of alpha,mybox,label=above:$\beta$] (beta) {4,5,6};\node[right=1cm of beta,mybox,label=above:$\gamma$] (gamma) {7,8,9};\end{tikzpicture}\end{document} |
Consider the following two variants of subset sum problem:
Variant 1
Given a set of $2n$ positive integer elements and a positive integer target $W$, is there a subset with size $n$, whose sum is $W$?
Variant 2
Given a set of $n$ positive integer elements and a positive integer target $W$, where $(\text{the maximum element})< 2\times(\text{the minimum element})$, is there a subset whose sum is $W$?
Variant 1 is NP-hard by a reduction from one-in-three 3-SAT (almost the same reduction from 3SAT to the normal subset sum problem, except that we do not need the $s$ and $t$ values, and the target is always $111\ldots1$).
Variant 2 is also NP-hard by a reduction from Variant 1. Given an instance of Variant 1, we add a large enough integer $M$ to each element so that the new elements satisfy the constraint in Variant 2. Now the answer to Variant 1 is "yes" if and only if there is a subset of the new elements whose sum is $W+nM$ (since $M$ is large enough, the subset must have size $n$).
Now we show a reduction from Variant 2 to (the decision version of) your problem, so your problem is NP-hard.
Given an instance of Variant 2, we construct $S_1=S_2$ as the same set as the given instance. Let $s$ denote the sum of all elements. We set $R_1=W$ and $R_2=s-W$. The gain of each element is the same as itself.
Now if there is a subset $O\subseteq S_1=S_2$ whose sum is $W$, then we can choose $O_1=O$ and $O_2=S_2\backslash O$, meaning there is a solution to your problem with $|O_1|+|O_2|+|O_1\cap O_2|\le n$.
On the other hand, if there is a solution to your problem with $|O_1|+|O_2|+|O_1\cap O_2|\le n$, then $2|O_1\cap O_2|\le n-|O_1\cup O_2|$. Let $\min$ and $\max$ be the minimum and maximum elements respectively. We have $$\sum_{e\in O_1\cap O_2}e\le |O_1\cap O_2|\max\le 2|O_1\cap O_2|\min\le(n-|O_1\cup O_2|)\min\le\sum_{e\notin O_1\cup O_2}e.$$
This means$$\sum_{e\notin O_1}e=\sum_{e\in O_2}e-\sum_{e\in O_1\cap O_2}e+\sum_{e\notin O_1\cup O_2}e\ge \sum_{e\in O_2}e\ge s-W.$$
Note $\sum_{e\in O_1}e\ge W$ and $\sum_{e\notin O_1}e+\sum_{e\in O_1}e=s$, we have $\sum_{e\in O_1}e=W$ and $\sum_{e\notin O_1}e =s-W$, which means there is a subset $O\subseteq S_1=S_2$ whose sum is $W$. |
In his celebrated paper "Conjugate Coding" (written around 1970), Stephen Wiesner proposed a scheme for quantum money that is unconditionally impossible to counterfeit, assuming that the issuing bank has access to a giant table of random numbers, and that banknotes can be brought back to the bank for verification. In Wiesner's scheme, each banknote consists of a classical "serial number" $s$, together with a quantum money state $|\psi_s\rangle$ consisting of $n$ unentangled qubits, each one either
$$|0\rangle,\ |1\rangle,\ |+\rangle=(|0\rangle+|1\rangle)/\sqrt{2},\ \text{or}\ |-\rangle=(|0\rangle-|1\rangle)/\sqrt{2}.$$
The bank remembers a classical description of $|\psi_s\rangle$ for every $s$. And therefore, when $|\psi_s\rangle$ is brought back to the bank for verification, the bank can measure each qubit of $|\psi_s\rangle$ in the correct basis (either $\{|0\rangle,|1\rangle\}$ or ${|+\rangle,|-\rangle}$), and check that it gets the correct outcomes.
On the other hand, because of the uncertainty relation (or alternatively, the No-Cloning Theorem), it's "intuitively obvious" that, if a counterfeiter who
doesn't know the correct bases tries to copy $|\psi_s\rangle$, then the probability that both of the counterfeiter's output states pass the bank's verification test can be at most $c^n$, for some constant $c<1$. Furthermore, this should be true regardless of what strategy the counterfeiter uses, consistent with quantum mechanics (e.g., even if the counterfeiter uses fancy entangled measurements on $|\psi_s\rangle$).
However, while writing a paper about other quantum money schemes, my coauthor and I realized that we'd never seen a rigorous proof of the above claim anywhere, or an explicit upper bound on $c$: neither in Wiesner's original paper nor in any later one.
So,
has such a proof (with an upper bound on $c$) been published? If not, then can one derive such a proof in a more-or-less straightforward way from (say) approximate versions of the No-Cloning Theorem, or results about the security of the BB84 quantum key distribution scheme?
Update: In light of the discussion with Joe Fitzsimons below, I should clarify that I'm looking for more than just a reduction from the security of BB84. Rather, I'm looking for an explicit upper bound on the probability of successful counterfeiting (i.e., on $c$)---and ideally, also some understanding of what the optimal counterfeiting strategy looks like. I.e., does the optimal strategy simply measure each qubit of $|\psi_s\rangle$ independently, say in the basis
$$\{ \cos(\pi/8)|0\rangle+\sin(\pi/8)|1\rangle, \sin(\pi/8)|0\rangle-\cos(\pi/8)|1\rangle \}?$$
Or is there an entangled counterfeiting strategy that does better?
Update 2: Right now, the best counterfeiting strategies that I know are (a) the strategy above, and (b) the strategy that simply measures each qubit in the $\{|0\rangle,|1\rangle\}$ basis and "hopes for the best." Interestingly, both of these strategies turn out to achieve a success probability of (5/8) n. So, my conjecture of the moment is that (5/8) n might be the right answer. In any case, the fact that 5/8 is a lower bound on c rules out any security argument for Wiesner's scheme that's "too" simple (for example, any argument to the effect that there's nothing nontrivial that a counterfeiter can do, and therefore the right answer is c=1/2).
This post has been migrated from (A51.SE)
Update 3: Nope, the right answer is (3/4) n! See the discussion thread below Abel Molina's answer. |
I have two points. (-1,-5) is one of the minimum points and (3.5,-4) is one of the maximum points. I'm not sure if this is a sin or a cosine function. Please write the equation to this problem.
.I literally just watched a video on how to do this, this is a learning process for me too
I am assuming that you learned the trig vocabulary for this as you asking these problems
________________________
Lets first do a cosine graph cuz its easier
This is the base equation for a cosine graph:
\(y=A\cos{b}(x-h)+c\)
________________________________________
So first find the amplitude, which is the height of the waves of the graph. (using y-values)
\(\text{Amplitude}=|\frac{\text{Max}-\text{Min}}{2}|\)
\(|\frac{-5-(-4)}{2}|\)
\(|\frac{-1}{2}|\)
\(\text{Amplitude}=\frac{1}{2}\)
Now we have the "A" value
\(y=\frac{1}{2}\cos{b}(x-h)+c\)
_______________________________________
Ok now we have to find the period
We first find the positive difference of the x values to find the horizontal distance.
\(|-1-3.5|=4.5\)
Then we double what we got
\(9\)
Then we solve for the b-value
\(9=\frac{2\pi}{b}\rightarrow9b=2pi\rightarrow{b}=\frac{2pi}{9}\)
Now we have
\(y=\frac{1}{2}\cos{\frac{2pi}{9}}(x-h)+c\)
_________________________________________
Now we have to find the phase shift (h-value)
Since it has shifted -1, ( coordinate (-1, -5) tells us that. )
We now have:
\(y=\frac{1}{2}\cos{\frac{2pi}{9}}(x+1)+c\)
___________________________________________
Now we have to find the vertical shift (c-value)
Formula for that is
\(c=\frac{\text{Maximum}}{2}\)
\(c=\frac{-5+(-4)}{2}\)
\(c=-4.5\)
____________________________________________
So the equation of the sinusoidal graph is
\(y=\frac{1}{2}\cos{\frac{2pi}{9}}(x+1)-4.5\)
This is the cosine graph.
