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## Seminars and Colloquia by Series

Wednesday, February 20, 2019 - 15:00 , Location: Skiles 006 , Steven Heilman , USC , Organizer: Galyna Livshyts

A single soap bubble has a spherical shape since it minimizes its surface area subject to a fixed enclosed volume of air. When two soap bubbles collide, they form a “double-bubble” composed of three spherical caps. The double-bubble minimizes total surface area among all sets enclosing two fixed volumes. This was proven mathematically in a landmark result by Hutchings-Morgan-Ritore-Ros and Reichardt using the calculus of variations in the early 2000s. The analogous case of three or more Euclidean sets is considered difficult if not impossible. However, if we replace Lebesgue measure in these problems with the Gaussian measure, then recent work of myself (for 3 sets) and of Milman-Neeman (for any number of sets) can actually solve these problems. We also use the calculus of variations. Time permitting, we will discuss an improvement to the Milman-Neeman result and applications to optimal clustering of data and to designing elections that are resilient to hacking. http://arxiv.org/abs/1901.03934

Wednesday, February 20, 2019 - 13:55 , Location: Skiles 005 , Steven Heilman , USC , Organizer: Galyna Livshyts

It is well known that a Euclidean set of fixed Euclidean volume with least Euclidean surface area is a ball. For applications to theoretical computer science and social choice, an analogue of this statement for the Gaussian density is most relevant. In such a setting, a Euclidean set with fixed Gaussian volume and least Gaussian surface area is a half space, i.e. the set of points lying on one side of a hyperplane. This statement is called the Gaussian Isoperimetric Inequality. In the Gaussian Isoperimetric Inequality, if we restrict to sets that are symmetric (A= -A), then the half space is eliminated from consideration. It was conjectured by Barthe in 2001 that round cylinders (or their complements) have smallest Gaussian surface area among symmetric sets of fixed Gaussian volume. We discuss our result that says this conjecture is true if an integral of the curvature of the boundary of the set is not close to 1. https://arxiv.org/abs/1705.06643 http://arxiv.org/abs/1901.03934

Series: Other Talks
Wednesday, February 20, 2019 - 12:00 , Location: 005 , Anna Kirkpatrick , Georgia Tech , , Organizer: Cvetelina Hill

Understanding the structure of RNA is a problem of significant interest to biochemists. Thermodynamic energy functions are often key to this pursuit, but it is well-established that these energy functions do not perform well when applied to longer RNA sequences. This work specifically investigates the branching properties of RNA secondary structures, viewed as plane trees. By employing Markov chain Monte Carlo techniques, we sample from the probability distributions determined by these thermodynamic energy functions. We also investigate some of the challenges in employing Markov chain Monte Carlo, in particular the existence of local energy minima in transition graphs. This talk will give background, share preliminary results, and discuss future avenues of investigation.

Wednesday, February 20, 2019 - 00:05 , Location: Skiles 006 , Jennifer Kloke , Ayasdi , Organizer: Kirsten Wickelgren

The Mapper algorithm constructs compressed representations of the underlying structure of data but involves a large number of parameters. To make the Mapper algorithm accessible to domain experts, automation of the parameter selection becomes critical. This talk will be accessible to graduate students.

Tuesday, February 19, 2019 - 16:30 , Location: Skiles 006 , Michael Wigal , Georgia Tech , Organizer: Xingxing Yu

Let $g(n) = \max_{|T| = n}|\text{Aut}(T)|$ where $T$ is a tournament. Goldberg and Moon conjectured that $g(n) \le \sqrt{3}^{n-1}$ for all $n \ge 1$ with equality holding if and only if $n$ is a power of 3. Dixon proved the conjecture using the Feit-Thompson theorem. Alspach later gave a purely combinatorial proof. We discuss Alspach's proof and and some of its applications.

Monday, February 18, 2019 - 14:00 , Location: Skiles 006 , Jen Hom , Georgia Tech , Organizer: Jennifer Hom

We show that the three-dimensional homology cobordism group admits an infinite-rank summand. It was previously known that the homology cobordism group contains an infinite-rank subgroup and a Z-summand. Our proof relies on the involutive Heegaard Floer package of Hendricks-Manolescu and Hendricks-Manolescu-Zemke. This is joint work with I. Dai, M. Stoffregen, and L. Truong.

Monday, February 18, 2019 - 13:55 , Location: Skiles 005 , , Rensselaer Polytechnic Institute , , Organizer: Wenjing Liao

Abstract: The Euclidean distance geometry problem arises in a wide variety of applications, from determining molecular conformations in computational chemistry to localization in sensor networks. Instead of directly reconstruct the incomplete distance matrix, we consider a low-rank matrix completion method to reconstruct the associated Gram matrix with respect to a suitable basis. Computationally, simple and fast algorithms are designed to solve the proposed problem. Theoretically, the well known restricted isometry property (RIP) can not be satisfied in the scenario. Instead, a dual basis approach is considered to theoretically analyze the reconstruction problem. Furthermore, by introducing a new condition on the basis called the correlation condition, our theoretical analysis can be also extended to a more general setting to handle low-rank matrix completion problems under any given non-orthogonal basis. This new condition is polynomial time checkable and holds for many cases of deterministic basis where RIP might not hold or is NP-hard to verify. If time permits, I will also discuss a combination of low-rank matrix completion with geometric PDEs on point clouds to understanding manifold-structured data represented as incomplete inter-point distance data. This talk is based on:1. A. Tasissa, R. Lai, “Low-rank Matrix Completion in a General Non-orthogonal Basis”, arXiv:1812.05786 2018. 2. A. Tasissa, R. Lai, “Exact Reconstruction of Euclidean Distance Geometry Problem Using Low-rank Matrix Completion”, accepted, IEEE. Transaction on Information Theory, 2018. 3. R. Lai, J. Li, “Solving Partial Differential Equations on Manifolds From Incomplete Inter-Point Distance”, SIAM Journal on Scientific Computing, 39(5), pp. 2231-2256, 2017.

Monday, February 18, 2019 - 13:00 , Location: Skiles 005 , Robert Walker , U Michigan , Organizer: Josephine Yu

We survey dissertation work of my academic sister Sarah Mayes-Tang (2013 Ph.D.). As time allows, we aim towards two objectives. First, in terms of combinatorial algebraic geometry we weave a narrative from linear star configurations in projective spaces to matroid configurations therein, the latter being a recent development investigated by the quartet of Geramita -- Harbourne -- Migliore -- Nagel. Second, we pitch a prospectus for further work in follow-up to both Sarah's work and the matroid configuration investigation.

Monday, February 18, 2019 - 11:15 , Location: Skiles 005 , Boris Kalinin , Penn State , Organizer: Rafael de la Llave

We will discuss the regularity of the conjugacy between an Anosov automorphism L of a torus and its small perturbation. We assume that L has no more than two eigenvalues of the same modulus and that L^4 is irreducible over rationals. We consider a volume-preserving C^1-small perturbation f of L. We show that if the Lyapunov exponents of f with respect to the volume are the same as the Lyapunov exponents of L, then f is C^1+ conjugate to L. Further, we establish a similar result for irreducible partially hyperbolic automorphisms with two-dimensional center bundle. This is joint work with Andrey Gogolev and Victoria Sadovskaya

Monday, February 18, 2019 - 10:10 , Location: Skiles 005 , Victoria Sadovskaya , Penn State , Organizer: Rafael de la Llave

We consider a hyperbolic dynamical system (X,f) and a Holder continuous cocycle A over (X,f) with values in GL(d,R), or more generally in the group of invertible bounded linear operators on a Banach space. We discuss approximation of the Lyapunov exponents of A in terms of its periodic data, i.e. its return values along the periodic orbits of f. For a GL(d,R)-valued cocycle A, its Lyapunov exponents with respect to any ergodic f-invariant measure can be approximated by its Lyapunov exponents at periodic orbits of f. In the infinite-dimensional case, the upper and lower Lyapunov exponents of A can be approximated in terms of the norms of the return values of A at periodic points of f. Similar results are obtained in the non-uniformly hyperbolic setting, i.e. for hyperbolic invariant measures. This is joint work with B. Kalinin.


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# Experiment in the Dark Fringe Raises Questions

I set up a single slit and produced this interference pattern on a distant wall.

I then placed another single slit in the deconstructive portion of the pattern. Here's a pic. The slit is between the two bright areas.

The dark fringe area produced no light after the second slit. Please consider my two questions.

1.If the dark areas are places where the waves are 180 out of phase then shouldn't the second slit disrupt the out of phase waves and create new wavelets that would the create a new interference pattern.

2.If the wave is not simply out of phase but destroyed. Where did the energy go? I put a beam splitter between the two slits and nothing returned to the source.

Here's a diagram of the setup.

• It is not clear to me what we are seeing in the 2nd image, because you say that "the dark fringe area produced no light after the slit." So what is the light that we are seeing? – sammy gerbil Oct 20 '16 at 17:57
• The second image shows the second slit positioned into a deconstructive zone between two constructive areas. The slit did not photograph very well. It is lost in the glare of the two constructive areas. – Lambda Oct 20 '16 at 18:27
• Does the 2nd image show only the light emerging through the 2nd slit? The one which you said "produced no light after the slit"? Or is the light in this image light which doesn't pass through the 2nd slit? – sammy gerbil Oct 20 '16 at 18:41
• It's the light that does not pass through slit 2. It's taken from the laser side of slit two. – Lambda Oct 20 '16 at 20:25

The question puts forward 2 hypotheses for what causes the dark fringes in a single-slit experiment. In (1), the hypothesis is that it's due to the cancellation of 2 waves which are 180-degrees out of phase. Whereas in (2), the hypothesis is that it's due to the wave simply getting "destroyed".

Neither of these are quite right, but (1) is much closer to the truth. It's not really that 2 particular waves are physically canceling each other, it's more that if you add together all possible paths that light could take from the slit to the wall, you end up with a pattern which is zero at certain points (the nodes / dark fringes) and non-zero at most points. The cancellation is between an infinite number of paths, but it's true that since most of them will cancel out anyway, you only really need to consider the light rays that come from either edge of the slit.

The particular pattern of intensity you end up with is called a sinc function ($\frac{\sin x}{x}$), or more specifically ${\rm sinc}^2 x$.

The intensity is only exactly zero at single isolated points. So no matter how small you make the second slit, there is no way to avoid some light going through. But as long as it is narrow enough and centered on a node, you can reduce the intensity of the light passing through to very close to zero. So likely, there is some minimal amount of light getting through the 2nd slit you made, you are just not seeing it because it's far too dim.

