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Deep Rectifier Networks as geometric decision trees

# Introduction:

In my previous article on deep rectifier networks I focused on their algebraic structure and showed that the latent space formed a Hilbert space which means that they function by de-correlating input signals $x \sim X$. Before continuing I must clarify that by deep rectifier networks I am referring to networks of arbitrary architecture with relu activations. Now, while the algebraic perspective is insightful it doesn’t clearly explain why deeper networks might be more expressive. If the latent space of a network with $N$ nodes has $2^N$ possible linear feature maps why would it matter whether it has five layers with $\lfloor N/5 \rfloor$ nodes per layer or one hidden layer with $N$ nodes? Why would depth make a difference?

A colleague suggested that I take a look ‘Tropical Geometry of Deep Neural Networks’ [2] which supposedly explains why deeper networks are exponentially more expressive but as I have little familiarity with tropical geometry I decided to try and figure out whether I could answer as much of the question myself before reading that publication. In fact, by following my own reasoning it occurred to me that a geometric interpretation of affine transformations was crucial and that it may be useful to identify deep rectifier networks with a particular kind of geometric decision tree.

In summary, here’s a short breakdown of my attempt to answer this question:

1. A deep rectifier network may be identified with an $N$ level decision tree for solving geometric problems in a sequential manner with lower layers of the network forming re-usable sub-expressions.

2. Each expression is an affine transformation and all geometric transformations in Euclidean space are affine. It follows that deep rectifier networks may be viewed as geometric programs where each sub-expression is a geometric operation.

3. At each layer of the network the candidate expressions are a subset of the power set of distinct nodes at that layer. Hence the importance of network width for versatile geometric reasoning.

4. Furthermore, each sub-expression at layer $i$ may be re-used at most $2^{N-\sum_{k=1}^i n_k}$ times by sub-expressions at higher levels of the decision tree. Montufar [1] gives a similar argument but he identifies expressions with ‘folds’ which is incorrect in my opinion.

5. The importance of the last statement is clear when we think in terms of geometric transformations. Deep rectifier networks permit an exponential number of possible sequences of geometric transformations of length $N$ and therefore the complexity of the set of possible transformations of the latent space is proportional to network depth.

In summary depth gives us complexity and width gives us versatility. Now, to better understand these ideas I propose applying deep rectifier networks to the problem of self-organising origami which involves designing fault-tolerant structures by folding matter in a self-organised manner, something nature does amazingly well [3]. In fact, I see the application of self-organising origami as a proof-of-principle that my explanation must be correct.

# Deep Rectifier Networks as N-level decision trees:

Let’s suppose that our deep rectifier network has $N$ layers and $n_i$ nodes at each layer $i$. Then I claim that a deep rectifier network may be identified with an $N$ level decision tree for solving geometric problems in a sequential manner. Let me explain:

1. At the ith layer there are $m \leq 2^{n_i}$ possible affine transformations that may be applied to the output of the (i-1)th layer.

2. It follows that the combinatorial versatility at layer $i$ is proportional to $2^{n_i}$ so network width is important.

3. Within the context of computational geometry(where algorithms and data structures are defined in terms of geometric objects), these affine transformations are sub-expressions may be re-used by higher levels of the decision tree.

The last insight is useful because it becomes immediately clear that each sub-expression at layer $i$ may be re-used at most:

$$2^{N-\sum_{k=1}^i n_k}$$

times by sub-expressions at higher levels of the decision tree. Now, we see that the tree analogy is actually quite good because the number of geometric sequences grows exponentially with network depth so by the time we have reached the end of the network we are in a real jungle with a astronomical number of options.

But, why would the composition of affine transformations lead to greater geometric reasoning ability? Why not simply use a single layer network with $N$ nodes?

# All geometric transformations in Euclidean space are affine transformations:

Let’s make the reasonable assumption that each layer $i$ has $n_i \geq 3$ nodes. How do sequences of affine transformations transform subspaces of $\mathbb{R}^{n_i \geq 3}$? To answer this question we shall first handle the case of $n_i = 3$ i.e. three dimensional space.

## Geometric transformations:

There are four elementary geometric transformations in $\mathbb{R}^{3}$ which I shall describe briefly below. The reader may check that these are all invertible transforms and therefore geometric transforms are elements of the affine group on $\mathbb{R}^{3}$, $Aff(\mathbb{R}^{3})$.

### Translation:

Given a vector $X \in \mathbb{R}^{3}$, the identity matrix $I \in \mathbb{R}^{3 \times 3}$ and a translation vector $b \in \mathbb{R}^{3}$ we may define translations as follows:

$$X \mapsto IX + b$$

## Shear:

This is simply the identity matrix with one of the six null elements replaced with a non-zero real number.

## Rotation matrices:

For these I actually have to type-set matrices in mathjax which I haven’t figured out how to do. In any case, there are three types of rotation matrices as you can guess and one interesting thing about rotations is that they aren’t commutative.

## Scaling:

This is similar to the identity matrix except that one or more of the diagonal entries has an absolute value which is neither zero nor one.

Now, the power of these elementary geometric transforms is due to the fact that any geometric transform is a composition of these elementary transforms.

## Geometric transformations in higher-dimensional Euclidean spaces:

Let’s suppose $n_{i} > 3$ so that we have to interpret the affine transformation at layer $i+1$ assuming that $n_{i+1} \geq n_{i} > 3$. Let’s note that there are ${n_i \choose 3}$ orthogonal projections of $\mathbb{R}^{n_i}$ onto $\mathbb{R}^{3}$. It follows that there is a direct interpretation of affine transforms on $\mathbb{R}^{n_i > 3}$. These are deformations of at most ${n_i \choose 3}$ three dimensional subspaces of the $n_i$ dimensional latent space where each geometric transform is followed by a feature-dependent translation which is a linear combination of the $n_i -3$ other values of the feature vector.

There is however the natural question of whether this interpretation is exactly correct. How many of the ${n_i \choose 3}$ column-wise projections of the $i+1$ layer affine transform are invertible? These are the ones that may be interpreted as geometric transformations.

## A kinematic view:

Let’s consider a finite set of transforms $\{x_i\}_{i=1}^n \subset Aff(\mathbb{R}^3)$. How can we interpret the sequence of geometric(and hence invertible) transformations:

$$T = x_n \circ x_{n-1} \circ … \circ x_1$$

Well, these may be understood as a motion through space with $n$ discrete time increments. For all the above reasons network depth makes a difference in terms of the affordable complexity of the geometric expressions. With one affine transform you have a line, with two a curve and with a large number you have arbitrarily complex trajectories i.e. deformations of space.

# Applying deep and wide rectifier networks to self-organised origami:

Origami, as the reader probably knows, is the art of creating new shapes via a sequence of elementary folds which correspond to rotations followed by translations and vice-versa. What may surprise the reader however is the Universality of Origami.

It may be demonstrated that every 3D shape can be approximated by a sequence of origami folds applied to a sufficiently large piece of paper. Similarly, the universality of rectifier networks implies that a sufficiently deep and wide rectifier network may be used to approximate any finite number of distinct sequences of geometric transformations.

This begs the question: can we use deep rectifier networks in combination with a planning algorithm to explore the space of possible origami shapes?

## The challenge of self-organised origami:

L. Mahadevan in ‘self-organised origami’ [3] introduces the challenge of self-organised origami in the following manner:

The controlled folding and unfolding of maps, space structures, wings, leaves, petals, and other foldable laminae is potentially complicated by the independence of individual folds; as their number increases, there is a combinatorial explosion in the number of folded possibilities’

This combinatorial explosion is what makes the search problem in neural network parameter space potentially really hard. But, the potential benefits of self-organised origami is that we will discover structures that are fault-tolerant in principle and it may prove to be a very useful tool for modeling morphogenesis. Structures that are commonly observed in nature must be robust by necessity.