This is the video I learned from, if you want to find the sine graph, follow the steps in the video.
I was on Khan Academy. I typed in the answer, and it turns out that y=0.5cos((2π/9)x−(7π/9))−4.5 is the answer. |
I have a problem about a general multivariate skew normal distribution. There is a $p\times 1$ vector, $\mathbf{y}=(\mathbf{y}_1',\mathbf{y}_2',\ldots,\mathbf{y}_n')',p>n$, which has the density as
$$f_p(\mathbf{y})\propto\phi_p(\mathbf{y};\boldsymbol{\mu},\Sigma)\prod_{i=1}^n\Phi(c_1+c_2\mathbf{1}'\mathbf{y}_i)=\phi_p(\mathbf{y};\boldsymbol{\mu},\Sigma)\Phi_n(c_1+c_2A\mathbf{y}),$$
where $\phi$ and $\Phi$ denote the pdf and cdf of the multivariate or univariate normal distribution. $c_1,c_2$ are constant scalars. $A$ is an $n\times p$ matrix, $A=\left(\begin{array}{cccc} 1,\ldots,1 & & & 0\\ & 1,\ldots,1\\ & & \ddots\\ 0 & & & 1,\ldots,1 \end{array}\right).$
The goal is to calculate the mean and variance of this distribution. Gupta et al. (2004) have derived them for the case $p=n$ and they seem tedious. I would like to know if there exists a simple solution for my problem. Any clue is welcome! Thanks! |
I analyzed an unknown Alkali Halide crystal using x-rays from a copper source. I need to find the d-spacing for this crystal, i.e. the interplanar distance between planes of atoms.
I found my $K_\alpha$ and $K_\beta$ peaks and used Bragg's Law to calculate the d-spacing. However, these values do not match. Is this an experimental error or could the non-matching values be correct somehow?
For $K_\alpha$:
$$d_{hkl}=\frac{λ}{(2 \sin(θ))}=\frac{(1.54184 Å )}{(2 \sin((28.3°)/2) )}=0.771 Å$$
For $K_\beta$:
$$d_{hkl}=\frac{λ}{(2 \sin(θ))}=\frac{(1.39222 Å )}{(2 \sin((25.7°)/2) )}=2.488 Å$$
My attempt: in both cases, I assumed that these were first order peaks. Could it be that the $K_\beta$ peak is actually a third order peak? If so, then $n=3$ and $d_{hkl}=2.313 Å$. This is reasonably close to the other value. But then what would that tell me about the crystal? |
Let $S$ be the set of $2^n$ binary $n$-bit strings. For every $x\in S$, let $f(x)$ is the maximal chain of bits 1 in $x$. So Can we find a good upper bound of $$F(n)=\frac{\sum_{x\in S}f(x)}{2^n}$$ Of course, $O(1)\le F(n) \le O(n)$. I think the upper bound is a constant or $O(\log n)$. Can anyone help me?
There is an easy way to get a good upper bound.
The probability that there is a streak of length $k$ is at most the expected number of streaks of $1$s length $k$, which is at most the expected number of all-$1$ substrings of length $k$ (which may overlap). It is easy to get the last expected value. There are $n-k+1$ possible substrings of length $k$, so the expected number of all-$1$ substrings of length $k$ is $(n-k+1)2^{-k} \lt n /2^k$. For $k = \lceil \log_2 n \rceil + c$ this gives us an upper bound of $1/2^c$ for the probability that there is a streak of length $\lceil \log_2 n \rceil + c$. So, the average excess over $\lceil \log_2 n \rceil$ is at most $1$, and the average length of the longest streak is at most $\lceil \log_2 \rceil + 1$.
Of course, it's not clear that this upper bound is good until you get a lower bound which is close to this. I think the value should be something like $(\log_2 n) -1$. (Actually, my guess is that the difference from $\log_2 n$ is not asymptotically constant, but fluctuates depending on the fractional part of $\log_2 n$.) |
Let's derive the proof. Consider you are short an option and long its self-financing delta hedge. You get the portfolio $\Pi$ whose $t$-value verifies$$ \Pi_t = \underbrace{- V_t}_{\text{Short option}} + \underbrace{\Delta_t S_t}_{\text{Long stocks}} + \underbrace{\frac{(V_t - \Delta_t S_t)}{B_t} B_t}_{\text{Residual cash position}} $$At time $t$, this position is worth zero: the option is perfectly replicated by the hedge. As soon as time will pass though, you will need to rebalance your hedge to remain delta neutral. Let's see what happens in-between two rebalancing dates when we keep the delta unchanged over say $[t,t+dt[$.
Because the strategy is self-financing, assuming the risk-free money market account verifies the ODE $dB_t = B_t r dt$ and the stock pays no dividends then we have that the replication error over $[t,t+dt[$ writes$$ d\Pi_t = - dV_t + \Delta_t dS_t + (V_t - \Delta_t S_t) r dt \tag{0} $$Let's expand the different terms at the lowest trivial order. If we consider a pure diffusion model (no jumps) this means order 1 in $dt$ and order 2 in $dS_t$. Assuming you're pricing the option $V_t$ under a BS framework with volatility $\sigma$, Itô's lemma gives\begin{align}dV_t &= \frac{\partial V}{\partial t} dt + \frac{\partial V}{\partial t} dS_t + \frac{1}{2} \frac{\partial^2 V}{\partial S^2} d\langle S \rangle_t \\&= \theta_t dt + \Delta_t dS_t + \Gamma_t d\langle S \rangle_t \end{align}Plugging this in $(0)$ yields$$ d\Pi_t = \underbrace{(-\theta_t - rS_t \Delta_t + rV_t)dt}_{\text{(a)}} - \Gamma d\langle S \rangle_t \tag{1} $$The first terms $(a)$ should look familiar. Indeed, if you're pricing under the BS framework, then $v_t = V(t,S_t)$ ought to verify the BS pricing PDE$$ \theta_t + r S_t \Delta_t + \frac{1}{2} \Gamma S_t^2 \sigma^2 - r V_t = 0$$Using this to rewrite $(1)$ yields the following replication error over the period $[t,t+dt[$\begin{align}d\Pi_t &= \frac{1}{2} \Gamma S_t^2 \sigma^2 dt - \Gamma d \langle S \rangle_t \\ &= \frac{1}{2} \Gamma S_t^2 \left( \sigma^2 - \frac{d \langle S \rangle_t}{(S_t)^2 dt} \right) dt \\ &= \frac{1}{2} \Gamma S_t^2 \left( \sigma^2 - \beta_t^2 \right) dt\end{align}where $\beta^2$ is the "realised" quadratic variation of the log returns, i.e. not the one postulated and priced in by your model but the manner in which the market truly behaves.
Now if you want to find out the total replication error up to the maturity $T$ as seen from current time $t$, all which is left to do is to integrate and discount the infinitesimal P\&L leaks described above$$ P\&L_t = \int_0^T e^{-r(T-t)} \frac{1}{2} \Gamma S_t^2 \left( \sigma^2 - \beta_t^2 \right) dt $$
This is a very interesting and well-known equation because it ties 3 important concepts:$$ \int_0^T e^{-r(T-t)} \frac{1}{2} \underbrace{\Gamma}_{\text{Instrument related}} S_t^2 \left( \underbrace{\sigma^2}_{\text{Model related}} - \underbrace{\beta_t^2}_{\text{Market related}} \right) dt $$
A naive interpretation of it would go like this. Suppose I sold a vanilla option by pricing in a future volatility of $\sigma$ at inception. If the realised volatility $\beta$ is always higher than $\sigma$, then I expect to lose money since this would amount to me having effectively underpriced the option (note that the $\Gamma$ of a vanilla option is always positive).
The trick here is to observe that selling an option and delta hedging it dynamically is not a pure volatility trade though. The gamma term in the P\&L equation above introduces a path dependence: only along paths where $\Gamma(t,S_t)$ is not zero will the discrepancy between the pricing and realised vol accumulate and P\&L crystallise. You can find out more info in this related question.
