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# *Seond Order Derivative*

1. Jun 13, 2008

### physixguru

Been doing calculus all my life , that's why i am at the Indian Institute of Technology:shy:..
but never faced a headache like this......

Prove that:

If x=cos theta and y = sin^3 theta

then [d^2y/dx^2] + [dy/dx]^2 = 3sin^2 theta [ 5 cos ^2 theta -1 ]

My side:

dx/d theta = -sin theta
dy/d theta = 3 sin^2 theta cos theta
>> {dy/dx}== -3 sin theta cos theta

now d^2y/dx^2= -3 [ cos^2 theta - sin ^2 theta ]* 1/-sin theta

moving with the give expression....

3cos^2 theta-3 sin^2 theta/sin theta + 9 sin^2 theta cos^2 theta

Spent a lot of time...cud not get anywhere...:grumpy:
Am i messing up the trigo functions?

This is not a homework question. I don't have to do homework anymore.I work in a MNC.

Last edited: Jun 13, 2008
2. Jun 13, 2008

### Ben Niehoff

You've never faced a "headache" like that? Use trig identities.

3. Jun 14, 2008

### physixguru

Please be a bit more clear.Even a moron knows that you gotta use trigo identities.Its no rocket science!!!!

4. Jun 14, 2008

### maze

Ok, I'm going to assume you didn't make any algebra/differentiation mistakes so far (I didn't check)
If that is the case, then you are starting here:
3cos^2 theta-3 sin^2 theta/sin theta + 9 sin^2 theta cos^2 theta

And you want to get here:
3sin^2 theta [ 5 cos ^2 theta -1 ]

I would write each expression in terms of complex exponentials, then multiply all the exponentials out completely, working both backwards and forwards simultaneously. Then when your results meet in the middle, rearrange your steps to go from where you started to where you finished.

Here is a small (easier) example of this strategy in action: Say we want to show [itex]cos^2\left(x\right) = 1-sin^2\left(x\right)[/tex]
Working forwards:
$$cos^2\left(x\right)=\left(\frac{e^{i x}+e^{-i x}}{2}\right)^2=\frac{e^{2 i x} + 2 + e^{-2 i x}}{4}$$
Now hold that thought while we work backwards...

Working backwards:
$$1-sin^2\left(x\right) = 1-\left(\frac{e^{i x}-e^{-i x}}{2 i}\right)^2=1-\frac{e^{2 i x}-2+e^{-2 i x}}{-4}=\frac{e^{2 i x} + 2 + e^{-2 i x}}{4}$$

Then to write the final proof, you would just rearrange the steps:
$$cos^2\left(x\right)=\left(\frac{e^{i x}+e^{-i x}}{2}\right)^2=\frac{e^{2 i x} + 2 + e^{-2 i x}}{4}=1-\frac{e^{2 i x}-2+e^{-2 i x}}{-4}=1-\left(\frac{e^{i x}-e^{-i x}}{2 i}\right)^2=1-sin^2\left(x\right)$$

Last edited: Jun 14, 2008
5. Jun 16, 2008

### physixguru

Dear maze, thanks a lot for your efforts and spending your precious time on the problem.
Your method is really impressive and shows your multidimensional thinking.But the problem is that the method you supplied is a lengthy one and can be extremely confusing in long trigo expressions like mine is the case.I would like a simple approach.


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# Bloomberg Treasuries PX_Last and daily returns

I tried to search for this specific question, although I didn’t found a conclusive answer. I have a dataset containing the yields of several T-Bills and T-Notes that were downloaded from a Bloomberg Terminal using the function px_last. In order to achieve the actual prices of such securities how I should set up the formula?

1. $$P=\frac{FV}{\left(1 + \frac{px_{last}}{100}\right)^{\frac{m}{12}}}$$
2. $$P=FV \times \left(1 - \left(\frac{px_{last}}{100} \right)^{\frac{m}{12}}\right)$$

where $$m$$ are the months up to maturity.

Then if I need to calculate daily returns, I would kindly ask you if it is correct to use a formula:

\begin{aligned} R =& \left(1+\frac{px_{last}.shift(1)}{100}\right)^{\frac{1}{365}} - 1 \times \frac{\partial P}{\partial px_{last}} \times \frac{px_{last} - px_{last}.shift(1)}{100} \\ &+ 0.5 \times \frac{\partial^2 P}{\partial px_{last}^2} \times \left( \frac{px_{last} - px_{last}.shift(1)}{100} \right)^2 \end{aligned}

or simply

$$R=\frac{P_{t+1}-P_t}{P_t}$$

where: $$P$$ represent the price formula, $$px_{last}.shift(1)$$ is intended as the yield of the previous day

• (3) The formula $R=(P_{t+1}-P_t)/P_t$ is the best way to calculate return. The other is (a) an approximation (b) requires additional complicated calculations to find the two derivatives, Duration and Convexity. – Alex C Dec 19 '19 at 16:39

format: plain


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# Illinois Geometry Lab

## IGL Projects, Fall 2017

Applications

To apply for one of these projects, use one of the following forms. Application deadline: Monday August 21, 2017.

• #### Complex dynamics and zeros of derivatives

##### Faculty Mentor: Aimo Hinkkanen

This is a continuation of the IGL project of the same name from Fall 2016. The team will produce graphics of Julia sets of suitable rational functions (these are fractals arising in complex dynamics) to illustrate the behavior of the zeros of the derivatives of a given function. The zeros of the first derivative of a function are poles of the associated Newton's method function, and the zeros of the second derivative of a function are among the zeros of the derivative of the Newton's method function, which, in turn, are often connected to certain components of the Fatou set (the complement of the Julia set) of the Newton's method function. For this reason, complex dynamics has been used to study the location of the zeros of the derivatives of a given function. Graphics will be produced to illustrate situations that can occur.

Weekly
• ##### Project Difficulty
Basic
Completion of Calculus 3
• #### Stability of Quasicrystal Frameworks in 3D

##### Faculty Mentor: George Francis

This project concerns the stability of a rod-and-pinion framework under unrestricted deformations. Such a framework is the 1-skeleton a Penrose cluster of 3D quasicrystals. We develop real-time interactive computer animations for experiments to generate conjectures, verify them experimentally, and then develop proofs of the theorem if possible.

Weekly
• ##### Project Difficulty
Intermediate
Some linear algebra, 3D vector geometry and programming experience are needed for the advanced component. But even if you have never programmed an animation you are welcome to join the novice subgroup. We will be using the (free) online textbook on computer graphics by David Eck at Hobart College.
• #### Visual Cliffs, Virtual Reality, and Movement Disorders

##### Faculty Mentors: Richard Sowers (Math, ISE) and Manuel Hernandez (Kinesiology)

Virtual reality (VR) is entering into the public consciousness. One area where it may have significant impact is the medical industry. In this project, we seek to create a VR component of a testbed for understanding responses to visual stimuli and its relation to Parkinson's disease and other movement disorders. We will likely develop on a Hololens (computer languages are likely to be C# and Unity) or a similar VR platform. Hopefully you will gain some expertise which is valuable in this exciting area. Solid coding ability is required. Professor Manuel Hernandez of the Dept. of Kinesiology & Community Health is a cofaculty mentor for this project. We intend to connect this project to the UIUC Health Care Engineering Systems Center.

• ##### Team Meetings
Every other week
• ##### Project Difficulty
Solid coding ability is required.
• #### Video as a Sensor

##### Faculty Mentors: Richard Sowers (Math, ISE) and Daniel Work (CEE)

This project is a continuation of the Spring 2017 "Geometry of Autonomous Vehicles" project. The near future promises a rise in sensor-equipped vehicles.. While machine learning is a core enabling technology, we believe that some models and context may provide additional benefits. Our goal is to compute some of the geometry of traffic surrounding the driver and understand how this affects driving decisions. Hopefully you will gain some expertise which is valuable in this exciting area (think Uber and Tesla). Solid coding ability is required. Professor Dan Work of Civil and Environmental Engineering is a co-faculty mentor for this project.

• ##### Team Meetings
Every other week
• ##### Project Difficulty
Solid coding ability is required.
• #### Surfaces moving under Hamiltonian flows

##### Faculty Mentors: Ely Kerman

In the study of Hamiltonian flows (classical mechanics) the motion of a point (particle) is determined by a single function (the energy). In this project we will study a new evolution equation for star-shaped surfaces that is defined in terms of Hamiltonian flows. To each starshaped surface one can associate a simple function which measures how far the surface is from a round sphere. This function defines a Hamiltonian flow which moves the surface, which changes the flow, which moves the surface, .... etc. The resulting evolution of the original surface is governed by a differential equation which in two dimensions equals the standard heat equation, and in higher dimensions resembles it. The hope is that, like solutions of the heat equation, the evolving surface quickly converges to an interesting equilibrium. The first goal of the project will be to model the relevant equation in order to study the convergence of the solutions and the special features of the equilibrium surfaces.