Here are a concrete examples of origami-inspired designs already used by engineers:

1. NASA’s star shade project: a sunflower shaped object that is designed to protect a space telescope. More information in this wonderful interview.
2. James Webb space telescope: The primary mirror is made from segments that can be folded inwards. More information in this BBC article.

The primary reason why space technology makes heavy use of origami structures is due to the rather small diameter of rockets ~5 m relative to the large structures that must be sent into space. Having said this, I think we have barely scratched the potential of self-organised origami from both the perspective of science and engineering.


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# Junior Faculty Research

This session invites research presentations by faculty who are early in their careers (example: untenured Assistant Professors).
Organizer:
Katharine Shultis and Kate Kearney
Contact:
• Christian Millichap*, Linfield College
Given two arbitrary knots (tangled up strings with their ends tied together), how can we (easily) tell if they are different or not? In general, this problem is extremely difficult to answer, and has led to the development of a variety of knot invariants. In this talk, we will examine geometric invariants of knots. These are knot invariants that arise from examining the (often hyperbolic) geometry of the space surrounding a knot. Historically, these invariants have been extremely useful in helping classify knots. However, it is possible to construct large sets of knots that are geometrically similar, that is, all the knots in such a set are different, yet these knots have a number of geometric invariants in common. We will give two examples of geometrically similar sets of knots and raise some interesting questions about geometric invariants.
Time: Saturday, April 2, 2016 - 10:15
Room Number: STAG 160
• Kate Kearney*, Gonzaga University
The relationship between embedded surfaces and their knotted boundaries has been one of the main topics of knot theory for much of the last half century. This talk focuses on a particular case, namely whether a given knot in the three-sphere can be the boundary of a Mobius band embedded in the four-ball, B^4. We will discuss a new example of a knot which does not bound a Mobius band in B^4, and describe how the d-invariant of Heegaard-Floer theory is used to obstruct this and other knots from bounding Mobius bands in B^4.
Time: Saturday, April 2, 2016 - 10:35
Room Number: STAG 160
• Masaki Ikeda*, Western Oregon University
In enumerative combinatorics, the study of permutation patterns blossomed in the 1980s with the Stanley-Wilf conjecture. In this talk, I will introduce the basic concept of permutation patterns and some approachable examples as well as the final result of my doctoral research.
Time: Saturday, April 2, 2016 - 10:55
Room Number: STAG 160
• Kyle Riggs*, Eastern Washington University
An abelian group is decomposable if it can be written as a direct sum of two (or more) nontrivial subgroups. Otherwise it is indecomposable. The only indecomposable torsion groups are cyclic groups of the form $\mathbb{Z}(p^n)$, where $p$ is prime (as well as Pr\"ufer groups, $\mathbb{Z}(p^\infty)$). Mathematicians have been unable to describe the class of decomposable torsion-free groups. As it turns out, this problem is analytic complete (it cannot be characterized by a first-order formula). We show this by comparing it to the problem of determining whether an infinitely branching tree in $\omega^{<\omega}$ has an infinite path.
Time: Saturday, April 2, 2016 - 11:15
Room Number: STAG 160
• Yajun An*, Pacific Lutheran University
Finite Difference (FD) schemes have been used widely in computing approximations for partial differential equations for wave propagation, as they are simple, flexible and robust. However, even for stable and accurate schemes, waves in the numerical schemes can propagate at different wave speeds than in the true medium. This phenomenon is called numerical dispersion error. Traditionally, FD schemes are designed by forcing accuracy conditions, and in spite of the advantages mentioned above, such schemes suffer from numerical dispersion errors. Traditionally, two ways have been used for the purpose of reducing dispersion error: increasing the sampling rate and using higher order accuracy. More recently, Finkelstein and Kastner (2007, 2008) propose a unified methodology for deriving new schemes that can accommodate arbitrary requirements for reduced phase or group velocity dispersion errors, defined over any region in the frequency domain. Such schemes are based on enforcing exact phase or group velocity at certain preset wavenumbers. This method has been shown to reduce dispersion errors at large wavenumbers. In this talk, we discuss the construction and behaviors of FD schemes designed to have reduced numerical dispersion error. We prove that the system of equations to select the coefficients in a centered FD scheme for second order wave equations with specified order of accuracy and exact phase velocity at preset wavenumbers can always be solved. Furthermore, from the existence of such schemes, we can show that schemes which reduce the dispersion error uniformly in an interval of the frequency domain can be constructed from a Remez algorithm. In these new schemes we propose, we can also specify wavenumbers where the exact phase or group dispersion relation can be satisfied. For an incoming signal consisting of waves of different wavenumbers, our schemes can give more accurate wave propagation speeds. Furthermore, when we apply our schemes in two dimensional media, we can obtain schemes that give small dispersion error at all propagation angles.
Time: Saturday, April 2, 2016 - 11:35
Room Number: STAG 160
• Corban Harwood*, George Fox University
In numerically solving linear partial differential equations, carefully formed matrices are powered up as the algorithm iterates the solution through time. Depending upon the eigenvalues, the solution either blows up to non-numerical values, stabilizes with bounded oscillations, or stabilizes free of oscillations. This talk presents comparisons of eigenvalue conditions through extended von Neumann stability analysis of standard numerical methods for foundational partial differential equations. Known lower eigenvalue bounds for stability are not optimal, so computational testing is necessary. We will compare several implementations to automate detection of solution behavior for testing theoretical conjectures. Along the way, we will pause to enjoy some surprisingly beautiful figures resulting from such analysis and computation.
Time: Saturday, April 2, 2016 - 11:55
Room Number: STAG 160

format: plain


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# Tag Info

3

Assuming the wheel moves at the same speed as the water (ie: neglecting and 'slip' of the wheel past the water), angular velocity is given by: $$\omega = v / r$$ where $r$ is the 'radial' distance from the centre of the wheel to the top of the water. In practice, there will be some water sliding past the wheel, depending upon the hydrodynamics of the ...

8

The reason that you get slip at even the smallest forces results not from the fact that the tire is slipping against the ground, but that the tire is elastic. There is no way to completely eliminate slip with an elastic tire. Let's see why this is. To measure the slip, lets put twenty little green splotches of die evenly spaced on the circumference of the ...

0

Both graphs have 0% force at 0% slip. That implies that no force is being applied to the tire. As applied force is increased until a certain point the increase in slip remains nearly linear because the coefficient of friction remains nearly constant. Thus effective force also increases nearly linearly. After that point the coefficient of friction gets ...

2

The tension in the rope should be different on the left than on the right - it is this difference that gives rise to the torque that accelerates the pulley. You seem to think that it should be the same: but if it was, then where would the torque to move the pulley come from? Annotate your diagram carefully: you did not show $T$ anywhere.

2

Your mistake is that the two tensions are different, because of the presence of a pulley with non-zero $I$. What you have missed is: 1) connect the two tensions to the torque: $(T_1-T_2)R=I\alpha$ and 2) link the accelerations $R\alpha=a_1=a_2$ NOTE (from comments): If the pulley had a zero $I$ (moment of inertia), then the two tensions would be be ...

0

I don't know how this is supposed to simulate planetary motion, but we can still look at what data you would expect. The force on the stopper is $\frac {mv^2}r$, which if there is no friction (good luck!) equals the gravitational force on the masses at the bottom. We would therefore expect that for given mass on the bottom, $r \propto v^2$, which doesn't ...

0

Assume that the origin of frame $r$ is moving with acceleration $\newcommand{\a}{\mathbf{a}}\a$, with respect to a fixed frame $f$, but that the coordinate axes of $r$ are aligned with $f$'s. Then if an object has an apparent acceleration $\a_r$ in the frame $r$, then it's actual acceleration $\a_f$ in the fixed frame is $\a_r + \a$. Now when your allow the ...