Of course as mentioned in the comments, this P\&L equation assumes no transaction costs and continuous trading (when dynamically rebalancing the Delta). Also you are right in practice this is used to monitor the daily evolution of a delta-hedged portfolio ex post. It is usually part of the explained P&L calculations for instance. |
removes tags created more than 6 months ago which have been used only a single time. It will run monthly.
Hake's Theorem, due to Heinrich Hake of Düsseldorf in 1921, says that an improper Henstock–Kurzweil integral (aka generalized Riemann integral, gauge integral, Perron integral, or Denjoy integral) on a bounded interval is already proper. That is, if $f$ is defined on a half-open interval $[a,c)$...
Question as in the title: Who was Hermann Künneth? Where can I find some biographical information beyond what is available on Wikipedia? The well-known Künneth formula, for example in the form of exactness of the sequence $$ 0 \to \bigoplus_{p+q = n} H_p(C) \otimes H_q(D) \to H_n(C \otimes D) ...
Since calculus originated long before our modern function concept, much of our language of calculus still focuses on variables and their interrelationships rather than explicitly on functions. For example, in the assertion "If $y=x^2$ then $\frac{dy}{dx}=2x$," the functions $f$ and $f'$ remain u...
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Background
For my thesis, I'm trying to wrap my head around a relatively small field within mathematics called "Stochastic Cooperative Game Theory". To that end, I've read a few papers on the subject, including
Cooperative games with stochastic payoffs (Suijs et al., 1999, link) and Convexity in stochastic cooperative situations (Timmer et al., 2005, link).
In classical cooperative game theory, the payoffs (rewards) to players are deterministic. The players know beforehand what the benefits are of forming coalitions with the other players. In stochastic cooperative game theory, however, uncertainty is introduced. This means players do not know for sure what payoff they will receive: the payoffs are modeled as random variables.
The players try to decide which payoffs based on random variables are most beneficial to them. However, there many ways to define "most beneficial". In order to determine which random variable is preferred by player $i \in N$, Timmer et al. introduce the notion of a preference relation $\succeq_{i}$ (on page 4). For some stochastic payoffs $X$ and $Y$ (where $X$ and $Y$ are nonnegative random variables with finite expectation), they write $X \succeq_{i} Y$ if player $i$ weakly prefers the stochastic payoff $X$ to receiving $Y$. Furthermore, $X \succ_{i} Y$ means that player $i$ strictly prefers $X$ to $Y$.
On page 5, the same authors give an example of such a preference relation. The preferences of player $i$ could be such that $X \succeq_{i} Y$ if and only if $E(x) \geq E(Y)$, where $E(X)$ is the expectation of $X$. I suppose I understand this type of preference. As a player, you'd want to maximize the expectation of your payoffs $(*)$.
However, I don't understand the second type of preference. It involves quantiles of random variables. They define $ u_{ \beta_{i} }^{X} = \sup \{t \in \mathbb{R} | \Pr \{ X \leq t \} \leq \beta_{i} \} $ be the $\beta_{i}$-quantile of $X$. Furthermore, they define the utility function $U_{i}$ by $U_{i} (X) = u_{ \beta_{i} }^{X} $. They say that a player has
quantile-preferences if $X \succeq_{i} Y$ if and only if $U_{i} (X) \geq U_{i} (Y)$. Questions Could you provide me with an intuitive explanation of the quantile preference relation? (Perhaps similar to the one I gave myself in $(*)$, but a bit longer/detailed?). Can you give an example with actual distributions?
Suijs et al. introduce a similar preference relation in their paper on p. 197. They add that agent $i$ is more
risk averse than agent $j$ if and only if $\beta_{i} < \beta_{j}$ Could you explain and motivate this definition of risk aversiveness? And how does it relate to this type of risk aversiveness?
Finally, I tried understanding the quantile preference relation by comparing two cumulative distribution functions. These are the cumulative exponential distribution with $\lambda = 0.5$ (distribution $X$) and $\lambda = 1.5$ (distribution $Y$), respectively. Suppose we take $\beta_{i} = 0.3$. As $E[X] = 1/0.5 = 2 > 0$ and $E[Y] = 1/1.5 = 2/3 > 0$, we need to find the utility with $1-\beta_{i} = 0.7$. Then we get the following picture:
(adapted from an image on Wikipedia). I suppose now we have $u_{ \beta_{i} }^{X} \approx 2.3$ and $u_{ \beta_{i} }^{Y} \approx 0.9$, so we have $U_{i} (X) \geq U_{i} (Y)$
How do we interpret this result? Is there an intuitive explanation? |
Is my solution correct?
Question: Find all \(z\), such that \((z+1)^7=z^7\).
My answer: Since \(z\ne 0\), we can divide by \( z^7\) on both sides to get \(\left(\frac{z+1}{z}\right)^7=1\). If we let \(w=\frac{z+1}{z}\), then \(w\) is equal to the \(7^{\text{th}}\) roots of unity. We can manipulate \(w=\frac{z+1}{z}\) by multiplying both sides by \(z\), then subtracting \(z\) from both sides to get \(1=zw-z\). We can then divide both sides by \(w-1\) to get \(z=\frac{1}{w-1}\).
The seven roots of unity are \(1\), \(e^{\frac{2i\pi}{7}}\), \(e^{\frac{4i\pi}{7}}\), \(e^{\frac{6i\pi}{7}}\), \(e^{\frac{-6i\pi}{7}}\), \(e^{\frac{-4i\pi}{7}}\), \(e^{\frac{-2i\pi}{7}}\). \(z=\frac{1}{1-1}\) if \(w=1\) is not defined, and we can see that it doesn't work for \((z+1)^7=z^7\), either, so the solutions for \(z\) are \(\frac{1}{e^{\frac{2i\pi}{7}}-1}\), \(\frac{1}{e^{\frac{4i\pi}{7}}-1}\), \(\frac{1}{e^{\frac{6i\pi}{7}}-1}\), \(\frac{1}{e^{\frac{-6i\pi}{7}}-1}\), \(\frac{1}{e^{\frac{-4i\pi}{7}}-1}\), and \(\frac{1}{e^{\frac{-2i\pi}{7}}-1}\), respectively. |
In theoretical population genetics, it is very common to have to assume a model of epistatic interaction. The two most common models are the additive model and the multiplicative model.
Additive model
Under the additive model the fitness $w$ of an individual carrying all $n$ mutations of effects $[s_1,s_2,...,s_n]$ is
$$w=1-\sum_{i=1}^n s_i$$
, which becomes $w=1-sn$, for the special case when $s=s_i \forall i \in [1,n] $
Multiplicative model
Under the multiplicative model the fitness $w$ of an individual carrying all $n$ mutations of effects $[s_1,s_2,...,s_n]$ is
$$w=\prod_{i=1}^n (1-s_i)$$
, which becomes $w=(1-s)^n$, for the special case when $s=s_i \forall i \in [1,n] $
Old observation
, where the additive $\left(w=1-sn\right)$ and multiplicative $\left(w=(1-s)^n \right)$ models correspond to the two lines that poorly fit the data. The line that best fit the data is a quadratic model of synergistic epistasis
$$w=1-sn-an^2$$
In all cases, $s$ and $a$ where estimated from these data. We probably have better data and better studies today.
Question What model of epistatic interaction would you recommend? Here are some clarifications if needed Consider $n$ mutations. The fitness of an individual carrying only the first mutation is $1-s_1$. The fitness of an individual carrying only the second mutation is $1-s_2$, etc... the fitness of an individual carrying the $n^{th}$ mutation is $1-s_n$. What is the fitness of the individual carrying all $n$ mutations? Of course, there is no single answer to this question and it will depend on the epistatic interactions between these loci. However, there is probably a general trend that a theoretician would like to consider in its assumptions. |
It might be thought that there is rather a limited amount that could be written about the geometry of a straight line. We can manage a few Equations here, however, (there are 35 in this section on the Straight Line) and we shall return for more on the subject in Chapter 4.
