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2.3. Tidal simulation in the Orkney island¶

2.3.1. Introduction¶

This example demonstrates how OpenTidalFarm can be used for simulating the tides in a realistic domain.

It shows how to:
• load a mesh from file;
• use realistic tidal forcing on the open boundaries;
• load the bathymetry from a NetCDF file;
• use a non-homogenous viscosity;
• solve a time-dependent shallow water solver and store the results to file.

We will be simulating the tides in the Pentland Firth, Scotland for 12.5 hours, starting at 14:40 am on the 18.9.2001. The flow result at the end of the simulation looks like:

This example requires some large data files, that must be downloaded separately by calling in the source code directory:

git submodule init
git submodule update

2.3.2. Implementation¶

We begin with importing the OpenTidalFarm module.

from opentidalfarm import *
set_log_level(ERROR)

We also need the datetime module for the tidal forcing.

import datetime

Next we define the UTM zone and band, as we we will need it multiple times later on.

utm_zone = 30
utm_band = 'V'

Next we create shallow water problem and attach the domain and boundary conditions

prob_params = SWProblem.default_parameters()

We load the mesh in UTM coordinates, and boundary ids from file

domain = FileDomain("../data/meshes/orkney/orkney_utm.xml")
prob_params.domain = domain

The mesh and boundary ids can be visualised with

#plot(domain.facet_ids, interactive=True)

Next we specify boundary conditions. We apply tidal boundary forcing, by using the TidalForcing class.

eta_expr = TidalForcing(grid_file_name='../data/netcdf/gridES2008.nc',
data_file_name='../data/netcdf/hf.ES2008.nc',
ranges=((-4.0,0.0), (58.0,61.0)),
utm_zone=utm_zone,
utm_band=utm_band,
initial_time=datetime.datetime(2001, 9, 18, 10, 40),
constituents=['Q1', 'O1', 'P1', 'K1', 'N2', 'M2', 'S2', 'K2'], degree=2)

bcs = BoundaryConditionSet()

# Apply a strong no-slip boundary condition. This can be changed to
# free slip (weakly enforced), by leaving out the Constant((0, 0))
# argument and changing bctype to "free_slip"
prob_params.bcs = bcs

Next we load the bathymetry from the NetCDF file.

bathy_expr = BathymetryDepthExpression(filename='../data/netcdf/bathymetry.nc',
utm_zone=utm_zone, utm_band=utm_band, domain=domain.mesh, degree=2)
prob_params.depth = bathy_expr

The bathymetry can be visualised with

#plot(bathy_expr, mesh=domain.mesh, title="Bathymetry", interactive=True)

Equation settings

For stability reasons, we want to increase the viscosity at the inflow and outflow boundary conditions. For that, we read in a precomputed function (generated by compute_distance) and

V = FunctionSpace(domain.mesh, "CG", 1)
dist = Function(V)
File("dist.xml") >> dist

With that we can define an expression that evaluates to a nu_inside value inside the domain and a nu_outside value near the in/outflow boundary.

class ViscosityExpression(Expression):
def __init__(self, *args, **kwargs):
self.dist_function = kwargs["dist_function"]
self.nu_inside = kwargs["nu_inside"]
self.nu_boundary = kwargs["nu_boundary"]
self.dist_threshold = kwargs["dist_threshold"]

def eval(self, value, x):
if self.dist_function(x) > self.dist_threshold:
value[0] = self.nu_inside
else:
value[0] = self.nu_boundary

Finally, we interpolate this expression to a piecewise discontinuous, constant function and attach it as the viscosity value to the shallow water problem.

W = FunctionSpace(domain.mesh, "DG", 0)
nu = ViscosityExpression(dist_function=dist, dist_threshold=1000, nu_inside=10., nu_boundary=1e3, degree=1)
nu_func = interpolate(nu, W)
prob_params.viscosity = nu_func

The other parameters are set as usual.

prob_params.friction = Constant(0.0025)
# Temporal settings
prob_params.start_time = Constant(0)
prob_params.finish_time = Constant(12.5*60*60)
prob_params.dt = Constant(5*60)
# The initial condition consists of three components: u_x, u_y and eta.
# Note that we set the velocity components to a small positive number, as some
# components of the Jacobian of the quadratic friction term is
# non-differentiable.
prob_params.initial_condition = Constant((DOLFIN_EPS, DOLFIN_EPS, 1))

# Now we can create the shallow water problem
problem = SWProblem(prob_params)

# Next we create a shallow water solver. Here we choose to solve the shallow
# water equations in its fully coupled form:
sol_params = CoupledSWSolver.default_parameters()
solver = CoupledSWSolver(problem, sol_params)

Now we are ready to solve and save the results to file.

f_u = XDMFFile("results/u.xdmf")
f_eta = XDMFFile("results/eta.xdmf")

# To save memory, we deactivate the adjoint model with annotate=False.
# We do not need the adjoint because we will not solve an optimisation problem
# or compute sensitivities
for sol in solver.solve(): #annotate=False):
print "Computed solution at time {}.".format(sol["time"])

# Write velocity and free-surface perturbation to file.
f_u.write(sol["u"], sol["time"])
f_eta.write(sol["eta"], sol["time"])

2.3.3. How to run the example¶

The code for this example can be found in examples/tidal-simulation/ in the OpenTidalFarm source tree, and executed as follows:

\$ mpirun -n 4 python orkney-coupled.py

where 4 should be replaced by the number of CPU cores available.

The results are stoed in the results directory and can be visualised with Paraview.


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# Use the guidelines of this section to sketch the curve.$y = \frac{x}{\sqrt{x^2 - 1}}$

## see solution

Derivatives

Differentiation

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### Discussion

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

So we know the graph is gonna look something like this because it's a rational function. But, um, first, let's find out the ass in totes so that we can know exactly where the graph is gonna be. First of all, the domain. Um, since there's a square on the bottom, we cannot. Well, since there's a square, yeah, we cannot have a negative under the radical. And also we cannot have the denominator to equal zero. So that means for the domain, it could be anything besides x zero so X can not equals here for the domain and ex cannot equal plus or minus one, but it can't exist anywhere else. Intercepts there are none. So substituting zero for extra Why we see that there are no intercepts. Symmetry. It's odd because F of negative X is equal to negative f of X. So it hasn't always symmetry ass until it's so first was finding vertical ascent oats. So were the denominator equal zero. We already said that was plus or minus one. So let's test out the limit as X approaches, plus or minus one. So let's do one first. Um, and if this equals infinity or negative infinity. We know it's a vertical lassitude. So substituting one, we see it's one over zero, so that's infinity. And if we do the same thing for a negative one, we just know it's gonna be negative. Infinity. So for vertical, ask himto we can say that X is plus or minus one. That's where Assam totes are, which is why the domain cannot be plus or minus one. All right, enough reverted for ah, horizontal as in two. The horizontal ask himto is where the limit approaches infinity or negative infinity. So the limit as X approaches infinity of our function X over swear ou of X squared minus one. So if we factor out in the denominator, we get the limit as X approaches Infinity of X over the square of X squared one minus one over X squared. And if we take that out of the square root. If you take this out of the square root, we get an X and X and X cancel. So we're left with the limit as X approaches. Infinity of one over the square root one minus one over X squared, won over. Affinity is zero. So one minus zero is one. So our limit is one see, and the limits as X approaches infinity or a negative infinity, we're gonna follow very similar steps. The only difference is that when we pull out because it's approaching negative infinity here. When we pull out the X squared out of the radical, we're just gonna make it negative X. So we're gonna be left with, um when they cancel out a negative on the bottom, that's the only difference. So the limit of negative square root of one minus one over X squared is negative one. So we have a horizontal ass in two. That why equals one and why equals negative one. And now we can graft with the information that we have. All right, so we're gonna put in the Assam totes and red, So here we know I should have been angry. So here we know X plus or minus one, one, minus one. That's our vertical asking totes. And we also know that why is one when why is one and negative one? We have horizontal ass in two, and that's where our function can lie. That's how you grow

#### Topics

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University of Nottingham

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# 3 – Latent Variables

Well, the answer is, we need one arrow for each pair document word. Since we have 500 documents and 1,000 words, the number of parameters is the number of documents times the number of words. This is 500 times 1,000, which is 500,000. This is too many parameters to figure out. Is there any way we can reduce this number and still keep most of the information? The answer is yes. We can reduce the number of parameters and still catch most of the information. We’ll do this by adding to the model the notion of a small set of topics or latent variables that actually drive the generation of words in each document. So, in this model, any document is considered to have an underlying mixture of topics associated with it. Similarly, a topic is considered to be a mixture of terms that is likely to generate. In here, we have say three topics: Science, politics, and sports. So now, there are two sets of parameters of probability distributions that we need to compute. The probability of a topic given a document d or P(z/d) and the probability of a term P given a topic z, or P(t/z). Our new probability of a document given a term can be expressed as a sum over the two previous probabilities. We’ll see more about this later. So, it’s time for another quiz. If we have 500 documents, 10 topics and 1,000 words, how many parameters does this new model have? Enter your answer below.


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# Bagging of xgboost

The extreme-gradient boosting algorithm seems to be widely applied these days. I often have the feeling that boosted models tend to overfit. I know that there are parameters in the algorithm to prevent this. Sticking to the documentation here the parameters subsample and colsample_bytree could (among others) prevent overfitting. But they do not serve for the same purpose as bagging xgboosted models would - right?

My question: would you apply bagging on top of xgboost to reduce the variance of the fit?

So far the question is statistical and I dare to add a code detail: in case bagging makes sense I would be happy about example code using the R package caret.

EDIT after the remark: if we rely on the parameters only to control the overfit, then how can we design the cross-validation best? I have approx. 6000 data points and apply 5-fold x-validation. What could improve the out-of-sample performance: going to something like 10-fold x-validation or doing repeated 5-fold x-validation? Just to mention: I use the package cartet where such strategies are implemented.

• Just a comment. You didn't mention the learning rate of boosted models explicitly, which is extremely important in preventing over-fitting. – Matthew Drury Mar 25 '16 at 16:29
• Could work, but ensembles of ensembles can grow quite big. It may be more efficient to find a appropriate set of training parameters not leading to over fitting for a given data set. – Soren Havelund Welling Mar 27 '16 at 13:43
• @SorenHavelundWelling please see my edit. – Ric Mar 29 '16 at 6:47
• link.springer.com/article/10.1186/1758-2946-6-10 If I were to publish some A-grade ML model I would go for the proposed Algorithm 3: repeated grid-search cross-validation for variable selection and parameter tuning. I don't use caret that much (I should). As I remember caret do not provide a outer cross-validation for a grid-search. I would feel comfortable by wrapping a caret grid search in a outer 5 or 10fold-CV loop and check if each fold optimal paramters close to the same. For final model, pick the typical parameter set from folds and use outer CV as error estimation. – Soren Havelund Welling Mar 29 '16 at 20:42
• @SorenHavelundWelling I opened up a discussion about overfitting here: stats.stackexchange.com/questions/204489/… in case you want to join. – Ric Mar 30 '16 at 8:03


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Outlook: Medicenna Therapeutics Corp. Common Shares is assigned short-term Ba1 & long-term Ba1 estimated rating.
Dominant Strategy : Hold
Time series to forecast n: 23 Feb 2023 for (n+1 year)
Methodology : Deductive Inference (ML)

## Abstract

Medicenna Therapeutics Corp. Common Shares prediction model is evaluated with Deductive Inference (ML) and Paired T-Test1,2,3,4 and it is concluded that the MDNA stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Hold

## Key Points

1. How do you decide buy or sell a stock?
2. What are the most successful trading algorithms?
3. What is the best way to predict stock prices?