You can do all of this using water waves instead of light waves if you prefer. Create a pool with a barrier which has a narrow slit near the middle. Then make some waves on one side of the pool, and see which waves end up getting through to the other side and crashing on the far wall. You can think of the waves as individual points in the pool moving up and down. Each time any point in the pool oscillates up and down, it send ripples out in all directions. Then all of the surrounding points are waving up and down, and those end up also sending out ripples in all directions. It doesn't make much of a difference whether there is a 2nd slit or not, if not many ripples are making it to a particular point on the wall, making a slit there is not going to change that. Not much light (or water waves) will get through a slit which you place near a dark fringe. Whereas a lot of light (or water waves) will get through a slit you place near a bright fringe.

At no point does anything like (2) happen where waves are simply destroyed and energy is lost.

• After reading your answer I spent some time investigating sinc functions. Your answer was helpful. It shifted my thinking from mechanical to probabilities. Given time and effort I will gain insight. For those like me trying to understand this here's a helpful video.youtube.com/watch?v=XaJnygMuT7M – Lambda Oct 20 '16 at 14:27

I do not understand the thinking behind your 1st question. In what way do you think the 2nd slit "disrupts the out-of-phase waves and create new wavelets that would create a new interference pattern"? Perhaps as follows :

We can think of the illumination at slit 2 as either (a) individual light waves from different parts of slit 1 arriving with approximately the same amplitude but different phases, or (b) darkness resulting from the collective destructive interference of the waves in (a). In case (b) it is obvious that there will be darkness behind slit 2, but what about case (a)?

The different parts of the 1st slit can be thought of as coherent sources (Huygens' Construction). Each of these sources produces its own diffraction pattern behind the 2nd slit, similar to that in your 1st image. Each diffraction pattern has its own phase - or more accurately, each point in the pattern has a different phase depending on which source created it. But just as the sources at slit 1 are out of phase and interfere destructively at slit 2, so the diffraction patterns behind slit 2 are also out of phase, and interfere destructively at the screen.

So there is darkness behind the 2nd slit whichever way you think of the incident light.

Your 2nd question is answered in What happens to the energy when waves perfectly cancel each other? and many similar questions such as How does interference move energy from destructive to constructive regions? These answers explain that energy which is "lost" at places of destructive interference "reappears" at places of constructive interference. You cannot have destructive interference at one place without having constructive interference somewhere else. So there is always conservation of energy.

But these answers don't seem to explain why or how this happens. The explanation may be that waves are not objects themselves but descriptions of the transfer of energy or probability or something else. This distinction is explicit in quantum physics, between wave-functions and the particles they describe. So individual photons do not annihilate at points where there is destructive interference of the wave-function. But interference also occurs in Newtonian physics - eg sound in air, ripples on water. If we were to examine the collisions between elements of air or water, we would see the mechanics by which energy is transferred away from or towards certain regions at which we say destructive or constructive interference is occurring.

The wave picture is a shorthand method of describing some more complex underlying mechanism. Although it makes some things easier to understand (such as the appearance of interference patterns), it makes other things more difficult to understand (such as the disappearance of energy from one point and its reappearance somewhere else).

• Thanks for your response. My thinking behind my first question was that the light energy was still present in the dark fringes but cancelled. Kind of like two generators that are running in parallel, when their waves are in sync then the difference of potential between the two is zero, even though they may be producing high voltage when compared to ground. I was thinking the second slit would shift the phase, by causing the waves to travel different distances and expose the "voltage." Obviously, my experiment showed that my theory was incorrect. – Lambda Oct 21 '16 at 3:06

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Podcast #128: We chat with Kent C Dodds about why he loves React and discuss what life was like in the dark days before Git. Listen now.

# Tag Info

23

Compare the following two images: The path planning is somewhat trivial. There's only one path: the rope. The motion planning on the other hand is not that easy. In a maze the path planning is hard and motion planning is easy: Of course both planning tasks can be easy or hard at the same time or anything in between. They are linked to one another in that ...

9

First, we need to define optimal. Since you do not say what you consider optimal, most people choose a quadratic expression. For example, suppose your current joint angles are given by the vector $\vec{\alpha}$. We can consider minimizing the movement required - with an error $\vec{x} = \vec{\alpha} - \vec{\alpha}_{start}$, you can define a cost function $J=\... 6 You are correct that kd-trees typically only work in small, Euclidean metric spaces. But, there is lots of work for general nearest neighbor applications in metric spaces (anywhere you can define a distance function essentially). The classic work is on ball trees, which then were generalized to metric trees. There is some newer work called cover trees ... 6 In order to answer my own question Configuration space and Joint space must be defined. Configuration space of a rigid body is a minimum set of parameters that can determine position of each point in that body or Configuration space is set of all possible configurations of that body. Configuration space of the end-effector is set of all possible positions ... 6 I implemented something like this in College: https://github.com/Auburn-Automow/au_automow_common/tree/master/automow_planning Basically we just passed the vertices of the boustrophedon path as goals to move_base. Here's a video of a bag file being played back: https://www.youtube.com/watch?v=R7nLgYquECg Here's the class paper we did for the planner: ... 6 I think you may be misunderstanding the nature of gimbal lock. It sounds like you may be trying to remove an actual term in a rotation matrix calculation , but this is incorrect because each axis is still able to rotate. What happens with gimbal lock is that one of the rotational degrees of freedom of the object you are rotating is removed. This happens when ... 6 What's the difference between turn-by-turn GPS and driving a car? GPS is path planning: high-level commands like, "turn right in 1 mile." Driving is motion planning, which means following a route established by path planning while at the same time taking care of the minutia: interfacing with the car, staying in lane, watching for pedestrians, obeying ... 5 In robotics the configuration space is exactly the joint space of the manipulator. Differently, to indicate the space where the forward kinematic law maps the joints configuration into, we use the terms task space and operational space, equivalently. 5 Kinematic constraints involves only constraint on the motion (kinematics means the study of motion without considering the force that causes it), which may involve configuration variables and their time derivatives (including higher order derivatives). In particular, no inertial parameters are involved in the constraint. The constraint is geometric in nature,... 5 Since the configuration space is the set of all possible configurations the link can have, i.e. all possible angles from 0° to 360°, shouldn't the c-space be a line rather than a circle? You are correct that a rotating link has one variable$\theta$that can attain any value in the range$[0, 2\pi]$(let's talk in radian). However, the configuration space ... 5 Obstacle padding/ robot padding. Suppose you are working in a 2D environment and that you have an obstacle of the size 2x2. When doing planning (graph search, etc.), you increase the size of the obstacle to, for example, 3x3. Then when you find a path, the path is guaranteed to be at least of the distance 1 away from the actual obstacle. As for a smoother ... 5 In general you need 6 parameters to describe the position and orientation of any joint with respect to a link coordinate frame. The DH parameterisation includes 2 constraints so only 4 parameters are required. The constraints are: axis$x_j$intersects axis$z_{j-1}$axis$x_j$is perpendicular to axis$z_{j-1}$(see Robotics, Vision & Control, ... 4 I believe it simply boils down to what your robot can do. If you are for some reason restricted to moving only in 4 directions, then you connect each grid cell to 4. If you can go in 8 directions, you connect each grid cell to 8. If you can go in 6 directions, you use a honeycomb grid and connect each grid cell to 6. There is no generic answer. If you ... 4 *Note,$a|b$is the concatenation of paths$a$and$b$. Then$c(\cdot)$defined as the minimum clearance implies$c(a|b)=min(c(a),c(b))$You refer to (in reference 1): Theorem 11: (Additivity of the Cost Function.) For all$\sigma_1$,$\sigma_2\in X_{free}$, the cost function c satisﬁes the following:$c(\sigma_1|\sigma_2) = c(\sigma_1) + c(\...

4

Classical control theory requires Linear Time Invariant systems. Most robots of interest actually have non-linear dynamics. Of course many optimal control techniques also require linear systems. LQR for instance stands for Linear Quadratic Regulator meaning it takes a quadratic cost function and gives the optimal control for a system with linear dynamics. ...

4

The state space is the space of possible values that the state can take. For any given system it depends on which variables you are taking into account. For instance it is common to consider the 2D position $(x, y)$ and the orientation $\theta$ of the system. For this example the state space consists of three dimensions where $x$ and $y$ could conceivable ...

4

Line following is a simple reactive behaviour. Before you get into planning to solve the obstacle avoidance problem - which can get quite complex - you should consider simpler solutions. What this simpler solution could be depends a bit on the environment and on your robot. But let me make a suggestion: while() follow line if( detect obstacle ) ...

4

Let me put homotopy into the context of planning algorithms Suppose you want to get from point A to point B. Clearly, the easiest way is to traverse a straight line. But if there is an obstacle in the way of this straight line path, what should you do? If you want to obtain the "optimal" path (e.g. traversing the least total distance to the goal), you ...

4

It's really easy. env = openravepy.Environment() env.StopSimulation() env.Load('_ArmFileName'_.xml') #Whatever .xml file you're using robot = env.GetRobots()[0] manip = robot.SetActiveManipulator('left_arm') #whatever arm you're using robot.SetDOFValues(joint_values, manip.GetArmIndices()) env.CheckCollision(robot) where joint_values is the an array of the ...

4

They are modeling the probability as a normal distribution with the given mean and variance.

4

I think that vector field histogram method should be a good solution here. It's a method of local motion planning (avoiding local obstacles while navigating to a global target). It involves mapping measurements into cartesian occupancy grid, and making a polar obstacle density histogram from that grid. Later the direction with lowest obstacle density and ...

4

If you are able to sense obstacles with a sensor pattern that is circular (eg laser scanner, contact sensors on a circular body, etc), and you can rotate the robot pose without translation, then you can satisfy the assumptions of the Bug algorithm. If you use a point model for your robot in the map, then you grow the obstacles by the radius of your robot. ...

4

Since the problem is one dimensional, you are actually asking to compute a velocity profile. (A velocity profile is the information of how a path is traversed with respect to time.) Now the problem is "How to travel for $B$ units within time $T$?" (Let's call the duration $T$ instead.) A velocity profile can be viewed as a curve in the $v$-$t$ (velocity vs ...

4

Many articles reference algorithms such as A*, PRM or RRT based planners to motion planning algorithms which seems unreasonable since it is still necessary to parametrize found path with time.I wonder, why? First of all, RRT, for example, can be used to plan trajectories directly. When the robot in question has $n$ DOFs, such a planning problem happens in a ...

3

There are several traps you might have stepped into, but it is difficult to tell without more information. The first issues that came to my mind: The equations you wrote down are for sampling from the velocity motion model. But then you write about the Kalman Gain approaching singularity, which only makes sense of you apply a Gaussian filter (EKF or UKF). ...

3

You have an interesting approach, but I think it's the wrong approach; you've painted yourself into a corner by trying to avoid some technical obstacles (instead of just tackling them). Based on the information you've provided, it sounds like the goal is for this robot to successfully mow an irregularly-shaped lawn while staying within an invisible ...

3

In case you're not already aware, the problem you are asking about is generally referred to as the two-point boundary value problem. For some systems, a closed form solution may be extremely difficult to calculate and may not even exist. As such it would help to know more about your dynamics. Your description would seem to imply that you are working with a ...