0

An object with corners or longer in one dimension than in the others, can fall on such a part and be in unstable equilibrium. E.g. an object touches the ground on one of its corners but the center is not on the vertical that passes through that corner. That generates a torque, and angular momentum appears that tends to be conserved.

2

If two eggs are given out of which one is hard boiled egg and the other is raw egg. When both of them are allowed to spin on a tabletop, the egg which spins slower must be the raw egg because the liquid inside it tries to get away from the axis of rotation increasing the value of moment of inertia (I). To remain the angular momentum L conserved in absence of ...

1

what does the commentator mean by concavity of the floor? He or she means that the surface you walk on is in fact the inside of a cylindrical surface. Like a very large version of the inside of a wedding-ring. The curvature of this surface can be measured.

1

That the floor is not flat but circular, an the centrifugal force acts radially, and is the same at the same radius. If you cover the concavity of the floor with a planar surface or either make the stations of planar segments (a polygon instead of a circle), then you will measure different accelerations at different points of the floor, because they are not ...

-1

I have yet to find a physics book that doesn't make this really confusing. If one has a vector fixed in inertial space, its components as viewed in a moving frame are obtained by the dot product of the vector with the moving unit triad fixed to the body but moving relative to inertial space. While the inertial frame would measure its components as constants ...

0

The key phrase to notice here is that you are being asked for the least horizontal force. If you use a different expression for torque, you should be able to find the least horizontal force by figuring out what the minimum angle needed is. Remember: $$\tau=\underline{\mathbf{r}}\times\underline{\mathbf{F}}$$

-1

If we look at the symmetry of the pear of mass $m$, we can see on the x axis that momentum is conserved because the derivative of its motion in space $x$ is the same from any point we measure it's motion from: $$p_x=m \frac{dx}{dt}$$ On the y axis however, the system is affected by an external field gravity, $g$ which pulls the pear downwards. This removes ...

-1

There are two types of vector.One is polar,the other is axial.Angular velocity is an axial vector.So,no displacement is needed along its direction.

Top 50 recent answers are included


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# diophantus

Hello, this is beta version of diophantus. If you want to report about a mistake, please, write to hello@diophantus.org

#### Number of moduli of irreducible families of plane curves with nodes and cusps

04 Apr 2007 math.AG arxiv.org/abs/0704.0618
Abstract. Consider the family S of irreducible plane curves of degree n with d nodes and k cusps as singularities. Let W be an irreducible component of S. We consider the natural rational map from W to the moduli space of curves of genus g=(n-1)(n-2)/2-d-k. We define the "number of moduli of W" as the dimension of the image of W with respect to this map. If W has the expected dimension equal to 3n+g-1-k, then the number of moduli of W is at most equal to the min(3g-3, 3g-3+\rho-k), dove \rho is the Brill-Neother number of the linear series of degree n and dimension 2 on a smooth curve of genus g. We say that W has the expected number of moduli if the equality holds. In this paper we construct examples of families of irreducible plane curves with nodes and cusps as singularities having expected number of moduli and with non-positive Brill-Noether number.

# Reviews

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# Alternative to high cost plexiglas and polycarbonate for large project

I am planning the construction of a large scale dodecahedron infinity mirror, and so far the largest cost by far are the glass (or similar clear, hard material) pentagons. Each panel would be screwed on to 3D printed frame parts, amounting to 6ft tilted on a vertex. Each pentagon would be 24 inches per side, the diagonal being 38.83 inches. Therefore at least a 40 X 40" clear sheet of 3mm width would be necessary to score the pentagons.

My question is, is there any material that could be economical? I was quoted each piece at around $150 (acrylic), so very pricey for 12 of them to make up the dodecahedron. I considered stretching flexible plastic sheets or mirror film alone, but that would inevitably be subject to wear and tear as well as a warped surface (i.e. funhouse mirror rather than flat and accurate). Hope this could be answered. Much thanks in advance. For reference to the project, it would be similar to this: https://www.youtube.com/watch?v=WDK5xYHuQf0 • Economical? Based on what? Bill Gates wallet? Or 19 year old on job seekers allowance? Sep 15, 2021 at 20:55 • Economical as in the concept of a number being lower than$150 per piece. Sep 15, 2021 at 22:04
• internet search indicates the panels cosing around \$60 each Sep 15, 2021 at 22:31
• use multiple pieces per if you can tolerate the defect. There are some sizes that are economical... check furniture stuff like ikea mirrors
– Abel
Sep 18, 2021 at 2:00
• @Abel yeah I did think about putting smaller pieces together, although I don't think it would look very good. Ikea may actually be a good idea though. Unfortunately it wouldn't work with just a regular mirror, as it needs to be a one way mirror you can see through. Sep 20, 2021 at 1:14


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# Proof Techniques to Prove System Secure

I have so far seen the proof by reduction technique to prove the security guarantee of cryptosystem where we consider breaking the cryptosystem as hard as solving a difficult mathematical problem. What are the other proof techniques used to prove the security of the Cryptosystem? It would be beneficial if the technique is briefly stated.


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# Homework Help: Inductance problem: mass dropped above magnetic field

1. May 8, 2015

### monkeymind

1. The problem statement, all variables and given/known data
"Consider a square metal picture frame of side length s, mass M, and total electrical resistance R. It is dropped from rest from a height H above a region of uniform magnetic field pointing into the page. The frame accelerates downward under the influence of gravity until reaching the magnetic field. It is observed that, while entering the magnetic field, the frame moves with constant velocity. What is the frame's speed when it begins to enter the magnetic field? What is the strength of the magnetic field? (Solve for B in terms of the given physical quantities (s, M, R, and H) and any constants you need.) After the frame has completely entered the magnetic field, what is the frame's acceleration? Justify your answer in two sentences or less."

2. Relevant equations
$${V_{f}}^2 = {V_{i}}^2 + 2 a x$$
$$F_{mag} = \frac{v l^2 B^2}{R}$$
$$F = m a$$

3. The attempt at a solution
First I found the velocity after the metal frame dropped a height H:

$$V_{f}^2 = V_{i}^2 + 2 a x$$
$$V_{f} = \sqrt{ 2 a x } = \sqrt{ 2 (9.8 \frac{m}{s^2 }) H} = 4.4 H \frac{m}{s}$$

Then I reasoned that the only portion of the square metal picture frame that would factor into the magnetic force was the bottom horizontal portion. I figured that the two sides were moving in the same direction as velocity, and therefore the cross-product would go to zero. I also reasoned that the top horizontal portion would not matter, because as soon as this portion reaches the magnetic field, the flux is no longer changing and there will be no induced magnetic field, and therefore no magnetic force.

I approached the next portion as a dynamic equilibrium problem. Since the velocity is constant, I summed the forces in the positive and negative y-direction. Downward I have $F_{gravity} = m g$, and upward $F_{B} = \frac{v s^2 B^2}{R}$. I then set them equal and solved for B, obtaining:

$$B = \sqrt{ \frac{M g R}{4.4 H s} }$$

After it completely enters the field, I assumed that its acceleration would return to $9.8 \frac{m}{s^2}$ because the flux is no longer changing, so no fields are being induced.

Is this at all on the right track?

2. May 9, 2015

### rude man

Vf and B are both wrong. Redo, leaving all parameters in terms of symbols rather than numbers until the very end. You are missing chances to check dimensions term-by-term as you go. You got all the right relations and then blew it by introducing numbers prematurely.

3. May 9, 2015

### PhysixChick

dude, that's almost what i got, except i left gravity as g, and for s, i considered the length of the whole perimeter to be 4s, then squaring I got 16s. are you in simpson's class? anyways, i used the conservation of energy equation for part A though: MgH = 1/2 Mv^2

Last edited: May 9, 2015
4. May 9, 2015

### Staff: Mentor

4.4√H

Hi monkeymind. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif [Broken]

You did a good job of explaining your working, which makes errors easy to spot.

rude man makes a good point about not being too hasty in substituting data values for symbols in your formulae.