## MDNA Target Price Prediction Modeling Methodology

We consider Medicenna Therapeutics Corp. Common Shares Decision Process with Deductive Inference (ML) where A is the set of discrete actions of MDNA stock holders, F is the set of discrete states, P : S × F × S → R is the transition probability distribution, R : S × F → R is the reaction function, and γ ∈ [0, 1] is a move factor for expectation.1,2,3,4

F(Paired T-Test)5,6,7= $\begin{array}{cccc}{p}_{a1}& {p}_{a2}& \dots & {p}_{1n}\\ & ⋮\\ {p}_{j1}& {p}_{j2}& \dots & {p}_{jn}\\ & ⋮\\ {p}_{k1}& {p}_{k2}& \dots & {p}_{kn}\\ & ⋮\\ {p}_{n1}& {p}_{n2}& \dots & {p}_{nn}\end{array}$ X R(Deductive Inference (ML)) X S(n):→ (n+1 year) $R=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$

n:Time series to forecast

p:Price signals of MDNA stock

j:Nash equilibria (Neural Network)

k:Dominated move

a:Best response for target price

For further technical information as per how our model work we invite you to visit the article below:

How do AC Investment Research machine learning (predictive) algorithms actually work?

## MDNA Stock Forecast (Buy or Sell) for (n+1 year)

Sample Set: Neural Network
Stock/Index: MDNA Medicenna Therapeutics Corp. Common Shares
Time series to forecast n: 23 Feb 2023 for (n+1 year)

According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Hold

X axis: *Likelihood% (The higher the percentage value, the more likely the event will occur.)

Y axis: *Potential Impact% (The higher the percentage value, the more likely the price will deviate.)

Z axis (Grey to Black): *Technical Analysis%

## IFRS Reconciliation Adjustments for Medicenna Therapeutics Corp. Common Shares

1. An entity that first applies IFRS 17 as amended in June 2020 after it first applies this Standard shall apply paragraphs 7.2.39–7.2.42. The entity shall also apply the other transition requirements in this Standard necessary for applying these amendments. For that purpose, references to the date of initial application shall be read as referring to the beginning of the reporting period in which an entity first applies these amendments (date of initial application of these amendments).
2. The credit risk on a financial instrument is considered low for the purposes of paragraph 5.5.10, if the financial instrument has a low risk of default, the borrower has a strong capacity to meet its contractual cash flow obligations in the near term and adverse changes in economic and business conditions in the longer term may, but will not necessarily, reduce the ability of the borrower to fulfil its contractual cash flow obligations. Financial instruments are not considered to have low credit risk when they are regarded as having a low risk of loss simply because of the value of collateral and the financial instrument without that collateral would not be considered low credit risk. Financial instruments are also not considered to have low credit risk simply because they have a lower risk of default than the entity's other financial instruments or relative to the credit risk of the jurisdiction within which an entity operates.
3. An entity is not required to incorporate forecasts of future conditions over the entire expected life of a financial instrument. The degree of judgement that is required to estimate expected credit losses depends on the availability of detailed information. As the forecast horizon increases, the availability of detailed information decreases and the degree of judgement required to estimate expected credit losses increases. The estimate of expected credit losses does not require a detailed estimate for periods that are far in the future—for such periods, an entity may extrapolate projections from available, detailed information.
4. A contractual cash flow characteristic does not affect the classification of the financial asset if it could have only a de minimis effect on the contractual cash flows of the financial asset. To make this determination, an entity must consider the possible effect of the contractual cash flow characteristic in each reporting period and cumulatively over the life of the financial instrument. In addition, if a contractual cash flow characteristic could have an effect on the contractual cash flows that is more than de minimis (either in a single reporting period or cumulatively) but that cash flow characteristic is not genuine, it does not affect the classification of a financial asset. A cash flow characteristic is not genuine if it affects the instrument's contractual cash flows only on the occurrence of an event that is extremely rare, highly abnormal and very unlikely to occur.

*International Financial Reporting Standards (IFRS) adjustment process involves reviewing the company's financial statements and identifying any differences between the company's current accounting practices and the requirements of the IFRS. If there are any such differences, neural network makes adjustments to financial statements to bring them into compliance with the IFRS.

## Conclusions

Medicenna Therapeutics Corp. Common Shares is assigned short-term Ba1 & long-term Ba1 estimated rating. Medicenna Therapeutics Corp. Common Shares prediction model is evaluated with Deductive Inference (ML) and Paired T-Test1,2,3,4 and it is concluded that the MDNA stock is predictable in the short/long term. According to price forecasts for (n+1 year) period, the dominant strategy among neural network is: Hold

### MDNA Medicenna Therapeutics Corp. Common Shares Financial Analysis*

Rating Short-Term Long-Term Senior
Outlook*Ba1Ba1
Income StatementCBaa2
Balance SheetBaa2Baa2
Leverage RatiosBaa2Baa2
Cash FlowBaa2Baa2
Rates of Return and ProfitabilityBaa2Baa2

*Financial analysis is the process of evaluating a company's financial performance and position by neural network. It involves reviewing the company's financial statements, including the balance sheet, income statement, and cash flow statement, as well as other financial reports and documents.
How does neural network examine financial reports and understand financial state of the company?

### Prediction Confidence Score

Trust metric by Neural Network: 72 out of 100 with 607 signals.

## References

1. Varian HR. 2014. Big data: new tricks for econometrics. J. Econ. Perspect. 28:3–28
2. Çetinkaya, A., Zhang, Y.Z., Hao, Y.M. and Ma, X.Y., Is FFBC Stock Buy or Sell?(Stock Forecast). AC Investment Research Journal, 101(3).
3. S. Devlin, L. Yliniemi, D. Kudenko, and K. Tumer. Potential-based difference rewards for multiagent reinforcement learning. In Proceedings of the Thirteenth International Joint Conference on Autonomous Agents and Multiagent Systems, May 2014
4. Athey S, Wager S. 2017. Efficient policy learning. arXiv:1702.02896 [math.ST]
5. Hastie T, Tibshirani R, Friedman J. 2009. The Elements of Statistical Learning. Berlin: Springer
6. Clements, M. P. D. F. Hendry (1996), "Intercept corrections and structural change," Journal of Applied Econometrics, 11, 475–494.
7. G. J. Laurent, L. Matignon, and N. L. Fort-Piat. The world of independent learners is not Markovian. Int. J. Know.-Based Intell. Eng. Syst., 15(1):55–64, 2011
Frequently Asked QuestionsQ: What is the prediction methodology for MDNA stock?
A: MDNA stock prediction methodology: We evaluate the prediction models Deductive Inference (ML) and Paired T-Test
Q: Is MDNA stock a buy or sell?
A: The dominant strategy among neural network is to Hold MDNA Stock.
Q: Is Medicenna Therapeutics Corp. Common Shares stock a good investment?
A: The consensus rating for Medicenna Therapeutics Corp. Common Shares is Hold and is assigned short-term Ba1 & long-term Ba1 estimated rating.
Q: What is the consensus rating of MDNA stock?
A: The consensus rating for MDNA is Hold.
Q: What is the prediction period for MDNA stock?
A: The prediction period for MDNA is (n+1 year)


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### Strings

We began by talking about a kind of collection I called strings. These are also sometimes called "ordered sequences" or "lists," or several other things. But I'm calling them strings because a paradigm case of these collections are linguistic expressions, like a word or phrase or sentence or novel. (Or gibberish. We're making no assumptions at this point about whether our strings are syntactically well-formed or have any semantic properties.)

One kind of string is the empty string. This could be written as "" or as $$\varepsilon$$.

Other kinds of strings are built out of an initial string and a final "letter". We should decide in advance what our "letters" are going to be. If we're thinking of the strings as words, we'll naturally take English letters to be our "letters." If on the other hand we're thinking of the strings as sentences, it might be more natural to take English words as the "letters." From the point of view of the string, all of its letters are atoms. The string is blind to any additional structure that its letters may have.

In class, I proposed we use as letters the following: $$\mathcal{A}, \mathcal{B}, \mathcal{C}, \textbf{:-)}, \textbf{:-(}$$. I will symbolize the operation by which a letter is appended to the end of an initial string with a caret, like this:

initial_string ^ new_letter

Hence, all of the following are strings of length 1:

$$\varepsilon\smalltriright\mathcal{A}$$

$$\varepsilon\smalltriright\mathcal{B}$$

$$\varepsilon\smalltriright\textbf{:-)}$$

And the following are strings of length 2:

$$(\varepsilon\smalltriright\mathcal{A}) \smalltriright\mathcal{B}$$

$$(\varepsilon\smalltriright\mathcal{A}) \smalltriright\mathcal{A}$$

$$(\varepsilon\smalltriright\textbf{:-)}) \smalltriright\mathcal{A}$$

$$(\varepsilon\smalltriright\textbf{:-)}) \smalltriright\textbf{:-(}$$

We'll adopt the convention that this caret symbol is to be understood as "associating to the left", so that:

$$\varepsilon\smalltriright\mathcal{A}\smalltriright\mathcal{B}$$

means the same as:

$$(\varepsilon\smalltriright\mathcal{A}) \smalltriright\mathcal{B}$$

Now we're in a position to give a recursive or inductive definition of a string. We do that like this:

1. $$\varepsilon$$ is a string.
2. if $$S$$ is a string and $$L$$ is a letter, then $$S \smalltriright L$$ is a string.
3. Nothing else is a string.

The second clause relies on the very notion we're defining, but that's okay because the definition has to bottom out at some point in the base clause. Alternatively, you can think of this definition as "generating" the class of strings, by first including the items picked out by any base clauses (here, $$\varepsilon$$), and then giving a rule that says how the class gets extended stage-by-stage. We don't place any limit on how many "stages" it might take to include in all the items we want to count as strings; but we do have to explicitly rule out that anything else is a string.

There's a technique for converting any recursive definition of this sort into an explicit, non-recursive definition. We'll talk about this later in the course.

Here's another recursive definition:

1. $$0$$ is a natural number.
2. If $$n$$ is a natural number, then the successor of $$n$$ is also a natural number.
3. Nothing else is a natural number.

One thing you might think about is why we had to put these definitions in the 1-2-3 form we did. Why couldn't we instead say:

$$n$$ is a natural number iff $$n$$ is $$0$$ or $$n$$ is the successor of a natural number.

Hint: does this inadequate definition exclude the possibility that $$\pi$$ is a natural number? If you think it does, try to explain why $$\pi$$ is excluded.