3

If both entrance and exit of the maze is at the edges of the maze, the left/right hand (wall following) algorithm should work. If you start following a wall that is connected to the exit, you could never get into a loop. If you want the robot to be able to start in the middle of a maze with loops then simple wall following is not enough. I would recommend ...

3

I'm not sure what the "standard techniques" are but here are a few a ideas. You could combine a standard jacobian based forward kinematics as your base and then apply a gradient search algorithm where you bias against things such as running out of slack on your wires or other constraints. Let me go into a bit more detail. Generally you would just give the ...

3

To expand on Alexandre's answer: Controlling the arm with the Jacobian along an end-effector trajectory is one way to do it. However, this will not give you obstacle avoidance. Although you could manually check for collisions at each time step, but i imagine this could get ugly. Additionally, this is a gradient technique, so you will be constrained to the ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# Speaker BJT Amplifier - very low gain

First of all this is not related to a HW or anything; I just wanted to build an audio amplifier and test my skills. I really don't see why my design is not working as intended, without the speaker load it amplifies the input signal just right but when I connect the speaker load (4 Ohm) the power of the speaker is limited to microwatts. Can anyone help? Sorry I didn't sketch the circuit in the editor since I already had drawn it in LTspice P.S. I actually have built the circuit but I can barely hear something coming from the speaker when I blow on the mic.

• Your values for R4 and R5, especially R5, limit the current available to the low resistance of the load, so the load power is very low. Ignoring C2 for a moment, the maximum voltage across the speaker is 24 x (4/2204), or about 44 mV, for a maximum peak power of 47 uW. – WhatRoughBeast Jun 16 '16 at 15:24
• Please space out your schematic, it's very hard to read. Also, you might consider linking us the LTSpice file so we can verify your simulation. Finally, the 1 mF capacitors look suspicious, are those the intended value? – uint128_t Jun 16 '16 at 15:25
• This circuit cannot be used as a POWER amplifier. Its a voltage amplifier but it has virtually no current driving capability. Connecting a 4 ohm speaker at the final stage is like adding a short circuit to the output. – JIm Dearden Jun 16 '16 at 16:15
• Did you even look at any audio power amplifier design before you started? – Brian Drummond Jun 16 '16 at 16:50
• Suggested reading: Nelson Pass Zen amplifier, or Douglas Self's various writings on audio amplifiers. Your output stage is not practical for a power amplifier. All the current that goes to the speaker has to pass through R5. – mkeith Jun 16 '16 at 16:54

## 2 Answers

The gain of the last stage is approximately equal to: $$Gain_3 = {{R8}\over{R4}} = {{4}\over{2200}}=0.001818$$ You must use a push-pull emitter follower. Transistors desirable to take a large gain. With such a supply voltage using two stage voltage gain enough ($Gain=1000 \div 5000$) output and push-pull emitter follower (gain approximately equal 1). The last stage can be done on a single-ended power transistor with the radiator cooling. But modal current in this case is very high.

When you take an output signal from the collector you are buying in to a circuit that has a signal gain that is governed by collector resistance divided by emitter resistance. This is because the signal on the base is also the same amplitude (within reason) as the signal on the emitter.

Because collector current largely equals emitter current it then becomes a triviality to say that gain = Rc/Re.

So, your final stage would have a gain of 1 because Rc = Re but you then put a 4 ohm speaker in parallel with Rc (via the output capacitor) and gain rockets down to a fraction of what it was.

Virtually all common-place amplifiers buffer the output (the sensitive colelctor node) with an emitter follower (push pull on more powerful amps).

Another way of looking at it is by saying the signal at the collector is sourced from an impedance equal to the collector resistor and of course if you have 1Vp-p being fed via 2k2 to a 4 ohm speaker your get 1.81 mVp-p.


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zbMATH — the first resource for mathematics

Subgraphs and well-quasi-ordering. (English) Zbl 0762.05093
Let $$K$$ be a class, a quasi-ordering $$\preccurlyeq$$ on $$K$$ is said to be a well-quasi-ordering (shortly a wqo) if for every infinite sequence $$x_ 1,x_ 2,\dots$$ of elements of $$K$$ there are $$i,j$$, $$i<j$$, such that $$x_ i\preccurlyeq x_ j$$. For the class $$G$$ of all finite simple graphs neither the subgraph relation $$\subseteq$$ nor the induced subgraph relation $$\preccurlyeq$$ is a wqo.
In the paper ideals $$H$$ of $$G$$ are studied such that $$\subseteq$$ or $$\preccurlyeq$$ is a wqo on $$H$$. The following theorem is proved: Let $$H$$ be an ideal of $$G$$, with respect to $$\subseteq$$, then the following are equivalent: (1) $$(H,\subseteq)$$ is a wqo; (2) $$(H,\preccurlyeq)$$ is a wqo; (3) $$H$$ contains only finitely many $$C_ n$$ and $$F_ n$$, where $$C_ n$$ is a circuit on $$n$$ vertices, $$F_ n$$ is the graph on $$\{y_ 1,y_ 2,z_ 1,z_ 2,x_ 1,x_ 2,\dots,x_ n\}$$ with the edge set $$\{y_ 1x_ 1,y_ 2x_ 1,z_ 1x_ n,z_ 2x_ n,x_ 1x_ 2,x_ 2x_ 3,\dots,x_{n-1}x_ n\}$$.
For $$\preccurlyeq$$ three wqo ideals are presented, one being the class of all graphs without the path on $$n$$ vertices as a subgraph, two others are classes of bipartite graphs.
Connections between the obtained results and digraphs are considered (although no necessary and sufficient condition for an ideal $$D$$ of digraphs for which $$(D,\subseteq)$$ is a wqo is known yet). Examples are constructed disproving some of the possible generalizations of the stated results.
Reviewer: M.Demlová (Praha)

MSC:
 05C99 Graph theory 06A06 Partial orders, general
Full Text:
References:
 [1] Damaschke, J. Graph Theory 14 pp 427– (1990) [2] Stable sets verus independent sets. Discrete Math. Accepted. [3] Algorithmic Graph Theory and Perfect Graphs. Academic Press, New York (1980). · Zbl 0541.05054 [4] Higman, Proc. London Math. Soc. 2 pp 326– (1952) [5] and , Graph minors–A survey. Surveys in Combinatorics, London Mathematics Society Lecture Note 103 (1985). · Zbl 0568.05025
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# How do you simplify (2r -5)(r + 10)?

Sep 22, 2015

$2 {r}^{2} + 15 r - 50$

#### Explanation:

To do this we use a technique called FOIL. It stands for firsts, outers, inners, and lasts.

$\left(2 r - 5\right) \left(r + 10\right)$
Here, $2 r$ and $r$ are the firsts, $2 r$ and $10$ are outers, $- 5$ and $r$ are inners, and $- 5$ and $10$ are lasts. Solving the problem only requires that we multiply these pairs together and combine them.

$\left(2 r - 5\right) \left(r + 10\right)$
$\left(2 r \cdot r\right) + \left(2 r \cdot 10\right) + \left(- 5 \cdot r\right) + \left(- 5 \cdot 10\right)$
$2 {r}^{2} + 20 r - 5 r - 50$
$2 {r}^{2} + 15 r - 50$


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# Balanced vs unbalanced superposition distinguisher

I've been looking at basic quantum algorithms such as the Deutsch-Jozsa algorithm that are able to characterize functions very well and I was wondering if similar approaches exist to characterize quantum states.

Consider the following example: Given a normalized state $$\sum_{i=1}^{N} a_i\vert x_i\rangle$$, is there some way to check if all the $$a_i = 1/\sqrt{N}$$ or if there exists some $$a_i$$ that are different from the others?

My initial lines of thinking were to apply a Hadamard on the state but for large $$N$$, it's not clear if that approach helps. Do you have any ideas on how to proceed, or is it the case that such problems are not things that quantum computers can help with?

EDIT: I should have emphasized some features of my question better

1) $$N$$ is large, yet the "unbalancedness" i.e. the variance among the $$a_i$$ can be small. Hence POVM measurements will succeed in giving me the answer with very low probability.

2) Methods that distinguish between two known non-orthogonal states seem slightly hard to apply since my question is subtly different. I ask if my state is of a specific kind (balanced) or not. Intuitively, my question seems easier to answer.

3) I should have stated it earlier but one can assume that one is allowed $$c\ll N$$ copies of the state if this helps.

It is not possible at an information-theoretic level to do what you want to do.

Let us suppose we have two pure states: $$|\phi\rangle$$ and $$|\psi\rangle$$, where $$|\phi\rangle = \frac{1}{\sqrt{N}}\sum_{i=1}^N |x_i\rangle$$ and $$|\psi\rangle$$ is similar to $$|\phi\rangle$$ but with a few of the coefficients tweaked in some way. Notice that we're fixing just two states, and even if you're promised that you're given one of these two states it will not be possible to determine which one you're given with high probability, under the assumption that $$|\phi\rangle$$ and $$|\psi\rangle$$ are close together. (As DaftWullie has suggested, a variant of this problem where $$|\phi\rangle$$ is fixed and $$|\psi\rangle$$ is not known ahead of time is certainly no easier than the case in which $$|\psi\rangle$$ is known ahead of time.)

To keep things simple, let us suppose that we're given $$|\phi\rangle$$ with probability $$1/2$$ and $$|\psi\rangle$$ with probability $$1/2$$, and we're aiming to maximize the probability of correctly determining which of the two states we were given. A theorem sometimes called the Holevo-Helstrom theorem tells us exactly what the optimal probability of a correct guess is: $$\frac{1}{2} + \frac{1}{4} \bigl\| |\phi\rangle\langle \phi| - |\psi\rangle\langle \psi| \bigr\|_1,$$ where the norm is the trace norm. Because we're working with pure states, this expression can be simplified to $$\frac{1}{2} + \frac{1}{2}\sqrt{1 - |\langle \psi | \phi \rangle|^2}.$$ This is for the optimal measurement; you cannot do any better than this, no matter what you try to do, assuming you start with one of the two states selected at random and by honest means do your best to determine which state you were given.

Now, the question suggests that the "unbalancedness," or variance among the coefficients of the states, is small. With that in mind we could define $$\varepsilon = 1 - |\langle \psi | \phi \rangle|^2$$ and regard $$\varepsilon$$ as a small (but nonnegative) real number. This means that our probability of a correct guess is $$\frac{1}{2} + \frac{\sqrt{\varepsilon}}{2},$$ which is not much better than just randomly guessing (which yields a correct answer with probability 1/2).