• ##### Team Meetings
once a week initially, biweekly eventually
• ##### Project Difficulty
Intermediate
Completion of Calculus 3 and familiarity with differential equations at the level of Math 285. Familiarity with software like Matlab and Python is also desired but not expected.
• #### Automata and Space-Filling Curves

##### Faculty Mentors: Philipp Hieronymi and Erik Walsberg

In the last few years novel connections between mathematical logic, automata theory and metric geometry have emerged. A question that often arises in this area is the following: let $r \in \mathbb{N}_{>1}$ and let $C \subseteq \mathbb{R}^n$ be a geometrically interesting set (often a fractal), can the set of all $r$-ary representations of elements of $C$ be recognized by a Büchi automaton? For example, let C be the usual middle-thirds Cantor set. Elements of C are precisely those real numbers in $[0,1]$ that have a ternary representation in which the digit 1 does not occur. Therefore it is not hard to see that the set of ternary representations of elements of $C$ can be recognized by such an automaton. The goal of this project to answer similar questions in the case that $C$ is the graph of a function. In particular, this project aims to determine whether graphs of space-filling curves can be recognized in this way.

• ##### Team Meetings
At least biweekly
• ##### Project Difficulty
Intermediate
Completion of MATH 347 and 400-level real analysis course are required. Completion of MATH 414 or CS 374 helpful, but knowledge of logic or automata theory is not a prerequisite.
• #### Poncelet theorem and complex analysis

##### Faculty Mentors: Jeremy Tyson

The goal of this project is to investigate a surprising connection between the geometry of conic sections and finite Blaschke products. Poncelet’s theorem states that if a polygon is inscribed in a conic section and circumscribes another conic section, then each point of the outer conic is a vertex of a polygon with the same property. Daepp, Gorkin and Mortini (2002) identified a surprising connection between Poncelet’s theorem and the geometry of finite Blaschke products. A finite Blaschke product is the product of finitely many automorphisms of the unit disc. Let B(z) be a Blaschke product of degree three with one zero at the origin and two other nonzero and distinct zeros. Each point on the unit circle has three preimages. The Daepp-Gorkin-Mortini result states that the (Euclidean) triangle formed by these preimages is of Poncelet type. In fact, the triangles formed by the three preimages of any point of the unit circle all circumscribe a fixed ellipse, whose foci lie at the remaining two zeros of B. In this project we’ll investigate various related questions. For instance, what happens if we replace “Euclidean triangle” by “hyperbolic triangle”? Alternatively, what is the locus spanned by all such triangles under other normalization for degree three Blaschke products? No background is needed in complex analysis; this project is accessible to students who have completed Math 241.

• ##### Team Meetings
weekly at first, then periodically
• ##### Project Difficulty
Basic
Completion of Calculus 3
• #### Calculus, Geometry, and Probability in n Dimensions

##### Faculty Mentors: A.J. Hildebrand

This project is part of an ongoing program that began in Fall 2012 and that is now in its tenth semester. The general goal of this program is to seek out and explore interesting problems in n-dimensional calculus and geometry that are accessible at the calculus level, motivated by applications in probability, statistics, economics, and other areas, and that have a broad appeal and are well-suited for creating interactive visualizations for presentation at outreach events, and for publication at the Wolfram Demonstrations website. During the past two semesters we focused on mathematical questions arising in game theory, and in particular on mathematical models of poker. Depending on the interests and backgrounds of the participants, we may continue with the "Mathematics of Poker" theme in the coming semester, or explore other problems of similar flavor at the interface of mathematics and economics. For further details, and reports on past projects, visit http://www.math.illinois.edu/~ajh/ugresearch/.

• ##### Team Meetings
Twice a week for two hours each time
• ##### Project Difficulty
Basic
Students should have taken Math 241, preferably in the honors version, with a grade of A- or better, or be concurrently enrolled in an honors section of Math 241. Programming experience with Mathematica and/or Python, would be a plus, but is not required. More important than any formal prerequisites are an enthusiasm about the subject and the project, an open mind, a willingness to work with like-minded students in a team, and a willingness and to make the necessary time commitment for this project to succeed.
• #### Chaotic maps, exotic number systems, and arithmetic coding schemes

##### Faculty Mentors: A.J. Hildebrand and Ken Stolarsky

The goal of this project is to explore surprising connections between two seemingly unrelated problems, one belonging to analysis, and the other belonging to number theory: The behavior of chaotic maps on the one hand, and the representation of numbers in "non-standard", or "exotic", number systems on the other hand. A chaotic map is, roughly speaking, a function that when iterated a large number of times behaves in unpredictable ways. We will focus on a particular class of such maps, namely piecewise linear functions from [0,1] to [0,1]. Examples include the "tent map", which consists of an up-slope followed by a down-slope, and maps of the form f(x)= {ax+b}, where the braces denote the fractional part. These maps lead in a natural way to representations, or "encodings", of real numbers with respect to generalized, or "exotic", number systems, a connection that has been used in recent years to design encryption and compression schemes. Despite the apparent simplicity of these maps, their behavior under iteration is still not well understood. In this project we will investigate these maps - from both the "chaotic map" and the "generalized number systems" angle - using computer experimentation and visualization, and hopefully also gain new theoretical insights.

• ##### Team Meetings
Twice a week
• ##### Project Difficulty
Intermediate


text:
# 4. Model development

Last time we used simulations as well as prior and posterior predictive checks to check and improve upon our model for Cepheids in a single galaxy. We then found the alpha and beta posteriors for each galaxy independently.

In order to use the period luminosity relation to extend the cosmic distance ladder, and eventually estimate the Hubble constant, the relation needs be universal, i.e. the same alpha and beta should hold for all galaxies, within uncertainties.

From our studies of individual galaxies, we should be able to see the similarities and differences between the results. But, how do we combine these results properly, and keep track of the uncertainties or correlations between parameters?

## Hierarchical models

A natural way to combine the inidividual galaxy results is to build a hierarchical model for all galaxies that we have data for. To understand how to do this, we can start with a simpler example of a simple normal model, like the one we used to understand Stan in the introduction notebook.

Imagine that $$\mu$$ is now sampled from some parent distribution, with mean $$\theta$$ and standard deviation $$\tau$$.

Single-level normal model:

$p(\mu, \sigma|x) \propto p(x|\mu, \sigma) p(\mu, \sigma).$

Hierarchical normal model:

$p(\theta, \tau | x) \propto p(x|\mu, \sigma) p(\sigma) p(\mu | \theta, \tau) p(\theta, \tau)$

Hierarchical models are ideal for describing populations of objects with similar properties, like our Cepheid-hosting galaxies. If we let alpha and beta for each galaxy come from parent distributions with means mu_X and standard deviations tau_X, we can then fit for the shape of the universal alpha and beta distributions!

We see that building up hierarchies in this way can quickly lead to a large number of free parameters, which may seem scary. However, Hamiltonian Monte Carlo is designed to cope with these kind of models, and this should be no problem for Stan.

## A joint hierarchical fit of all galaxies

Exercise 1 (10 points): Sketch the graphical model for our hierarchical model for Cepheids over all galaxies

A simplified Stan model for the full hierarchy can be found in cepheid_v3.stan. Note that this is based on cepheid_v1.stan, and you should also incorporate the changes that you made during the exercises in the model checking notebook in order to achieve a good fit to the data. I will use cepheid_v3.stan here to demonstrate a few important points about hierarchical modelling.

[1]:

!cat stan/cepheid_v3.stan

/**
* Hierarchical model for Cepheid variables
* - Multiple galaxies
* - Shared sigma_m
* - Weakly informative priors
**/

data {

/* usual inputs */
int Ng; // number of Galaxies
int Nt; // sum(Nc_g)

int gal_id[Nt]; // galaxy id for each entry [1 - 9]
vector[Nt] m_obs; // obs apparent mag.
real sigma_m; // mag uncertainty
vector[Nt] log10P; // log10(periods/days)
vector[Ng] z; // redshift of single galaxy

}

transformed data {

vector[Ng] dL;

/* luminosity distance */
dL = (3.0e5 * z) / 70.0; // Mpc

}

parameters {

/* parameters of the parent distributions */
real mu_alpha;
real<lower=0> tau_alpha;

real mu_beta;
real<lower=0> tau_beta;

/* individual galaxy parameters */
vector[Ng] alpha;
vector[Ng] beta;

}

transformed parameters {

vector[Nt] M_true;
vector[Nt] m_true;

/* P-L relation */
M_true = alpha[gal_id] + beta[gal_id] .* log10P;

/* convert to m */
m_true = M_true + 5 * log10(dL[gal_id]) + 25;

}

model {

/* priors */
mu_alpha ~ normal(0, 10);
mu_beta ~ normal(-5, 5);
tau_alpha ~ cauchy(0, 2.5);
tau_beta ~ cauchy(0, 2.5);

/* connection to latent params */
alpha ~ normal(mu_alpha, tau_alpha);
beta ~ normal(mu_beta, tau_beta);

/* connection to data */
m_obs ~ normal(m_true, sigma_m);

}

A few things worth noting:

• The easiest way to deal with a ragged array structure (different number of Cepheids in each galaxy) in Stan is to flatten the array, and pass a vector with the galaxy IDs.