In some cases, the "nothing else" clause is guaranteed by the other clauses, or by other assumptions you're making, and doesn't need to be explicitly added; we'll see an example shortly. But in these definitions it is needed.

Okay, we've discussed the operation by which you take a string and a letter and make a new string. A different sort of operation would be one where you take two strings and concatenate them, possibly getting a new string. I say "possibly" because it may be that one of your starting strings was $$\varepsilon$$, and when you concatenate that to another string, the result is just the other string. But for instance:

the concatenation of $$\varepsilon\smalltriright\mathcal{A}\smalltriright\mathcal{B}$$ with $$\varepsilon\smalltriright\mathcal{B}\smalltriright\mathcal{C}$$ will be the string $$\varepsilon\smalltriright\mathcal{A}\smalltriright\mathcal{B}\smalltriright\mathcal{B}\smalltriright\mathcal{C}$$

I will use the symbol $$\curl$$ to represent concatenation.

In class, we figured out how to define concatenation in terms of operations we already knew. We came up with another recursive definition for this. We assume that $$S$$ and $$T$$ are strings. Then there are just two cases, depending on the structure of $$T$$:

1. If $$T$$ is $$\varepsilon$$, then $$S \curl T$$ is $$S$$.
2. Else $$T$$ has the form $$T_\latin{start} \smalltriright L$$, where $$T_\latin{start}$$ is a string and $$L$$ is a letter. In this case, $$S \curl T$$ is $$(S \curl T_\latin{start})\smalltriright L$$.

Strings illustrate a few important ideas. For one thing, they gave us an opportunity to familiarize ourselves with recursive definitions. For another thing, they give us one example of a collection: a kind of thing that (may) contain other things. There is a third important idea they illustrate, which we'll come to a bit further down the road.

Some authors talk about strings in such a way that the individual letters are identified with what I would call the length-1 strings that contain them. In certain contexts, this won't get you into trouble. But in other contexts it will, so I've tried to keep these two notions conceptually distinct. Consider: I haven't placed any constraints on what can count as letters. What if we permit strings to themselves be letters? Consider the letter which is the length-2 string $$\varepsilon\smalltriright\mathcal{A}\smalltriright\mathcal{B}$$. Make a string out of that one letter: what is its length? 1 or 2? If you append the length-2 string $$\varepsilon\smalltriright\mathcal{C}\smalltriright\mathcal{A}$$ to the end of the string you just made, do you get a length-2 string, each of whose letters is itself a length-2 string---as I'd naturally expect? Or do you get a length-1 string, whose single letter is a string of length-4? Or a length-3 string? Or something else? How should the $$\smalltriright$$ operation tell where the last letter of its left-hand operand is?

If we distinguish between strings and the letters they're made of, even when the string is of length 1, then this all becomes unproblematic.

The first HOMEWORK EXERCISE I assigned was: (1a) give a recursive definition of the factorial function. And (1b) prove in a rigorous way (presumably with a mathematically inductive argument) that the factorial of every $$n \ge 2$$ is even.

In this class, it will be common to give mathematically inductive arguments.

• If we want to prove things about every formula in a language, then we'll "induce on the structure" of the formulas. That is, our base clauses will talk about the atomic formulas, and our inductive steps will concern all the different ways in which compound formulas can be built up out of things already counted as formulas.

• If we want to prove things about all provable formulas, then our base clauses will concern axioms, and our inductive steps will concern all the different ways in which formulas can be proven from other formulas already counted as provable.

### Other collections

There are a variety of collections that a mathematician or logician might work with. Strings pay attention to the order of their elements; they also pay attention to the multiplicity of their elements. The string $$\varepsilon\smalltriright\mathcal{A}$$ is different from the string $$\varepsilon\smalltriright\mathcal{A}\smalltriright\mathcal{A}$$. Other collections make other choices. Sets, for example, don't pay attention to either the order or the multiplicity of their elements. The set $$\set{a, a, b}$$ is the same as the set $$\set{b, a}$$. We will be working with sets a lot this term.

Another collection, less often discussed, is a multiset. This pays attention to the multiplicity of its elements but not their order. So a multiset containing the elements $$a, a,\latin{ and }b$$ will be the same as a multiset containing the elements $$a, b,\latin{ and }a$$, but different from a multiset containing only $$b$$ and $$a$$ (once).

A bit later, we'll discuss graphs and trees, which are collections having more complex structures.

### Sets

I'll try to stick to the convention of naming sets with capital greek letters, like $$\Delta$$ and $$\Gamma$$. But I may sometimes slip; and there is no uniform practice. (And sometimes ☺ I will use capital Greek letters like $$\Alpha, \Beta, \dots$$)

As we said, sets are collections that may have elements or members. Their members may, but need not, also be sets. When the members aren't sets, they're called urelements or individuals or atoms. When $$d$$ is a member of set $$\Delta$$, we write $$d \in \Delta$$. The empty set, written as $$\emptyset$$ or $$\set{}$$, has no members.

For a given set of things $$\Delta$$, there is exactly one empty set of $$\Delta$$s, which contains no $$\Delta$$s. Where $$\Delta$$s and $$\Gamma$$s are sets of things that are fundamentally different in certain ways, it's less clear and more contentious whether the empty set of $$\Delta$$s and the empty set of $$\Gamma$$s are the same entity. But we will just ignore this debate.

Sets can have lots of members, and they can contain sets which have lots of members, but on the standard picture, no set can have itself as a member. Nor can there be an infinite descending chain of sets, each of whom is a member of the next outermost set. That is, no sets like this:

$$\set{0, \set{1, \set{2, \set{3, \dots}}}}$$

These claims aren't indisputable, and alternative set theories have been developed which don't respect them. But those kinds of controversies aren't going to be our topic here. We're just going to assume we're working with well-behaved sets, we're not going to worry about Russell's paradox, and so on.

Sometimes mathematicians or logicians will specify a set by listing all its members:

$$\set{0, 16, 128}$$

Other times, they'll specify the set using set-builder notation, which can look like this:

$$\set{y \where y\in\N \latin{ and }\exists x\in \N ( y = x^2 ) }$$

$$\set{y\in\N \where \exists x\in \N ( y = x^2 )}$$

$$\set{x^2 \where x\in\N}$$

All three of these expressions specify the same set. Sometimes this is instead written like this: $$\set{x^2 : x\in\N}$$.

In the informal metalanguage you're reading now, I will always specify what domain I'm drawing from when I use a quantifier. Thus I write $$\exists x\in \N$$ and not just $$\exists x$$. In the formal languages we'll begin talking about in a few weeks, we'll at first only have the second, "unrestricted" quantifier. But we'll talk later about introducing the first, "restricted" kind of quantifier into the languages we're studying.

I'll assume you're familiar with basic set-theoretic relations and operations:

$$\Gamma \union \Delta$$

$$\Gamma \intersect \Delta$$

If $$\Gamma_0, \Gamma_1, \dots \Gamma_n$$ are sets, then another way to write $$\Gamma_0 \union \Gamma_1 \union \dots \union \Gamma_n$$ is $$\Union_{i=0}^n \Gamma_i$$. Similarly for $$\Intersect_{i=0}^n \Gamma_i$$.

If $$\Delta$$ is a set of sets, then $$\Union\Delta$$ is the union of all the members of $$\Delta$$. Or you could write $$\Union_{d\in\Delta} d$$. Similarly for $$\Intersect\Delta$$.

The "difference" between $$\Gamma$$ and $$\Delta$$---that is, the set of all members of $$\Gamma$$ that aren't also members of $$\Delta$$---is sometimes written $$\Gamma \setminus \Delta$$, but I prefer to write it just as $$\Gamma - \Delta$$

Sometimes people write $$\overline{\Delta}$$ or $$\overline{\Delta \union \Gamma}$$. Here they are assuming some specific larger set $$\Omega$$ that $$\Delta$$ and $$\Gamma$$ are subsets of, and these are just alternative ways of specifying $$\Omega - \Delta$$ and $$\Omega - (\Delta \union \Gamma)$$.

$$\Delta$$ is a subset of $$\Omega$$ when everything that's a member of $$\Delta$$ is also a member of $$\Omega$$. We write this relation as $$\Delta \se \Omega$$. $$\Omega$$ may have additional members besides the things that are also in $$\Delta$$. Or it may not. If it doesn't, then $$\Omega$$ has exactly the same members as $$\Delta$$. And as we understand sets, that means that they are exactly the same set. So among the sets that $$\Delta$$ is a subset of is itself.

If you want to say that $$\Delta$$ is a subset of and not identical to some other set, then you say that $$\Delta$$ is a proper subset of the other set, and you write that like this: $$\Delta \subset \Gamma$$. For the most part this term, we'll be working with the $$\se$$ relation and rarely if ever with the $$\subset$$ relation.

More HOMEWORK EXERCISES: (2) Let $$\Gamma = \set{1}$$ and $$\Delta = \set{1,\set{1}}$$. Then: (a) Why is $$\Gamma \in \Delta$$? (b) Why is $$\Gamma \se \Delta$$? (c) What is $$\Gamma \intersect \Delta$$? (d) Is it always the case when $$\Alpha \se \Beta$$ that $$\Alpha \intersect \Beta = \Beta$$? (This last question contains a mistake. Fix the mistake and then answer the corrected question.)

As we said, sets are the same when they have exactly the same members. For collections like multisets, which care about the multiplicity of their elements, then the collections will be the same when they have all the same elements to the same multiplicity. For collections like strings, which care about the order of their elements, then the collections will be the same when they have all the same elements in the same order. For collections with other sorts of structure, the rules will be analogous. There's a general idea here, that the identity of a collection depends upon the identity of its elements and possibly on the structure by which the collection organizes them, and on nothing more. You can't have two collections with all the same elements organized (in the way that collection cares about) in the same way, yet those collections be numerically distinct. This general idea is called extensionality. (The root term "extensional" also has other meanings you will be familiar with, and that we'll come back to later in the term. These meanings are conceptually related, but don't expect that you could derive specifically the one meaning from the other. We'll encounter this kind of situation fairly often in our studies.)

When mathematicians or logicians work with a collection, they'll generally be understanding the collection to be one that's extensional in the way I described. But in computer science contexts, this can't be taken for granted. There it can be useful to introduce notions of structures that aren't extensional, where we can have two numerically distinct but indiscernible collections, with the same elements organized in the same way. We're not going to get into those contexts here. I'm just flagging that though the idea of an extensional collection is very natural and useful to work with, it's not conceptually mandatory, and in some formal settings we can make good sense of, and it's useful to work with, collections that aren't extensional.

The power set of a set $$\Omega$$ is the sets that are subsets of $$\Omega$$. This includes $$\Omega$$ itself, and it includes the empty set $$\emptyset$$. That's because all of the empty set's members (all none of them) are also members of $$\Omega$$. Sometimes the power set of $$\Omega$$ is written $$\varpow{\Omega}$$, but I will write it as $$\pow{\Omega}$$.

When two sets have members in common, we say that the sets overlap. When the sets fail to have any members in common, we say that they're disjoint. (The empty set is an edge case: should we define "disjointness" in a way to count it as disjoint from other sets, because their intersection is empty? Or should we count it as non-disjoint, since after all, it's a subset of those other sets?) When two sets fail to be numerically identical, then---regardless of whether they overlap or are disjoint---we say that the sets are distinct.