If you assume that you are given $$k$$ copies of $$|\phi\rangle$$ or $$k$$ copies of $$|\psi\rangle$$, the calculation is exactly the same, except replacing the states with $$|\phi\rangle^{\otimes k}$$ and $$|\psi\rangle^{\otimes k}$$, also as DaftWullie has suggested. We can see how this will affect the optimal correctness probability: it now becomes $$\frac{1}{2} + \frac{1}{2}\sqrt{1 - |\langle \psi | \phi \rangle|^{2k}}.$$ Notice that we're placing no constraints on the measurements for distinguishing the two cases -- they could be arbitrarily correlated across the $$k$$ copies of the states.

We can compare this optimal correctness probability with the one-shot case by using the inequality $$1 - |\langle \psi | \phi \rangle|^{2k} \leq k \bigl(1 - |\langle \psi | \phi \rangle|^2\bigr).$$ To prove this inequality, notice that it is trivial if $$|\langle \psi | \phi \rangle| = 1$$, and otherwise we can use the crude estimate $$\frac{1 - |\langle \psi | \phi \rangle|^{2k}}{1 - |\langle \psi | \phi \rangle|^2} = 1 + |\langle \psi | \phi \rangle|^2 + \cdots + |\langle \psi | \phi \rangle|^{2(k-1)} < k$$ for the case $$|\langle \psi | \phi \rangle| < 1$$. We find that the optimal correctness probability is upper-bounded by $$\frac{1}{2} + \frac{\sqrt{k\varepsilon}}{2}.$$ So, if $$k$$ is considered to be much smaller than $$N$$, and $$\varepsilon$$ is on the order of $$1/N$$, then we're still not doing much better than randomly guessing.

There are many different variants depending on what it is precisely that you want to achieve (note, this was written before recent edits, although I think there is still value/relevance in this more general answer).

The closest analogy to the Deutsch-Jozsa algorithm is probably to say that you're given one of two states, and you want to know which you've been given (with maximum probability). If those two states are orthogonal, you can perfectly distinguish them just by using projective measurements, two of which project onto those particular states. Beyond the orthogonality assumption, there are ways of selecting the measurement to maximise the probability of distinguishing them. For more details, see this answer and this one.

On the other hand, if I take your question literally (which one certainly should given the edits to the question), then probably the best strategy is to define the projectors $$E_1=|\psi\rangle\langle\psi|^{\otimes c}\qquad E_2=\mathbb{I}-E_1,$$ where $$|\psi\rangle$$ is the uniform superposition state ($$a_i=1/\sqrt{N}$$), and $$c$$ copies are allowed. If you get the measurement result $$E_1$$, you learn essentially nothing. However, if you get the answer $$E_2$$ (which may well happen with low probability if the $$a_i$$s are close to uniform), you know that the state definitely was not $$|\psi\rangle$$, so some of the amplitudes were different.

• Thank you very much for the answer. I apologize for not emphasizing the features of my question better but I have edited the question to make it clearer. POVMs would succeed with very low probability in my case and I'm not sure how non-orthogonal distinguishing measurements work (since in my case the unbalanced state can be unbalanced in many ways - it isn't one specific unbalanced state). Your answer is, of course, completely correct but I will leave it open for a bit to see if a more efficient solution is possible. Thank you – user1936752 Nov 6 '18 at 10:26
• @user1936752 I believe that, for a single copy, this is the best that you can do. For multiple copies, you just update $E_1=|\psi\rangle\langle\psi|^{\otimes c}$. This is quite effective at improving your success probability. – DaftWullie Nov 6 '18 at 10:33
• @user1936752 I suppose one thing to take away from the first part of my answer is that, even if you fix just one state (which is certainly easier than your stated problem), the probability of success is very low. – DaftWullie Nov 6 '18 at 10:36


text:
# Math Help - Probability questions

1. ## Probability questions

Ok, I completely useles at probability and stats and don't understand these questions.

Computer parts are exponentially distributed and have a mean lifetime of 1000 hours, and are independent of each other.

What proportion are in excess of 100 hours?

Boxed in packages of 10, each box is to have at least 9 in excess of 100 hours. Calculate the percentage?

Also gold coins are found 1 per 1km sq. Jim has a metal detector and searches 10km sq.

Let N be the number of coins in that area. State the distribution X.

And the probability of finding a coin is 0.6. Let X be the number of coins found and N=n, what is the distribution of X. The calculate the mean of this distribution.

Is it a Poisson distribution?

2. Originally Posted by Happy Dancer
Ok, I completely useles at probability and stats and don't understand these questions.

Computer parts are exponentially distributed and have a mean lifetime of 1000 hours, and are independent of each other.

What proportion are in excess of 100 hours?

Boxed in packages of 10, each box is to have at least 9 in excess of 100 hours. Calculate the percentage?

[snip]
The pdf of X is $f(x) = 0.001 e^{-0.001 x}$.

(a) $\Pr(X > 1100) = \int_{1100}^{+\infty} 0.001 e^{-0.001 x}$.

(b) Makes no sense to me. Is that exactly how the question was asked?

Originally Posted by Happy Dancer

[snip]
Also gold coins are found 1 per 1km sq. Jim has a metal detector and searches 10km sq.

Let N be the number of coins in that area. State the distribution X. Mr F says: Do you mean N, not X here?

And the probability of finding a coin is 0.6. Mr F says: Finding a coin where? In 1 km sq? 10 km sq?

Let X be the number of coins found and N=n, what is the distribution of X. The calculate the mean of this distribution.

Is it a Poisson distribution?
I'm sorry but I cannot make sense of most of these questions the way you have posted them. You will need to post them more carefully and exactly as they are written in the book (or wherever it is you got them from) if you want to get help with them.


text:
flexspin compiler for P2: Assembly, Spin, BASIC, and C in one compiler - Page 15 — Parallax Forums

flexspin compiler for P2: Assembly, Spin, BASIC, and C in one compiler

• Posts: 4,885
Rayman wrote: »
If you do this with "OBJ", then you must need some kind of Spin "Start" function to start the assembly cog, right?

Sounds like you're saying to just do it the same way as with P1...

Yes. The P1 way of doing objects seems to work well, and provides good namespace isolation and modularity. Did you have something else in mind?
• Posts: 11,962
Maybe it's OK. I'll have to try it out and see...
• Posts: 8,690
Are there version of FastSpin, Spin2CPP, and Spin2gui available for or planned for Linux?
• Posts: 10,907
Ah, yep, that's the native development environ. I see mention of Debian and Ubuntu explicitly.
• Posts: 4,885
kwinn wrote: »
Are there version of FastSpin, Spin2CPP, and Spin2gui available for or planned for Linux?

There are no pre-packaged binaries, but the default for the source code is to build for Linux. I used to provide pre-built binaries for Linux, but nobody was downloading them. I think most Linux users prefer to build code themselves.

The build process for fastspin/spin2cpp is very straightforward:
  git clone https://github.com/totalspectrum/spin2cpp
cd spin2cpp
make

Binaries are placed in the "build" directory.

• Posts: 337
I've been working on getting my USB "boot protocol" mouse/keyboard host/drive converted to a spin object. I've got the USB controlRead/controlWrite transfer routines working and I can enumerate a device. Currently the code is all cog/lut execution, but I'm out of space. To get a device configured and transferring mouse/keyboard data will require hub execution too, and my first attempts at implementing it haven't gone well.

Is "native" p2 hub execution in a DAT block supported by fastspin?

I'm sure my woes are due to not fully understanding which addressing mode to use when calling hub code from cog/lut, but trying to determine the "right" way has me scratching my head::

CALL #hubsub fails.
CALL #\hubsub fails.
CALL #@hubsub works.

Aren't #2 and #3 essentially the same?

#2 compiles to:
01690     BC 0B A0 FD |                 call    #\testsub

and #3:
01690     D0 21 A0 FD |                 call    #@testsub

I haven't found anything in the fastspin docs regarding hub execution in a DAT block of a spin object. If it is doable, IMO a "Cliff Notes" regarding JMP/CALL addressing "gotchas" for cog/lut/hub regions would be a welcome addition.

Any insight would be greatly appreciated!
Thanks!
• Posts: 4,885
The four calls
   call #testsub
call #\testsub
call #@testsub
call #\@testsub

are potentially all different. "\" forces the call to be compiled as an absolute address rather than PC relative, and "@" forces the assembler to use the HUB address of testsub rather than the default address. The default address is the same as the HUB address after an ORGH, but is not the same after ORG. For example:
   ORGH $400 ORG 0 call #label1 call #\label1 call #@label1 call #\@label1 label1 ret  There "label1" is located in HUB memory at address$410, but its COG address is 4 (COG memory is addressed by longwords). So this is basically assembled the same as:
   call #4    ' will be assembled as a relative call in COG memory
call #\4   ' will be assembled as an absolute call in COG memory
call #$410 ' will jump to HUB memory (absolute call if we start in COG) call #\$410 ' always an absolute jump to HUB memory

All this is the same in both fastspin and PNut, by the way.

If you want to use hubexec in a Spin object, which is only available in fastspin, then I would recommend (a) using the most recent version of fastspin (3.9.25), and (b) always putting @ in front of labels used for HUB calls. (b) isn't always necessary in pure PASM code: if the last ORG type directive before a label was an ORGH then the label will have its HUB value by default, so label and @label are the same. But in Spin objects the compiler may not compute the HUB address of "label" correctly if there's no @ in front of it. It's not clear yet what the "official" Spin2 compiler will do in this case.
• Posts: 11,962
I think the backslash means absolute address.
Don’t think that can work with spin...
• Posts: 337
edited 2019-04-28 00:24
ersmith wrote: »
All this is the same in both fastspin and PNut, by the way.

If you want to use hubexec in a Spin object, which is only available in fastspin, then I would recommend (a) using the most recent version of fastspin (3.9.25), and (b) always putting @ in front of labels used for HUB calls. (b) isn't always necessary in pure PASM code: if the last ORG type directive before a label was an ORGH then the label will have its HUB value by default, so label and @label are the same. But in Spin objects the compiler may not compute the HUB address of "label" correctly if there's no @ in front of it. It's not clear yet what the "official" Spin2 compiler will do in this case.
Thanks, your info has removed the scales from mine eyes . It's been at least a decade since I last did anything meaningful in spin, and my aging brain is taking longer than I'd like to get back in the groove

Thanks again for the quick response!
• Posts: 17,701
IIRC there is details of the / @ etc label prefixes in the Tricks and Traps Reference thread and followup discussion in the Tricks and Traps Discussion thread. There are some gotchas here where the compilers seem to do different things. There is also reference to a further thread where its' discussed in more detail. I suggest you look at the listing to see what has been used where it really matters.
• Posts: 673
@ersmith Here's a bug submission for SELECT CASE in BASIC.