Most readers will be familiar with the equation for a straight line:
\[y = mx + c \label{2.2.1} \tag{2.2.1}\]
The slope (or gradient) of the line, which is the tangent of the angle that it makes with the \(x\)-axis, is \(m\), and the intercept on the \(y\)-axis is \(c\). There are various other forms that may be of use, such as
\[\frac{x}{x_0} + \frac{y}{y_0} = 1 \label{2.2.2} \tag{2.2.2}\]
\[\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1} \label{2.2.3} \tag{2.2.3}\]
which can also be written
\begin{array}{| c c c | c}
x & y & 1 \\ x_1 & y_1 & 1 & =0 \\ x_2 & y_2 & 1 \\ \label{2.2.4} \tag{2.2.4} \end{array}
\[x \cos \theta + y \sin \theta = p \label{2.25} \tag{2.2.5}\]
The four forms are illustrated in figure \(\text{II.1}\).
\(\text{FIGURE II.1}\)
A straight line can also be written in the form
\[Ax + By + C = 0. \label{2.2.6} \tag{2.2.6}\]
If \(C = 0\), the line passes through the origin. If \(C ≠ 0\), no information is lost, and some arithmetic and algebra are saved, if we divide Equation \ref{2.2.6} by \(C\) and re-write it in the form
\[ax + by = 1. \label{2.2.7} \tag{2.2.7}\]
Let \(P (x , y)\) be a point on the line and let \(P_0 (x_0 , y_0 )\) be a point in the plane not necessarily on the line. It is of interest to find the perpendicular distance between \(P_0\) and the line. Let \(S\) be the square of the distance between \(P_0\) and \(P\). Then
\[S = (x-x_0)^2 + (y-y_0)^2 \label{2.2.8} \tag{2.2.8}\]
We can express this in terms of the single variable \(x\) by substitution for \(y\) from Equation \(\ref{2.2.7}\). Differentiation of \(S\) with respect to \(x\) will then show that \(S\) is least for
\[x = \frac{a+b(bx_0 - ay_0)}{a^2 + b^2} \label{2.2.9} \tag{2.2.9}\]
The corresponding value for \(y\), found from Equations \(\ref{2.2.7}\) and \(\ref{2.2.9}\), is
\[y = \frac{b+a(ay_0 - bx_0)}{a^2 + b^2}. \label{2.2.10} \tag{2.2.10}\]
The point \(\text{P}\) described by Equations \(\ref{2.2.9}\) and \(\ref{2.2.10}\) is the closest point to \(\text{P}_0\) on the line. The perpendicular distance of \(\text{P}\) from the line is \(p = √S\) or
\[p = \frac{1-ax_0 - by_0}{\sqrt{a^2+b^2}}. \label{2.2.11} \tag{2.2.11}\]
This is positive if \(\text{P}_0\) is on the same side of the line as the origin, and negative if it is on the opposite side. If the perpendicular distances of two points from the line, as calculated from Equation \(\ref{2.2.11}\), are of opposite signs, they are on opposite sides of the line. If \(p = 0\), or indeed if the numerator of Equation \(\ref{2.2.11}\) is zero, the point \(\text{P}_0 (x_0 , y_0 )\) is, of course, on the line.
Let \(\text{A}(x_1 , y_1 ), \ \text{B}(x_2 , y_2 )\) and \(\text{C}(x_3 , y_3 )\) be three points in the plane. What is the area of the triangle \(\text{ABC}\)? One way to answer this is suggested by figure \(\text{II.2}\).
\(\text{FIGURE II.2}\)
We see that
area of triangle \(\text{ABC}\) = area of trapezium \(\text{A}^\prime \text{ACC}^\prime\) (see comment*)
+ area of trapezium \(\text{C}^\prime \text{CBB}^\prime\) − area of trapezium \(\text{A}^\prime \text{ABB}^\prime\).
\[ = \frac{1}{2} (x_3 - x_1) (y_3 + y_1) + \frac{1}{2} (x_2 - x_3)(y_2 + y_3) - \frac{1}{2} (x_2 - x_1) (y_2 + y_1) \]
\[ = \frac{1}{2} [x_1(y_2 - y_3) + x_2 (y_3 -y_1) + x_3(y_1 - y_2)]\]
\begin{array}{l r | c c c |}
& & x_1 & x_2 & x_3 \\ = & \frac{1}{2} & y_1 & y_2 & y_3 \\ & & 1 & 1 & 1 \\ \label{2.2.12} \tag{2.2.12} \end{array}
* Since writing this section I have become aware of a difference in U.S./British usages of the word "trapezium". Apparently in British usage, "trapezium" means a quadrilateral with two parallel sides. In U.S. usage, a trapezium means a quadrilateral with no parallel sides, while a quadrilateral with two parallel sides is a "trapezoid". As with many words, either British or U.S. usages may be heard in Canada. In the above derivation, I intended the British usage. What is to be learned from this is that we must always take care to make ourselves clearly understood when using such ambiguous words, and not to assume that the reader will interpret them the way we intend.
The reader might like to work through an alternative method, using results that we have obtained earlier. The same result will be obtained. In case the algebra proves a little tedious, it may be found easier to work through a numerical example, such as: calculate the area of the triangle \(\text{ABC}\), where \(\text{A}\), \(\text{B}\), \(\text{C}\) are the points (2,3), (7,4), (5,6) respectively. In the second method, we note that the area of a triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\). Thus, if we can find the length of the side BC, and the perpendicular distance of \(\text{A}\) from \(\text{BC}\), we can do it. The first is easy:
\[(\text{BC})^2 = (x_3 - x_2)^2 + (y_3 - y_2)^2 . \label{2.2.13} \tag{2.2.13}\]
To find the second, we can easily write down the equation to the line \(\text{BC}\) from Equation \(\ref{2.2.3}\), and then re-write it in the form \(\ref{2.2.7}\). Then Equation \(\ref{2.2.11}\) enables us to find the perpendicular distance of \(\text{A}\) from \(\text{BC}\), and the rest is easy.
If the determinant in Equation \(\ref{2.2.12}\) is zero, the area of the triangle is zero. This means that the three points are collinear.
The angle between two lines
\[y = m_1 x + c_1 \label{2.2.14} \tag{2.2.14}\]
and \[ y = m_2 x + c_2 \label{2.2.15} \tag{2.2.15}\]
is easily found by recalling that the angles that they make with the \(x\)-axis are \(\tan^{-1} \ m_1\) and \(\tan^{-1} \ m_2\) together with the elementary trigonometry formula \(\tan( A − B) = (\tan A − \tan B) / (1+ \tan A \tan B)\). It is then clear that the tangent of the angle between the two lines is
\[\frac{m_2 - m_1}{1+ m_1 m_2} . \label{2.2.16} \tag{2.2.16}\]
The two lines are at right angles to each other if
\[m_1 m_2 = -1 \label{2.2.17} \tag{2.2.17}\]
The line that bisects the angle between the lines is the locus of points that are equidistant from the two lines. For example, consider the two lines
\[-2x + 5y = 1 \label{2.2.18} \tag{2.2.18}\]
\[30x - 10y = 1 \label{2.2.19} \tag{2.2.19}\]
Making use of Equation \(\ref{2.2.11}\), we see that a point \((x , y)\) is equidistant from these two lines if
\[\frac{1+2x-5y}{\sqrt{29}} = \pm \frac{1-30x+10y}{\sqrt{1000}}. \label{2.2.20} \tag{2.2.20}\]
The significance of the \(\pm\) will become apparent shortly. The + and − choices result, respectively, in
\[-8.568x + 8.079y = 1 \label{2.2.21} \tag{2.2.21}\]
and \[2.656x + 2.817y = 1. \label{2.2.22} \tag{2.2.22}\]
The two continuous lines in figure \(\text{II.3}\) are the lines \(\ref{2.2.18}\) and \(\ref{2.2.19}\). There are two bisectors, represented by Equations \(\ref{2.2.21}\) and \(\ref{2.2.22}\), shown as dotted lines in the figure, and they are at right angles to each other. The choice of the + sign in Equation \(\ref{2.2.20}\) (which in this case results in Equation \(\ref{2.2.21}\), the bisector in figure \(\text{II.3}\) with the positive slope) gives the bisector of the sector that contains the origin.