In mereology, the formal theory meant to model our intuitive understanding of part-whole relations, they also use the term "overlapping." But there instead of "disjoint," they talk about things with no parts in common as being wholly "discrete." Don't confuse these vocabularies.

(There's also another use of "discrete" that comes from topology; this is the sense in which the integers count as discrete relative to the reals.)

### Partitions and Covers

A partition of a set is a division of the set into one or more (non-empty) "cells", where everybody from the original set gets to be in one of the cells, and none of the cells overlap. So, for example, here is one partition of the set $$\set{1,2,3}$$:

$$\set{1} \qquad \set{2,3}$$

The partition is itself understood to be a set---the set containing the two sets just listed. But I'll write the partition as I did above, to make the intuitive idea more obvious. Another partition of the set $$\set{1,2,3}$$ is:

$$\set{1,3} \qquad \set{2}$$

Another HOMEWORK EXERCISE: (3) How many different partitions of the set $$\set{1,2,3}$$ are there?

A cover is similar to a partition, except that now we're allowed to have some overlap. It's allowed (but not required) that some of the members of the original set get to be in multiple cells. So both of the above partitions of $$\set{1,2,3}$$ also count as covers of that set. But so too does this:

$$\set{1,3} \qquad \set{2,3}$$

Covers come up sometimes in linguistics. I don't think we'll need to work with them in anything we'll be doing.

### Tuples (Ordered pairs, triples, $$\dots$$)

Sometimes we want to talk about several items taken from several sets. We do this using the notion of an ordered pair (or triple, or $$\dots$$) An ordered pair $$(a, b)$$ has a "first" element $$a$$ and a "second" element $$b$$. Two ordered pairs count as the same just in case they have the same first elements and the same second elements. The two elements might both be the same kind of thing (they might both be members of the set of natural numbers $$\N$$, for example). Or they might be different kinds of things (maybe $$a$$ is a number but $$b$$ is a person). When $$\Alpha$$ and $$\Beta$$ are two, not-necessarily distinct, sets, we describe the set of all ordered pairs whose first element comes from $$\Alpha$$ and whose second element comes from $$\Beta$$ as $$\Alpha \times \Beta$$. We can write this as:

$$\Alpha \times \Beta = \set{(a,b) \where a\in\Alpha \latin{ and } b\in\Beta}$$

This is called the Cartesian product of $$\Alpha$$ and $$\Beta$$. The ordered triple of $$\Alpha$$, $$\Beta$$, and $$\Gamma$$ we write as:

$$\Alpha \times \Beta \times \Gamma = \set{(a,b,c) \where a\in\Alpha \latin{ and } b\in\Beta \latin{ and } c\in\Gamma}$$

You might wonder is there something like a notion of "adding" two sets, in the way that this is a way of taking their "product"? There is. Whereas with the product of two sets, we get an element from the first set and an element from the second set, with the sum of the sets we'll get an element from the first set or an element from the second set. This may remind you of the operation of taking the union of the two sets.

However, there's a wrinkle. What if we're taking the product or the sum of two overlapping sets, or of the very same set? With $$(a_1, a_2)$$, a member of the product $$\Alpha \times \Alpha$$, we have an element from $$\Alpha$$ in its role as the left-hand product and an element from $$\Alpha$$ in its role as the right-hand product. And we can recover which role each $$a$$ came from $$\Alpha$$ as playing: that is, we can tell the difference between $$(a_1, a_2)$$ and $$(a_2, a_1)$$. Similarly, a member of the sum of $$\Alpha$$ and $$\Alpha$$ will either come from $$\Alpha$$ as playing the role of the left-hand sum operand, or from it playing the role of the right-hand operand. But how can we tell which? To get the closest parallel to the notion of a Cartesian product, we should be able to keep track of this.

The way this is usually done is by taking not the union of the summed sets, themselves, but rather the union of their-members-with-a-tag-attached indicating whether the element belongs to the sum by virtue of coming from the left-hand or from the right-hand operand. Sums understood in this way are called tagged or disjoint unions. If we model the tagging in terms of an ordered pair, and help ourselves to two primitive objects $$\mathsf{Left}$$ and $$\mathsf{Right}$$, then we can define the tagged union of $$\Alpha$$ and $$\Beta$$---with $$\Beta$$ allowed to be identical to, or a subset of, or to overlap, or to be completely disjoint from $$\Alpha$$---as: $$\set{(\mathsf{Left},a) \where a\in\Alpha} \union \set{(\mathsf{Right},b) \where b\in\Beta}$$.

This summing operation is neither more nor less "correct" than the more familiar operation $$\union$$. For some purposes (the ones that will dominate our attention this semester) the latter is more appropriate; for other purposes, the former is.

I've seen each of the following notations used for disjoint union: $$\uplus$$, $$\udot$$, $$\sqcup$$, $$\amalg$$, and $$+$$. Since we won't be employing this notion, we don't need to pick a notation for it.

The notion of an ordered pair's "order" can be misleading. Really the important thing is just that we keep track of which element comes from which set, or which element comes from a single set playing the role of one operand rather than another. Instead of the "first" element and the "second" element, we could instead talk about the "west" element and the "east" element. And for ordered quatruples, about the "north", "south", "east" and "west" elements. If someone then took it in mind to ask whether the west element comes before or after the east element, this question wouldn't have any established sense.

Similarly, if you take the ordered quatruple $$(a,b,c,d)$$, we might call it "increasing" iff $$a \le b \le c \le d$$. But I could just as easily define another notion, call it "ascending", which holds iff $$a \le d \le b \le c$$. And I submit there's no sense in which one of these two notion is more intrinsically natural or less gruesome than the other.

With strings, on the other hand, the collection does have a more natural intrinsic ordering. It's genuinely more natural to count the letter $$\mathcal{B}$$ as coming "between" the letters $$\mathcal{A}$$ and $$\mathcal{C}$$ in $$\varepsilon\smalltriright\mathcal{A}\smalltriright\mathcal{B}\smalltriright\mathcal{C}$$ than it is to count $$\mathcal{C}$$ as coming "between" $$\mathcal{A}$$ and $$\mathcal{B}$$, because of way the string is built up.

Now in many mathematical and logical texts, you will see authors taking a very casual attitude towards these differences. It's also common to see authors propose reductions of some of these notions to the others, in ways that trample on the distinctions. For example, it's common to see the ordered triple $$(a,b,c)$$ defined as the pair $$(a,(b,c))$$ whose second element is itself a pair. But there might be settings where I need to differentiate between pairs that contain pairs and triples. In many computer science settings, for example, where we pay close attention to the types of things we're working with, conflating these can be disastrous.

So I'm going to downplay these kinds of reductions in this class. It's a philosophically substantial question whether any of these identifications or reductions are correct, and usually there are even good mathematical reasons to resist them, albeit reasons that come from taking a look at broader applications than the author who proposed the reduction was considering. Even if in the end we decide to accept some such reductions, I don't think it's right to be offering them in the guise of introductory definitions of the notions in question.

Whether these reductions are to be accepted or resisted is not central to the material we're aiming to learn this semester. It makes no fundamental difference to the mathematics. So I won't spend much time on it. I will myself tend to refrain from reductions like $$(a,b,c) = (a,(b,c))$$ that you see other authors going in for. The other authors may say that's what a triple is, or how it's defined. I'll tend to say instead that the author is mathematically modelling or representing some notion (a notion we have some intuitive, somewhat pre-theoretical grasp on) in terms of $$(a,(b,c))$$. In some settings it's worth keeping track of the difference. That's why I'm flagging these things for you. I also think you'll have better conceptual hygiene if you're more aware of the choices to be made here. But for the material we'll be focusing on, it won't matter in any direct way.

As I mentioned in class, some of the important respects in which strings and tuples differ are that:

• strings are length-insensitive, that is, if an operation is defined to work on strings of one length, typically you should expect the operation to make intuitive sense, and be defined for, shorter and longer strings too
• similarly, strings are homogenous, that is, their letters will always come from a single set or type
• so we should think of a length-2 string of numbers as itself being the same type of thing as a length-3 string of numbers. And we should only expect to find strings whose letters are a mix of numbers and people if we think of the string as generated from a larger letter set, that contains both numbers and people. And if we're going to allow such strings, we should be equally willing to work with strings that contain people and numbers in different orders.

On the other hand:

• tuples are length-sensitive. An operation that's defined to work on ordered pairs might in some cases be sensibly extensible to an operation that works on ordered triples, but there should be no presumption that it will be. If you're expecting pairs as arguments, there's no reason you should be able to deal with triples as well.
• tuples may be homogenous, but may also be heterogenous. We'll very often be working with things like pairs of numbers and people (with no expectation that the operations we're defining will also work on pairs of people and numbers, or pairs of people and people).

These are just rough gestures; but I hope they're enough to give you some intuitive sense of why these notions might be worth keeping separate.

Instead of $$(a,b)$$, some authors will write $$\tuple{a,b}$$. I think this notation tends even more strongly to be correlated with ignoring the differences between tuples and strings.

Another sort of reduction you will often encounter in math and logic texts is one that identifies ordered pairs with certain kinds of sets. There are different ways to do this. The most widespread proposal is that:

$$(a,b) = \set{\set{a},\set{a,b}}$$

As with the preceding reductions, I take this in the spirit of "Okay, let's suspend judgment about whether these different kinds of abstract collections really do essentially coincide in this way, or whether the nature of sets is really more fundamental than that of a pairing. I'll just assume that present purposes the author proposes to model or represent pairs in terms of sets in this way. Let's go along with that for the time being." And in fact, it almost never makes any mathematical difference whether pairs can be reduced to sets in this way. Perhaps it makes a difference to the philosophy of mathematics; but that is not our topic here.

Sometimes ordered pairs, triples, and so on---the general class of things I will call n-tuples or just tuples---are referred to as "ordered sets." Avoid this usage; it will be too confusing when we look at a different notion of ordered set, a bit later in the class.

When you're working with triples from the set $$\Alpha \times \Alpha \times \Alpha$$, it's customary to also write this as $$\Alpha^3$$. But then that raises the questions whether:

$$\Alpha \times (\Alpha \times \Alpha)$$, that is, $$\Alpha \times \Alpha^2$$, should $$= \Alpha^3$$ as well?

Those who identify $$(a,b,c)$$ with $$(a,(b,c))$$ will say sure, because that's just the same thing as $$\Alpha \times \Alpha \times \Alpha$$. But then should it also be the case that:

$$(\Alpha \times \Alpha) \times \Alpha = \Alpha^3$$

and so that $$((a,b),c)$$ always $$= (a,(b,c))$$? This would be disastrous. The $$\Alpha^3$$ notation papers over this issue. Nonetheless, I will sometimes go along with the notation because it's very prevalant.

Another question posed by that notation is what is the relation between $$\Alpha^1$$ and $$\Alpha$$? Should we take these to be the same? Many authors will just uncritically say yes, but in fact there are many interesting complexities to this. There's a strong case for acknowledging two kinds of tuple, one of which verifies this identification, and the other of which falsifies it. Many computing languages primarily work with the latter notion.