Code written in this manner only runs the code in ELSE, despite similar code listed in the basic.pdf manual.
Issue: It will always just print "Something Other Than A or B Was Pressed" despite sending an A or B keypress over serial.
    PRINT "PRESS A or B"
K$= INPUT$(1)
KEYPRESS = ASC(K$) PRINT " "; KEYPRESS SELECT CASE K$
CASE "A"
PRINT "KEY A Was Pressed"
CASE "B"
PRINT "KEY B Was Pressed"
CASE "a"
PRINT "KEY a Was Pressed"
CASE "b"
PRINT "KEY b Was Pressed"
CASE ELSE
PRINT "Something Other Than A or B Was Pressed"
END SELECT

PRESS A or B
C 67
Something Other Than A or B Was Pressed
PRESS A or B
B 66
Something Other Than A or B Was Pressed
PRESS A or B
A 65
Something Other Than A or B Was Pressed
PRESS A or B

• Posts: 4,885
whicker wrote: »
@ersmith Here's a bug submission for SELECT CASE in BASIC.

Thanks for the bug report! The code wasn't at all doing what I wanted it to. I think I see how to fix it though, so it should work properly in the next release.

• Posts: 4,885
There are new versions available now of (3.9.26) of spin2gui and fastspin. The changes are:

- Added DEF FNA(X)=X form to BASIC
- Fixed declarations of functions inside other C functions
- Fixed SELECT CASE in BASIC when the test is on strings or floats
- Fixed an optimizer bug that affected local arrays
- Fixed space allocation for local arrays in C functions
- On P2, default to 160MHz and 230400 baud if no explicit clkset is given
- Made BASIC read functions support base 16 and base 2 numbers
- Implemented struct assignment for C
- Experimental -z option to compress the code (produces smaller but slower code)

The -z flag is only supported for P1 now, and is still quite buggy, so not ready for production, but eventually it will allow for smaller but slower code (like PropGCC's CMM mode).

On P2 if no explicit clkset() is given before any output, we default to 160 MHz with 230400 baud. This makes porting programs much simpler (no need to add in the clkset and _setbaud calls).

The C compiler support in fastspin is quite a bit better now, and is able to compile the dhrystone, xxtea, and fftbench benchmarks properly. I hope to update the C compiler benchmarks threads later.
• Posts: 4,885
New versions (3.9.27) of fastspin and spin2gui are available now. The main change is that Spin CASE statements (and BASIC SELECT and C switch) use a jump table now if they can, and the CASE_FAST keyword is supported to force use of a jump table and to produce an error if a jump table cannot be used. There is also better support for old style C function declarations, and a CAST operator has been added to BASIC to allow easy conversion between pointer and integer types.
• Posts: 10,907
Eric,
Found another minor missing error message. NOP isn't permitted to have conditional execution, or a _ret_. When applying such, there is no error reported.
• Posts: 4,885
Thanks Evan. That's fixed in github now (adding any kind of condition to a NOP produces an error).
• Posts: 3,344
edited 2019-06-18 12:26
found some misbehavior

outa, ptra++ results in

error: unknown operator in constant expression 281

same for dirh / dirl ...
Enjoy!

Mike
• Posts: 4,885
msrobots wrote: »
found some misbehavior

outa, ptra++ results in

error: unknown operator in constant expression 281

same for dirh / dirl ...
Enjoy!

Mike

It sounds like you're trying to compile P2 code for P1?
• Posts: 3,344
edited 2019-06-18 12:59
no it is my current project

sorry typo

outh, ptra++ does not work and should, doesn't it?

confused

Mike
• Posts: 10,907
edited 2019-06-18 13:37
That one isn't legal Mike. ptra++ will only work for the RD/WRxxxx hub access instructions.
outh	ptra++

Fastspin gives an error, it's not an ordinary error but this occurs:
error: unknown operator in constant expression 281

EDIT: Doh! I see you already posted the error message.
• Posts: 4,885
Ah, I see -- fastspin wasn't specifically checking for bad use of ptra++ and such, it was just finding it out in a later phase of compilation when those expressions turned out to be invalid. I've checked in a fix for this. Thanks for the bug report Mike (and for the clarification Evan).
• Posts: 3,344

is it not valid to use ptra++ as source for a outl command?

I currently just use another long in COG ram and increment it by myself.

confused

Mike
• Posts: 4,885
Auto-incrementing (using ++ after ptra or ptrb) is only valid in the source for a read or write instruction (rdlong, wrlong, rdbyte, wrbyte, etc.). It is not valid for any other instructions.

What did you want "outl ptra++" to do? Did you think it would fetch the data pointed at by ptra and use that for the outl? You can accomplish that by:
   rdlong x, ptra++
outl x

Or, if you just wanted to increment ptra and use it for the out value you can do:
   outl ptra
add  ptra, #1  ' or whatever size increment you wanted here

• Posts: 10,907
msrobots wrote: »
is it not valid to use ptra++ as source for a outl command?
Correct. If you look through the instruction set spreadsheet you'll see just a few that have a /P in the S-operand column of the Assembly Syntax.
• Posts: 10,907
Eric,
I'm in the process of getting up to speed with one of Mikes many projects that is written in fastspin. He's used the CASE statement and clearly hasn't had any issues. I suspect I'm using a newer version of fastpin. I'm using fastspin version 3.9.28-beta- from a couple days back.

For me, the case is always taking the "other" branch. But when I edit the statement to be case_fast instead, it behaves correctly again.

• Posts: 4,885
evanh wrote: »
I'm in the process of getting up to speed with one of Mikes many projects that is written in fastspin. He's used the CASE statement and clearly hasn't had any issues. I suspect I'm using a newer version of fastpin. I'm using fastspin version 3.9.28-beta- from a couple days back.

For me, the case is always taking the "other" branch. But when I edit the statement to be case_fast instead, it behaves correctly again.

Could you give the most current version in github a try? Someone else ran into a case bug which I fixed yesterday, I hope it's the same as what you're seeing.

Regards,
Eric
• Posts: 10,907
Yep, fixed, thanks.
• Posts: 10,907
edited 2019-06-30 11:16
Eric,
I've gone and done something dumb without thinking, I tried to CMP two constants (each with a #)! ... and Fastspin didn't complain. It wasn't until I tested with Pnut that I realised my mistake.

Basically, I am sort of doing a selective build but without the code saving. Conditional compiles would be better but I don't think that's an option in Pasm only code. Or is it?

• Posts: 4,885
evanh wrote: »
Eric,
I've gone and done something dumb without thinking, I tried to CMP two constants (each with a #)! ... and Fastspin didn't complain. It wasn't until I tested with Pnut that I realised my mistake.

Whoops, that's definitely a bug! I must have broken something back when I added code to support immediates in the dest field (which some P2 instructions support). Thanks for the bug report, I'll look into it.
Basically, I am sort of doing a selective build but without the code saving. Conditional compiles would be better but I don't think that's an option in Pasm only code. Or is it?

fastspin is fine with conditional compiles in PASM code. It only supports #ifdef and #ifndef though, not full featured #if.

Regards,
Eric
• Posts: 10,907
ersmith wrote: »
fastspin is fine with conditional compiles in PASM code. It only supports #ifdef and #ifndef though, not full featured #if.
Thanks.


text:
Tag: marketing

# Source & Medium: A Medium Sized Dilemma

Subtitle: Source, Medium, Attribution, Stale Information, and The Future of Data

Here’s our situation – we want to be able to slice reporting and dashboards by a number of dimensions, including source and medium.

MARDAT (the team I’m lucky enough to be working with) is working to make this kind of thing a simple exercise in curiosity and (dare I say) wonder. It’s really interesting to me, and has become more and more clear over the last year or so, how important enabling curiosity is. One of the great things that Google Analytics and other business intelligence tools can do is open the door to exploration and semi-indulgent curiosity fulfillment.

You can imagine, if you’re a somewhat non-technical member of a marketing or business development team, you’re really good at a lot of things. Your experience gives you a sense of intuition and interest in the information collected by and measured by your company’s tools.

If the only way you have access to that information is by placing a request, for another person to go do 30 minutes, two hours, three hours of work, that represents friction in the process, that represents some latency, and you’re going to find yourself disinclined to place that kind of request if you’re not fairly certain that there’s a win there – it pushes back on curiosity. It reduces your ability to access and leverage your expertise.

That’s a little bit of a digression – let’s talk about Source and Medium. Source and Medium are defined pretty readily by most blogs and tools: these are buckets that we place our incoming traffic in. People who arrive at our websites, where ever they were right before they arrived at our websites, that’s Source and Medium.

We assign other things too – campaign name, keyword, all sorts of things. My dilemma here actually applies to the entire category of things we tag our customers with, but it’s quicker to just say, Source and Medium.

Broadly, Source is the origin (Google, another website, Twitter, and so forth) and Medium is the category (organic, referral, etc) – if this is all new to you I recommend taking a spin through this Quora thread for a little more context.

What I am struggling with, is this: for a site like WordPress.com, where folks may come and go many times before signing up, and they may enjoy our free product for a while before making a purchase, at what point do you say, “OK, THIS is the Source and Medium for this person!”

Put another way:  when you make a report, say, for all sales in May, and you say to the report, “Split up all sales by Source and Medium,” what do you want that split to tell you?

Here are some things it might tell you:

• The source and medium for the very first page view we can attribute back to that customer, regardless of how long ago that page view was.
• The source and medium for a view of a page we consider an entry page (landing pages, home page, etc), regardless of how long ago that page view was.
• The source and medium for the very first page view, within a certain window of time (7 days, 30 days, 1 year)
• The source and medium for the first entry page (landing page, homepage) within a certain window of time (7 days, 30 days, 1 year)
• The source and medium for the visit that resulted in a signup, rather than the first ever visit.
• The source and medium for the visit that resulted in a conversion, rather than the first ever visit.
• The source and medium for an arrival based on some other criteria (first arrival of all time OR first arrival since being idle for 30 days, something like that)

It feels like at some point Source and Medium should go bad, right? If someone came to the site seven years ago, via Friendster or Plurk or something, signed up for a free site, and then came back last week via AdWords, we wouldn’t want to assign Friendster | Referral to that sale, right?

Maybe we have to create more dynamic Source/Medium assignation: have one for “First Arrival,” one for “Signup,” one for “Purchase.” Maybe even something like Source/Medium for “Return After 60+ Days Idle”

In the long run, it feels like having a sense of what sources are driving each of those behaviors more or less effectively would be helpful, and could help build insights – but I also feel a little crazy: does no one else have this problem with Source and Medium?

# Cogitating on Return on Ad Spend – AKA ROAS

I’m still pretty new to this whole marketing thing: I’ve been a part of Automattic’s marketing efforts for just over a year, and I feel like I’m still learning: the pace of education hasn’t slowed down even a bit.

One of the things that was a real challenge for me was getting to understand the language of the work, especially given our interactions with a number of outside vendors and agencies: the number of acronyms, shorthand and unusual usage of otherwise common words is a huge part of the advertising world, and it serves many purposes.