An equation of the form
\[ax^2 + 2hxy + by^2 = 0 \label{2.2.23} \tag{2.2.23}\]
can be factored into two linear factors with no constant term, and it therefore represents two lines intersecting at the origin. It is left as an exercise to determine the angles that the two lines make with each other and with the \(x\) axis, and to show that the lines
\[x^2 + \left( \frac{a-b}{h} \right) xy - y^2 = 0 \label{2.2.24} \tag{2.2.24} \]
are the bisectors of \(\ref{2.2.23}\) and are perpendicular to each other.
\(\text{FIGURE II.3}\)
Given the equations to three straight lines, can we find the area of the triangle bounded by them? To find a general algebraic expression might be a bit tedious, though the reader might like to try it, but a numerical example is straightforward. For example, consider the lines
\[x - 5y + 12 = 0, \label{2.2.25} \tag{2.2.25}\]
\[3x + 4y - 9 = 0, \label{2.2.26} \tag{2.2.26}\]
\[3x - y - 3 = 0. \label{2.2.27} \tag{2.2.27}\]
By solving the Equations in pairs, it is soon found that they intersect at the points (−0.15789, 2.36842), (1.4, 1.2) and (1.92857, 2.78571). Application of Equation \(\ref{2.2.12}\) then gives the area as 1.544. The triangle is drawn in figure \(\text{II.4}\). Measure any side and the corresponding height with a ruler and see if the area is indeed about 1.54.
But now consider the three lines
\[x - 5y + 12 = 0, \label{2.2.28} \tag{2.2.28}\]
\[3x + 4y - 9 = 0 , \label{2.2.29} \tag{2.2.29}\]
\[3x + 23y - 54 = 0. \label{2.2.30} \tag{2.2.30}\]
\(\text{FIGURE II.4}\)
By solving the equations in pairs, it will be found that all three lines intersect at the same point (please do this), and the area of the triangle is, of course, zero. Any one of these equations is, in fact, a linear combination of the other two. You should draw these three lines accurately on graph paper (or by computer). In general, if three lines are
\[A_1 x + B_1 y + C_1 = 0 \label{2.2.31} \tag{2.2.31}\]
\[A_2 x + B_2y + C_2 = 0 \label{2.2.32} \tag{2.2.32}\]
\[A_3 x + B_3 y + C_3 = 0 \label{2.2.33} \tag{2.2.33}\]
they will be concurrent at a single point if
\begin{array}{| c c c | c}
A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 & = 0. \\ A_3 & B_3 & C_3 \\ \label{2.2.34} \tag{2.2.34} \end{array}
Thus the determinant in Equation \(\ref{2.2.12}\) provides a test of whether three points are collinear, and the determinant in Equation \(\ref{2.2.34}\) provides a test of whether three lines are concurrent.
Finally - at least for the present chapter - there may be rare occasion to write the equation of a straight line in polar coordinates. It should be evident from figure \(\text{II.5}\) that the Equations
\[r = p \csc (\theta - \alpha) \ \text{or} \ r = p \csc (\alpha - \theta) \label{2.2.35} \tag{2.2.35}\]
describe a straight line passing at a distance \(p\) from the pole and making an angle \(\alpha\) with the initial line. If \(p = 0\), the polar Equation is merely \(\theta = \alpha\).
\(\text{FIGURE II.5}\) |
As @nimbus3000 mentions, the shape of the vol curve differs by markets so I won't comment on that here. I'll restrict my comments to the Black(-Scholes) vs. Bachelier section of the question.
You can approximate Normal (Bachelier) vols from Black vols by (there is a second order effect related to the product of the square of the Black vol and the maturity but ignored here):
$$\sigma_N = \sigma_B \sqrt{F\times K}$$
Where $F$ and $K$ are the forward and strike, respectively. Since you're interested in moneyness, consider $K = F\times k$ for some %-moneyness $k$. Then $$\sigma_N = \sigma_B \times F \sqrt{k}$$
From this I have 2 observations:
Bachelier vols are not independent of the level of the underlying (unlike Black vols). The transformation is almost linear in F, so the shape of the Bachelier vol skew for a given maturity will roughly mimic the shape of the Black curve. |
First, the term "propagator" is usually defined as the Green's function of the first type, not the second type, i.e. as a solution to the diffential equation $\hat L G = \delta$.
At any rate, those definitions are ultimately equivalent – when the details are correctly written down – because the Green's function defined as the correlator in the second definition obeys the first differential equation.
The differential operator $\hat L$ is what appears in the linearized equations of motion for the field, in this case $\psi(x_1,t_1)$, and it only acts on $\psi(x_1,t_1)$, not $\psi^\dagger (x_2,t_2)$.
The time-ordering operator $T$ may be written in terms of the step function$$ T(\psi \psi^\dagger) = \psi \psi^\dagger\cdot \theta(t_1-t_2) - \psi^\dagger \psi\cdot\theta(-t_1+t_2)$$where $\psi$ is always at $x_1,t_1$ and $\psi^\dagger$ is at $x_2,t_2$. Now, ask what happens when you act with $\hat L$ on the right hand side of the displayed equation above.
By the Leibniz rule, there are the terms with $\hat L \psi = 0$. It vanishes by the equations of motion. But there are extra terms where $\hat L$ acts on the step functions.
The operator $\hat L$ contains the term that differentiates with respect to $t_1$ multiplied by a coefficient $C$. This turns $\theta(t_1-t_2)$ to $\delta(t_1-t_2)$. The same occurs in the next term, but with the opposite sign which cancels the sign that was already there. So the extra terms are$$ \hat L T(\psi \psi^\dagger) = C\delta(t_1-t_2) (\psi \psi^\dagger + \psi^\dagger \psi)$$I got two terms because there were two terms. However, these two terms exactly combine to the anticommutator of $\psi$ and $\psi^\dagger$ which only needs to be evaluated for $t_1=t_2$, the equal-time anticommutator, and the result is $D\cdot \delta(x_1-x_2)$.
That's why the action of $\hat L$ on the correlator ends up being $CD\delta(t_1-t_2)\delta(x_1-x_2)$ where the constants $C,D$ are mostly just factors of $i$ etc.
For bosonic fields, $\hat L$ has the second derivative with respect to time. One of the derivatives has the fate as above, the other one turns the other $\phi$, which plays the role of $\psi^\dagger$, into $\partial_t \phi$ which is the canonical momentum, and it's the right variable that has the $\delta$-function-like commutator. Also, the intermediate sign is the opposite one but the result is the same, some $CD\cdot\delta\cdot\delta$. |
I'm reading "Antennas and Wave Propagation" by Harish and Sachidananda. They introduce the phasor form of Maxwell's equations:
$$\nabla\times E=-j\omega\mu H$$ $$\nabla\times H = j\omega\epsilon E + J$$ $$\nabla\cdot D = \rho$$ $$\nabla\cdot B = 0$$
First of all they observe that by taking the curl of the first equation and rewriting the right hand side with the expression for $\nabla\times H$, we can show that $E$ satisfies a wave equation: $\nabla^2E+aE=bJ$, for some constants $a$ and $b$. They also note that in the same way we can show that $H$ satisfies the wave equation.
Then they introduce the notion of vector and scalar potential, and propose to find $V$, the scalar potential of $E$, and $A$, the vector potential of $H$. The motivation for doing this seems to be that finding $A$ and $V$ is easier because they both satisfy... the wave equation.
Why solve the wave equation to get the potentials, and then work out $E$ and $H$ from there? Why not just solve the wave equations for $E$ and $H$ directly? |
First of all credits:
Latex people for making this beauty happen
Dave at xyloid.org for his vblatex package
The makers of mhchem package
Tormod for being the most unappreciated person who created and manages this chaos every day, also for being a great man and a father
Tormod's family for being the people who appreciate what he does more then anyone else in the world (or at least i hope they do)
Mercedes Bebzene for being annoying
Jay-qu for being supportive
Buffy for being really cool
Irish for being the coolest
Users of Hypography for using Hypography
Linus Torvalds
Richard Stallaman for defining open-source and freedom
The open-source community
Steve Jobs
The guy that took the picture that i use as my background
My friends and family (especially here at Hypography)
Aspectradio for providing the music when i wanted to smash something because something didn't work
Microsoft for pissing me off so much and making me switch to open-source
AT&T for being a horrible internet provider, as well as for spying on me for the NSA (don't know what I'd do without you guys)
His majesty, the constitutional monarch of The Kingdom of Norway; Harald V
Her majesty the Queen of the United Kingdom of Great Britain
and Northern Ireland; Queen Elisabeth II
The Supreme Pontiff of the Vatican City
(excuse me if i have left anyone out, just don't have enough time)
Ok so what is new with this setup?