Last edited by a moderator: May 7, 2017
5. May 9, 2015

Oh?

6. May 9, 2015

### PhysixChick

oops, sorry... 16s^2... is that wrong rude man ? I figured it would be right, because this is the whole length in which current can flow through.

7. May 9, 2015

### rude man

That's the area, not the length. The length is 4s. But I'm not sure why you are interested in it.
Monkeymind did a good job with his reasoning, you should look at it. Just remember not to switch to numbers until the end.

8. May 9, 2015

### PhysixChick

@rude man , I did the same thing monkeymind did, but i didn't plug in numbers. and I didn't use the area, I used the length of the whole picture frame (if it was straightened out, it would be 4s). the formula for magnetic force is F = V(SB)^2/R = V S^2 B^2 / R so instead of plugging in just s, I plugged in 4s.

9. May 9, 2015

### rude man

Problem is, @PhysicsChick, that only one of the four sides gets the mag. force in the vertical direction.
.

10. May 9, 2015

### PhysixChick

OK, thanks for the help, I really appreciate it! That actually makes sense now that I think about it lol.

11. May 9, 2015

### monkeymind

I see that I made two mistakes (hopefully that's it!). One when calculating vf and one when calculating B.

Correcting, this gives me:

$$v_{f} = 4.4 \sqrt{H} \frac{m}{s}$$
$$B = \sqrt{\frac{MgR}{(4.4\sqrt{H})s^2}}$$

Is that better?

12. May 9, 2015

### PhysixChick

I got $$B = \sqrt{\frac{MgR}{(\sqrt{2gH})s^2}}$$

13. May 9, 2015

### rude man

That is right, and is probably what monkeyminbd would have gotten had he/she stuck to symbols to the end as you did.

14. May 9, 2015

### PhysixChick

I'm sure it's the same thing.

15. May 9, 2015

### monkeymind

Thanks for your help, rude man! I will definitely start solving symbolically from now on.

16. May 10, 2015

### rude man

I don't want to over-dramatize, but that is one of the most important things to learn in your physics studies.
So now try also to check units in each term. I prefer the term "dimensions". Like, force would have dimensions {F} = MLT-2, M for mass, L for length, T for time. When you get into elec. & magnetism you have to add Q for charge. You can deduce the dimensions of B for example by the familiar formula
F = iLB so B = F/iL and so {B} = {F}/{i}(L} = MLT-2/QT-1 = MLT-2Q-1TL-1 = MT-1Q-1.
Or, from ampere's law B = μi/L, {μi/L} = {B} which means that {μ} = {B}{L}/{i} = MLQ-2.

Last edited: May 10, 2015
17. May 10, 2015

### PhysixChick

@rude man that's what we're taking right now! lol. Thanks for your help! Much appreciated =]


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## Van't Hoff Eq.

$\ln K = -\frac{\Delta H^{\circ}}{RT} + \frac{\Delta S^{\circ}}{R}$

Christina_F_3F
Posts: 21
Joined: Fri Jul 22, 2016 3:00 am

### Van't Hoff Eq.

For the Van't Hoff equation, we are given two different versions of the same equation:
-ln(K2/K1) = -(deltaH*/RT2) + (deltaH*/R) where one assumes delta S naught is constant
-ln (K2/K1) = -deltaH*/R(1/T2-1/T1) where one assumes delta H naught is constant

My question is under what conditions are delta S naught and delta H naught not constant? If they are under standard conditions my understanding would be that they would be constant no matter what, because they are under a set of constant conditions.

Thanks!

Christian Hardoy 3F
Posts: 41
Joined: Fri Jul 15, 2016 3:00 am

### Re: Van't Hoff Eq.

From what I understand, we aren't being given 2 Van't Hoff equations. When the course reader says, "assume delta S naught is constant" and, "assume delta H naught is constant" it is not because there are two equations where one is constant and the other is not. The course reader says this because each line is a step in writing the Van't Hoff equation in the final form that is on the last line of page 42 in the course reader. In other words, I think the assumptions are part of rewriting and explaining the Van't Hoff equation, not two distinct equations with different assumptions.

Alexia Joseph 2B
Posts: 56
Joined: Thu Jul 27, 2017 3:01 am

### Re: Van't Hoff Eq.

From what we learned in class today, I'm assuming we only need to know how to use the equation when delta S and delta H are held constant.

Nicole 1F
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### Re: Van't Hoff Eq.

Can someone please explain what it means for ∆S°to be constant? I think Dr. Lavelle was explaining this at the end of lecture (something to do with the difference not changing between the two ∆S° values), but I didn't quite catch it all. Thanks!

Manvir2K
Posts: 32
Joined: Fri Sep 29, 2017 7:05 am

### Re: Van't Hoff Eq.

∆S°is constant because the difference between the two ∆S°values is the same. This is because no matter what two values are taken, the difference for both values will be the same.

Michael Lee 2I
Posts: 51
Joined: Fri Sep 29, 2017 7:05 am

### Re: Van't Hoff Eq.

Both delta S and delta H remain constant from what Lavelle told us in class today.


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Toronto Math Forum

APM346-2018S => APM346--Lectures => Topic started by: Ioana Nedelcu on January 24, 2018, 12:29:49 AM

Title: Uniqueness/ continuity of solutions
Post by: Ioana Nedelcu on January 24, 2018, 12:29:49 AM
I'm still trying to understand how the initial and boundary conditions affect the existence and types of solutions.

For example, in one of the tutorials the solution to $$xu_{t} + u_{x} = 0$$ $$u(x, 0) = f(x)$$ is not uniquely determined. What exactly determines this?

(It's also possible I understood the question completely wrong but I'm generally confused about uniqueness)
Title: Re: Uniqueness/ continuity of solutions
Post by: Victor Ivrii on January 24, 2018, 03:27:49 AM
Also existence problems at $(0,0)$.

Draw integral lines and consider their intersections with $\{(x,t)\colon t=0\}$.
Title: Re: Uniqueness/ continuity of solutions
Post by: Jingxuan Zhang on January 24, 2018, 06:01:01 AM
But should there be issue with existence? I think there at the origin we have this parabolic characteristic tangent to it and what's wrong with that?
Title: Re: Uniqueness/ continuity of solutions
Post by: Victor Ivrii on January 24, 2018, 06:48:43 AM
Look, at the picture. In the "yellow" domain initial condition $u(x,0)=f(x)$ does not define solution (why?).
In  the white domain initial condition could be impossible to satisfy (why?)
Title: Re: Uniqueness/ continuity of solutions
Post by: Jingxuan Zhang on January 24, 2018, 07:13:23 AM
Look, at the picture. In the "yellow" domain initial condition $u(x,0)=f(x)$ does not define solution (why?).
In  the white domain initial condition could be impossible to satisfy (why?)
But you mentioned in another post that existence problem is at $(0,0)$. For the white domain indeed you have to have even initial function to meet the symmetrical curves, but $t-x^{2}=0$ does not require this, and so shouldn't it rather be no issue at origin?
Title: Re: Uniqueness/ continuity of solutions
Post by: Victor Ivrii on January 24, 2018, 07:43:34 AM
It means "in the arbitrarily small vicinity of the origin"
Title: Re: Uniqueness/ continuity of solutions
Post by: Ioana Nedelcu on January 24, 2018, 05:40:54 PM
For the yellow domain, is it because when t=0 there are infinite possibilities for f(x)? So then there is no unique solution there

Impossible solution in the white domain because the characteristic curve doesn't cross the x-axis so there is no function of x, ie solution doesn't exist

Because the solution u(x, t) is constant on/ determined by the characteristic curves, do we essentially just look at how those curves behave at the initial and boundary conditions?
Title: Re: Uniqueness/ continuity of solutions
Post by: Victor Ivrii on January 25, 2018, 04:22:51 AM
1) In yellow domain characteristics do not intersect $\{t=0\}$ (non-unicity)
2) In white domain they intersect $\{t=0\}$ twice (possible non-existence)