In some theoretical settings, again ones closer to computer science than the material we'll be working with, it's useful to also introduce a notion $$\Alpha^0$$, that is, the ordered tuple that has zero elements. There is only one of these (at least, up to isomorphism). Sometimes the set of these tuples (all one of them) is called $$1$$. That should remind you of my practice of sometimes using the symbol $$2$$ to refer to the set of truth-values.

Don't confuse $$\Alpha^2$$ and $$\pow{\Alpha}$$!

Another HOMEWORK EXERCISE: (4) What is the difference between $$\Alpha^2$$ and $$\pow{\Alpha}$$? Give examples of some things that belong to one but not the other.

### More HOMEWORK EXERCISES

1. For each of the following, true or false?

1. If $$a \in \Alpha$$ and $$\Alpha \se \Beta$$ then $$a \in \Beta$$.
2. If $$\Alpha \se \Beta$$ and $$\Beta \se \Gamma$$ then $$\Alpha \se \Gamma$$.
3. $$a \in \Alpha$$ iff $$\set{a} \se \Alpha$$.
4. $$a \in \Alpha$$ and $$b \in \Alpha$$ iff $$\set{a,b} \se \Alpha$$.
5. If $$\Alpha \se \Beta$$ and $$\Beta \se \Alpha$$ then $$\Alpha = \Beta$$.
2. Are the following true or false for arbitrary sets $$\Gamma$$?

1. $$\Gamma \in \set{\Gamma}$$
2. $$\set{\Gamma} \in \set{\Gamma}$$
3. $$\set{\Gamma} \se \set{\Gamma}$$
3. What is the set whose only member is $$\set{\Gamma}$$?

4. Here is an example proof that $$\Alpha \union \Beta = \Beta \union \Alpha$$. Suppose that $$a \in \Alpha \union \Beta$$. By the definition of union, this means that $$a \in \Alpha$$ or $$a \in \Beta$$. But then of course it follows that $$a \in \Beta$$ or $$a \in \Alpha$$ (the commutativity of disjunction in our metalanguage is not questioned). So $$a \in \Beta \union \Alpha$$. So we've proved that every member of $$\Alpha \union \Beta$$ is also a member of $$\Beta \union \Alpha$$. The proof of the converse goes analogously.

Prove the following:

1. If $$\Alpha \se \Beta$$ and $$\Beta \se \Gamma$$ then $$\Alpha \se \Gamma$$.
2. $$\Alpha \se \Alpha \union \Beta$$
3. $$\Alpha \intersect \Beta \se \Alpha$$
4. $$\Alpha \union \Alpha = \Alpha$$
5. $$\Alpha \intersect \Alpha = \Alpha$$
6. $$\Alpha \union \emptyset = \Alpha$$
7. $$\Alpha \intersect \emptyset = \emptyset$$
8. $$\Alpha \union (\Alpha \intersect \Beta) = \Alpha$$
9. $$\Alpha \intersect (\Alpha \union \Beta) = \Alpha$$
10. $$\Alpha \se \Beta$$ iff $$\Alpha \union \Beta = \Beta$$
11. $$\Alpha \se \Beta$$ iff $$\Alpha \intersect \Beta = \Alpha$$
12. $$\Alpha - (\Alpha-\Beta) = \Alpha \intersect \Beta$$
13. $$(\Alpha-\Beta) \intersect \Beta = \emptyset$$
5. Prove the following:

1. $$\Alpha \times (\Beta \intersect \Gamma) = (\Alpha \times \Beta) \intersect (\Alpha \times \Gamma)$$
2. $$\Alpha \times (\Beta \union \Gamma) = (\Alpha \times \Beta) \union (\Alpha \times \Gamma)$$
3. $$\Alpha \times (\Beta-\Gamma) = (\Alpha \times \Beta) - (\Alpha \times \Gamma)$$
6. Let $$T1$$ be a set, and let:

$$T2 = \set{T1}$$

$$T3 = \set{\set{T1}}$$

$$T4 = \set{T1, \set{T1}}$$

Questions:

1. Which of $$T1$$ -- $$T4$$ are members of $$T4$$?
2. Which of $$T1$$ -- $$T4$$ are subsets of $$T4$$?
7. Let $$\Alpha = \set{a,b,c}$$, $$\Gamma = \set{c,d}$$, and $$\Delta = \set{d,e,f}$$. Then:

1. List $$\pow{\Alpha}$$.
2. What is $$\Alpha \union \Gamma$$?
3. What is $$\Gamma \union \emptyset$$?
4. What is $$\Alpha \intersect \Gamma$$?
5. What is $$\Alpha - \Gamma$$?
6. What is $$\Alpha \union (\Gamma \intersect \Delta)$$?
7. What is $$\Alpha \intersect (\Gamma \intersect \Delta)$$?

1. Understand the symmetric difference of sets $$\Alpha$$ and $$\Beta$$ to be the set $$(\Alpha \union \Beta) - (\Alpha \intersect \Beta)$$. I've seen this operation expressed using each of the following notations: $$\triup \quad \ominus \quad \oplus$$ and $$+$$ (note that $$+$$ is also sometimes used instead to express disjoint union, explained above). For the purposes of this question, let's express symmetric difference using $$+$$. Prove that $$\Alpha + \Beta = (\Alpha-\Beta) \union (\Beta-\Alpha)$$.


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# How can I improve traction on a wheeled vehicle yet reduce friction?

This is inspired by a project I undertook for my Physics class last year; I'd like to apply it in the future.

Background: For the project, I and a few classmates were required to build a small car powered by any of a number of sources; we opted for a mousetrap that turned two rear wheels on a four-wheel car as it was released. Our group's goal was to make the car go as far as possible.

One problem was that the power source was not constant. The lever arm of the trap lifted in an arc for about five seconds, then returned to its resting position. During this time, it propelled the car. The car then coasted for the rest of the way.


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Acta Arithmetica

2013 | 157 | 4 | 357-364
Tytuł artykułu

Nonreciprocal algebraic numbers of small Mahler's measure

Autorzy
Treść / Zawartość
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EN
Abstrakty
EN
We prove that there exist at least cd⁵ monic irreducible nonreciprocal polynomials with integer coefficients of degree at most d whose Mahler measures are smaller than 2, where c is some absolute positive constant. These polynomials are constructed as nonreciprocal divisors of some Newman hexanomials $1 + x^{r₁} + ⋯ + x^{r₅}$, where the integers 1 ≤ r₁ < ⋯ < r₅ ≤ d satisfy some restrictions including $2r_{j} < r_{j+1}$ for j = 1,2,3,4. This result improves the previous lower bound cd³ and seems to be closer to the correct value of this function in d than the best known upper bound which is exponential in d.
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357-364
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2013
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autor
• Department of Mathematics and Informatics, Vilnius University, Naugarduko 24, Vilnius LT-03225, Lithuania
autor
• Department of Mathematics and Informatics, Vilnius University, Naugarduko 24, Vilnius LT-03225, Lithuania
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## Book Store

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## Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
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`Class 10` `Class 12`
How can the rate of a fast reaction be determined by Flash Photolysis?

Flash photolysis: When we pass a powerful flash of short duration or laser beam, through the reaction mixture to initiate a reaction the atoms, ions or free radicals are formed.
These atoms, ions or free radicals formed can be identified by passing a second, Flash of light through the mixture immediately after the first flash.
For this the absorption spectrum of the mixture is monitored continuously at small regular intervals after first flash and the changes in the spectrum with time indicate the various processes occurring in the system.
It can also be studied by observing some other property like electrical conductance or magnetic property of the reaction mixture.

126 Views

The conversion of molecules X to Y follows second order kinetics. If concentration of X is increased to three times how will it affect the rate of formation of Y?

Let the reaction is X →Y

This reaction follows second order kinetics.
So that, the rate equation for this reaction will
Rate, R = k[X]2 .............(1)
Let initial concentration is x mol L−1,
Plug the value in equation (1)
Rate, R1 = k .(a)2
= ka2
Given that concentration is increasing by 3 times so new concentration will 3a mol L−1
Plug the value in equation (1) we get
Rate, R2 = k (3a)2
= 9ka2
We have already get that R1 = ka2 plus this value we get
R2 = 9 R1
So that, the rate of formation will increase by 9 times.
Rate = k[A]2
If concentration of X is increased to three times,
Rate = k[3A]2
or Rate = 9 k A2
Thus, rate will increase 9 times.

972 Views

For the reaction R → P, the concentration of a reactant changes from 0.03 M to 0.02 M is 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.

Given that
Initial concentration, [R1] = 0.03
Final concentration, [R2] = 0.02
Time taken ∆t = 25 min = 25 × 6 0 = 1500 sec (1 min = 60 sec )
The formula of average rate of change

(i) Average rate

(ii) Average rate

1717 Views

In a reaction 2A → Products, the concentration of A decreases from 0.5 mol to 0.4 mol L–1 in 10 minutes. Calculate the rate during this interval.

Given that
Initial concentration [A1] =0.5
Final concentration [A2] =0.4
Time is  = 10 min

Rate of reaction = Rate of disappearance of A.

Rate of reaction = $-\frac{1}{2}\frac{∆\left[\mathrm{A}\right]}{∆\mathrm{t}}\phantom{\rule{0ex}{0ex}}$

2088 Views

A first order reaction has a rate constant 1.15 x 10–3 s–1. How long will 5 g of this reactant take to reduce to 3 g?

Given that
Initial quantity, [R]o= 5 g
Final quantity, [R] = 3 g
Rate constant, k = 1.15 x 10−3 s−1
Formula of 1st order reaction,
We know that
$\mathrm{t}=\frac{2.303}{\mathrm{k}}\mathrm{log}\frac{{\left[\mathrm{R}\right]}_{0}}{\left[\mathrm{R}\right]}$
or

1299 Views

For a reaction, A + B → Product; the rate law is given by, r = k[ A]1/2 [B]2. What is the order of reaction?

The order of the reaction is sum of the powers on concentration.
So that sum will

r = k[A]
1/2[B]2

Order of reaction =

1504 Views


text:
# Tag Info

5

There is Monte Carlo Simulation and there is Monte Carlo Simulation. If you are referring to a simple question like simulating dice or calculation of $\pi$ or even vanilla option price calculation, it is one thing and "concisely" available. I recommend get a gist of small examples from CS books and then get on with finance. But if you are referring ...

5

Yes, this technique is called moment matching variance reduction and it may indeed lead to a form of variance reduction. The first and second order moments correspond to the mean and the variance of the distribution. You can extend to higher order moments, which is of course more difficult to implement and creates some extra overhead. The mean can be adjust ...