The import of accessible language is probably something I should save for its own post: I think that, especially in highly interdependent company like Automattic, opaque language, complex jargon, and inscrutable acronyms are more of a hindrance than a help, and in fact likely do us harm, given the way that we, as humans, myself included, want to feel smart, and powerful, and it can be very attractive to nod along rather than ask hard questions.

If you’ve been following this blog for a little while, you know that measurement and the implications of measurement are things that I think about – here’s a piece about metrics generally.

My broad position on metrics is, they’re reductive, necessarily and usefully so, and need to be understood as means rather than as ends.

All that to say, we should also be careful not to treat our metrics as being perhaps more reductive than they really are, or to behave as though what we are measuring is simple, when in fact it is not simple at all.

Taking something complex and making it simple enough to be useful – that’s the essential core of all measurement. Taking something complex and acting like it is something simple is another thing entirely, and a very easy way to increase your overall Lifetime Error Rate.

This brings us to Return on Ad Spend, sometimes shortened to ROAS. Return on Ad Spend can be calculated like this:

$\dpi{150} \fn_jvn \LARGE \frac{r}{S}$

…with being revenue and being spend. Generally the output is represented either by a ratio like 3:1, where for every dollar you spend on advertising, you get three dollars worth of revenue, or with a percentage – 3:1 would be represented as 300%.

It looks pretty simple. It’s generally referred to as being very simple, or easy, that kind of thing. Which, well, it is, at least on the face of it.

(The rest of this Post is about the sometimes hidden complexities of ROAS. If you want to learn more about using the metric in a tactical way, John at Ignite Visibility has a great write up on how to calculate and break out ROAS, as well as some wrinkles about attribution, which I recommend if that’s what you’re looking for. Here’s a link)

Let’s talk about this metric: ROAS. The name holds a lot of promise, right? Return on Ad Spend: something everyone who spends money on ads wants to learn, the dream of marketers everywhere. How much are we taking back in, for the amount we are putting out?

The trick of ROAS is, we have built in a set of assumptions: specifically, that the numbers we put in represent the whole of each of those categories. The trouble here is that there are only very specific parts of the marketing spend where that is a safe assumption: low-funnel type tactics, especially for e-commerce companies shipping physical products.

In these situations, for these companies, ROAS tends to be a clean metric: you have a very clear picture of where you are spending money, and each transaction has a straightforward, static revenue.

The trick is,

For SaaS companies, ROAS can become much more complicated: imagine your company sells a single product, some type of Helpful Business Software, and it retails for \$100 / year. If you run some numbers, you find that you spend on average \$50 in ads to get a customer – this looks good, right? We can say we have 200% ROAS and call it a day.

Of course, one of the great advantages of having Data is that we are able to record it, and then see how it changes over time, and try to do the sorts of things in our business that move the needle in our desired direction.

For a SaaS company, two of the metrics that you live or die by are Customer Lifetime Value (sometimes called CLV or LTV) and the dreaded Churn Rate – astute readers will note that these two metrics are inextricably linked. Briefly: LTV is the amount of revenue that your business can expect to make from a given customer, and the dreaded Churn Rate is the expected number of customers (generally at a rate out of 100, represented as a percent, like: “Our Dreaded Churn Rate is a spooky 13%!” )

A saavy SaaS marketing analyst will use the expected lifetime value of a customer in the top of the fraction up there, to determine Return on Ad Spend – for two great reasons. FIRST, because it is more accurate: if you’re looking to determine the total return it makes more sense to use LTV than simply the ticket price. SECOND, because it will make her look better in her reporting.

Consider: for this same sale of our Helpful Business Software, our expected LTV isn’t \$100, which is the annual cost of our product, but rather, \$200. This doubles our ROAS. This is great news!

(It’s not really news at all though, right? We’re not actually improving either our ads or our product, we just used a more accurate number. Metrics are means!)

One wrinkle, though, is that now we’re not really using that equation above anymore – we’re using something more like:

$\dpi{200} \frac{LTV}{S}$

If you’ve ever spent any time trying to calculate your customers’ lifetime value, you know that this has suddenly become a much more complicated metric.

What happens once we start to bring in more complicated ingredients into our ROAS pie here, things like LTV, is that ROAS moves from being a static sort of snapshot into a metric that is much more dependent on other parts of the business to be successful.

In the above example, imagine if your company has had a disastrous year, and your Dreaded Churn Rate has skyrocketed, driving your LTV down to below \$100 (due to let’s say sweeping customer refunds and growing customer support costs) – now our ROAS is below 100%, even though literally nothing has changed on the advertising side. In this situation, ROAS becomes a larger aggregate metric, telling us something about the business at large.

This brings to mind a larger question: do we want ROAS to be a heartbeat metric, an indicator of the business overall? Or do we want it to be what it was about a thousand words ago, a simple snapshot of how our advertising efforts are going?

As we move away from direct retail e-commerce businesses into more complex companies, and up what’s called the advertising funnel, ROAS becomes additionally tricky, not because the equation itself becomes more complicated, but because we start to introduce uncertainty, and even worse than that, we introduce unequal uncertainty.

Generally, you know how much you’ve spent. This is true even for less measurable marketing efforts, things like event sponsorships, branding, and so forth. What you decide to include is a little bit of a wrinkle: do you include agency fees? Payroll?

The uncertainty comes into play in the revenue piece, and this is why ROAS as a metric starts to break down as we move up the funnel, because the lower part of your fraction, your spend, stays certain, while the upper part, the revenue, becomes increasingly uncertain, which makes the output more and more difficult to use in a reliable way.

This is a problem that crops up a lot in marketing metrics, and something I’ve been thinking on quite a lot: we often will compare or do arithmetic on numbers which have wildly different underlying levels of base uncertainty, sometimes to our detriment, maybe sometimes to our advantage.

I’ve been working with ROAS quite a lot, and trying to really get my teeth into it, and my brain around its under-the-surface complexity. For most businesses today, ROAS is useful, but it is not as simple as it looks.

This is where I ask you to add something in the comments! What metrics are stuck in your craw this week? Do you think I spend too much time trying to become certain about uncertainty? Let me know!


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# Calculate [Cr^3+] in equilibrium with Cr(OH4)- when 0.010 mol of Cr(NO3)3 is dissolved in a liter of solution buffered at pH 10.0

I tried setting up a RICE chart with the initial conditions being:

$$\begin{array}{c|c} \hline \text{Species} & \text{Concentration / M} \\ \hline \ce{Cr^3+} & 0.010 \\ \ce{OH-} & 10^{-4.0} \\ \ce{Cr(OH)4-} & 0 \\ \hline \end{array}$$

I'm assuming I should use the given $K_\mathrm{f}$ value for $\ce{Cr(OH)4-}$ ($8 \times 10^{29}$).

However, when I solve for the change I end up with a negative value for the concentration of $\ce{Cr^{3+}}$? The correct answer is supposed to be $10^{-16}$.

• You probably mean $\ce{Cr^{3+}}$ and not $\ce{Cr3+}$. Here is a link on how to properly typeset chemical equations and other formulae using MathJax. Also, please write down the balanced reaction and we can proceed from there. – Todd Minehardt Jan 31 '17 at 2:07
• I forgot to mention that you do not have nitrate in your query. You'll need to account for that in the overall reaction. – Todd Minehardt Jan 31 '17 at 2:13
• You don't have to worry about the nitrate. The answer is the same as if you threw some $\ce{Cr(OH)3}$ into a solution buffered at pH 10. Until you know what you are doing don't ski[p steps by doing some of this in your head. Write out reaction, then the equilibrium equation, solve the equilibrium equation for $\ce{[Cr^{3+}]}$, and then do the RICE chart. – MaxW Jan 31 '17 at 4:25
• I've written out the full RICE chart and tried setting it up multiple ways, with no luck. – khajiit Feb 9 '17 at 22:30
• Could you please add the source of the question? (Book title, page number or question number) – orthocresol Mar 26 '17 at 22:57

I know this problem and the answer the book gives, and unfortunately the book is in error. Let me explain. Adding 0.010 mol of Cr(NO3)3 will also give you 0.010 mol of $\ce{Cr^3+}$, as we can assume the $\ce{Cr^3+}$ will completely dissociate from the $\ce{NO^3-}$. The $\ce{OH^-}$ concentration can be found from the pH, and it is 0.0001 $\frac{moles}{L}$.

The "before reaction starts" gets set up accordingly:

$\ce{Cr^3+ + 4OH- <=> Cr(OH)4-}$

0.010 M .0001 M 0 M

Notice, however, that there is a 4:1 ratio of $\ce{OH^-}$ to $\ce{Cr^3+}$. It takes 4 $\ce{OH^-}$ for every 1 $\ce{Cr^3+}$, and $\ce{OH^-}$ is quite obviously the limiting reactant. So after the reaction, what is the concentration of $\ce{Cr^3+}$? Let's make it 1:1, then.

$\frac{.0001 M}{4}$ = $\ce{2.5 x 10^{-5}}$. And this is essentially the [$\ce{Cr^3+}$] and [$\ce{Cr(OH)4-}$]. However, this is not what the book says. They meant to put the concentration of $\ce{Cr(OH)4-}$ as .010 M, not $\ce{Cr^3+}$. Let's try it that way, and see what we get!

Cr(OH)4- <=> Cr3+ + 4OH-

Here, in the equation, the concentrations are:

$\ce{Cr(OH)4-}$ = .010 M

$\ce{Cr^3+}$ = 0

4$\ce{OH^-}$ = .0001 M

-x +x +4x

.010 - x +x .0001 + 4x

Because .010 - x and .0001 + 4x are so small, let .010 - x = .010 and .0001 + 4x = .0001.

The $\ce{K_f}$ = 8 x $10^{29}$, but that is for the equation when it looks like:

$$\ce{Cr(OH)4- <=> Cr3+ + 4OH-}$$

We switched it around to find out how much $\ce{Cr(OH)4-}$ would dissolve to make $\ce{Cr^3+}$, so we have to use $\frac{1}{Kf}$, which = 1.25 x $10^{-30}$.

$\ce{K_f}$ = $\frac{[{Cr^3+}][{OH-}]^4}{[\ce{Cr(OH)4-}]}$

$\ce{K_f}$ = $\frac{(x)(.0001)^4}{(.010)}$

1.25 x $10^{-30}$ = $\frac{1 x 10^{-16}x}{.010}$

You can take it from there. The answer will be 1.25 x $10^{-16}$ M.


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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$k = 4~N/m$
When all the energy in the system is in the form of kinetic energy, the kinetic energy is 8 J. Therefore, the total energy in the system is 8 J. When the spring is stretched or compressed by 2 meters, all the energy in the system is in the form of potential energy. We can find the spring constant. $U_s = E$ $\frac{1}{2}kA^2 = E$ $k = \frac{2E}{A^2}$ $k = \frac{(2)(8~J)}{(2~m)^2}$ $k = 4~N/m$


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# How do you find all the zeros of y=x^2-3x-4 with its multiplicities?