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location:  Publications → journals
Search results

Search: MSC category 11G09 ( Drinfel'd modules; higher-dimensional motives, etc. [See also 14L05] )

 Expand all        Collapse all Results 1 - 3 of 3

1. CMB 2009 (vol 53 pp. 95)

Ghioca, Dragos
 Towards the Full Mordell-Lang Conjecture for Drinfeld Modules Let $\phi$ be a Drinfeld module of generic characteristic, and let X be a sufficiently generic affine subvariety of $\mathbb{G_a^g}$. We show that the intersection of X with a finite rank $\phi$-submodule of $\mathbb{G_a^g}$ is finite. Keywords:Drinfeld module, Mordell-Lang conjectureCategories:11G09, 11G10

2. CMB 2000 (vol 43 pp. 282)

Boston, Nigel; Ose, David T.
 Characteristic $p$ Galois Representations That Arise from Drinfeld Modules We examine which representations of the absolute Galois group of a field of finite characteristic with image over a finite field of the same characteristic may be constructed by the Galois group's action on the division points of an appropriate Drinfeld module. Categories:11G09, 11R32, 11R58

3. CMB 1997 (vol 40 pp. 385)

Bae, Sunghan; Kang, Pyung-Lyun
 Elliptic units and class fields of global function fields Elliptic units of global function fields were first studied by D.~Hayes in the case that $\deg\infty$ is assumed to be $1$, and he obtained some class number formulas using elliptic units. We generalize Hayes' results to the case that $\deg\infty$ is arbitrary. Categories:11R58, 11G09


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• 研究报告: 植物多样性 •

### 内蒙古西部自然植被的物种多样性及其影响因素

1. 1 .地表过程与资源生态国家重点实验室, 北京师范大学地理科学学部, 北京 100875
2 .内蒙古大学生态与环境学院, 呼和浩特 010021
• 收稿日期:2017-05-12 接受日期:2017-12-01 出版日期:2017-12-20
• 通讯作者: 黄永梅 E-mail:ymhuang@bnu.edu.cn
• 基金项目:
科技基础性工作专项“华北地区自然植物群落资源综合考察” (2011FY110300)

### Biodiversity of natural vegetation and influencing factors in western Inner Mongolia

Chongyao Yang1, Engui Li1, Huiying Chen1, Jinghui Zhang2, Yongmei Huang1, *()

1. 1 State Key Laboratory of Earth Surface Processes and Resource Ecology, Faculty of Geographical Science, Beijing Normal University, Beijing 100875
2 School of Ecology and Environment, Inner Mongolia University, Hohhot 010021
• Received:2017-05-12 Accepted:2017-12-01 Online:2017-12-20
• Contact: Huang Yongmei E-mail:ymhuang@bnu.edu.cn

To determine biodiversity, 74 plant community plots were surveyed in western Inner Mongolia between 2012 and 2014. These samples were divided into four vegetation types, including temperate steppe, temperate shrub, meadow, and desert. Four biodiversity indices (Patrick index, Simpson index, Shannon-Wiener index and Pielou index) were calculated to show biodiversity differences among the four vegetation types. Climate and soil data were also collected over the course of the survey. Pearson correlation analysis was conducted to reveal the relationship between biodiversity and environmental factors and stepwise regression analysis was used to determine dominant environmental factors. Results indicated that: (1) In western Inner Mongolia, biodiversity indices were different among the four vegetation types. Temperate steppe and meadow had similar values for biodiversity indices, and were also the highest amongst all vegetation types, while the lowest biodiversity was found in the desert. (2) Pearson correlation analysis showed that the Patrick index, Shannon-Wiener index, and Simpson index were all negatively and linearly correlated with mean annual temperature and potential evapotranspiration. (3) Among all chemical properties of the soil, pH, organic carbon and total nitrogen all influenced biodiversity indices. (4) Stepwise regression analysis showed that potential evapotranspiration and soil organic carbon were the most influential factors affecting the Patrick index and Shannon-Wiener index. The only dominant factor for the Simpson index was soil organic carbon while soil total nitrogen was most important factor for the Pielou index.


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# Suffix Automaton Study Note

This is a review & note for studying suffix automaton.

### What is a Suffix Automaton?

In computer science, a suffix automaton is the smallest partial deterministic finite automaton that recognizes the set of suffixes of a given string(computer_science).

The above definition is a short excerpt from Wikipedia about suffix automaton. From the definiton of deterministic finite automaton, we let $\mathrm{trans(s,ch)}$ be the state transferred from current state by adding a new character $\mathrm{ch}$; we let $\mathrm{null}$ be a empty state; we let $\mathrm{trans(s,str)}$ be the state transferred from current state by adding a new string $\mathrm{str}$. Then, we know that we can obtain $\mathrm{trans(s,str)}$ by adding $\mathrm{ch_i}\in\mathrm{str}$ one by one into the automaton.

From the definition on Wikipedia, it is obvious that a suffix automaton can recognize all suffixes of a given string. The word recognize is defined as a existence of a path in the automaton that the word formed by the path, by adding the characters on the path, is a suffix. Meanwhile, the suffix automaton can also be used to recognize all substrings.

### Important Properties of Suffix Automaton

A obvious way is to insert all of the suffixes into a Trie tree. But sometime it may take $\mathrm{O(N^2)}$ space. Lets consider how to simplify the states (reduce the number of vertices). We denote $\mathrm{ST(str)}$ the state / vertex $\mathrm{trans(init,str)}$, where $\mathrm{init}$ is the initial state. Obviously, if $\mathrm{str}$ is not a substring of the original string, then $\mathrm{ST(str)} = \mathrm{null}$, since no suffix will be formed by adding character after the vertex.

Shall we build nodes for all substrings? The anwser is also NO, since the total number of states will also be $O(N^2)$. Lets denote the set of end position, which is the occurence index of substrings in the original string, for each vertex as $\mathrm{right(v)}$. Then, we can see that if $\mathrm{right(a)} = \mathrm{right(b)}$, then $\mathrm{ST(a)} = \mathrm{ST(b)}$, this simplifies the overall states. Let the length of substrings in the vertex be in range $[\min(\mathrm{v}),\max(\mathrm{v})]$. Consider two different state $\mathrm{a},\mathrm{b}$ and their right set $\mathrm{Ra},\mathrm{Rb}$. Then, the following property holds:

Property 1: If $\mathrm{Ra},\mathrm{Rb}$ are intersecting then one set is another’s proper subset.

Lets define a parent tree. where $\mathrm{fa} = \mathrm{Parent(v)}$. Let $\mathrm{right(fa)} = \bigcup_{\mathrm{v}\in \mathrm{child(fa)}}\mathrm{right(v)}$. We also have following properties.

Property 2: $\max(\mathrm{Parent(v)}) = \min(\mathrm{v}) – 1$.

Property 3: $\mathrm{trans(a,c) = b}$, then an edge from $\mathrm{a}\rightarrow \mathrm{b}$ marked with character $\mathrm{c}$.

Property 4: If $\mathrm{ST(s)}\neq \mathrm{null}$, then $\mathrm{s}$ must be a substring of the original string. Otherwise, $s$ is not a substring.

Property 5: The string denoted by $\mathrm{Parent(v)}$ is a suffix of the string denoted by $\mathrm{v}$.

Property 6: If $\mathrm{trans(v,ch)}\neq \mathrm{null}$, then $\mathrm{trans(Parent(v),ch)} \neq \mathrm{null}$.

Property 7: Within a node $\mathrm{v}$, the lengths of strings denoted by the vertex are in range $[\min(\mathrm{v}),\max(\mathrm{v})]$, and all of the lengths of strings denoted by the vertex within such range exist.


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# How to get a more intuitive/and motivated understanding of a solution?