4

You have typo "vol^2", but it should be "vol". Its $$\sqrt{\sigma^2T}=\sigma\sqrt{T}$$

3

The Black-Scholes price of this option is approximately $14.8$. When I run a Monte Carlo simulation with $10000$ paths and "exact" time stepping, I get results very close to this value. You are simulating the terminal asset price with the first-order Euler approximation over multiple time steps: $$S(t+\Delta t)= S(t) + rS(t)\Delta t + \sigma ... 3 You have the correct approach. (1) The simulation generates sampled portfolio values, P_1,P_2, \dots, P_n at time t=T. VaR is specified as a left-tail percentile. Order the sample as$$P_{(1)} \leq P_{(2)} \leq \dots \leq P_{(n)}.$$If you are considering VaR_\alpha at the 100(1-\alpha) \%  confidence level , then choose the smallest integer k ... 3 Your question is too general because Monte Carlo methods differ quite a bit. It's driven more by the problem you are trying to solve, significant result sets, etc, etc. You would either have to provide more details to what you're trying to solve or; try programming some Monte Carlo simulations yourself. My first experience with them was trying to solve ... 3 The error is, you are not storing the random numbers for the same path at the end: xbefore = x + c*tau + sigma*sqrt(tau)*randn() A = muA + sigmaA*randn(); xafter = xbefore + A; But then at end you set a different path here by creating a new random number: xT = log(S0)+(c+muA*lambda)*T+sqrt((sigma^2+(muA^2+sigmaA^2)*lambda)*T)*randn(); randn() ... 2 Historical returns are not to be used 'untreated' for the calculation of option prices. The expectation that you will be using in Monte Carlo will take the form$$ C(K,T) = E^Q\{D(T)\ \max[0, S_T-K, 0]\} $$where T is the maturity, K is the strike price, S is the stock price and D is the discount factor. But the expectation is taken under the 'risk ... 2 This depends on your method to generate the normal random numbers. The problem with normal cdf is that the direct inverse \phi^{-1}(Z) is hard to solve for directly. There are some other methods to generate N(\mu, \sigma^2) from U(0,1). Two notable methods are: Box-Muller method Marsaglia polar method For most purposes you can use the above methods ... 2 I agree with @encor that it isn't an issue to include some logic to avoid errors. I imagine that most non-uniform random number generators already include that. I don't think I've used a pseudo-random number generator that's given a 0. I'm not an expert on the topic, but a common implementation (linear congruential generator) relies on modular arithmetic. ... 2 Indeed for computational purposes, best you can do is use a uniform distribution on another interval [10^{-10},1-10^{-10}], or just discard all occurences of Z=0,1. Discarding Z=0,1 is justified, since for continuous distributions P(Z=0)=P(Z=1)=0. 2 LSM is very fiddly. The most important things in my view are 1) don't believe anyone who says that the choice of basis functions doesn't matter. 2) implement an upper bounder, eg Andersen--Broadie (2003) or Joshi-Tang (2014) so you can tell if your prices are good 3) do two passes, one to build the strategy, one to price, if they give very different ... 2 You should look at confidence interval. Normally, your confidence interval size is proportional to the standard deviation, looking something like: with probability p your value will be in the interval:$$[\bar{S} - k*StdDev, \bar{S} + k*StdDev]$$Then, getting back to your simulation, we can say that your time step is very big (1 year) and you simulate ... 2 I'd recommend M. Joshi and T. Leung "Using Monte Carlo simulation and importance sampling to rapidly obtain jump-diffusion prices of continuous barrier options". Though it assumes jump-diffusion process for the returns it is straightforward to obtain the scheme for a diffusion process. Also Paul Glasserman's [book][2] [2]: ... 2 I am a professor too and I did work with Siemens Corporate Technology which provides the quantitative technology for their copper and electricity trading (Siemens being one of the biggest players in this area in Europe). They are mainly using sophisticated neural networks. We also published a paper together, see my answer here: What types of neural networks ... 2 This sounds correct, however step 2 is a little vague, so I will try to restate the steps here for you. The assets in your portfolio must be priced with respect to a set of risk factors (e.g. interest rate curve). Each scenario consists of a value for each of your risk factors. Given the value of your risk factors you can price your portfolio. You want ... 2 To keep things simple let's assume you have a perfect random number generator (i.e. I will discuss only the statistics not the numerics of the problem). I will also focus on the practical matter and gloss over some mathematical details. From a practical perspective "convergence" means that you will never get an exact answer from Monte-Carlo but ... 2 the output of an MC simulation depends on the random numbers used and if the distribution used is not too weird, after 10,000 runs you will get an answer that is distributed$$ \mu + \frac{\sigma}{\sqrt{n}} Z,  with $Z$ a standard normal. Here $n=10,000.$ With $\mu$ the quantity you want and $\sigma$ the standard deviation. So you won't get precisely the ...

2

You should write some kernel functions in CUDA (Nvidia language) for your matlab code. Arrayfun is quite restrictive and not appropriate. Look at this link http://fr.mathworks.com/help/distcomp/run-cuda-or-ptx-code-on-gpu.html for more details about matlab and parallel computing.

2

First, please make sure that when you resimulate sample paths, you are keeping your underlying random samples constant, as in this answer. For your delta, vega and rho there is some ambiguity in the definition of the greeks. Consider the simple case of delta in the presence of a skew $\sigma(K/S)$, and say that the underlying price right now is $S_0$. We ...

1

Since there is a closed form in the BS case for continuous barrier options, you probably won't find a huge amount of work on this since it's not needed. In the discrete case, I did a paper with Tang: http://ssrn.com/abstract=1441142 Pricing and Deltas of Discretely-Monitored Barrier Options Using Stratified Sampling on the Hitting-Times to the Barrier

1

There are some restrictions to using arrayfun. You can read the restrictions here. Judging from the error, you cannot use indexes the way you are. You probably have to create separate GPU arrays for $V_{t+1}$ and $V_t$. I suggest that you find similar examples in Matlab's website and try to replicate its functionality. Here is an article with ...

1

I though about this one more time: method of moments means that you do the following: calculate some statistics (i.e. the moments) on the sample express the moments of the distribution that you want to fit in terms of the parameters of this distribution solve the resulting system of equations. If you estimate $E[S^n]$ by averaging the ...

1

as you post 3 questions on this topic and after reading them: this is homerwork/study material- right? So for comparing Fast Fourier, MC and Panjer there are tons of publications out there. For the formulas for the momemts of $S$ look here or google "moments in the collective risk model". You should notice that: If you know the distribution of $N$ and $X$ ...

1

You can do the following: For each $i$ in $1$ to number of Mont-Carlo runs $K$ simulate the number of losses $N_i$ simulate $N_i$ many loss-sizes $X_{i,1},\ldots,X_{i,N_i}$ calculate $L_i = \sum_{j=1}^{N_i} X_{i,j}$ Doing this you get a sample of losses $L_1,\ldots,L_K$ and you can do all sorts of hisograms, density fits, VaR, ES calculations on it. ...

1

You can use: Antithetic variates and; Control variates. Both are variance reduction techniques which will allow you to use fewer paths/simulations. Usually antithetic variates are very efficient on their own. Combining both can be a bit tricky. You could start by simulating the value of a plain vanilla call. Then include antithetic variates and/or ...

1

You are generating a price series, in time steps $p_{dt},p_{2dt},p_{3dt},...$. I assume? So if you do $\text{Median}\{p_{dt},p_{2dt},p_{3dt},..\}$ then you will get a value biased towards $0$ (for the parameters you gave, and the sample size you have). If you want to look at the distributional characteristics of the price, then you have to do so per each ...

1

The GBM is a continuous model, so using large integer time steps naturally leaves large discretization error (which vanishes when you increase the number of steps). Use small time step 0.001: paths(j + 1,i) = paths(j,i) * exp((mu - vol^2/2)*0.001 + vol * 0.001^0.5*shocks_ant(j,i)); Then the mean is almost exactly 100 as expected.

1

Giuseppe Bruno, Bank of Italy, did some interesting work in R showing that the use of Quasi Random Numbers in Monte Carlo simulations was superior to Pseudo-Random. Here is an abstract of what he presented at useR! 2014: Pricing Credit Risk Derivative with R

1

We set out a general scheme for doing this sort of thing in our paper http://ssrn.com/abstract=1401094 and its sequel http://ssrn.com/abstract=1437847 Whilst the case studied is different, the techniques are the same. I also discuss in detail the whole process in a chapter of More Mathematical Finance. The adjoint method when it applies is generally ...

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# Problem:  K a for acetic acid is 1.8 x 10-5. What is the first digit of the pK a value for acetic acid?

🤓 Based on our data, we think this question is relevant for Professor Bindell's class at UCF.

###### FREE Expert Solution

$\overline{){{\mathbf{pK}}}_{{\mathbf{a}}}{\mathbf{=}}{\mathbf{-}}{\mathbf{log}}\mathbf{\left(}{\mathbf{K}}_{\mathbf{a}}\mathbf{\right)}}$

###### Problem Details

K a for acetic acid is 1.8 x 10-5. What is the first digit of the pK a value for acetic acid?


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# According to MOT, two atomic orbitals overlap resulting in the formation of molecular orbital formed. Number of atomic orbitals overlapping together is equal to the molecule orbital formed. The two atomic orbital thus formed by LCAO (linear combination of atomic orbital) in the phase or in the different phase are known as bonding and antibonding molecular orbitals respectively. The energy of bonding molecular orbital is lower than that of the pure atomic orbitals by an amount Delta. This known as the stabilization energy. The enerby of antibonding molecular orbital in increased by Delta' (destabilisation energy). <br> Q. How many nodal plane is present in sigma_(s and p) bonding molecular orbital ?

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# Inner product and orthogonality in non orthogonal basis

According to the definition of orthogonality (on finite vector spaces),

Given an inner product space, two vectors are orthogonal if their inner product is zero.

So as an example, assuming the inner product is the "the standard" Euclidean inner product, two vectors (1,0) and (0,1), in $\mathbb{R}^2$, are obviously orthogonal in standard basis. However, if we change the basis to another basis of two linearly independent vectors which are not orthogonal in standard $\mathbb{R}^2$ basis, then these new basis vectors would be also represented as, (1,0) and (0,1) in the new basis. So keeping the same Euclidean inner product, they would be orthogonal in the new basis, although, they are not orthogonal in standard $\mathbb{R}^2$ basis. This would mean that orthogonality depends on the basis chosen, which seems "strange" to me?

Can you please explain what is wrong with in the above reasoning (assuming it is wrong).

Thanks.

Note, I have read the thread with the very similar title, change of basis and and inner product in non orthogonal basis but that doesn't answer my question

The key point to understand here is that you really are dealing with two $\mathbb R^2$ here, although it's not that obvious when using the standard basis.

The first $\mathbb R^2$ is your vector space. Let's write this vector space and everything in it in blue. This first $\color{blue}{\mathbb R^2}$ is equipped with a vector space structure and additionally with the dot product $\color{blue}{\mathbf x\cdot\mathbf y = x_1y_1+x_2y_2}$.

Now as soon as you choose a basis $\{\color{blue}{\mathbf b_1},\color{blue}{\mathbf b_2}\}\subset\color{blue}{\mathbb R^2}$, you can write every vector $\color{blue}{\mathbf x}\in\color{blue}{\mathbb R^2}$ in an unique way as $\color{blue}{\mathbf x}=\color{red}{\xi_1}\color{blue}{\mathbf b_1}+\color{red}{\xi_2}\color{blue}{\mathbf b_2}$. Note that $\color{red}{\xi_1}$ and $\color{red}{\xi_2}$ are not the components of the vector in $\color{blue}{\mathbb R^2}$, but are base dependent.