Well it uses latex rendering engine itself to convert the tex code to png images, its slower, but it yields better results, produces a better quality pictures and supports plugins (also has better fonts and is more configurable)
Note: the old rendering engine is still up and running, i want people to use the new one though, it is better trust me!
Old Vs. New:
First of all the new version does waaay better arrays, take a look at this code
left[ begin{array}{ c c } 1 & 2 \ 3 & 4 end{array} right]perfectly legitimate array code, but in our old package it renders as
Old:
[math]
\left[
\begin{array}{ c c }
1 & 2 \\
3 & 4
\end{array} \right]
[/math]
while the new one does a better job and renders it as such:
New:
[math]
\left[
\begin{array}{ c c }
1 & 2 \\
3 & 4
\end{array} \right]
[/math]
here is an example of how much cleaner the rendering is in this version:
old:
[math]\frac{\frac{1}{x}+\frac{1}{y}}{y-z}[/math]
new:
[math]\frac{\frac{1}{x}+\frac{1}{y}}{y-z}[/math]
old:
[math]\Delta =\sum_{i=1}^N w_i (x_i - \bar{x})^2[/math]
new:
[math]\Delta =\sum_{i=1}^N w_i (x_i - \bar{x})^2[/math]
in addition to the new render engine, a new mode is now available for displaying chemical equations
lets try to display a simple chemical formula (not a chemist, merely a hypothetical formula):
frac{1}{2}^{277}_{90}Th^+
[math]\frac{1}{2}^{277}_{90}Th^+[/math]
looks clumsy and hard to understand, does the job, but kinda hard to write
now the new method
[FONT=monospace]1/2^{277}_{90}Th+[/FONT][ce]{1/2^{277}_{90}Th+}
[/ce]
does this not look better?
or another one H2O2
H_{2}O_{2}
[math]H_{2}O_{2}[/math]
vs
H2O2
[ce]{H2O2}[/ce]
Also, if you are typing a response, instead of taking the whole function and having to rewrite it from scratch (in a reverse engineering fashion), you can click the image and voila, it shows you exactly what the person typed in, correct it and repost it! (saves lots of time)
ok, so now how do you use this new rendering engine, and as you guessed latex tags stay for the old engine (at least until i make sure that everything written for the old engine will render in the new one:
there are 3 new tags:
[math][/math] - renders mathematical equations
[imath][/imath]- renders inline math
[ce]{}[/ce] - renders chemical equations
notes, for ce, you need a set of curly brackets, otherwise the equation will be taken as math or something like that...
more about all that later, please post any problems you may find here or pm me directly about them
also feel free to try the new engine out here, its sort of still in testing (there are some minor differences between mimetex and this package)
there is a 400 character limit on the math, 200 on imath and 200 on ce, if that becomes an issue i can alsways increase the limit, but as i said, any comments or problems you may find, please post here or PM me so they get solved...
enjoy
Play around with it a little, see what you can make of it yourself
oh so i dont forget this is really cool, an online wyswig latex editor:
LaTeX Equation Editor
its not a bad thing to play around with and the code it produces should be compliant with our engine
More references:
Latex Math Symbols |
Is there an example of an eigenfunction of a linear time invariant (LTI) system that is
not a complex exponential? Justin Romberg's Eigenfunctions of LTI Systems says such eigenfuctions do exist, but I am not able to find one.
All eigenfunctions of an LTI system can be described in terms of complex exponentials, and complex exponentials form a complete basis of the signal space. However, if you have a system that is
degenerate, meaning you have eigensubspaces of dimension >1, then the eigenvectors to the corresponding eigenvalue are all linear combination of vectors from the subspace. And linear combinations of complex exponentials of different frequencies are not complex exponentials anymore.
Very simple example: The identity operator 1 as an LTI system has the whole signal space as eigensubspace with eigenvalue 1. That implies ALL functions are eigenfunctions.
I thought I had worded my response clearly---apparently not :-). The original question was, "Are there eigensignals besides the complex exponential for an LTI system?". The answer is, if one is given the fact that the system is LTI but nothing else is known, then the only confirmed eignensignal is the complex exponential. In specific cases, the system may have additional eigensignals as well. The example I gave was the ideal LPF with sinc being such an eigensignal. Note that the sinc function is not an eigensignal of an arbitrary LTI system. I gave the LPF and the sinc as an example to point a non-trivial case---x(t) = y(t) will satisfy a mathematician but not an engineer :->. I am sure one can come up with other specific non-trivial examples that have other signals as eigensignals besides the complex exponential. But these other eigensignals will work for those specific examples only.
Also, cos and sin are not, in general, eigensignals. If cos(wt) is applied and the output is A cos(wt + theta), then this output cannot be expressed as a constant times the input (except when theta is 0 or pi, or A=0), which is the condition needed for a signal to be an eigensignal. There may be conditions under which cos and sin are eigensignals, but they are special cases and not general.
CSR
For any arbitrary LTI sytem, the complex exponential is, to the best of my knowledge, the only known eigensignal. On the other hand, consider the ideal LPF. The $\operatorname{sinc}$ function: $$\operatorname{sinc}(t) \triangleq \frac{\sin(\pi t)}{\pi t}$$ can easily be seen to be an eigen signal. This points to the existence of LTI systems (such as the ideal LPF) having signals other than complex exponentials as eigen signals ($\frac{\sin(\pi t)}{\pi t}$ in this case).
Maybe spatially invariant multidimensional objects like lenses with circular symmetry. It is called the Fourier Bessel expansion. There is no T for time but the convolution frequency domain relations hold |
Let $A$, $B$ be two smooth vector bundles of finite rank over a smooth manifold $M$. Let $Diff(A,B)$ be the space of differential operators from $A$ to $B$. Can I talk about "the space of smooth maps from $[0,1]$ to $Diff(A,B)$?
You could use the bijective correspondence between differential operators $L: \Gamma(A) \to \Gamma(B)$ and vector bundle homomorphisms $\tilde{L}: J^k A \to B$ to topologize your space $Diff(A,B)$. Then $C^1$-curves in $Diff(A,B)$ are defined as usual, that is, the differential quotient has to exist and be continuous. Iterating yields the notion of smooth curves.
One way to introduce the space you are interested in is: $$ \widetilde{\textrm{Diff}}(A,B):=C^\infty([0,1])\otimes_{\mathbb{R}}\textrm{Diff}(A,B). $$ I'm sure this approach is equivalent to the topological one, suggested by @Tobias, at least for operators of bounded order.
You can also characterize it as a distinguished subspace $$ \widetilde{\textrm{Diff}}(A,B)\subseteq \textrm{Diff}(p^*A,p^*B).\quad\quad\quad (*) $$ Indeed, if $\{\alpha^i\}$ (resp., $\{b_j\}$) is a basis of $\Gamma(A)^*$ (resp., $\Gamma(B)$), then $$ L=L_{i\, I\, l}^j\cdot b_j\otimes \frac{\partial^{|I|+l}}{\partial x^I\partial t^l}\circ \alpha^i $$ is the local expression of a generic element $L$ belonging to the right--hand side of $(*)$, which sits in the left--hand side if and only if $L_{i\, I\, l}^j=0$ for $l>0$, i.e., as you said, $L$ is of order 0 along the fibers of $p$.
P.S. I used coordinates $x^i$ on $M$ and $t$ on $[0,1]$; $I$ is a multi-index of length $|I|\leq k$, and $L_{i\, I\, l}^j\in C^\infty(M\times [0,1])$; I tacitly used the same bases for $A$ and $B$ and their pull-backs (it is only the algebra of scalars which grows bigger). |
This is not a complete answer, just a start as you indicate that you don't even know where to begin.