• We introduce mu and tau parameters for alpha and beta, with sensible priors.

• We assume the parent alpha and beta distributions are independent, as a starting point.

• In a hierarchical model, it is not so clear how to separate the priors and likelihood, the model is more a series of conditional steps.

Now, let’s see how the model performs when fitting our dataset.

[2]:

import numpy as np
import arviz as av
from cmdstanpy import CmdStanModel

[3]:

# load data

# compile data from all galaxies
galaxies = [int(ngc_no) for ngc_no in galaxy_info[:, 0]]
data = {int(x[0]):{"z":x[1]} for x in galaxy_info}

for g in galaxies:
j = np.where(cepheid_info[:, 0] == g)[0]
data[g]["gal"] = np.array([int(i) for i in cepheid_info[j, 0]])
data[g]["Nc"] = len(data[g]["gal"])
data[g]["m_obs"] = cepheid_info[j, 1]
data[g]["sigma_m"] = cepheid_info[j, 2]
data[g]["P"] = cepheid_info[j, 3]
data[g]["delta_logO_H"] = cepheid_info[j, 4]
data[g]["log10P"] = np.log10(data[g]["P"])

[4]:

# Select out two galaxies
sel = 2

# prepare Stan input
input_data = {}
input_data["Ng"] = len(galaxies[0:sel])
input_data["Nt"] = sum([data[g]["Nc"] for g in galaxies[0:sel]])
input_data["m_obs"] = np.concatenate([data[g]["m_obs"] for g in galaxies[0:sel]])
input_data["log10P"] = np.concatenate([data[g]["log10P"] for g in galaxies[0:sel]])
input_data["gal_id"] = np.concatenate([np.tile(i, data[g]["Nc"])
for i, g in enumerate(galaxies[0:sel])]) + 1
input_data["z"] = galaxy_info.T[1][0:sel]
input_data["sigma_m"] = 0.5

[5]:

stan_model = CmdStanModel(stan_file="stan/cepheid_v3.stan")

INFO:cmdstanpy:found newer exe file, not recompiling
INFO:cmdstanpy:compiled model file: /Users/fran/projects/BayesianWorkflow/src/notebooks/stan/cepheid_v3

[6]:

fit = stan_model.sample(data=input_data, chains=4, iter_sampling=1000)

INFO:cmdstanpy:start chain 1
INFO:cmdstanpy:start chain 2
INFO:cmdstanpy:start chain 3
INFO:cmdstanpy:start chain 4
INFO:cmdstanpy:finish chain 3
INFO:cmdstanpy:finish chain 1
INFO:cmdstanpy:finish chain 2
INFO:cmdstanpy:finish chain 4

We should always check the sampler diagnostics first!

[7]:

fit.diagnose();

INFO:cmdstanpy:Processing csv files: /var/folders/8d/cyg0_lx54mggm8v350vlx91r0000gn/T/tmpomivi5y3/cepheid_v3-202109111349-1-i76uz3kv.csv, /var/folders/8d/cyg0_lx54mggm8v350vlx91r0000gn/T/tmpomivi5y3/cepheid_v3-202109111349-2-sdcvnxp4.csv, /var/folders/8d/cyg0_lx54mggm8v350vlx91r0000gn/T/tmpomivi5y3/cepheid_v3-202109111349-3-mp1dna27.csv, /var/folders/8d/cyg0_lx54mggm8v350vlx91r0000gn/T/tmpomivi5y3/cepheid_v3-202109111349-4-ywx326xl.csv

Checking sampler transitions treedepth.
Treedepth satisfactory for all transitions.

Checking sampler transitions for divergences.
208 of 4000 (5.2%) transitions ended with a divergence.
These divergent transitions indicate that HMC is not fully able to explore the posterior distribution.
Try increasing adapt delta closer to 1.
If this doesn't remove all divergences, try to reparameterize the model.

Checking E-BFMI - sampler transitions HMC potential energy.
E-BFMI satisfactory.

Effective sample size satisfactory.

Split R-hat values satisfactory all parameters.

Processing complete.

[8]:

fit.summary()

[8]:

Mean MCSE StdDev 5% 50% 95% N_Eff N_Eff/s R_hat
name
lp__ -450.0 0.1100 2.600 -450.00 -450.00 -440.0 560.0 340.0 1.0
mu_alpha -1.3 0.0630 1.800 -3.90 -1.30 1.6 800.0 490.0 1.0
tau_alpha 2.4 0.0890 2.500 0.61 1.70 6.4 800.0 490.0 1.0
mu_beta -4.2 0.0340 0.980 -5.70 -4.10 -2.9 820.0 500.0 1.0
tau_beta 1.0 0.0680 1.900 0.06 0.47 3.4 760.0 470.0 1.0
... ... ... ... ... ... ... ... ... ...
m_true[158] 28.0 0.0038 0.150 28.00 28.00 29.0 1508.0 925.0 1.0
m_true[159] 25.0 0.0014 0.068 25.00 25.00 25.0 2472.0 1516.0 1.0
m_true[160] 25.0 0.0014 0.068 25.00 25.00 25.0 2445.0 1499.0 1.0
m_true[161] 27.0 0.0025 0.100 27.00 27.00 27.0 1707.0 1046.0 1.0
m_true[162] 24.0 0.0018 0.074 24.00 24.00 24.0 1625.0 996.0 1.0

333 rows × 9 columns

Hmmm, it seems something is off. Let’s check the trace and pair plots…

[9]:

av.plot_trace(fit, var_names=["mu_alpha", "mu_beta", "alpha", "beta"]);

To make a sensibly-sized pair plot and view divergent transitions, we have to pass some more information to arviz. Let’s focus on the high level parameters now, seeing as these are the new things we have introduced. The Ng and Nt indicies are also specified, in case you want to use arviz to explore some other cases.

[10]:

fit_a = av.from_cmdstanpy(fit, coords={"Ng" : np.arange(input_data["Ng"]),
"Nt" : np.arange(input_data["Nt"])},
dims={"alpha": ["Ng"], "beta": ["Ng"], "M_true":["Nt"]})
av.plot_pair(fit_a, var_names=["mu_alpha", "mu_beta", "tau_alpha", "tau_beta"],
divergences=True, coords={"Ng": np.array([0, 1]), # Ng coords
"Nt": np.array([0])}); # Nt coords

The geometry of a hierarchical model with its many latent parameters is much more complex than that of a single model, and divergent transitions can occur. Fortunately, these can be used to help us understand and fix our model.

Here, we can see that the misspecification of the model causes problems, although these issues become less servere if we add data from more galaxies, as I’ll now demonstrate. It is important to remember that the geometry the sampler sees depends on both the model and the data.

Exercise 2 (5 points): Update the above Stan code with your improvements from the single galaxy case that you carried out in the model checking notebook and display the code. Verify that you can run the above demo to fit all galaxies with no divergent transitions.

Note: To start with, you can experiment with adding the reparametrisation and/or individual sigma_m into the model. If you complete the homework in the model checking notebook, you’ll also figure out some other updates that you could use to get a better fit to the data.

[13]:

# to be completed...

Exercise 3 (10 points): Perform posterior predictive checks for your improved fit, and visualise the posterior predictive distribution of your fit. Can you fit the data well?

[12]:

# to be completed...

Homework exercise 1 (10 points): Compare the galaxy-level alpha and beta distributions to those found from individual fits that you completed as part of the model checking notebook. What differences can you see?

[ ]:

# to be completed...

Homework exercise 2 (20 points): Make a plot to show how the constraints on the parent distributions for alpha and beta change depending on the number of galaxies used in the sample. What happens when we only use one or two galaxies?

[ ]:

# to be completed...

## Estimation of the Hubble constant

So far we have assumed a value for H0 in our fits in order to convert between apparent and absolute magnitude. This is necessary to fit for alpha and beta. Let’s think about how we can extend the model to make an estimate of H0, rather than assuming it.

Exercise 4 (5 points): What happens if we simply leave H0 as a free parameter in our existing hierarchical model? Why is this?

[ ]:

# to be completed...

Exercise 5 (5 points): What happens if we change the value of H0 assumed in our analysis? Which results change, and which stay the same?