Math is a deductive science, that one starts from basic premises to prove more complicated results. From doing questions and reading solutions (at least at the elementary level eg: Olympiad questions), I find that there is always a dichotomy between a 'solution as it is presented' and 'a solution about how it is motivated', and sometimes when I am reading a solution I find myself saying to myself 'yes, this is true' without understanding how it is motivated until the end.

In short, I am wondering how to get a more intuitive/and motivated understanding of a solution. So far, I have found that playing with examples and then trying to generalize the result is helpful.

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What do you mean by "motivated"? –  Warren P May 27 '12 at 16:09
Some writers are good at explaining the idea behind the proof, or how one can take an intuitive idea and turn it into a proof (evading the pitfalls along the way). Other writers are not, or prefer to present the polished jewels rather than the raw material. Often, you have to find your own way of "how to think about the result" and understand the solution, and that comes from practice and reflection. A good exercise is to take a proof of a theorem and see exactly how and where each hypothesis comes into it, and try to come up with examples where the argument fails if the hypothesis fails. –  Arturo Magidin May 27 '12 at 16:19
Well, it might happen that even the author of the proof does not recognize the idea behind it. It had happened for me (more than once) that I reinvented some well-known methods, but only a few months/years later, when introduced to whole theory, I was able to grasp the bigger picture and understand my own proofs. Of course, there probably are even better perspectives that puts all this into new contexts, however, let us not dissolve into philosophy, I do like using a trash can ;-) –  dtldarek May 27 '12 at 17:00

In addition to the great comments above (I especially agree with Arturo's):

Let us suppose for a moment that we are considering math at the level of olympiad-style questions. Then the single most important method of developing intuition and motivation for such problems that can think of is to attempt, and I mean really attempt, to do the problem first.

I bring this up largely because of your phrasing with respect to 'readng solutions.' One should not, perhaps, just read one solution after another. There is a large amount to be gained from struggling with a problem. This is said every now and then, but it is really true.

So let's say that you are working on developing olympiad inequality intuition. Then what do you do? Find a set of olympiad inequality problems, and start to try them. Maybe you'll have no idea what to do. That's alright - push along. Try things. Play, struggle, move to other problems. Maybe you'll make progress on one that might give you ideas on others. If you're lucky, or perhaps brilliant, then you will motivate some of the solutions on your own. But it's likely that some will elude you.

And in that case, the time you spent struggling on the problem will have given you a certain roadmap of the problem, so that when (if) you start to parse a solution, you might be able to see exactly what obstacles different aspects of the solution overcame, or at least start to develop such an understanding.

Yet still, this does not give a full intuition. It is when you repeat this, trying more problems, struggling, etc., and when you are able to push a bit further, modify your previous methods of attacking the problems, and see how ideas of previous problems present themselves in present problems that the intuition starts to settle.

The fact is that it's hard to see what bits are important, or how to apply them. So you must struggle in order to see what different things are useful, struggle some more to see how they are useful, and struggle yet more to make it worthwhile for you to learn it.

It reminds me of the language teacher who lamented the prevalence of easy, online dictionaries. The idea is that when a student has to look through a paper dictionary for words all the time (which takes a bit, and a bit of effort), they remember the words more than when they are able to instantly look up anything with just a few effortless keystrokes.

Now let's leave behind elementary or olympiad-style questions, and consider the question again. It is an unfortunate aspect of the math publishing world that people often do not explain how they came up with the ideas behind their papers, or how they thought to cross whatever gap they crossed. (their motivation, to use your word). There are many causes - one that's perhaps undermentioned is that often the birth of an idea is not the clear, clean, seemingly divinely inspired flash of insight, but instead something that comes out of a very messy, circuitious, roundabout jumble of work. And the act of cleaning up the idea into something presentable manages to obfuscate the origin of the idea.

This aspect of math is very hard to overcome, I think. In a certain sense, struggling with similar problems and trying to generalize are still a good idea.

Two examples come to mind. One might show that $\displaystyle \sum_{p \; \text{prime}} \frac{1}{p}$ diverges by many different elementary techniques. Dirichlet's result on the infinitude of primes in arithmetic progressions can be recast in a similar statement about the divergence of $\displaystyle \sum_{\underset{(a,b) = 1}{an + b \; \text{prime}}} \frac{1}{an + b}$. Of course, the proofs are radically different. But the heart of the idea is sort of there.

A bit easier to fathom is Sophie Germain's identity. For a long time, I never saw how someone might come across it. But then one day I realized that it comes sort of naturally if you complete a square $a^2 + b^2 = (a+b)^2 - 2ab$ and wonder what it would take to use $x^2 - y^2 = (x-y)(x+y)$. You might quickly ratonalize that having $a^4 + 4b^4 = (a^2 + 2b^2)^2 - 4a^2b^2$ lets you proceed, and leads immediately to Sophe Germain's identity. Somehow, this thought process gives me peace.

-
Training for Olympiad is more reducing the problem to pre-existing methods rather than blindly approach the problem. Quoting from Terrence Tao's Solving Mathematical Problem: "The journey of a thousand miles begins with one step." Terrence Tao: As long as one steps in a firm, clear direction, with long strides and sharp vision, one would need far, far less than the millions of steps needed to journey a thousand miles. And of course, one should endeavor to solve the problem before reading the solution. –  Mat.S May 27 '12 at 17:38


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What are some examples of problems in R but not in PR?

Also are there any R complete problems? These would be the hardest decidable problems to my knowledge. Sorry if this isn't specific enough.

PR class is associated with primitive recursive functions but not every total recursive function is a primitive recursive function. The most famous example of such functions is the Ackermann function. If you need to specify a decision problem then you can use e.g. the language $$L_{A} =\{ \langle m,n,k\rangle \mid m,n,k \in \mathbb{N}, A(m,n) \le k \} \subset \{0,1\}^{*}$$, (A - Ackermann function and e.g. $$\langle m,n,k\rangle = 1^{m}0^{n}1^{k}$$ or use any other encoding scheme). In this way you can build other examples including partial recursive functions (assuming that $$\bot < 0$$) which are also not in PR. Class R is closed under $$\le_{m}$$ reduction but any non-trivial language in R is R-complete under this reduction so you need to use some stronger reduction to find a specific set of the hardest problems in R.


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How do you simplify 45/(10a-10) ?

Sep 22, 2016

$\frac{9}{2 \left(a - 1\right)}$

Explanation:

In algebraic fractions, check first if you can find factors.