This quadratic has zeros $x = 4$ and $x = - 1$, both with multiplicity $1$.
$y = {x}^{2} - 3 x - 4 = \left(x - 4\right) \left(x + 1\right)$
Hence zeros $x = 4$ and $x = - 1$, both with multiplicity $1$.


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vif {car} R Documentation

Variance Inflation Factors

Description

Calculates variance-inflation and generalized variance-inflation factors (VIFs and GVIFs) for linear, generalized linear, and other regression models.

Usage

vif(mod, ...)

## Default S3 method:
vif(mod, ...)

## S3 method for class 'lm'
vif(mod, type=c("terms", "predictor"), ...)

## S3 method for class 'merMod'
vif(mod, ...)

## S3 method for class 'polr'
vif(mod, ...)

## S3 method for class 'svyolr'
vif(mod, ...)

Arguments

 mod for the default method, an object that responds to coef, vcov, and model.matrix, such as a glm object. type for unweighted lm objects only, how to handle models that contain interactions: see Details below. ... not used.

Details

If all terms in an unweighted linear model have 1 df, then the usual variance-inflation factors are calculated.

If any terms in an unweighted linear model have more than 1 df, then generalized variance-inflation factors (Fox and Monette, 1992) are calculated. These are interpretable as the inflation in size of the confidence ellipse or ellipsoid for the coefficients of the term in comparison with what would be obtained for orthogonal data.

The generalized VIFs are invariant with respect to the coding of the terms in the model (as long as the subspace of the columns of the model matrix pertaining to each term is invariant). To adjust for the dimension of the confidence ellipsoid, the function also prints GVIF^{1/(2\times df)} where df is the degrees of freedom associated with the term.

Through a further generalization, the implementation here is applicable as well to other sorts of models, in particular weighted linear models, generalized linear models, and mixed-effects models.

Two methods of computing GVIFs are provided for unweighted linear models:

• Setting type="terms" (the default) behaves like the default method, and computes the GVIF for each term in the model, ignoring relations of marginality among the terms in models with interactions. GVIFs computed in this manner aren't generally sensible.

• Setting type="predictor" focuses in turn on each predictor in the model, combining the main effect for that predictor with the main effects of the predictors with which the focal predictor interacts and the interactions; e.g., in the model with formula y ~ a*b + b*c, the GVIF for the predictor a also includes the b main effect and the a:b interaction regressors; the GVIF for the predictor c includes the b main effect and the b:c interaction; and the GVIF for the predictor b includes the a and c main effects and the a:b and a:c interactions (i.e., the whole model), and is thus necessarily 1. These predictor GVIFs should be regarded as experimental.

Specific methods are provided for ordinal regression model objects produced by polr in the MASS package and svyolr in the survey package, which are "intercept-less"; VIFs or GVIFs for linear and similar regression models without intercepts are generally not sensible.

Value

A vector of VIFs, or a matrix containing one row for each term, and columns for the GVIF, df, and GVIF^{1/(2\times df)}, the last of which is intended to be comparable across terms of different dimension.

Author(s)

John Fox jfox@mcmaster.ca and Henric Nilsson

References

Fox, J. and Monette, G. (1992) Generalized collinearity diagnostics. JASA, 87, 178–183.

Fox, J. (2016) Applied Regression Analysis and Generalized Linear Models, Third Edition. Sage.

Fox, J. and Weisberg, S. (2018) An R Companion to Applied Regression, Third Edition, Sage.

Examples

vif(lm(prestige ~ income + education, data=Duncan))
vif(lm(prestige ~ income + education + type, data=Duncan))
vif(lm(prestige ~ (income + education)*type, data=Duncan),
type="terms") # not recommended
vif(lm(prestige ~ (income + education)*type, data=Duncan),
type="predictor")

[Package car version 3.1-0 Index]


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Groupe d’études et de recherche en analyse des décisions

# Using Size for Bounding Expressions of Graph Invariants

## Jelena Sedlar, Damir Vukicevic et Pierre Hansen

With the help of the AutoGraphiX system, we study relations of the form

$\underline{b}_m \le i_1(G) \oplus i_2(G) \le \overline{b}_m$

where $i_1(G)$ and $i_2(G)$ are invariants of the graph $G$, $\oplus$ is one of the operations $-, +, /, \times$, $\underline{b}_m$ and $\overline{b}_m$ are best possible lower and upper bounding functions depending only one the size m of G. Specifically, we consider couples of indices where $i_1(G)$ is a measure of distance, i.e., diameter, radius or average eccentricity, and $i_2(G)$ is a measure of connectivity, i.e., minimum degree, edge connectivity and vertex connectivity. Conjectures are obtained and then proved in almost all cases.

, 20 pages


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# Math Help - Finding the Degree

1. ## Finding the Degree

I used sine law to find 31.44 and 55.1 degrees but i'm not sure it's right now i'm confused. what should i do? please help.

2. So everything EXCEPT 31.44 and 55.1 was on the original problem given to you? Can you post the exact problem as it was given?

3. ## Original Problem

.

4. Originally Posted by megs_world
I used sine law to find 31.44 and 55.1 degrees but i'm not sure it's right now i'm confused. what should i do? please help.
Your answers are correct:

$\frac{\sin \alpha}{6} = \frac{\sin 73^0}{11} \Rightarrow \alpha = 31.44^0$.

$\frac{\sin \beta}{11} = \frac{\sin 31.44^0}{7} \Rightarrow \beta = 55.1^0$.

Originally Posted by mollymcf2009
So everything EXCEPT 31.44 and 55.1 was on the original problem given to you? Can you post the exact problem as it was given?
The question as originally posted by the OP is in no need of clarification.


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# How do I open a certain file location in Powershell

Using Windows Powershell is there a shorter way of opening a program other than:

cd c:\program files
cd program folder
start program.exe

-
Like start "c:\program files\program.exe"  – 50-3 Sep 10 '13 at 3:12

You can use the Invoke-Item cmdlet and specify the full path as opposed to changing location multiple times to get to the directory containing your target executable

Invoke-Item "C:\Program Files\Program Folder\Program.exe"

-

You could skip the cd steps by specifying the whole path, e.g c:\path\to\program.exe

If it's a program you launch with some regularity, you could also create an alias. With an alias, you could have PowerShell launch c:\path\to\program.exe with just myapp, for example.

To create an alias, use the following command:

Set-Alias MyApp "C:\Path\To\Program.exe"

Finally, you could also add the folder location to your Windows PATH environment variable. Here's a link to a post detailing how to do this for Python, but you should be able to follow the same instructions for your folder. http://stackoverflow.com/questions/6318156/adding-python-path-on-windows-7

-


text:
# Bond order

Last updated

In chemistry, bond order, as introduced by Linus Pauling, is defined as the difference between the number of bonds and anti-bonds.

## Contents

The bond order itself is the number of electron pairs (covalent bonds) between two atoms. [1] For example, in diatomic nitrogen N≡N, the bond order between the two nitrogen atoms is 3 (triple bond). In acetylene H–C≡C–H, the bond order between the two carbon atoms is also 3, and the C–H bond order is 1 (single bond). In carbon monoxide C≡O+, the bond order between carbon and oxygen is 3. In thiazyl trifluoride N≡SF3, the bond order between sulfur and nitrogen is 3, and between sulfur and fluorine is 1. In diatomic oxygen O=O the bond order is 2 (double bond). In ethylene H2C=CH2 the bond order between the two carbon atoms is also 2. The bond order between carbon and oxygen in carbon dioxide O=C=O is also 2. In phosgene O=CCl2, the bond order between carbon and oxygen is 2, and between carbon and chlorine is 1.

In some molecules, bond orders can be 4 (quadruple bond), 5 (quintuple bond) or even 6 (sextuple bond). For example, potassium octachlorodimolybdate salt (K4[Mo2Cl8]) contains the [Cl4Mo≣MoCl4]4− anion, in which the two Mo atoms are linked to each other by a bond with order of 4. Each Mo atom is linked to four Cl ligands by a bond with order of 1. The compound (terphenyl)–CrCr–(terphenyl) contains two chromium atoms linked to each other by a bond with order of 5, and each chromium atom is linked to one terphenyl ligand by a single bond. A bond of order 6 is detected in ditungsten molecules W2, which exist only in a gaseous phase.

Bond order gives an indication of the stability of a bond. Isoelectronic species have the same bond order. [2]

In molecules which have resonance or nonclassical bonding, bond order may not be an integer. In benzene, the delocalized molecular orbitals contain 6 pi electrons over six carbons, essentially yielding half a pi bond together with the sigma bond for each pair of carbon atoms, giving a calculated bond order of 1.5 (one and a half bond). Furthermore, bond orders of 1.1 (eleven tenths bond), 4/3 (or 1.333333..., four thirds bond) or 0.5 (half bond), for example, can occur in some molecules and essentially refer to bond strength relative to bonds with order 1. In the nitrate anion (NO3), the bond order for each bond between nitrogen and oxygen is 4/3 (or 1.333333...). Bonding in dihydrogen cation H+2 can be described as a covalent one-electron bond, thus the bonding between the two hydrogen atoms has bond order of 0.5. [3]

## Bond order in molecular orbital theory

In molecular orbital theory, bond order is defined as half the difference between the number of bonding electrons and the number of antibonding electrons as per the equation below. [4] [5] This often but not always yields similar results for bonds near their equilibrium lengths, but it does not work for stretched bonds. [6] Bond order is also an index of bond strength and is also used extensively in valence bond theory.

bond order = number of bonding electrons - number of antibonding electrons/2

Generally, the higher the bond order, the stronger the bond. Bond orders of one-half may be stable, as shown by the stability of H+2 (bond length 106 pm, bond energy 269 kJ/mol) and He+2 (bond length 108 pm, bond energy 251 kJ/mol). [7]

Hückel molecular orbital theory offers another approach for defining bond orders based on molecular orbital coefficients, for planar molecules with delocalized π bonding. The theory divides bonding into a sigma framework and a pi system. The π-bond order between atoms r and s derived from Hückel theory was defined by Charles Coulson by using the orbital coefficients of the Hückel MOs: [8] [9] [ clarification needed ]

${\displaystyle p_{rs}=\sum _{i}n_{i}c_{ri}c_{si}}$,

Here the sum extends over π molecular orbitals only, and ni is the number of electrons occupying orbital i with coefficients cri and csi on atoms r and s respectively. Assuming a bond order contribution of 1 from the sigma component this gives a total bond order (σ + π) of 5/3 = 1.67 for benzene, rather than the commonly cited bond order of 1.5, showing some degree of ambiguity in how the concept of bond order is defined.