But of course you need always two of them, and when doing vector addition and multiplication with scalar, you'll find they behave exactly like the components of a vector should behave. Therefore it does make sense to consider them as part of a $\color{red}{\mathbb R^2}$ which however is a different $\mathbb R^2$ than the original $\color{blue}{\mathbb R^2}$ we started with. In particular, the coordinate $\color{red}{\mathbb R^2}$ is not pre-equipped with an inner product.

The basis then defines a linear map $\beta$ from the coordinate $\color{red}{\mathbb R^2}$ to the original $\color{blue}{\mathbb R^2}$ given by $$\beta(\color{red}{\boldsymbol\xi})=\color{red}{\xi_1}\color{blue}{\mathbf b_1} + \color{red}{\xi_2}\color{blue}{\mathbf b_2}.$$

Now remember that I said that the coordinate $\color{red}{\mathbb R^2}$ is not pre-equipped with an inner product. That doesn't mean we cannot give it one. but we want to do it in a way that the product is preserved by the map $\beta$, that is, you want to have $$\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle} = \beta(\color{red}{\boldsymbol\xi})\color{blue}{\cdot}\beta(\color{red}{\boldsymbol\eta})$$ where $\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle}$ denotes the inner product in the coordinate $\color{red}{\mathbb R^2}$. By inserting the explicit formula of $\beta$, one easily sees that $$\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle} = \sum_{j,k=1}^2 (\color{blue}{\mathbf b_j\cdot\mathbf b_k})\color{red}{\xi_j\eta_k}.$$

Now quite obviously, if $\{\color{blue}{\mathbf b_1},\color{blue}{\mathbf b_2}\}$ is not an orthogonal basis, then $\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle}\ne\color{red}{\xi_1\eta_1}+\color{red}{\xi_2\eta_2}$, and indeed, the inner product on the coordinate $\color{red}{\mathbb R^2}$ explicitly depends on the chosen basis $\{\color{blue}{\mathbf b_1},\color{blue}{\mathbf b_2}\}$. But that is not really surprising, because the vector in $\color{blue}{\mathbb R^2}$ those coordinates describe does depend on the basis chosen, and of course different vectors in general have different inner products.

Note that by definition of the inner product, with $\beta(\color{red}{\boldsymbol\xi})=\color{blue}{\mathbf x}$ and $\beta(\color{red}{\boldsymbol\eta})=\color{blue}{\mathbf y}$ it is of course still true that $$\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle} = \color{blue}{\textbf x\cdot\textbf y} = \color{blue}{x_1y_1} + \color{blue}{x_2y_2}.$$ But in general, $\color{blue}{x_1y_1} + \color{blue}{x_2y_2} \ne \color{red}{\xi_1\eta_1}+\color{red}{\xi_2\eta_2}$.

However if you chose the standard basis $\color{blue}{\mathbf b_k}{\mathbf e_k}$ then you obviously have $\color{red}{\xi_k}=\color{blue}{x_k}$ and $\color{red}{\langle \boldsymbol\xi,\boldsymbol\eta\rangle} = \color{blue}{x_1y_1+x_2y_2} = \color{red}{\xi_1\eta_1}+\color{red}{\xi_2\eta_2}.$ This is why it is so easy not to see the fact that you are really working with two different $\mathbb R^2$ when using the standard basis.

You seem to be confusing the notion of an inner product and a basis.

A (real) inner product on a vector space $V$ is a bilinear function $\langle \cdot | \cdot \rangle\colon V\times V\to {\bf R}$ satisfying the standard axioms of positive definiteness and symmetry. It is this inner product that we use to determine whether a given tuple of vectors is orthogonal or orthonormal, not any basis.

Now, if $V$ is finite dimensional and $(v_1,\ldots, v_n)$ is a basis of $V$, there is a unique inner product such that $(v_1,\ldots, v_n)$ is an orthonormal basis. But the converse is far from truth: given any inner product on a space of dimension at least two, there are infinitely many orthonormal bases.

Now, if you have a space $V$ with a fixed inner product $\langle \cdot | \cdot \rangle_1$, you can take a basis $(v_1,\ldots, v_n)$ of $V$ and find the unique inner product $\langle \cdot | \cdot \rangle_2$ such that $(v_1,\ldots,v_n)$ are orthonormal with respect to it, but unless $(v_1,\ldots,v_n)$ are orthonormal with respect to $\langle \cdot | \cdot \rangle_1$, the two inner products are distinct, so it should come as no surprise that the notions of "orthonormal" differ (precisely because of what I have mentioned in the last paragraph about the inner product being unique given a basis which we want to be orthonormal).

Interestingly, it is actually true that any two inner products have a common orthogonal basis, but nonetheless, the notions of orthogonality do not coincide for distinct inner products, either. In other words, while there are special bases which are orthogonal jointly for both inner products, there are always bases which are orthogonal in one, but not in the other basis (for distinct inner products).

$\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$ and $\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$ are the coordinates of the two basis vectors, not the actual vectors .

So $\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$ represents $\vec{e_1}$ and $\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$ represents $\vec{e_2}$ .

In an orthonormal basis we have the dot product of basis vectors is zero : $\vec{e_1} \cdot \vec{e_2} =0$.

But in some other basis it might be that : $\vec{e'_1} \cdot \vec{e'_2} \neq 0$ even though the coordinates of the basis vectors are the same in both cases.

In other words : in the $\left\{ \vec{e'_1}, \vec{e'_2}\right\}$ basis , $\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right) \cdot \left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right) \neq 0$ . In terms of coordinates you could say that this is because the matrix for the dot product is not : $\left( \begin{array}{c , c} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$ in that basis.

Update :

The question is about an inner product space and how it is affected by a change of basis.

As an example we first look at the normal 2-D Euclidean space with orthonormal basis. The basis vectors now are represented in coordinates by these two matrices:

$\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$ and $\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$

The inner product is represented by this matrix : $\left( \begin{array}{c , c} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$ so that the inner product of $\left( \begin{array}{c} a \\ b \\ \end{array} \right)$ and $\left( \begin{array}{c} c \\ d \\ \end{array} \right)$ becomes : $\left( \begin{array}{c , c} 1 & 0 \\ 0 & 1 \\ \end{array} \right)\left( \begin{array}{c} a \\ b \\ \end{array} \right)\left( \begin{array}{c} c \\ d \\ \end{array} \right) = \left( \begin{array}{c} a & b\\ \end{array} \right)\left( \begin{array}{c} c \\ d \\ \end{array} \right)=ac+bd$

Now we express this same system with respect to a new basis : Let's say our new basis vectors become (expressed in terms of the old orthonormal basis vectors):

$\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$ and $\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)$ . These are their coordinates with respect to the 'old' $\left\{ \vec{e_1}, \vec{e_2}\right\}$ basis .

But the SAME two new basis vectors expressed in terms of the new basis are of course : $\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)$ and $\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$ , so these are their coordinates with respect to the new $\left\{ \vec{e'_1}, \vec{e'_2}\right\}$ basis . The old $\vec{e_1} , \vec{e_2}$ vectors with respect to the new coordinate system ($\left\{ \vec{e'_1}, \vec{e'_2}\right\}$) become: $\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)$ and $\left( \begin{array}{c} -1 \\ 1 \\ \end{array} \right)$

Changing the basis also means that coordinates of all other vectors and matrices in the old system need to be transformed to the new $\left\{ \vec{e'_1}, \vec{e'_2}\right\}$ system :

A : Old system $\left\{ \vec{e_1}, \vec{e_2}\right\}$ :

$\vec{e_1}=\vec{e'_1}$
$\vec{e_2}=\vec{e'_2}-\vec{e'_1}$

$\vec{a}= \left( \begin{array}{c} a_1 \\ a_2 \\ \end{array} \right)$ , $\vec{b}= \left( \begin{array}{c} b_1 \\ b_2 \\ \end{array} \right)$

The dot product : $\vec{a}\cdot\vec{b}=a_1b_1\vec{e_1}\cdot\vec{e_1}+a_2b_2\vec{e_2}\cdot\vec{e_2}$ $\enspace$ , $\enspace$ matrix form dot product : $\left( \begin{array}{c , c} 1 & 0 \\ 0 & 1 \\ \end{array} \right)$

B : New system $\left\{ \vec{e'_1}, \vec{e'_2}\right\}$ :

$\vec{e'_1}=\vec{e_1}$
$\vec{e'_2}=\vec{e_1}+\vec{e_2}$

$\vec{a}= \left( \begin{array}{c} a_1' \\ a_2' \\ \end{array} \right) = a_1\vec{e'_1} + a_2(\vec{e'_2}-\vec{e'_1}) \implies a_1'=a_1-a_2 \enspace , \enspace a_2' = a_2$
$\vec{b}= \left( \begin{array}{c} b_1' \\ b_2' \\ \end{array} \right) = b_1\vec{e'_1} + b_2(\vec{e'_2}-\vec{e'_1}) \implies b_1'=b_1-b_2 \enspace ,\enspace b_2' = b_2$

The dot product : $\vec{a}\cdot\vec{b}=a_1b_1+a_2b_2=(a_1'+a_2')(b_1'+b_2')+a_2'b_2' =a_1'b_1'+a_1'b_2'+ a_2'b_1'+ 2a_2'b_2'= \left( \begin{array}{c} a_1'+a_2' & a_1'+2a_2' \\ \end{array} \right)\left( \begin{array}{c} b_1' \\ b_2' \\ \end{array} \right) \implies$

Transformed matrix for the dot product becomes : $\left( \begin{array}{c , c} 1 & 1 \\ 1 & 2 \\ \end{array} \right)$

So in the new coordinates $\vec{e_1}\cdot\vec{e_2}$ is still zero : $\left( \begin{array}{c , c} 1 & 1 \\ 1 & 2 \\ \end{array} \right)\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)\left( \begin{array}{c} -1 \\ 1 \\ \end{array} \right)=\left( \begin{array}{c} 1 & 1 \\ \end{array} \right)\left( \begin{array}{c} -1 \\ 1 \\ \end{array} \right) =0$

And $\vec{e_1'}\cdot\vec{e_2'}$ is : $\left( \begin{array}{c , c} 1 & 1 \\ 1 & 2 \\ \end{array} \right)\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right)=\left( \begin{array}{c} 1 & 1 \\ \end{array} \right)\left( \begin{array}{c} 0 \\ 1 \\ \end{array} \right) =1$

Compare this last result to $\vec{e_1'}\cdot\vec{e_2'}$ in the old system: $\left( \begin{array}{c , c} 1 & 0 \\ 0 & 1 \\ \end{array} \right)\left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right)\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)=\left( \begin{array}{c} 1 & 0 \\ \end{array} \right)\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right) =1$

We conclude that when we transform the coordinates of the dot product matrix to the new basis $\left\{ \vec{e'_1}, \vec{e'_2}\right\} = \left\{ \left( \begin{array}{c} 1 \\ 0 \\ \end{array} \right), \left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)\right\}$ we get the new dot product coordinates : $\left( \begin{array}{c , c} 1 & 1 \\ 1 & 2 \\ \end{array} \right)$ and the dot products of all vectors remain unchanged as they should .