[ ]:

# to be completed...

The Cepheid variable experts that we talked to earlier are impressed with our careful analysis of the period-luminosity relation. They share a new dataset of Cepheid variable stars that are in our own Galaxy, the Milky Way. The distances to these stars are well-measured using geometric methods.

This dataset is given in milkyway_cepheids.dat. It similar to that we have seen before, except that each Cepheid has a distance measurement in kpc. We are also told that the uncertainty in the distance measurements is 150 pc, and that on the apparent magnitudes is 0.1, for all Cepheids. Any effect due to metallicity can be assumed to be negligible here.

Let’s take a look at the data:

[ ]:

mw_cepheids = np.loadtxt("data/milkyway_cepheids.dat")
d = mw_cepheids.T[0] # distance in kpc
m_obs = mw_cepheids.T[1] # apparent mag
P = mw_cepheids.T[2] # period in days

Homework exercise 3 (20 points): Use this new data set to make an independent estimate of alpha and beta, without assuming H0, what do you find?

[ ]:

# to be completed...

Homework exercise 4 (20 points): Update the hierarchical model that you have to include key information from the calibration data into your joint fit. If done correctly, this should allow you to break the degeneracy in the model and directly estimate the Hubble constant. Try to include all relevant uncertainties.

Hint: Assume that the mean alpha inferred from the Milky Way dataset is fair estimate of mu_alpha.

Homework exercise 5 (20 points): What factors affect the uncertainties in your estimate? How could you get a more constrained result?

[ ]:


text:
Unlimited PS Actions, graphics, videos & courses! Unlimited asset downloads! From \$16.50/m

by

This tutorial will show you how to layer a couple of grunge and watercolor brushes, then use them with some textures to create a wall with a real vintage look. After that, we'll combine the text and some dripping brushes to add the writing on that wall. Finally I'll give you some tips on adjusting the lighting so that you can end up with an intense, vivid design.

This effect is inspired by the promotion images for the new season of American Horror Story: Freak Show.

## Tutorial Assets

The following assets were used during the production of this tutorial.

## 1. Modifying the Pattern

### Step 1

Open the 35.jpg image from the French Wallpaper Patterns 1 pack, then go to Image > Adjustments > Levels. Click the Sample in image to set white point icon, then click the peachy orange colored part in the image. This will set it along with any lighter points to white.

### Step 2

Go to Image > Adjustments > Hue/Saturation, click the Colorize box, and then change the Hue to 32, the Saturation to 18, and the Lightness to 40.

### Step 3

Go to Edit > Define Pattern, and type in a name for the pattern.

## 2. Patching the Crack Texture

### Step 1

Create a new 1280 x 720 px document, then place the Crack 5 texture on top of the Background layer. Resize it as needed, but for a result similar to the one in this tutorial, make sure to keep the top and bottom parts open at the ends of the document as shown in the screenshot below.

In order to get a neater result, we will need to get rid of the hole extending to the right of the image, which is what we'll be doing next.

### Step 2

Pick the Stamp Tool, and a soft round 25 px brush tip. Press and hold the Option/Alt key, then click at the beginning of the part that you want to use as an edge for the hole to the right.

### Step 3

Release all, then click and drag to use the part you sampled earlier to patch the hole. You can undo and resample a couple of times until you get a result you like. Don't fill the whole part yet, just make sure to create the edge.

### Step 4

Now pick the Patch Tool, make sure that the Patch option in the Options bar is set to Normal, and that the Source button is clicked. Then, click and drag to create a selection around the remaining part of the hole.

### Step 5

Click inside the selection and drag it down until you find a part of the wall you like, then release. This will patch that part pretty nicely.

### Step 6

You can also use the Spot Healing Brush Tool to get rid of any other spots or parts you don't like, by simply painting over them.

### Step 7

When done, go to Filter > Noise > Reduce Noise, and change the Strength to 10, the Preserve Details value to 37, the Reduce Color Noise value to 84, and the Sharpen Details value to 75, and make sure that the Remove JPEG Artifact box is unchecked.

### Step 8

Finally, go to Image > Adjustments > Hue/Saturation, and change the Hue to -64 and the Saturation to -70.

## 3. Masking Out the Crack's Center and Adding the Concrete Texture

### Step 1

Pick the Magnetic Lasso Tool, then click at the beginning of one of the crack's sides to start creating a selection. Drag slowly at the edges, and if you need to force it to add a point, click once where you want the point to be added, then continue dragging.

You can play around with the settings in the Options bar, like increasing the Frequency value if necessary.

### Step 2

When you're done, you can use the Polygonal Lasso Tool, and click the Add to Selection icon in the Options bar, to add any left-out parts at the top and bottom of the image.

When the selection is complete, go to Select > Inverse.

### Step 3

Click the Add layer mask icon at the bottom of the Layers panel. This will add a layer mask that will remove the center of the crack.

### Step 4

Place the ConcreteBare0183 image right below the crack texture layer, and rename its layer to Concrete, then resize it as needed, making sure not to leave any empty areas.

### Step 5

Go to Image > Adjustments > Hue/Saturation, and change the Hue to -12 and the Saturation to -20.

### Step 6

Double-click the crack texture layer to apply a simple Drop Shadow effect using the values below:

• Opacity: 25%
• Distance: 10
• Size: 10

## 4. Enhancing the Crack's Color, Adding the Pattern, and Modifying the Edges

### Step 1

Create a new layer on top of all layers, call it Fill Color, and change its Blend Mode to Color. Then set the Foreground color to #968978, and use the Paint Bucket Tool to fill the layer with that color.

### Step 2

Create another new layer, call it Pattern, and change its Blend Mode to Multiply. Then with the Paint Bucket Tool still active, choose the Pattern fill from the Options bar's dropdown menu, and fill the new layer with the pattern you defined at the beginning of this tutorial.

### Step 3

Command/Control-Click the crack texture's layer mask thumbnail to create a selection.

Select the Fill Color layer, then click the Add layer mask icon to apply the same layer mask to it. Repeat that for the Pattern layer as well.

### Step 4

Place the PlasterWhite0080 texture on top of all layers, rename its layer to Edges, click the Add layer mask icon, and fill the layer mask with Black.

Pick the Eraser Tool, and make sure that the Foreground color is set to Black, and that the layer mask is still selected.

With a small soft round brush tip, start erasing some parts along the edges to show the masked texture back for those parts.

Not all the edges should be erased. A couple of areas around the crack should be just fine. This will make the final result look more realistic and interesting.

## 5. Grunging Up the Wallpaper

### Step 1

Create a new layer on top of all layers, and call it Border 1. Change its Blend Mode to Multiply and its Opacity to 70%, set the Foreground color to #524741, and then use any of the Grunge brushes to darken up the document's edges.

### Step 2

Create another new layer, call it Border 2, change its Blend Mode to Linear Burn, and use a soft round brush to add another darkening layer to the grungy edges.

### Step 3

Create one more layer, and call it Grunge Overlay. Change its Blend Mode to Multiply and its Opacity to 50%, then scatter some of the Grunge brushes randomly over the wallpaper (the part that has the pattern).

### Step 4

Create a new layer and call it Highlight. Change its Blend Mode to Soft Light, set the Foreground color to #acacac, and use a soft round brush tip to highlight some areas of the crack's edges. Then, add the same previous layer mask to the layer just like you did for the Fill Color and Pattern layers before.

### Step 5

You can also create another layer and call it Shadow, change its Blend Mode to Multiply, and use the same Foreground color to darken up any areas that need darkening along the edges. Don't forget to add a layer mask for this layer too.

## 6. Intensifying the Crack Edges' Shadow and Staining the Concrete Center

### Step 1

Create a new layer on top of the Concrete layer, and call it Center Shadow 1. Change its Blend Mode to Multiply and its Opacity to 50%, set the Foreground color to Black, then use a big soft round brush tip to add some shadowing to the edges in order to create more depth.

### Step 2

Create another new layer on top of the Center Shadow 1 layer, and call it Center Shadow 2. Change its Blend Mode to Linear Burn, and set the Foreground color to #342915.

The first part of this step is to pick a round brush tip with a Hardness value of 70%, then create some dark, intense shadow areas below a few parts of the crack's edges. You can use the Eraser Tool to help you shape those shadows to match the shape of the edges that are supposed to be casting them.

The second part is to pick the Eraser Tool, and use the brush 24 from the Watercolours 3 PS 7.0 pack to erase parts of the outer edges of the shadows you created.

### Step 3

Next, start creating new layers between the crack and the Center Shadow 2 layers, and change each of those layers' Blend Modes to Multiply, then choose shades of green and brown colors to add stains to the concrete center using the Watercolours 3 PS 7.0 brushes. Place those layers in a group and call it Center.