$\frac{45}{10 a - 10}$

=$\frac{45}{10 \left(a - 1\right)} \text{ } \leftarrow$ cancel by 5

=${\cancel{45}}^{9} / \left({\cancel{10}}^{2} \left(a - 1\right)\right)$

=$\frac{9}{2 \left(a - 1\right)}$


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When we have assets for multiple holding periods, it is necessary to aggregate the returns into one overall return. Portfolio Management 1. ©AnalystPrep. Asset classes could â¦ What are the arithmetic mean and geometric mean, respectively, of an investment which returns 8%, -2% and 6% each year for three years? In general, returns are presented pre-tax and with no adjustment for the effects of inflation. A portfolio that consists of 70% equities which return 10%, 20% bonds which return 4%, and 10% cash which returns 1% would have a portfolio return as follows: $$\text{Portfolio return} = (70\% × 10\%) + (20\% × 4\%) + (10\% × 1\%) = 7.9\%$$. (read difference between saving & investment) Investment is about deferring your present consumption for future goals with expectation of security of amount & getting returns. β or Beta describes the relationship between the stock’s return and the Market index return. Meaningful diversification is one which involves holding of stocks of more than one industry so that risks of losses occurring in one industry are counterbalanced by gains from the other industry. It is important to be able to compute and compare different measures of return to properly evaluate portfolio performance. This website includes study notes, research papers, essays, articles and other allied information submitted by visitors like YOU. It is the percentage change in the price of the stock regressed (or related) to the percentage changes in the market Index. Risk avoidance and risk minimization are the important objectives of portfolio management. Why an investor does has so many securities in his portfolio? The last column of weighted squared deviations are derived by multiplying the weights of probabilities with squared deviations. 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Types of PMS There are three types of Portfolios in PMS, you can choose any of them-Discretionary PMSâ Discretionary Portfolio provides the service provider a right to make decisions on behalf of the client, whether he wants to sell or buy the shares. An investor considering investment in securities is faced with the problem of choosing from among alarge number of securities. Diversification of risk in portfolio management - Portfolio should be diversified to the degree at which specific risk has virtually been eliminated. If there is no relation between them, the coefficient of correction can be zero. Return and standard deviation and variance of portfolios can also be compared in the same manner. investments in high quality, blue-chip stocks . If the security “ARC” gives the maximum return why not he invest in that security all his funds and thus maximise return? Passive portfolio management Passive portfolio management is a process where the portfolio manager creates a fixed portfolio aligning it with current trends in the market. If an investor makes use of derivative instruments within a portfolio or borrows money to invest, then leverage is introduced into the portfolio. It is a long-term strategy for index investing whose purpose is to generate â¦ For example, if a share has returned 15%, 10%, 12% and 3% over the last four years, then the arithmetic mean is as follows: $$\text{Arithmetic mean} = \frac {15\% + 10\% + 12\% + 3\%} {4} = 10\%$$. Another measure is the standard deviation and variance-. Disclaimer Copyright, Share Your Knowledge Before publishing your Articles on this site, please read the following pages: 1. So the portfolio manager according to the risk-taking capacity and the kind of returns calculated provides a portfolio structured in tandem with that. Headline stock market indices typically report on price appreciation only and do not include the dividend income unless the index specifies it is a “total return” series. In doing so, active managers would buy stocks with lower value and would sell it once it climbs above the norm. A net return is the return post-deduction of fees. 1. Dow â¦ Share Your Word File Each security in a portfolio contributes returns in the proportion of its investment in security. Arithmetic returns tend to be biased upwards unless the holding period returns are all equal. Average investors are risk averse. Portfolio Management Involves,-Investing and divesting different, -investment Risk management,- Monitoring and analyzing returns Portfolio management is a process encompassing many activities aimed at optimizing the investment Active Portfolio Management: This type of portfolio management aims to make better returns in contrast to what the market hypothesizes. A geometric return provides a more accurate representation of the portfolio value growth than an arithmetic return. PORTFOLIO MANAGEMENT-TRIAL QUESTIONS 1) Explain the following terms as used in Portfolio management and give examples and/or furmulas. Thus, one measure of risk is absolute deviations of returns under column 6. For example, a year relative to a 20-month holding period is a fraction of 12/20. The different types of portfolio investment are as follows: Risk-Free Portfolios â Risk-free portfolios are the ones that have investment securities regarding treasury bonds and such where the risk is almost nil but low returns. When the return of company XYZ is compared with that of company ABC, we have to take into account, their expected return and standard deviation of the return. Portfolio management is a process of choosing the appropriate mix of investments to be held in the portfolio and the percentage allocation of those investments. Using the same annual returns of 15%, 10%, 12% and 3% as above, we compute the geometric mean as follows: $$\text{Geometric mean} = [(1+15\%) × (1+10\%) × (1+12\%) × (1+3\%)]^{1/4} – 1 = 9.9\%$$. Return on Portfolio: Each security in a portfolio contributes returns in the proportion of its investment in security. x̅ is the mean or the weighted average return. Thus, the portfolio expected return is the weighted average of the expected returns, from each of the securities, with weights representing the proportionate share of the security in the total investment. On the other hand, the 7-stock portfolio might represent a number of different industries where returns might show low correlation and, hence, low portfolio returns variability. First, financial assets may provide some income in the form of dividends or interest payments. Thus (.04)2 = .0016 and this multiplied by the probability of it 0.25 gives .0004 and so on. 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(Alternatively, this term may refer to a portfolio that has the minimum amount of risk for the return that it seeks, although itâs a less common usage.) There are other measures of returns that need to be taken into account when evaluating performance. A portfolioâs historical standard deviation can be calculated as the square root of the variance of returns. Portfolio optimization should result in what investors call an âefficient portfolioâ. Portfolio management helps an individual to decide where and how to invest his hard earned money for guaranteed returns in the future. Portfolio management thus refers to investment of funds in such combination of different securities in which the total risk of portfolio is minimized while expecting maximum return from it. We examine the cash flows from the perspective of the investor where amounts invested in the portfolio are seen as cash outflows and amounts withdrawn from the portfolio by the investor are cash inflows. In order to compute this, we weight the returns of the underlying assets by the amounts allocated to them. An aggressive portfolio takes on great risks in search of great returns. Rho or Correlation Coefficient (Covariance): Rho measures the correlation between two stocks say i and j. 2. Covariance (or correlation) denotes the directional relationship of the returns from any two stocks. To annualize a return earned for a period shorter than one year, the return must be compounded by the number of periods in the year. When we talk about the returns from a financial asset, we can broadly classify them into two types. Investment Analysis & Portfolio Management (FIN630) VU general types: those that are pervasive in nature, such as market risk or interest rate risk, and This means itâs generating the highest possible return at your established risk tolerance . There shall be bifurcation across these five units of the total corpus provided. There are different types of returns, it can be either short-term return or long term, it must go according to the goals of the corporate entity etc. PORTFOLIO MANAGEMENTWhat is portfolio management? But when you want to calculate the expected volatility, you must include the covariance or correlation of each asset. Capital Asset Pricing Model When an asset needs to be added to an already well diversified portfolio, Capital Asset Pricing Model is used to calculate the assetâs rate of profit or rate of return (ROI). Our mission is to provide an online platform to help students to discuss anything and everything about Economics. The answers to this question lie in the investor’s perception of risk attached to investments, his objectives of income, safety, appreciation, liquidity and hedge against loss of value of money etc. The summation of all deviations from the mean and then squared will give the variance. The IRR provides the investor with an accurate measure of what was actually earned on the money invested but does not allow for easy comparison between individuals. A weekly return of 2%, when annualized, is as follows: $$\text{Annualized return} = (1+2\%)^{52} – 1 = 180\%$$. Computing a geometric mean follows a principle similar to the computation of compound interest. Defensive Portfolio : Defensive portfolio comprises of stocks with low risk and stable earnings â e.g. Beta measures the systematic market related risk, which cannot be eliminated by diversification. It can normally vary from – 1 to 4- 1. Understanding Active Return Active return refers to the portion of returns (profit or loss) in an investment portfolio that can be directly attributed to the active management decisions made by the portfolio manager Portfolio Manager Portfolio managers manage investment portfolios using a six-step portfolio management process. If the Beta is minus one, it indicates a negative relationship with the Market Index and if the market goes up by a + 1%, the stock price will fall by 1%. Privacy Policy3. Tax considerations like capital gains tax and tax on interest or dividend income will need to be deducted from the investment to determine post-tax returns. Real returns are useful for comparing returns over different time periods as inflation rates vary over time. And one of the things you should know in order to complete one is to know the various types of portfolio management. The previous year’s returns are compounded to the beginning value of the investment at the start of the new period in order to earn returns on your returns. Passive Portfolio: When the An Example for Covariance and Correlation: Given the standard deviation of X and Y as 13.23 and 9.75 in the portfolio of two securities the covariance between them has to be estimated. Assuming that investor puts his funds in four securities, the holding period return of the portfolio is described in the table below: The above describes the simple calculation of the weighted average return of a portfolio for the holding period. You will notice the workflow logic is the same for each object type. If we had a return of 12% for 20 months, then the annualized return is as follows: $$\text{Annualized return} = (1+12\%)^{12/20} – 1 = 7\%$$. The leverage amplifies the returns on the investor’s capital, both upwards and downwards. If the period during which the return is earned is not exactly one year, we can annualize the return to enable an easy comparative return. Returns are typically presented in nominal terms which consist of three components: the real risk-free return as compensation for postponing consumption, inflation as compensation for the loss of purchasing power and a risk premium. You These are as follows: A gross return is earned prior to the deduction of fees (management fees, custodial fees, and other administrative expenses). Active portfolio management strategy is normally dependent on the style of management and analysis that can generate healthy returns and beat the stock market as well. As returns and prices of all securities do not move exactly together, variability in one security will be offset by the reverse variability in some other security. The IRR is the discount rate applied to determining the present value of the cash flows such that the cumulative present value of all the cash flows is zero. If on the other hand, Rho is -1, the direction of the movement will be opposite as between the stock price and market index. These are briefly discussed below. These theories can be classified into different categories as depicted in figure 6.1. The expected return depends on the probability of the returns and their weighted contribution to the risk of the portfolio. Employees of both private and public companies often save and invest for retirement... Risk budgeting is focused on implementing the risk tolerance decisions taken at a... 3,000 CFA® Exam Practice Questions offered by AnalystPrep – QBank, Mock Exams, Study Notes, and Video Lessons, 3,000 FRM Practice Questions – QBank, Mock Exams, and Study Notes. If the portfolio is efficient, Beta measures the systematic risk effectively. The Company with a higher return of 11.5% has a higher risk than the other company. Financial market assets generate two different streams of return: income through cash dividends or interest payments and capital growth through the price appreciation of the asset. A holding period return is a return earned from holding an asset for a specified period of time. This means we look at the return as a composite of the price appreciation and the income stream. It measures the unsystematic risk of the company. Companies can be ranked in the order of return and variance and select into the portfolio those companies with higher returns but with the same level of risk. Portfolio management involves selecting and managing an investment policy that minimizes risk and maximizes return on investments. Reducing the Risk of Certain Asset Classes When the holding period is longer than one year, we need to express the year as a fraction of the holding period and compound using this fractional number. Now, the variance is sum of the weighted squared deviations, namely, .0008. This can be positive and negative. Traditional Approach: 1. If α = 0, then the regression line goes through the origin and its return simply depends on the Beta times the market return. CFA® and Chartered Financial Analyst® are registered trademarks owned by CFA Institute. A monthly return must be compounded 12 times, a weekly return 52 times and a daily return 365 times. ADVERTISEMENTS: Portfolio theories guide the investors to select securities that will maximize returns and minimize risk. A risk averse investor always prefer to â¦ Content Guidelines 2. Portfolio management is the art of selecting the right investment tools in the right proportion to generate optimum returns with a balance of risk from the investment made. We note the geometric return is slightly less than the arithmetic return. On the other hand alpha and Epsilon (α + ∑) measures the unsystematic risk, which can be reduced by efficient diversification. The total risk of a portfolio (as measured by the standard deviation of returns) consists of two types of risk: unsystematic risk and systematic risk. A money-weighted return is similar in computation methodology to an internal rate of return (IRR) or a yield to maturity. Second, some financial assets may also offer capital appreciation as the prices of these securities change themself. Investors typically have one or more types of portfolios among their investments and seek to achieve a balanced return on investment over time. In this article we will discuss about the analysis of return and risk on portfolio of a company. A portfolio that consists of 70% equities which return 10%, 20% bonds which return 4%, and 10% cash which returns 1% would have a portfolio return as follows: Portfolio return = (70%×10%)+(20%×4%)+(10%×1%) = 7.9% Portfolio return = ( 70 % × 10 %) + ( â¦ I. The basic equation for calculating risk can be formulated as a regression equation-, Where, y = Return in the security in a given period and, α = The intercept where the regression line crosses the y-axis. Each investor has his own risk profile, but there is a possibility that he does not have the relevant investment security that matches his own risk profile.