Another way to calculate the coordinates for the transformed dot product matrix is realizing that the dot product consists of combinations of two 'row'-basisvectors.

Row vectors and column vectors :

$\vec{e_{i}}$ is column vector , $\vec{e^{i}}$ is row vector.

Column vectors and row vectors together form numbers : $\vec{e^{i}}(\vec{e_{j}}) = \vec{e_{i}}(\vec{e^{j}}) = \vec{e'_{i}}(\vec{e'^{j}})=\vec{e'^{i}}(\vec{e'_{j}})=\delta_{ij}$ where $\delta_{ij}$ is the Kronecker delta. This combination of column and row vectors to form a number is NOT the dot product and it IS coordinate independent.

Because column and row vectors together must produce coordinate independent numbers we can derive their transformation properties :

$\vec{e^1}=a\vec{e'^1} +b\vec{e'^2}$
$\vec{e^2}=c\vec{e'^1} +d\vec{e'^2}$

$1=\vec{e^2}(\vec{e_2})=\vec{e^2}(\vec{e'_2}-\vec{e'_1}) = (c\vec{e'^1} +d\vec{e'^2})(\vec{e'_2}-\vec{e'_1}) =-c+d$

$1=\vec{e^1}(\vec{e_1})= \vec{e^1}(\vec{e'_1}) = (a\vec{e'^1} +b\vec{e'^2})(\vec{e'_1}) =a$

$0=\vec{e^2}(\vec{e_1})= \vec{e^2}(\vec{e'_1}) = (c\vec{e'^1} +d\vec{e'^2})(\vec{e'_1})=c$

$0=\vec{e^1}(\vec{e_2})= \vec{e^1}(\vec{e'_2}-\vec{e'_1}) = (a\vec{e'^1} +b\vec{e'^2})(\vec{e'_2}-\vec{e'_1}) =-a+b$

It follows that :

$\vec{e^1}= \vec{e'^1} + \vec{e'^2}$
$\vec{e^2}= \vec{e'^2}$

The dot product ($DP$) expressed in the old row-basisvectors is :

$DP=\vec{e^{1}}\otimes \vec{e^{1}}+\vec{e^{2}}\otimes \vec{e^{2}}$

In terms of the new basis vectors this becomes :

$DP=(\vec{e'^1} + \vec{e'^2})\otimes (\vec{e'^1} + \vec{e'^2})+\vec{e^{2}}\otimes \vec{e^{2}} \implies$

$DP= \vec{e'^1}\otimes \vec{e'^1} + \vec{e'^1}\otimes \vec{e'^2} +\vec{e'^2}\otimes \vec{e'^1} + 2\vec{e'^2}\otimes \vec{e'^2} = \left( \begin{array}{c , c} 1 & 1 \\ 1 & 2 \\ \end{array} \right)$

• But according to the definition of orthogonality Two vectors in some inner product space V are orthogonal if their inner product is zero. This is the definition of orthogonality given in Axler's Linear Algebra book (as well as my Croatian math textbook). Now both the dot product and Euclidean inner product of two vectors $x, y \in \mathbb{R}^n$ are given by $x_1y_1+ \cdots + x_ny_n$. Therefore (1,0) and (0,1) should be orthogonal irrespective of the "actual vectors" they represent (which obviously depends on the basis). Definition above doesn't mention anything about basis. – ivan zg Apr 15 '17 at 13:58
• @ivanzg Added one last update, hope this explains it. – Rutger Moody Apr 17 '17 at 7:41


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# Infinite sums of logs.

1. Jun 8, 2013

### cp255

So I was trying to see if $\Sigma$ln($\frac{n}{n+1}$) diverges or converges. To see this I started writing out [ln(1) - ln(2)] + [ln(2) - ln(3)] + [ln(4) - ln(5)] ...

I noticed that after ln(1) everything must cancel out so I reasoned that the series must converge on ln(1) which equals ZERO. However, Wolfram Alpha says the series is divergent. I tried looking it up how to do this problem and I read that the last term also does not cancel out which is why the series diverges. However, shouldn't there always be a term bigger than the last so everything must cancel.

2. Jun 8, 2013

### Infrared

Write an expression for $\sum_{k=1}^{n} ln(\frac{k}{k+1})$. What is the limit of this expression as n tends to infinity?


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# Anabolism

Anabolism (from Greek αναβολισμός , anabolismós , " Aufwurf " ) is material building process. The opposite of anabolism is catabolism , namely the degradation of substances . Catabolism and anabolism, are parts of metabolism.

definitions

The precise definitions in the literature are similar, but are different.

Widely used is the definition of all the anabolic metabolic reactions. After that , strictly speaking , some exergonic conversion processes included for the purpose of energy production , as far as in them complicated from simpler substances arise , such as the reduction of carbon dioxide CO2 with hydrogen H2 to EssigsäureCH3COOH (a form of Homoacetatgärung ) and the oxidation of hydrogen H2 with oxygen O2 to water H2O ( oxyhydrogen reaction) in some bacteria. In general, technical parlance , however, be such conversion processes that serve the energy , not referred to as anabolism.

Part of this definition is restricted to the construction of endogenous substances , which means that any substance reactions in which the products are excreted excluded , such as the formation of antibiotics and the above-mentioned hydrogen-oxygen reaction . In a further variation , only the combination of simple molecules, complex molecules, is referred to as anabolism and thus excluding the participation of atoms and monatomic ions. Since atoms and monatomic ions can not be products of a structure and atoms as starting materials of construction reactions are not known, only monatomic ions are excluded as starting materials with this definition , this exclusion is not justified.

Other definitions describe the totality of the endergonic ( energy-consuming ) conversion processes as anabolism. Thus, this is also the formation of substances that are excreted included. Variations of this definition be restricted to the endergonic structure of these materials, or to the endergonic structure of molecules. It is likely that this endergonic conversion processes which are only partial reaction a reaction sequence which is exergonic in the sum and the energy used ( catabolism) , should be excluded , such as the formation of sugar phosphates to initiate glycolysis .
energy demand

Most energy is needed for the construction of fabrics. In chemotrophic organisms they extracted from chemical , energy- releasing ( exergonic ) nutrient cycling , beiphototrophen organisms from light. The energy from these sources is first converted into a short-term energy storage and transfer agents , namely inAdenosintriphosphat (ATP). ATP is used in the construction of energy-consuming reactions as a direct energy source. Also is einReduktionsmittel If required by the building of endogenous substances , serves to NADPH , which is formed in phototrophs by the light reaction in chemotrophic by the oxidation of substances.
Structure of molecules

The simplest molecules , which result from the anabolic process of photosynthesis , are simple sugars . By further anabolic processes these sugar molecules to be rebuilt. This results in , among other amino acids, isoprenoids and nucleotides. From these precursors then complex molecules such as proteins, fats , carbohydrates, genetic material or lignin can be formed.

Quellen:

1.  Lexikon der Biologie. Herder, Freiburg u. a. O. 1983, S. 166.
2.  Brockhaus ABC Biologie. Brockhaus, Leipzig 1986, S. 861.
3.  Meyers Taschenlexikon Biologie. 2. Auflage. B.I.-Taschenbuchverlag, Mannheim, Wien, Zürich 1988.
4.  Gertrud Scherf: Wörterbuch Biologie. DTV, München 1997, S. 25.
5.  Lexikon der Biologie. Spektrum, Heidelberg 1999, S.310.
6.  Kompaktlexikon der Biologie. Spektrum, Heidelberg 2001.
7.  Peter H. Raven, George B. Johnson, Jonathan B. Losos, Susan R. Singer: Biology. 7. Auflage. McGraw-Hill, Boston u. a. O. 2005, Glossar S. G-1.
8.  Eike Libbert: Allgemeine Biologie. UTB 1197, 7. Auflage. Gustav Fischer, Jena 1991, S. 33, 149.
9.  Daniel N. Lapedes (Hrsg.): McGraw-Hill Dictionary of the Life Science. McGraw-Hill, New York u. a. O. 1976, S. 45-46.
10.  William K. Purves, David Sadava, Gordon H. Orians, H. Craig Heller: Biologie. 7. Auflage. Übersetzung ins Deutsche, Spektrum Elsevier, München 2006 (Original: 2004), S. 128.
11.  Peter H. Raven, George B. Johnson, Jonathan B. Losos, Susan R. Singer: Biology. 7. Auflage. McGraw-Hill, Boston u. a. O. 2005, S. 155 (hier wird an anderer Stelle auch die erstgenannte Definition verwendet)
12. ↑ a b Neil A. Campbell, Jane B. Reece: Biologie. 8. Auflage. Übersetzung ins Deutsche, Pearson Studium, München 2009.
13.  G. Czihak, H. Langer, H. Ziegler (Hrsg): Biologie – Ein Lehrbuch. 6. Auflage. Springer, Berlin u. a. O. 1996, S. 87.M.W.G. de Bolster: Glossary of Terms Used in Bioinorganic Chemistry: Anabolism. International Union of Pure and Applied Chemistry. 1997. Abgerufen am 10. Mai 2012.
14. http://de.wikipedia.org/wiki/Anabolismus


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# Mathematical Sciences Research Institute

Home » Workshop » Schedules » Models for homotopical higher categories

# Models for homotopical higher categories

## Introductory Workshop: Algebraic Topology January 27, 2014 - January 31, 2014

January 28, 2014 (03:30 PM PST - 04:30 PM PST)
Speaker(s): Julie Bergner (University of California)
Location: MSRI: Simons Auditorium
Video
Abstract

Models for homotopical categories, known as $(\infty,1)$-categories, are now fairly well-developed. Comparisons are in place between them, and there is a wide variety of applications ranging from topology to algebra to algebraic geometry. Generalizing them to models for homotopical higher categories, or $(\infty,n)$-categories, requires substantially greater technicality. Much work has been done in this direction, due to recent work of many authors, but the full picture is still very much work in progress. In this talk we will review different models for $(\infty,1)$-categories and discuss the number of ways they are being generalized to models for $(\infty,n)$-categories, as well as the known and conjectured comparisons between them

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# Thread: Shortcut calculation for tax savings on capital expenditure

1. ## Shortcut calculation for tax savings on capital expenditure

Hi

Say,

Machinery = $400,000 Useful life = 4 years Capital Allowance Rate = 25% Corporation Tax Rate = 35% Scrap Value =$5,000

Do you know any short cut that being used to calculate the total tax savings amount?

I only know till certain steps. ie 40,000 x 0.25 x 0.35 x 0.75 x 0.75 x....?

Will appreciate the help

Thanks

2. Originally Posted by mingali
Machinery = $400,000 Useful life = 4 years Capital Allowance Rate = 25% Corporation Tax Rate = 35% Scrap Value =$5,000

Do you know any short cut that being used to calculate the total tax savings amount?

I only know till certain steps. ie 40,000 x 0.25 x 0.35 x 0.75 x 0.75 x....?