### Step 4

The brushes and colors used in this tutorial from bottom to top are:

• Layer - Color - Brush
• 1 - #998675 - 28
• 2 - #a6a497 - 30
• 3 - #a6a497 - 29
• 4 - #998675 - 27
• 5 - #342915 - 29
• 6 - #a6a497 - 09
• 7 - #342915 - 22

### Step 1

Create the text using the font Nightbird, the font Size 99 pt, the color #993133, and set the Tracking value to -25.

### Step 2

Go to Edit > Transform > Rotate, rotate the text slightly, then hit Return/Enter to accept the changes.

### Step 3

Create a new layer on top of the text layer, call it Drips, and set the Foreground color to #993133. Then add some of the Drips 1 Brush Pack for Photoshop or Gimp brushes to a couple of the letters you have. You can combine more than one brush for the same letter as well.

### Step 4

Select both the text and the Drips layers, then go to Layer > Merge Layers, rename the merged layer to Text, and change its Blend Mode to Multiply.

### Step 5

You can use the Watercolours 3 PS 7.0 brushes to erase some parts of the text if you like.

### Step 6

Go to Image > Adjustments > Hue/Saturation, and change the Saturation to 10.

### Step 7

Create a new layer on top of the Center Shadow 2 layer, call it Stain, change its Blend Mode to Linear Burn, set the Foreground color to #b3897c, and add the brush 28 from the Watercolours 3 PS 7.0 pack behind the text. You can modify some of its settings if you like, as shown in the next step.

### Step 8

Open the Brush panel (Window > Brush), change the Size to 400, and check the Flip Y box, then add the brush.

## 8. Lighting and Final Touches

### Step 1

Click the Create new fill or adjustment layer icon at the bottom of the Layers panel and choose Levels. Make sure to place its layer on top of all layers, then change the Highlights value to 245.

### Step 2

Click the Create new fill or adjustment layer icon again, but this time choose Brightness/Contrast, and change the Contrast value to 10.

### Step 3

Double click the Fill Color layer to apply a Gradient Overlay effect using the following values.

• Blend Mode: Overlay
• Opacity: 35%
• Scale: 150%
• Check the Reverse box
• Use the Black to White gradient fill.

This will make the effect look more vivid.

### Step 4

Pick the Dodge Tool, choose a soft round brush tip, and set the Range to Midtones and the Exposure to 50% in the Options bar. Select the crack texture's layer, then start brightening up its edges a little bit.

## Congratulations! You're Done

In this tutorial, we patched, modified, and masked a crack texture to create the base of a pattern-filled wallpaper.

Then, we added the fill color and pattern to the crack texture, as well as a couple of brushes to make it look more vintage and grungy. We used other brushes to add highlights and shadows, creating more depth and dimension.

Then we created the text, added some drips, and merged and blended both layers with the design.

Finally, we enhanced the lighting, and added some finishing touches to bring the design to life and make it look more vivid.


text:
# Cardinality of the continuum

In set theory, the cardinality of the continuum is the cardinality or "size" of the set of real numbers ${\displaystyle \mathbb {R} }$, sometimes called the continuum. It is an infinite cardinal number and is denoted by ${\displaystyle {\mathfrak {c}}}$ (lowercase fraktur "c") or ${\displaystyle |\mathbb {R} |}$.[1]

The real numbers ${\displaystyle \mathbb {R} }$ are more numerous than the natural numbers ${\displaystyle \mathbb {N} }$. Moreover, ${\displaystyle \mathbb {R} }$ has the same number of elements as the power set of ${\displaystyle \mathbb {N} .}$ Symbolically, if the cardinality of ${\displaystyle \mathbb {N} }$ is denoted as ${\displaystyle \aleph _{0}}$, the cardinality of the continuum is

${\displaystyle {\mathfrak {c}}=2^{\aleph _{0}}>\aleph _{0}\,.}$

This was proven by Georg Cantor in his uncountability proof of 1874, part of his groundbreaking study of different infinities. The inequality was later stated more simply in his diagonal argument in 1891. Cantor defined cardinality in terms of bijective functions: two sets have the same cardinality if, and only if, there exists a bijective function between them.

Between any two real numbers a < b, no matter how close they are to each other, there are always infinitely many other real numbers, and Cantor showed that they are as many as those contained in the whole set of real numbers. In other words, the open interval (a,b) is equinumerous with ${\displaystyle \mathbb {R} .}$ This is also true for several other infinite sets, such as any n-dimensional Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ (see space filling curve). That is,

${\displaystyle |(a,b)|=|\mathbb {R} |=|\mathbb {R} ^{n}|.}$

The smallest infinite cardinal number is ${\displaystyle \aleph _{0}}$ (aleph-null). The second smallest is ${\displaystyle \aleph _{1}}$ (aleph-one). The continuum hypothesis, which asserts that there are no sets whose cardinality is strictly between ${\displaystyle \aleph _{0}}$ and ${\displaystyle {\mathfrak {c}}}$, means that ${\displaystyle {\mathfrak {c}}=\aleph _{1}}$.[2] The truth or falsity of this hypothesis is undecidable and cannot be proven within the widely used Zermelo–Fraenkel set theory with axiom of choice (ZFC).

## Properties

### Uncountability

Georg Cantor introduced the concept of cardinality to compare the sizes of infinite sets. He famously showed that the set of real numbers is uncountably infinite. That is, ${\displaystyle {\mathfrak {c}}}$ is strictly greater than the cardinality of the natural numbers, ${\displaystyle \aleph _{0}}$:

${\displaystyle \aleph _{0}<{\mathfrak {c}}.}$

In practice, this means that there are strictly more real numbers than there are integers. Cantor proved this statement in several different ways. For more information on this topic, see Cantor's first uncountability proof and Cantor's diagonal argument.

### Cardinal equalities

A variation of Cantor's diagonal argument can be used to prove Cantor's theorem, which states that the cardinality of any set is strictly less than that of its power set. That is, ${\displaystyle |A|<2^{|A|}}$ (and so that the power set ${\displaystyle \wp (\mathbb {N} )}$ of the natural numbers ${\displaystyle \mathbb {N} }$ is uncountable). In fact, one can show[citation needed] that the cardinality of ${\displaystyle \wp (\mathbb {N} )}$ is equal to ${\displaystyle {\mathfrak {c}}}$ as follows:

1. Define a map ${\displaystyle f:\mathbb {R} \to \wp (\mathbb {Q} )}$ from the reals to the power set of the rationals, ${\displaystyle \mathbb {Q} }$, by sending each real number ${\displaystyle x}$ to the set ${\displaystyle \{q\in \mathbb {Q} :q\leq x\}}$ of all rationals less than or equal to ${\displaystyle x}$ (with the reals viewed as Dedekind cuts, this is nothing other than the inclusion map in the set of sets of rationals). Because the rationals are dense in ${\displaystyle \mathbb {R} }$, this map is injective, and because the rationals are countable, we have that ${\displaystyle {\mathfrak {c}}\leq 2^{\aleph _{0}}}$.
2. Let ${\displaystyle \{0,2\}^{\mathbb {N} }}$ be the set of infinite sequences with values in set ${\displaystyle \{0,2\}}$. This set has cardinality ${\displaystyle 2^{\aleph _{0}}}$ (the natural bijection between the set of binary sequences and ${\displaystyle \wp (\mathbb {N} )}$ is given by the indicator function). Now, associate to each such sequence ${\displaystyle (a_{i})_{i\in \mathbb {N} }}$ the unique real number in the interval ${\displaystyle [0,1]}$ with the ternary-expansion given by the digits ${\displaystyle a_{1},a_{2},\dotsc }$, i.e., ${\displaystyle \sum _{i=1}^{\infty }a_{i}3^{-i}}$, the ${\displaystyle i}$-th digit after the fractional point is ${\displaystyle a_{i}}$ with respect to base ${\displaystyle 3}$. The image of this map is called the Cantor set. It is not hard to see that this map is injective, for by avoiding points with the digit 1 in their ternary expansion, we avoid conflicts created by the fact that the ternary-expansion of a real number is not unique. We then have that ${\displaystyle 2^{\aleph _{0}}\leq {\mathfrak {c}}}$.

By the Cantor–Bernstein–Schroeder theorem we conclude that

${\displaystyle {\mathfrak {c}}=|\wp (\mathbb {N} )|=2^{\aleph _{0}}.}$

The cardinal equality ${\displaystyle {\mathfrak {c}}^{2}={\mathfrak {c}}}$ can be demonstrated using cardinal arithmetic:

${\displaystyle {\mathfrak {c}}^{2}=(2^{\aleph _{0}})^{2}=2^{2\times {\aleph _{0}}}=2^{\aleph _{0}}={\mathfrak {c}}.}$

By using the rules of cardinal arithmetic, one can also show that

${\displaystyle {\mathfrak {c}}^{\aleph _{0}}={\aleph _{0}}^{\aleph _{0}}=n^{\aleph _{0}}={\mathfrak {c}}^{n}=\aleph _{0}{\mathfrak {c}}=n{\mathfrak {c}}={\mathfrak {c}},}$

where n is any finite cardinal ≥ 2, and

${\displaystyle {\mathfrak {c}}^{\mathfrak {c}}=(2^{\aleph _{0}})^{\mathfrak {c}}=2^{{\mathfrak {c}}\times \aleph _{0}}=2^{\mathfrak {c}},}$

where ${\displaystyle 2^{\mathfrak {c}}}$ is the cardinality of the power set of R, and ${\displaystyle 2^{\mathfrak {c}}>{\mathfrak {c}}}$.