For more elaborate forms of molecular orbital theory involving larger basis sets, still other definitions have been proposed. [10] A standard quantum mechanical definition for bond order has been debated for a long time. [11] A comprehensive method to compute bond orders from quantum chemistry calculations was published in 2017. [6]

## Other definitions

The bond order concept is used in molecular dynamics and bond order potentials. The magnitude of the bond order is associated with the bond length. According to Linus Pauling in 1947, the bond order between atoms i and j is experimentally described as

${\displaystyle s_{ij}=\exp {\left[{\frac {d_{1}-d_{ij}}{b}}\right]}}$

where d1 is the single bond length, dij is the bond length experimentally measured, and b is a constant, depending on the atoms. Pauling suggested a value of 0.353 Å for b, for carbon-carbon bonds in the original equation: [12]

${\displaystyle d_{1}-d_{ij}=0.353~{\text{ln}}(s_{ij})}$

The value of the constant b depends on the atoms. This definition of bond order is somewhat ad hoc and only easy to apply for diatomic molecules.

## Related Research Articles

A chemical bond is a lasting attraction between atoms or ions that enables the formation of molecules and crystals. The bond may result from the electrostatic force between oppositely charged ions as in ionic bonds, or through the sharing of electrons as in covalent bonds. The strength of chemical bonds varies considerably; there are "strong bonds" or "primary bonds" such as covalent, ionic and metallic bonds, and "weak bonds" or "secondary bonds" such as dipole–dipole interactions, the London dispersion force and hydrogen bonding. Strong chemical bonding arises from the sharing or transfer of electrons between the participating atoms.

A covalent bond is a chemical bond that involves the sharing of electrons to form electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs. The stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding. For many molecules, the sharing of electrons allows each atom to attain the equivalent of a full valence shell, corresponding to a stable electronic configuration. In organic chemistry, covalent bonding is much more common than ionic bonding.

In theoretical chemistry, a conjugated system is a system of connected p-orbitals with delocalized electrons in a molecule, which in general lowers the overall energy of the molecule and increases stability. It is conventionally represented as having alternating single and multiple bonds. Lone pairs, radicals or carbenium ions may be part of the system, which may be cyclic, acyclic, linear or mixed. The term "conjugated" was coined in 1899 by the German chemist Johannes Thiele.

In chemistry, aromaticity is a chemical property of cyclic (ring-shaped), typically planar (flat) molecular structures with pi bonds in resonance that gives increased stability compared to saturated compounds having single bonds, and other geometric or connective non-cyclic arrangements with the same set of atoms. Aromatic rings are very stable and do not break apart easily. Organic compounds that are not aromatic are classified as aliphatic compounds—they might be cyclic, but only aromatic rings have enhanced stability. The term aromaticity with this meaning is historically related to the concept of having an aroma, but is a distinct property from that meaning.

The octet rule is a chemical rule of thumb that reflects the theory that main-group elements tend to bond in such a way that each atom has eight electrons in its valence shell, giving it the same electronic configuration as a noble gas. The rule is especially applicable to carbon, nitrogen, oxygen, and the halogens; although more generally the rule is applicable for the s-block and p-block of the periodic table. Other rules exist for other elements, such as the duplet rule for hydrogen and helium, or the 18-electron rule for transition metals.

In chemistry, resonance, also called mesomerism, is a way of describing bonding in certain molecules or polyatomic ions by the combination of several contributing structures into a resonance hybrid in valence bond theory. It has particular value for analyzing delocalized electrons where the bonding cannot be expressed by one single Lewis structure.

In chemistry, polarity is a separation of electric charge leading to a molecule or its chemical groups having an electric dipole moment, with a negatively charged end and a positively charged end.

In chemistry, a lone pair refers to a pair of valence electrons that are not shared with another atom in a covalent bond and is sometimes called an unshared pair or non-bonding pair. Lone pairs are found in the outermost electron shell of atoms. They can be identified by using a Lewis structure. Electron pairs are therefore considered lone pairs if two electrons are paired but are not used in chemical bonding. Thus, the number of electrons in lone pairs plus the number of electrons in bonds equals the number of valence electrons around an atom.

In chemistry, molecular orbital theory is a method for describing the electronic structure of molecules using quantum mechanics. It was proposed early in the 20th century.

In chemistry, valence bond (VB) theory is one of the two basic theories, along with molecular orbital (MO) theory, that were developed to use the methods of quantum mechanics to explain chemical bonding. It focuses on how the atomic orbitals of the dissociated atoms combine to give individual chemical bonds when a molecule is formed. In contrast, molecular orbital theory has orbitals that cover the whole molecule.

In organic chemistry, Hückel's rule predicts that a planar ring molecule will have aromatic properties if it has 4n + 2 π electrons, where n is a non-negative integer. The quantum mechanical basis for its formulation was first worked out by physical chemist Erich Hückel in 1931. The succinct expression as the 4n + 2 rule has been attributed to W. v. E. Doering (1951), although several authors were using this form at around the same time.

In chemistry, orbital hybridisation is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory. For example, in a carbon atom which forms four single bonds the valence-shell s orbital combines with three valence-shell p orbitals to form four equivalent sp3 mixtures in a tetrahedral arrangement around the carbon to bond to four different atoms. Hybrid orbitals are useful in the explanation of molecular geometry and atomic bonding properties and are symmetrically disposed in space. Usually hybrid orbitals are formed by mixing atomic orbitals of comparable energies.

In organic chemistry, a bent bond, also known as a banana bond, is a type of covalent chemical bond with a geometry somewhat reminiscent of a banana. The term itself is a general representation of electron density or configuration resembling a similar "bent" structure within small ring molecules, such as cyclopropane (C3H6) or as a representation of double or triple bonds within a compound that is an alternative to the sigma and pi bond model.

A quintuple bond in chemistry is an unusual type of chemical bond, first reported in 2005 for a dichromium compound. Single bonds, double bonds, and triple bonds are commonplace in chemistry. Quadruple bonds are rarer but are currently known only among the transition metals, especially for Cr, Mo, W, and Re, e.g. [Mo2Cl8]4− and [Re2Cl8]2−. In a quintuple bond, ten electrons participate in bonding between the two metal centers, allocated as σ2π4δ4.

The Hückel method or Hückel molecular orbital theory, proposed by Erich Hückel in 1930, is a simple method for calculating molecular orbitals as linear combinations of atomic orbitals. The theory predicts the molecular orbitals for π-electrons in π-delocalized molecules, such as ethylene, benzene, butadiene, and pyridine. It provides the theoretical basis for Hückel's rule that cyclic, planar molecules or ions with π-electrons are aromatic. It was later extended to conjugated molecules such as pyridine, pyrrole and furan that contain atoms other than carbon and hydrogen (heteroatoms). A more dramatic extension of the method to include σ-electrons, known as the extended Hückel method (EHM), was developed by Roald Hoffmann. The extended Hückel method gives some degree of quantitative accuracy for organic molecules in general and was used to provide computational justification for the Woodward–Hoffmann rules. To distinguish the original approach from Hoffmann's extension, the Hückel method is also known as the simple Hückel method (SHM).

In chemical bonding theory, an antibonding orbital is a type of molecular orbital that weakens the chemical bond between two atoms and helps to raise the energy of the molecule relative to the separated atoms. Such an orbital has one or more nodes in the bonding region between the nuclei. The density of the electrons in the orbital is concentrated outside the bonding region and acts to pull one nucleus away from the other and tends to cause mutual repulsion between the two atoms. This is in contrast to a bonding molecular orbital, which has a lower energy than that of the separate atoms, and is responsible for chemical bonds.

In chemistry, Bent's rule describes and explains the relationship between the orbital hybridization of central atoms in molecules and the electronegativities of substituents. The rule was stated by Henry A. Bent as follows:

Atomic s character concentrates in orbitals directed toward electropositive substituents.

A molecular orbital diagram, or MO diagram, is a qualitative descriptive tool explaining chemical bonding in molecules in terms of molecular orbital theory in general and the linear combination of atomic orbitals (LCAO) method in particular. A fundamental principle of these theories is that as atoms bond to form molecules, a certain number of atomic orbitals combine to form the same number of molecular orbitals, although the electrons involved may be redistributed among the orbitals. This tool is very well suited for simple diatomic molecules such as dihydrogen, dioxygen, and carbon monoxide but becomes more complex when discussing even comparatively simple polyatomic molecules, such as methane. MO diagrams can explain why some molecules exist and others do not. They can also predict bond strength, as well as the electronic transitions that can take place.

Water is a simple triatomic bent molecule with C2v molecular symmetry and bond angle of 104.5° between the central oxygen atom and the hydrogen atoms. Despite being one of the simplest triatomic molecules, its chemical bonding scheme is nonetheless complex as many of its bonding properties such as bond angle, ionization energy, and electronic state energy cannot be explained by one unified bonding model. Instead, several traditional and advanced bonding models such as simple Lewis and VSEPR structure, valence bond theory, molecular orbital theory, isovalent hybridization, and Bent's rule are discussed below to provide a comprehensive bonding model for H
2
O
, explaining and rationalizing the various electronic and physical properties and features manifested by its peculiar bonding arrangements.

In theoretical chemistry, the bonding orbital is used in molecular orbital (MO) theory to describe the attractive interactions between the atomic orbitals of two or more atoms in a molecule. In MO theory, electrons are portrayed to move in waves. When more than one of these waves come close together, the in-phase combination of these waves produces an interaction that leads to a species that is greatly stabilized. The result of the waves’ constructive interference causes the density of the electrons to be found within the binding region, creating a stable bond between the two species.

## References

1. IUPAC , Compendium of Chemical Terminology , 2nd ed. (the "Gold Book") (1997). Online corrected version: (2006) " Bond number ". doi : 10.1351/goldbook.B00705
2. Dr. S.P. Jauhar. Modern's abc Chemistry.
3. Clark R. Landis; Frank Weinhold (2005). Valency and bonding: a natural bond orbital donor-acceptor perspective. Cambridge, UK: Cambridge University Press. pp. 91–92. ISBN   978-0-521-83128-4.
4. Jonathan Clayden; Greeves, Nick; Stuart Warren (2012). Organic Chemistry (2nd ed.). Oxford University Press. p. 91. ISBN   978-0-19-927029-3.
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# Ex.1.2 Q6 Rational Numbers Solution - NCERT Maths Class 8

## Question

Write five rational numbers greater than $$–2.$$

Video Solution
Rational Numbers
Ex 1.2 | Question 6

## Text Solution

What is known?

Rational numbers.

What is unknown?

The rational numbers between given rational numbers.

Reasoning:

We can find infinitely many rational numbers between any two given rational numbers by taking the mean of the two rational numbers. Another method: We can make the denominator same for the two given rational numbers.

Steps:

Some of the five rational numbers greater than $$– 2$$ are \begin{align}- 1,\;\;0,\;\;\frac{1}{2},\;\;\frac{1}{3}, \end{align} and \begin{align}\frac{1}{4} \end{align}

[There can be more such rational numbers]

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