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# Mean value theorem!

oks...im having problems applying the mean value theorem...i understand the concept behind it but whenever i try 2 do a question i have no idea wat 2 do...heres 1 question i've been trying:

Showing all your work, apply the Mean Value Theorem to show
that the function arctan x - x is equal to zero only at x = 0 .

where do i start??? thx.

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HallsofIvy
Homework Helper
The mean value theorem says that if f is differentiable on (a,b) then for some c between a and b, f'(c)= (f(b)- f(a))/(b-a).

Certainly tan(x) is differentiable for x between -pi/2 and pi/2 so also is tan(x)- x.

Suppose tan(b)- b= 0 for some b (between -pi/2 and pi/2). Apply the mean value theorem between 0 and b.
There must be a point c between 0 and b so that f'(c)= (f(b)- f(a))/(b-a) = 0. But for f(x)= tan(x)- x, f'(x)= sec2(x)- 1.
f'(x)= 0 means sec2(x)= 1 or sec(x)= +/- 1. That only happens at 0 and pi/2 NOT between 0 and 1.

NateTG
Homework Helper
Huh. That's not the mean value theorem that I'm familiar with. (In fact, it's not true unless the derivative is contiuous.)

The version I'm familiar with is more like:

If $$f$$ is continuous on $$[a,b]$$ then there is some point $$c; a \leq c \leq b$$ so that $$f(c)=\frac{f(a)+f(b)}{2}$$

Now, you know that
$$f(x)=\arctan{x}-x$$
is continuous.
If you can show that it is always less than zero for $$x>0$$ and alwauys greather than zero for $$x<0$$, and that there must be some point where it is zero (mean value theorem might be usefull here), then you're done.

Hurkyl
Staff Emeritus
Gold Member
I think you're thinking of the intermediate value theorem:

If $f(x)$ is continuous on the interval $[a, b]$, then for any $x$ between $f(a)$ and $f(b)$, there exists a $c \in (a, b)$ such that $f(c) = x$.

Last edited:
NateTG


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# Rate Equations¶

Recall, a rate equation looks like this: $$\frac{dy}{dt} = f(y,t).$$

• $y$ is the dependent variable and is a function of $t$, that is $y(t)$.
• $t$ is the independent variable.
• $f(y,t)$ is some function of $y$ and $t$; that is, some formula written in terms of $y$ and possibly $t$.
• We call this function the rate or the right hand side function (RHS).
• There is also an initial condition $$y(0)=y_0.$$

$$\frac{dC}{dt} = -\frac{C}{\tau}.$$

The exact solution is $$C(t) = C_0e^{-t/\tau},$$ where $C_0$ is the initial concentration.

### Pretest: odeint¶

• New function odeint
• from scipy.integrate import odeint
• Questions:
• What arguments will this need?
• What will it return?
• Will you need to define a function and what form will it have?
• What does the solution look like? (Like, what do you get as an answer?)
• What were the things we needed/did in Excel?
• Try to solve the above ODE with $\tau=1$, $C_0=1$ and $t_{end}=5$.

#### Numerical Solution:¶

$$\frac{dC}{dt} = -\frac{C}{\tau}.$$

1. define the function f(y,t)
2. set the desired solution times t
3. set the initial condition C0
4. call odeint as C=odeint(f, C0, times)
• from scipy.integrate import odeint
5. plot the results

Solve this problem with $\tau=1$, $C_0=1$, and t_end=5.

In [2]:
import numpy             as np
import matplotlib.pyplot as plt
%matplotlib inline
from scipy.integrate     import odeint

In [3]:
#------ Step 1, define the function

def rhsf(C, t, τ):
return -C/τ

#------ Step 2, set times to get the solution at

t = np.linspace(0, 5, 1000)

#------ Step 3, set initial condition

C0 = 1

#------ Step 4, solve the problem

τ=1
C = odeint(rhsf, C0, t, args=(τ,))

#------ Step 5 plot the answer

plt.rc('font', size=14)
plt.plot(t,C)
plt.xlabel('t')
plt.ylabel('C(t)');

#------ Step 6: does the answer make sense? Analyse it.

### Exercise¶

Solve the rate equation for $v(t)$ at discrete points $t$ and plot the result:

$$\frac{dv}{dt} = g - cv^2.$$
• $g=9.81$, $c=1$.
• $v_0 = 0$.
• Solve to $t=2$.
In [5]:
def rhsf(v, t):
g = 9.81
c = 1
dvdt = g - c*v**2
return dvdt

v0 = 0.0
t = np.linspace(0,2,200)
v = odeint(rhsf, v0, t)

plt.plot(t,v)
plt.xlabel('t')
plt.ylabel('v(t)')

Out[5]:
Text(0, 0.5, 'v(t)')

## Multiple equations¶

What is we have multiple rate equations in multiple variables?