### Alternative explanation for ${\displaystyle {\mathfrak {c}}=2^{\aleph _{0}}}$

Every real number has at least one infinite decimal expansion. For example,

1/2 = 0.50000...
1/3 = 0.33333...
π = 3.14159....

(This is true even in the case the expansion repeats, as in the first two examples.)

In any given case, the number of digits is countable since they can be put into a one-to-one correspondence with the set of natural numbers ${\displaystyle \mathbb {N} }$. This makes it sensible to talk about, say, the first, the one-hundredth, or the millionth digit of π. Since the natural numbers have cardinality ${\displaystyle \aleph _{0},}$ each real number has ${\displaystyle \aleph _{0}}$ digits in its expansion.

Since each real number can be broken into an integer part and a decimal fraction, we get:

${\displaystyle {\mathfrak {c}}\leq \aleph _{0}\cdot 10^{\aleph _{0}}\leq 2^{\aleph _{0}}\cdot {(2^{4})}^{\aleph _{0}}=2^{\aleph _{0}+4\cdot \aleph _{0}}=2^{\aleph _{0}}}$

where we used the fact that

${\displaystyle \aleph _{0}+4\cdot \aleph _{0}=\aleph _{0}\,.}$

On the other hand, if we map ${\displaystyle 2=\{0,1\}}$ to ${\displaystyle \{3,7\}}$ and consider that decimal fractions containing only 3 or 7 are only a part of the real numbers, then we get

${\displaystyle 2^{\aleph _{0}}\leq {\mathfrak {c}}\,.}$

and thus

${\displaystyle {\mathfrak {c}}=2^{\aleph _{0}}\,.}$

## Beth numbers

The sequence of beth numbers is defined by setting ${\displaystyle \beth _{0}=\aleph _{0}}$ and ${\displaystyle \beth _{k+1}=2^{\beth _{k}}}$. So ${\displaystyle {\mathfrak {c}}}$ is the second beth number, beth-one:

${\displaystyle {\mathfrak {c}}=\beth _{1}.}$

The third beth number, beth-two, is the cardinality of the power set of ${\displaystyle \mathbb {R} }$ (i.e. the set of all subsets of the real line):

${\displaystyle 2^{\mathfrak {c}}=\beth _{2}.}$

## The continuum hypothesis

The famous continuum hypothesis asserts that ${\displaystyle {\mathfrak {c}}}$ is also the second aleph number, ${\displaystyle \aleph _{1}}$.[2] In other words, the continuum hypothesis states that there is no set ${\displaystyle A}$ whose cardinality lies strictly between ${\displaystyle \aleph _{0}}$ and ${\displaystyle {\mathfrak {c}}}$

${\displaystyle \nexists A\quad :\quad \aleph _{0}<|A|<{\mathfrak {c}}.}$

This statement is now known to be independent of the axioms of Zermelo–Fraenkel set theory with the axiom of choice (ZFC), as shown by Kurt Gödel and Paul Cohen.[3][4][5] That is, both the hypothesis and its negation are consistent with these axioms. In fact, for every nonzero natural number n, the equality ${\displaystyle {\mathfrak {c}}}$ = ${\displaystyle \aleph _{n}}$ is independent of ZFC (case ${\displaystyle n=1}$ being the continuum hypothesis). The same is true for most other alephs, although in some cases, equality can be ruled out by König's theorem on the grounds of cofinality (e.g., ${\displaystyle {\mathfrak {c}}\neq \aleph _{\omega }}$). In particular, ${\displaystyle {\mathfrak {c}}}$ could be either ${\displaystyle \aleph _{1}}$ or ${\displaystyle \aleph _{\omega _{1}}}$, where ${\displaystyle \omega _{1}}$ is the first uncountable ordinal, so it could be either a successor cardinal or a limit cardinal, and either a regular cardinal or a singular cardinal.

## Sets with cardinality of the continuum

A great many sets studied in mathematics have cardinality equal to ${\displaystyle {\mathfrak {c}}}$. Some common examples are the following:

• the real numbers ${\displaystyle \mathbb {R} }$
• any (nondegenerate) closed or open interval in ${\displaystyle \mathbb {R} }$ (such as the unit interval ${\displaystyle [0,1]}$)
• the irrational numbers
• the transcendental numbers We note that the set of real algebraic numbers is countably infinite (assign to each formula its Gödel number.) So the cardinality of the real algebraic numbers is ${\displaystyle \aleph _{0}}$. Furthermore, the real algebraic numbers and the real transcendental numbers are disjoint sets whose union is ${\displaystyle \mathbb {R} }$. Thus, since the cardinality of ${\displaystyle \mathbb {R} }$ is ${\displaystyle {\mathfrak {c}}}$, the cardinality of the real transcendental numbers is ${\displaystyle {\mathfrak {c}}-\aleph _{0}={\mathfrak {c}}}$. A similar result follows for complex transcendental numbers, once we have proved that ${\displaystyle \left\vert \mathbb {C} \right\vert ={\mathfrak {c}}}$.
• the Cantor set
• Euclidean space ${\displaystyle \mathbb {R} ^{n}}$[6]
• the complex numbers ${\displaystyle \mathbb {C} }$ We note that, per Cantor's proof of the cardinality of Euclidean space,[6] ${\displaystyle \left\vert \mathbb {R} ^{2}\right\vert ={\mathfrak {c}}}$. By definition, any ${\displaystyle c\in \mathbb {C} }$ can be uniquely expressed as ${\displaystyle a+bi}$ for some ${\displaystyle a,b\in \mathbb {R} }$. We therefore define the bijection
{\displaystyle {\begin{aligned}f\colon \mathbb {R} ^{2}&\to \mathbb {C} \\(a,b)&\mapsto a+bi\end{aligned}}}
• the power set of the natural numbers ${\displaystyle {\mathcal {P}}(\mathbb {N} )}$ (the set of all subsets of the natural numbers)
• the set of sequences of integers (i.e. all functions ${\displaystyle \mathbb {N} \rightarrow \mathbb {Z} }$, often denoted ${\displaystyle \mathbb {Z} ^{\mathbb {N} }}$)
• the set of sequences of real numbers, ${\displaystyle \mathbb {R} ^{\mathbb {N} }}$
• the set of all continuous functions from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} }$
• the Euclidean topology on ${\displaystyle \mathbb {R} ^{n}}$ (i.e. the set of all open sets in ${\displaystyle \mathbb {R} ^{n}}$)
• the Borel σ-algebra on ${\displaystyle \mathbb {R} }$ (i.e. the set of all Borel sets in ${\displaystyle \mathbb {R} }$).

## Sets with greater cardinality

Sets with cardinality greater than ${\displaystyle {\mathfrak {c}}}$ include:

• the set of all subsets of ${\displaystyle \mathbb {R} }$ (i.e., power set ${\displaystyle {\mathcal {P}}(\mathbb {R} )}$)
• the set 2R of indicator functions defined on subsets of the reals (the set ${\displaystyle 2^{\mathbb {R} }}$ is isomorphic to ${\displaystyle {\mathcal {P}}(\mathbb {R} )}$ – the indicator function chooses elements of each subset to include)
• the set ${\displaystyle \mathbb {R} ^{\mathbb {R} }}$ of all functions from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} }$
• the Lebesgue σ-algebra of ${\displaystyle \mathbb {R} }$, i.e., the set of all Lebesgue measurable sets in ${\displaystyle \mathbb {R} }$.
• the set of all Lebesgue-integrable functions from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} }$
• the set of all Lebesgue-measurable functions from ${\displaystyle \mathbb {R} }$ to ${\displaystyle \mathbb {R} }$
• the Stone–Čech compactifications of ${\displaystyle \mathbb {N} }$, ${\displaystyle \mathbb {Q} }$ and ${\displaystyle \mathbb {R} }$
• the set of all automorphisms of the (discrete) field of complex numbers.

These all have cardinality ${\displaystyle 2^{\mathfrak {c}}=\beth _{2}}$ (beth two).