Let $f\in\Bbb{Z}[x]$ monic. If its image $f_p$ in $\Bbb{F}_p[x]$ is separable and factors as $f_p=\prod_{i=1}^kg_k$, then $\operatorname{Gal}(f)$ contains an element of cycle type $(\deg g_1,\ldots,\deg g_k)$. So it makes sense to start from an irreducibe polynomial $h\in\Bbb{F}_7[x]$ as then any lift $\tilde{h}\in\Bbb{Z}[x]$ already has an element of order $8$ in $\operatorname{Gal}(\tilde{h})$. An easy first candidate is$$h_7=x^8+x+3\in\Bbb{F}_7[x].$$To make sure we also have a transpotion in $\operatorname{Gal}(f)$, we choose a lift $\tilde{h}\in\Bbb{Z}[x]$ that factors into one quadratic and six linear factors mod $2$. A bit of fiddling around yields, for example$$\tilde{h}=x^8+x+3+7(x^7+x^6+x+1)\in\Bbb{Z}[x],$$so that$$h_2=x^8+x^7+x^6=x^6(x^2+x+1)\in\Bbb{F}_2[x].$$which shows that $\operatorname{Gal}(\tilde{h})$ contains a transposition. This doesn't quite give you that $\operatorname{Gal}(\tilde{h})\cong S_8$, but gets you on the right track.
EDIT: As pointed out in the comments below $h_2$ is not separable, so this choice of $\tilde{h}$ doesn't quite work. I have no doubt the argument can be salvaged, but the result will likely not be as pretty. I'll give it some thought tomorrow.
UPDATE: One way to salvage the argument is to take a larger prime $p$, so that a lift $\tilde{h}$ splits into six distinct linear factors and one irreducible quadratic factor. As $\deg\tilde{h}=8$ this requires $p\geq6$, hence $p=11$ is the smallest prime that might work. And indeed, surprisingly little fiddling around shows that the lift$$\tilde{h}=x^8+x+3+7(x+8)\big((x+3)(x+4)+6(x+1)(x+2)(x+5)(x+10)\big)\in\Bbb{Z}[x],$$satisfies$$h_p=(x+1)(x+2)(x+3)(x+5)(x+8)(x+10)(x^2+4x+5)\in\Bbb{F}_p[x],$$which shows that $\operatorname{Gal}(\tilde{h})$ contains a transposition. |
In "Classical Mechanics" book by Goldstein, Poole & Safko they develop the formulas of scattering to due to central-force problem.
In page 88, after receiving the expression for the differential cross section
$$ \sigma(\theta) =\frac{s}{sin \theta} |\frac{ds}{d\theta}|$$
They said that the relation between $s$(impact parameter), $\theta$ (deflection angle) can be achieved from the central-force problem formulas.
Questions:
How they have chosen such integration boundaries?
How did they use the formula: how the angles and radius match, for example: why does $r=\infty$ corresponds to $\theta=\pi$? |
Consider a probe approaching a planet for a hyperbolic flyby manoeuvre;
The eccentricity of the hyperbolic trajectory can be calculated using the following formula:
$$e = 1 + \frac{r_pv_\infty^2}{\mu_1}$$ $v_\infty$ refers to the probe's hyperbolic excess velocity. $\mu_1$ refers to the standard gravitational parameter $GM$ of the primary. $r_p$ refers to the periapsis radius.
In the event that eccentricity and periapsis radius are both unknown, however, this formula is insufficient. Does there exist an alternative formula that would allow the eccentricity and the periapsis radius to be worked out separately, or by use of simultaneous equations?
Assume for now that all other necessary values are known, excluding $e$ and $r_p$.
I suspect it could be possible using a known aiming radius $∆$. Where:
$$ \Delta= r_p\sqrt{1+\frac{2\mu_2}{r_pv_\infty^{2} }}, $$
But I unfortunately can't quite get my head around the maths. |
Let $T: \R^n \to \R^m$ be a linear transformation.Suppose that the nullity of $T$ is zero.
If $\{\mathbf{x}_1, \mathbf{x}_2,\dots, \mathbf{x}_k\}$ is a linearly independent subset of $\R^n$, then show that $\{T(\mathbf{x}_1), T(\mathbf{x}_2), \dots, T(\mathbf{x}_k) \}$ is a linearly independent subset of $\R^m$.
Let $A$ be the matrix given by\[A=\begin{bmatrix}-2 & 0 & 1 \\-5 & 3 & a \\4 & -2 & -1\end{bmatrix}\]for some variable $a$. Find all values of $a$ which will guarantee that $A$ has eigenvalues $0$, $3$, and $-3$.
Let\[A=\begin{bmatrix}8 & 1 & 6 \\3 & 5 & 7 \\4 & 9 & 2\end{bmatrix}.\]Notice that $A$ contains every integer from $1$ to $9$ and that the sums of each row, column, and diagonal of $A$ are equal. Such a grid is sometimes called a magic square.
Define two functions $T:\R^{2}\to\R^{2}$ and $S:\R^{2}\to\R^{2}$ by\[T\left(\begin{bmatrix}x \\ y\end{bmatrix}\right)=\begin{bmatrix}2x+y \\ 0\end{bmatrix},\;S\left(\begin{bmatrix}x \\ y\end{bmatrix}\right)=\begin{bmatrix}x+y \\ xy\end{bmatrix}.\]Determine whether $T$, $S$, and the composite $S\circ T$ are linear transformations.
Let\[\mathbf{v}_{1}=\begin{bmatrix}1 \\ 1\end{bmatrix},\;\mathbf{v}_{2}=\begin{bmatrix}1 \\ -1\end{bmatrix}.\]Let $V=\Span(\mathbf{v}_{1},\mathbf{v}_{2})$. Do $\mathbf{v}_{1}$ and $\mathbf{v}_{2}$ form an orthonormal basis for $V$?
Let $A$ be an $m \times n$ matrix.Suppose that the nullspace of $A$ is a plane in $\R^3$ and the range is spanned by a nonzero vector $\mathbf{v}$ in $\R^5$. Determine $m$ and $n$. Also, find the rank and nullity of $A$.
Suppose that a set of vectors $S_1=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\}$ is a spanning set of a subspace $V$ in $\R^5$. If $\mathbf{v}_4$ is another vector in $V$, then is the set\[S_2=\{\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3, \mathbf{v}_4\}\]still a spanning set for $V$? If so, prove it. Otherwise, give a counterexample.
For a set $S$ and a vector space $V$ over a scalar field $\K$, define the set of all functions from $S$ to $V$\[ \Fun ( S , V ) = \{ f : S \rightarrow V \} . \]
For $f, g \in \Fun(S, V)$, $z \in \K$, addition and scalar multiplication can be defined by\[ (f+g)(s) = f(s) + g(s) \, \mbox{ and } (cf)(s) = c (f(s)) \, \mbox{ for all } s \in S . \]
(a) Prove that $\Fun(S, V)$ is a vector space over $\K$. What is the zero element?
(b) Let $S_1 = \{ s \}$ be a set consisting of one element. Find an isomorphism between $\Fun(S_1 , V)$ and $V$ itself. Prove that the map you find is actually a linear isomorpism.
(c) Suppose that $B = \{ e_1 , e_2 , \cdots , e_n \}$ is a basis of $V$. Use $B$ to construct a basis of $\Fun(S_1 , V)$.
(d) Let $S = \{ s_1 , s_2 , \cdots , s_m \}$. Construct a linear isomorphism between $\Fun(S, V)$ and the vector space of $n$-tuples of $V$, defined as\[ V^m = \{ (v_1 , v_2 , \cdots , v_m ) \mid v_i \in V \mbox{ for all } 1 \leq i \leq m \} . \]
(e) Use the basis $B$ of $V$ to constract a basis of $\Fun(S, V)$ for an arbitrary finite set $S$. What is the dimension of $\Fun(S, V)$?
(f) Let $W \subseteq V$ be a subspace. Prove that $\Fun(S, W)$ is a subspace of $\Fun(S, V)$. |
Tagged: abelian group
Abelian Group Problems and Solutions.
The other popular topics in Group Theory are:
Problem 616
Suppose that $p$ is a prime number greater than $3$.