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1
JEE Main 2021 (Online) 24th February Morning Shift
+4
-1
If Y, K and $$\eta$$ are the values of Young's modulus, bulk modulus and modulus of rigidity of any material respectively. Choose the correct relation for these parameters.
A
$$Y = {{9K\eta } \over {3K - \eta }}N/{m^2}$$
B
$$Y = {{9K\eta } \over {2\eta + 3K}}N/{m^2}$$
C
$$\eta = {{3YK} \over {9K + Y}}N/{m^2}$$
D
$$K = {{Y\eta } \over {9\eta - 3Y}}N/{m^2}$$
2
JEE Main 2020 (Online) 6th September Evening Slot
+4
-1
A fluid is flowing through a horizontal pipe of varying cross-section, with
speed v ms–1 at a point where the pressure is P pascal. At another point where pressure is $${P \over 2}$$ Pascal its speed is V ms–1. If the density of the fluid is $$\rho$$ kg m–3 and the flow is streamline, then V is equal to :
A
$$\sqrt {{P \over {2\rho }} + {v^2}}$$
B
$$\sqrt {{P \over \rho } + {v^2}}$$
C
$$\sqrt {{{2P} \over \rho } + {v^2}}$$
D
$$\sqrt {{P \over \rho } + {v}}$$
3
JEE Main 2020 (Online) 6th September Morning Slot
+4
-1
An object of mass m is suspended at the end of a massless wire of length L and area of crosssection A. Young modulus of the material of the wire is Y. If the mass is pulled down slightly its frequency of oscillation along the vertical direction is :
A
$$f = {1 \over {2\pi }}\sqrt {{{YA} \over {mL}}}$$
B
$$f = {1 \over {2\pi }}\sqrt {{{mL} \over {YA}}}$$
C
$$f = {1 \over {2\pi }}\sqrt {{{YL} \over {mA}}}$$
D
$$f = {1 \over {2\pi }}\sqrt {{{mA} \over {YL}}}$$
4
JEE Main 2020 (Online) 5th September Evening Slot
+4
-1
In an experiment to verify Stokes law, a small spherical ball of radius r and density $$\rho$$ falls under gravity through a distance h in air before entering a tank of water. If the terminal velocity of the ball inside water is same as its velocity just before entering the water surface, then the value of h is proportional to :
(ignore viscosity of air)
A
r
B
r4
C
r3
D
r2
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# PCA/ordination/Correspondence analysis for data with two groupings

I am working with a data set on plant presence-absence within 40 plots. For two years, the presence-absence of all plants were noted in each plot. I would now like to investigate whether the plant community is more alike within a given year or within a given plot.

My immediate thought is to run a PCA or other kind of ordination in R. But I am unsure of what kind of PCA/ordination is the most suitable when I have not one, but two grouping variables along with my presence absence data? I have loooked at the PCoA analysis and the MCA analysis. My goal in using the PCA is to give a visual interpretation of potential shift en species within and between plots.

My goal is to be able to create a plot where I can color according to year and link the dots corresponding to the same plot. I have tried drawing what I envision and inserted it below:

My data has the following structure:

Species 1; Species 2; Species 3; ...; Species n; Year; Plot

1;         0;         1; ...;          1; 2017; 1

0;         0;         1; ...;          0; 2018; 1

1;         1;         1; ...;          1; 2017; 2

1;         0;         0; ...;          1; 2018; 2


text:
## souvik one year ago thermodynamics : An idea gas undergoes a process in which its internal energy U is related to its volume U=aV^b ,where a,b are constants..Show that the work done by the gas and the quantity of heat transferred to it to increase its internal energy by $$\delta$$U are respectively.....(see in the comments)

1. souvik

1.$\frac{ \Delta U (\gamma-1)}{ b }$ 2.$\frac{ \Delta U\{1+(\gamma-1)b\} }{ b }$

2. souvik

the relation between U & V is$U=aV ^{b}$

3. souvik

@Vincent-Lyon.Fr

4. Vincent-Lyon.Fr

Sorry, it must be quite simple, but I'm not getting through.

5. IrishBoy123

.