## References

1. ^ "Transfinite number | mathematics". Encyclopedia Britannica. Retrieved 2020-08-12.
2. ^ a b Weisstein, Eric W. "Continuum". mathworld.wolfram.com. Retrieved 2020-08-12.
3. ^ Gödel, Kurt (1940-12-31). "Consistency of the Continuum Hypothesis. (AM-3)". doi:10.1515/9781400881635. {{cite journal}}: Cite journal requires |journal= (help)
4. ^ Cohen, Paul J. (December 1963). "THE INDEPENDENCE OF THE CONTINUUM HYPOTHESIS". Proceedings of the National Academy of Sciences. 50 (6): 1143–1148. doi:10.1073/pnas.50.6.1143. ISSN 0027-8424.
5. ^ Cohen, Paul J. (January 1964). "THE INDEPENDENCE OF THE CONTINUUM HYPOTHESIS, II". Proceedings of the National Academy of Sciences. 51 (1): 105–110. doi:10.1073/pnas.51.1.105. ISSN 0027-8424.
6. ^ a b


text:
How to compute convolution fast for real-time applications?

## Introduction

Sound engineering and Virtual Reality audio demand real-time performance. Convolutions are inherently embedded in their inner workings, e.g., in the form of finite impulse response (FIR) filtering for artificial reverberation. Convolution evaluated according to the definition is too slow to keep up and introduces audible latency. To mitigate this we reach out to the fast convolution methods.

But speed itself is not enough. In the above-mentioned applications we need to process sound in a block-wise fashion. Methods allowing this are called partitioned convolution techniques.

In this article, we first show why the naive approach to the convolution is inefficient, then show the FFT-based fast convolution. What follows is a description of two of the most popular block-based convolution methods: overlap-add and overlap-save. Finally, we show how to deal with very long filters (over 1 second long) when using partitioned convolution.

Note: An amazing source about fast convolution techniques is [1]. I highly encourage you to check it out especially if you would like to read more on the topic.

## Notation

In software, the signals are always of finite length. Thus, we will denote a signal $x[n]$ of length $N_x$ by a fixed-size vector $\pmb{x} = [x[0], x[1], \dots, x[N_x-1]]$.

## Naive implementation of convolution

We could implement the convolution between discrete-time signals $x$ and $h$ by implementing the definition directly [1]

$y[n] = x[n] \ast h[n] = \sum_{k=\max\{0,n-N_h+1\}}^{\min\{n, N_x-1\}} x[k] h[n - k], \\ \quad n \in \{0, \dots, N_x + N_h - 1\}. \quad (1)$

This translates to the following code

Listing 1. Naive linear convolution.

## FFT-based implementation

As we know from the article on circular convolution, multiplication in the discrete-frequency domain is equivalent to circular convolution in the discrete-time domain. Element-wise multiplication of 2 vectors has time complexity $O(N)$. This is superior to $O(N^2)$ in case of the naive approach. The bottleneck of frequency-based convolution is the transformation to that domain, but it can be achieved in $O(N \log N)$ time, which is still better than $O(N^2)$. And so arises the FFT-based fast convolution (Figure 1).

Figure 1. FFT-based fast convolution. Source: [1].

In order to make circular convolution correspond to linear convolution we need to pad the input signals with sufficiently many zeros so that the frequency-domain representation has enough capacity to represent the result of the linear convolution. This means, that each signal should be extended to have length $K = Nx + Nh - 1$.

We can obtain it using function pad_zeros_to().

Listing 2.

If our signals are sufficiently long we can compute their discrete Fourier transforms (DFTs) using the Fast Fourier Transform (FFT) algorithm. Thanks to the FFT, the transformation from the time domain to the frequency domain can be computed in $O(N \log N)$ time.

In the frequency domain we can multiply our signals element-wise and then come back to the time domain via inverse FFT.

FFT is known to be most computationally efficient if the transformed signal’s length is a power of 2. We can compute the exponent of the next largest power of 2 with respect to an integer $n > 0$ using the formula

$q_{\text{next}} = \log2(n - 1) + 1. \quad (2)$

The above equation allows us to implement another helper function

Listing 3.

Wrapping it all together

Listing 4. FFT-based fast convolution.

Remember that the above algorithm is fast algorithmically. I am not claiming this code is maximally optimized 😉. It is provided for understanding and as a possible baseline implementation.

## Block-based convolution

FFT-based fast convolution is sufficiently fast for offline computation, but musical and virtual reality applications require real-time performance. In these applications, sound is usually processed in blocks of length 64, 128, 256, or more samples. If the signals to be convolved are longer than the block length, we need to adapt our methods so that we can convolve and output only partial results and handle incoming input.

In this section, we will assume that the input signal is an infinite stream (e.g., a looped source replaying a sound file) that comes in blocks $\pmb{x}$ of $B$ samples, and $\pmb{h}$ is a FIR filter of length $N$. Such a set-up is called unified input partitioning because the input comes in blocks and the blocks are of equal size.

The first idea to process the input in blocks is to convolve each incoming block with the full filter using FFT-based convolution, store the results of appropriately many past convolutions, and output sums of their subsequent parts in blocks of the same length as the input signal we process. This scheme can be seen in Figure 2.

Figure 2. Overlap-Add convolution scheme. Source: [1].

Some important remarks concerning this methodology:

• $K$ (transform length) must be sufficiently large to ensure that the circular convolution is equivalent to the linear convolution (no samples are time-aliased). Lengthening the convolved signals is achieved through zero-padding.

• Overlap-Add operation to form the output block is much more difficult than it may seem from the scheme. One needs to store all the convolution results and then sum appropriate indices. This may incur high memory and time cost.

### Overlap-Save Scheme

A significant drawback of the Overlap-Add scheme is the necessity to store and sum the computed partial convolutions. Can we do better?

Another approach is called Overlap-Save. It is based on storing appropriately many input blocks rather than output blocks, which are shorter than the calculated convolutions. The input blocks are stored in a buffer of length $K$ in a First-In-First-Out manner, i.e., each new block of input samples shifts all previously stored samples in the buffer, discarding the oldest $B$ samples. We then perform a $K$-point FFT-based convolution with the zero-padded filter coefficients.

Inherently, the result of this operation is a circular convolution (because the input signal is not zero-padded). It means that $K$ has to be sufficiently large so as to have the last $B$ samples correspond to the linear convolution. We can then output these $B$ valid samples and discard the rest. This results in a potentially longer transform length than in the Overlap-Add scheme, but removes the need to store and sum the outputs.

The entire scheme can be seen in Figure 3.

Figure 3. Overlap-Save convolution scheme. Source: [1].

#### Implementation

Listing 6. Overlap-Save convolution.

## What If the Filter Is Long Too?

If the filter we want to convolve the input with has also significant length (e.g., when it represents an impulse response of a large hall with a significant reverberation time), we may have to partition both, the input and the filter. These methods are called partitioned convolution.

Describing partitioned convolution algorithms is beyond the scope of this article, but the main idea is presented in Figure 4. It shows a non-uniformly partitioned convolution scheme, which is generally faster than uniformly partitioned versions [1]. It also involves the use of frequency-domain delay lines (FDLs), which allow computing the FFT of the input blocks just once (in contrast to time-delaying the input blocks and calculating their FFTs for each new output block).

Non-uniformly partitioned convolution is the current state of the art in artificial reverberation using FIR filters for real-time auralization in games and Virtual Reality.

Figure 4. Non-uniformly partitioned convolution scheme. Note the presence of frequency-domain delay lines. Source: [1].


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Planck's law describes the spectral radiance of electromagnetic radiation at all wavelengths from a black body at temperature T. As a function of frequency ?, Planck's law is written as
$I(\nu,T) = \frac{2 h\nu^3 }{c^2} \frac{1}{e^{h\nu/kT}-1}$ or  [img]http://upload.wikimedia.org/math/9/7/4/974887ba0f7030f31b7b27b619afde87.png[/img]

It can be converted to an expression for I'(?,T) in wavelength units by substituting ? by c / ? and evaluating

$I'(\lambda,T) = I(\nu,T)\left|\frac{d\nu}{d\lambda}\right|$ or [img]http://upload.wikimedia.org/math/2/7/a/27aa3105194d765b91f845f50968e7cf.png[/img]

$\frac{2 h\nu^3 }{c^2}=\frac{2 h(c\lambda)^3 }{c^2}=\frac{2hc}{\lambda^3}$
and from $\nu= c\lambda$, we have
$d\nu=-\frac{c}{\lambda^2}d\lambda$
so
$I(\lambda,T)=\frac{2hc^2}{\lambda^5}\frac{1}{e^{hc/\lambda kT}-1}$ or [img]http://upload.wikimedia.org/math/1/0/c/10c96a8f2af7d9631a0f4dde5b817b62.png[/img]
The above equation is energy per unit wavelength per unit solid angle.