Consider the multiplicative group $G=(\Zmod{p})^*$ of order $p-1$. (a) Prove that the set of squares $S=\{x^2\mid x\in G\}$ is a subgroup of the multiplicative group $G$. (b) Determine the index $[G : S]$.
Add to solve later
(c) Assume that $-1\notin S$. Then prove that for each $a\in G$ we have either $a\in S$ or $-a\in S$. If a Half of a Group are Elements of Order 2, then the Rest form an Abelian Normal Subgroup of Odd Order Problem 575
Let $G$ be a finite group of order $2n$.
Suppose that exactly a half of $G$ consists of elements of order $2$ and the rest forms a subgroup. Namely, suppose that $G=S\sqcup H$, where $S$ is the set of all elements of order in $G$, and $H$ is a subgroup of $G$. The cardinalities of $S$ and $H$ are both $n$.
Then prove that $H$ is an abelian normal subgroup of odd order.Add to solve later
Problem 497
Let $G$ be an abelian group.
Let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively. Prove that there exists an element $c$ in $G$ such that the order of $c$ is the least common multiple of $m$ and $n$.
Also determine whether the statement is true if $G$ is a non-abelian group.Add to solve later
Problem 434
Let $R$ be a ring with $1$.
A nonzero $R$-module $M$ is called irreducible if $0$ and $M$ are the only submodules of $M$. (It is also called a simple module.) (a) Prove that a nonzero $R$-module $M$ is irreducible if and only if $M$ is a cyclic module with any nonzero element as its generator.
Add to solve later
(b) Determine all the irreducible $\Z$-modules. Problem 420
In this post, we study the
Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem.
Add to solve later
Problem. Let $G$ be a finite abelian group of order $n$. If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic to the cyclic group $Z_n=\Zmod{n}$ of order $n$. Problem 343
Let $G$ be a finite group and let $N$ be a normal abelian subgroup of $G$.
Let $\Aut(N)$ be the group of automorphisms of $G$.
Suppose that the orders of groups $G/N$ and $\Aut(N)$ are relatively prime.
Then prove that $N$ is contained in the center of $G$. |
This question already has an answer here:
Equation of state of a rubber band 1 answer
Which of these two equations of state are valid?
$$S_1 = L_0 \gamma (\theta E/L_0)^{1/2} - L_0\gamma\left[\frac{1}{2} \left(\frac{L}{L_0}\right)^2 + \frac{L_0}{L} - \frac{3}{2}\right]$$
$$S_2 = L_0 \gamma e^{\theta n E/L_0} - L_0\gamma\left[\frac{1}{2} \left(\frac{L}{L_0}\right)^2 + \frac{L_0}{L} - \frac{3}{2}\right]$$
where $\gamma$, $\theta$ are constants, $L_0$ is a function of $n$ (which is the mole number). $S$ has to be monotonically increasing and linear as a function of $E$ since it's extensive. Clearly $S_2$ can't fit that description, but $S_1 \propto \sqrt{E}$ which is not linear either, and it also scales that way for the system length too.
This is exercise 1.2 from David Chandler's introduction to statistical mechanics book. |
Answer
$\frac{2x-1}{2x}, x\ne0$
Work Step by Step
Canceling common terms and using exponent rules, we find: $\frac{\sqrt x-\frac{1}{2\sqrt x}}{\sqrt x}=\frac{\frac{2x-1}{2\sqrt x}}{\sqrt x}=\frac{2x-1}{2x}$
You can help us out by revising, improving and updating this answer.Update this answer
After you claim an answer you’ll have
24 hours to send in a draft. An editorwill review the submission and either publish your submission or provide feedback. |
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Mike Spivey
I am a math professor at the University of Puget Sound. My background is in operations research, and I teach typical OR courses such as optimization, modeling, and probability, as well as calculus, statistics, and differential equations.
My math blog,
A Narrow Margin, includes (among other things) discussion of some of my favorite posts - of mine and of others - from math.SE.
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19311 gold badge22 silver badges99 bronze badges View network profile → Top network posts 166 Why is $1^{\infty}$ considered to be an indeterminate form 116 Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$ 102 Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? 94 Ways to evaluate $\int \sec \theta \, \mathrm d \theta$ 90 Intuition behind using complementary CDF to compute expectation for nonnegative random variables 89 Probability density function vs. probability mass function 87 Sample Standard Deviation vs. Population Standard Deviation View more network posts → |
It is quiet easy to prove whether C4 has any effect on total capacitance when C2=C3=C5=C6. Here's your circuit:
simulate this circuit – Schematic created using CircuitLab
I will calculate total capacitance Ct of this circuit and if the final result does not contain Cx than it can be omitted, it has no effect. But if the final result contains Cx that it has some effect and cannot be omitted.
First step is to use \$ \Delta - Y \$ transformation:
simulate this circuit
\$C1=(Ca\cdot Cb+Cb\cdot Cc+Ca\cdot Cc)/Ca\$
\$C2=(Ca\cdot Cb+Cb\cdot Cc+Ca\cdot Cc)/Cb\$ \$C3=(Ca\cdot Cb+Cb\cdot Cc+Ca\cdot Cc)/Cc\$
So after transformation our circuit will look like this:
simulate this circuit
where
\$C1=(Cx\cdot C+C\cdot C+Cx\cdot C)/Cx=(2\cdot C\cdot Cx+C\cdot C)/Cx\$\$C2=(Cx\cdot C+C\cdot C+Cx\cdot C)/C=2\cdot Cx+C\$\$C3=(Cx\cdot C+C\cdot C+Cx\cdot C)/C=2\cdot Cx+C\$
After this transformation we can easily calculate total capacitance of this circuit.
Capacitance of C3 and C in series: \$C3c=\frac{C3\cdot C}{C3+C}=\frac{C(2\cdot Cx+C)}{2(Cx+C)}\$ Capacitance of C2 and C in series: \$C2c=\frac{C2\cdot C}{C2+C}=\frac{C(2\cdot Cx+C)}{2(Cx+C)}\$
simulate this circuit
C3c and C2c are in parallel so it is easy to join:
\$C32=C3c+C2c=\frac{C(2\cdot Cx+C)}{2(Cx+C)}+\frac{C(2\cdot Cx+C)}{2(Cx+C)}=\frac{C(2\cdot Cx+C)}{(Cx+C)}\$
And now the grand finale, the total capacitance (C1 and C32 in series):
\$Ct=\frac{C1\cdot C32}{C1+C32}=\frac{\frac{2\cdot C\cdot Cx+C\cdot C}{Cx}\cdot \frac{C(2\cdot Cx+C)}{(Cx+C)}}{\frac{2\cdot C\cdot Cx+C\cdot C}{Cx}+\frac{C(2\cdot Cx+C)}{(Cx+C)}}=\frac{\frac{C^2(2\cdot Cx+C)^2}{Cx(Cx+C)}}{\frac{C(2\cdot Cx+C)^2}{Cx(Cx+C)}}=C\$
So, as we can see, the capacitance (Ct) of the circuit is represented only by value of C, and there is no Cx, so it is proved that Cx has no effect on that circuit at all and can be omitted. |
For the symmetrized quantum field theory Lagrangian of the free dirac field
$$\mathcal{L} = i[\overline{\psi}_a,({\partial_\mu}\gamma^\mu \psi)^a] -m[\overline{\psi}_a,\psi^a ]$$
the terms are symmetrized since $\overline{\psi}_a$ and $\psi_a$ do not anti-/commute. Is this the correct symmetrized Lagrangian? Now I tried to order the terms in the Lagrangian such that
$$\mathcal{L} = i\overline{\psi}{\partial_\mu}\gamma^\mu \psi +A -m \overline{\psi}\psi +B $$
Can I determine $A,B$ with the canonical commutation relations since I do not know something about the commutation relations about $\overline{\psi}$ and ${\partial_\mu}\gamma^\mu \psi$? I hope to finde that
$$\delta \mathcal{L} = \delta (i\overline{\psi}{\partial_\mu}\gamma^\mu \psi) -\delta(m \overline{\psi}\psi)$$
with $\delta A = 0 = \delta B$ where $\delta$ is the variation. |
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