This applets will show six black cureves of blackbody radiation curve betwen Tmin and Tmax.
Another curve in red is also shown (it's temperature can be adjusted with left slider bar)
Maximum wavelength shown can be adjusted with right slider bar.
You can use it for study the intensity for blackbody radiation.
If you want to study different temperature range, You can change Tmin and Tmax, to change the temperature range,too.
The wavelength unit in the simulation is Å (angstrom).
[hide][/hide]


text:
Uncertain Attitudes

Posted in Bad Statistics, Politics with tags , , , , on May 28, 2014 by telescoper

It’s been a while since I posted anything in the bad statistics file, but an article in today’s Grauniad has now given me an opportunity to rectify that omission.
The piece concerned, entitled Racism on the rise in Britain is based on some new data from the British Social Attitudes survey; the full report can be found here (PDF). The main result is shown in this graph:

The version of this plot shown in the Guardian piece has the smoothed long-term trend (the blue curve, based on a five-year moving average of the data and clearly generally downward since 1986) removed.

In any case the report, as is sadly almost always the case in surveys of this kind, neglects any mention of the statistical uncertainty in the survey. In fact the last point is based on a sample of 2149 respondents. Suppose the fraction of the population describing themselves as having some prejudice is $p$. For a sample of size $n$ with $x$ respondents indicating that they describe themselves as “very prejudiced or a little prejudiced” then one can straightforwardly estimate $p \simeq x/n$. So far so good, as long as there is no bias induced by the form of the question asked nor in the selection of the sample…

However, a little bit of mathematics involving the binomial distribution yields an answer for the uncertainty in this estimate of p in terms of the sampling error:

$\sigma = \sqrt{\frac{p(1-p)}{n}}$

For the sample size given, and a value $p \simeq 0.35$ this amounts to a standard error of about 1%. About 95% of samples drawn from a population in which the true fraction is $p$ will yield an estimate within $p \pm 2\sigma$, i.e. within about 2% of the true figure. This is consistent with the “noise” on the unsmoothed curve and it shows that the year-on-year variation shown in the unsmoothed graph is largely attributable to sampling uncertainty; note that the sample sizes vary from year to year too. The results for 2012 and 2013 are 26% and 30% exactly, which differ by 4% and are therefore explicable solely in terms of sampling fluctuations.

I don’t know whether racial prejudice is on the rise in the UK or not, nor even how accurately such attitudes are measured by such surveys in the first place, but there’s no evidence in these data of any significant change over the past year. Given the behaviour of the smoothed data however, there is evidence that in the very long term the fraction of population identifying themselves as prejudiced is actually falling.

Newspapers however rarely let proper statistics get in the way of a good story, even to the extent of removing evidence that contradicts their own prejudice.


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# Monty Corporation sells computers under a 2-year warranty contract that requires the corporation to replace defective...

###### Question:

Monty Corporation sells computers under a 2-year warranty contract that requires the corporation to replace defective parts and to provide the necessary repair labor. During 2020, the corporation sells for cash 427 computers at a unit price of $2,730. On the basis of past experience, the 2-year warranty costs are estimated to be$166 for parts and \$196 for labor per unit. (For simplicity, assume that all sales occurred on December 31, 2020.) The warranty is not sold separately from the computer. (a) - Your answer is partially correct. Record any necessary journal entries in 2020. (If no entry is required, select "No Entry" for the account titles and enter O for the amounts. Credit account titles are automatically indented when the amount is entered. Do not indent manually.) Account Titles and Explanation Debit Credit Cash 1165710 Sales Revenue 1165710 Warranty Expense 154574 Warranty Liability 154574

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# Coat-hanger Bubbles Experiment

2,129 views

As advised by Mr. Fullerton, I did the Coat-hanger bubbles experiment to further understand flux!

Pre-experiment preparation:

First, in my closet I found a nice metal coat-hanger suitable for the trial. After attempting to reshape the coat-hanger, I learned that my hangers are very strong, or that I lack strength; so, I went to my brother's toolbox and grabbed pliers to help bend the wire into a slinky-like shape. My coil ended up having four turns. I then ventured into my kitchen to fill the sink with soapy water. With the bubbly solution complete, I was ready to start the experiment.

The experiment:

I dipped my wire coil into the water, and slowly pulled it out. I found that the bubbles didn't form well to the structure. So, I compressed the coil by pushing the turns closer together. When I tried again with the compressed coil, the bubbles formed nicely between each turn and along the outside of the coil. The formation of the bubbles between each turn demonstrated how the number of turns matter when calculating flux. Therefore, the more turns, the greater the flux. Hence, the equation for magnetic flux is:

$\phi_m= NBA\cos \theta$

N=number of turns

A=area within one loop

B=magnetic field

$\theta$=angle between magnetic field and positive normal direction

Everyone should try this experiment before the test on Wednesday!

• 2

Glad to hear you were able to get that coathanger bent and see the continuous shape that the solenoid makes with the soap bubbles!

Yay coat hanger! I hope you don't mind, I posted on this topic too but cited your blogpost in it. Nice work here

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# ASCII Puzzler 2: Arrows Depicting Movement, Things, or Pointing Out Objects

This is a sequel to ASCII Puzzler 1: Part 1 and a prequel to ASCII Puzzler 3: The Trees and https://puzzling.stackexchange.com/questions/32439/ascii-puzzler-4-not-a-rebus.

Here is the second rebus I have for you to figure out:

 |
V
---        O <---
B --|--> -|-
---        /\
^
|

Here is another rebus I made. The main theme of this is arrows.

Hint 1: (if you reword it, this is a major hint)

After that the B was flat.

Hint 2:

A hint to Hint 1: strike out 1, 2, 3, 5.

• Why are there downvotes? – haykam May 8 '16 at 23:30
• what are we supposed to do with this? – Jasen May 9 '16 at 6:07
• "See sharp or be flat" :) – KeyboardWielder May 9 '16 at 20:58
• @KeyboardWielder nope! – haykam May 9 '16 at 21:09
• @KeyboardWielder Drop your piano down the mineshaft to get the answer a tone flat. – boboquack Nov 24 '16 at 3:02

I hope this is wrong, but the answer could conceivably just be

B flat minor

on the grounds that

the person depicted is a miner (perhaps that nice round head is actually a miner's helmet) and he's got a flattened B attached to him. (And, er, "miner" and "minor" sound alike and "B flat minor" is a musical key.)

It seems unsatisfactory because

there is nothing in the puzzle that actually indicates that the person is a miner (or, I guess, a minor -- i.e., someone not yet legally adult), and all those arrows (featured so prominently in the title) end up not really signifying anything much

so I hope there's a better solution.

• This is one of the most disappointing correct answers I've seen on the PSE... (disappointing to be correct). -1 for question and +1 for answer. – oleslaw Nov 23 '16 at 10:56
• For the latecomers who wonders why +500: This answer is so, so exemplary, that it obviously reserves a +500 (but not from me, from Lukas Rotter's rep bank) – user31076 Nov 23 '16 at 15:00
• (The actual reasons are indicated in the open-bounty notice. I think that goes away once the bounty is actually awarded, so here's the story. An anonymous user set up a sockpuppet called "Mushroom Bang Boom", there was some speculation about who it was, and Lukas Rotter offered 500 rep for the first correct guess made in a particular form. Lots of people had the same suspicion but I was the only one who stated it in the requested form. Clearly that was an act of genius richly deserving of a 500-point bounty.) – Gareth McCaughan Nov 23 '16 at 15:36

My guess is that the rebus is referring to

Acupuncture

via the phrase

"A sharp stick to the body goes right to the head"

Reasoning

B flat is equivalent to 'A sharp'. The vertical bar, |, is also called a stick. The arrow from the stick points to the body of the stick figure and the arrow 'going right' on the right points to the head. This may be referring to the apparent placebo effect associated with acupuncture.

Other possibilities

'The injection of drugs' or 'voodoo'

• Can you take the "other possibilities" and make them into other answers? – haykam May 11 '16 at 22:18
• It's clearly going left to the head :) – paste May 12 '16 at 2:39
• @paste Haha, what a dummy I am! You'd think being able to tell left from right would be a fundamental requirement for trying to solve a visual puzzle. – hexomino May 12 '16 at 10:21
• @Peanut Hmmm, does that mean I'm on the right track? Essentially, all the possibilities are derived from the phrase in the middle. I'd hate to clog up your question's page with answers only to delete all the wrong ones later on. Although, if you're happy for me to do this... – hexomino May 12 '16 at 10:24
• @hexomino I was just joking around; I hope I didn't offend (?) – paste May 12 '16 at 14:13

Perhaps

Be with your crush, and hold hands and kiss??

Because

B = Be
The B is getting crushed, and we also have the hint "After that the B was flat."

The rest is a bit of a stretch, but the arrows are pointing to the arms (/hand) and head (/mouth). And usually holding hands, kissing (and more ( ͡° ͜ʖ ͡°) ) is what you want to do with your crush.

• Nice try, but that is not the correct answer. – haykam May 9 '16 at 10:42