If we have multiple rate equations in multiple unknowns, we solve the problem in the same way, but using vector notation.

• This is similar to what we do for fsolve.

### Example¶

Solve for the velocity of a falling object: \begin{align} \frac{dv}{dt} &= g - cv^2, \\ \frac{dx}{dt} &= v. \end{align}

Here, g is gravitational acceleration, x is position, v is velocity, and c is a drag coefficient.

We first cast this in the form $$\frac{d\vec{y}}{dt} = \vec{f}(\vec{y},t).$$

This is equivalent to \begin{align} \frac{dy_0}{dt} &= f_0(y_0,y_1,t) \\ \frac{dy_1}{dt} &= f_1(y_0,y_1,t) \end{align}

For our example problem $v\equiv y_0$, $x\equiv y_1$, $g-cv^2=f_0(v,x,t)$, and $v=f_1(v,x,t)$

• Note, similiar to fsolve:
• Pass in an array of variables,
• Recover variables from the array,
• Compute each rate in terms of the variables,
• Return an array of the rates.

#### Numerical solution¶

\begin{align} \frac{dv}{dt} &= g - cv^2, \\ \frac{dx}{dt} &= v. \end{align}
• c = 1 m$^{-1}$
• g = 9.81 (m/s$^2$)
• t_end = 2 (s)
• $v_0 = 0$ (m/s)
• $x_0 = 0$ (m)
In [7]:
#------ Step 1, define the function

def rhsf(vx, t):
v = vx[0]
x = vx[1]

g = 9.81
c = 1

dvdt = g - c*v*v
dxdt = v

return np.array([dvdt, dxdt])

#------ Step 2, set times to get the solution at

t = np.linspace(0,2,1001)

#------ Step 3, set initial condition

vx0 = np.array([0, 0])

#------ Step 4, solve the problem, recover v, x arrays

vx = odeint(rhsf, vx0, t)

#------ Step 5 plot the answer

plt.plot(t, vx[:,0], label='v')
plt.plot(t, vx[:,1], label='x')
plt.xlabel('t')
plt.ylabel('v(t), x(t)')
plt.legend(frameon=False);

#------ Step 6: does the answer make sense? Analyse it.

### Exercise¶

Solve the following rate equation for a coal particle in a furnace. $$\frac{dT}{dt} = \left(\frac{hA}{m c_p}(T_f - T) + \frac{\sigma A}{m c_p}(T_f^4-T^4)\right).$$

The initial particle temperature is $T_0=500$ K.

Use D=100 $\mu$m, tend = 0.05 s.

The following data is given:

Variable Value Units
$\rho_p$ 1000 kg/m$^3$
$c_p$ 1380 J/kg$\cdot$K
$k$ 0.1 W/m$\cdot$K
$Nu$ 2 --
$\sigma$ 5.67E-8 W/m$^2$K$^4$
$T_f$ 1500 K

Also, the area, mass and Nusselt number $Nu$ are given, respectively, by: $$A = \pi D^2,$$ $$m = \frac{\pi}{6}D^3\rho_p,$$ $$Nu = \frac{hD}{k}.$$

In [ ]:

### Exercise¶

We are performing a chemical reaction as follows. $$\mbox{Rxn 1: }\phantom{xxx} A+B\rightarrow C$$ $$\mbox{Rxn 2: }\phantom{xxx} B+C\rightarrow D$$ Here, symbols $A$, $B$, $C$, $D$ denote species concentrations in mol/L.

The initial concentrations are $A_0=1$, $B_0=1$, $C_0=0$, $D_0=0$. Also, $k_1=1\,L/mol\cdot s$, and $k_2=1.5\,L/mol\cdot s$.

The concentrations obey the following rate equations:

\begin{align} \frac{dA}{dt} &= -k_1AB, \\ \frac{dB}{dt} &= -k_1AB - k_2BC, \\ \frac{dC}{dt} &= k_1AB - k_2BC, \\ \frac{dD}{dt} &= k_2BC. \end{align}

Solve for the concentrations of $A$, $B$, $C$, and $D$ as functions of time to $t=3$ s.

Plot the concentrations of A, B, C, and D as functions of time on the same plot. Label the axes as "time (s)" and “concentration (mol/L)”.

In [ ]:

format: plain


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## Result of Only Adding One of the Reactants

Moderators: Chem_Mod, Chem_Admin

Joon Chang 2F
Posts: 59
Joined: Fri Sep 28, 2018 12:25 am

### Result of Only Adding One of the Reactants

Can someone explain what will happen if only one of the reactants is added to a chemical equation?
For example, in the equation 4NH3(g) + 5O2(g) $\rightleftharpoons$ 4NO(g) + 6H2O(g), if only NH3 is added to the equation, will equilibrium shift toward the products?

David S
Posts: 54
Joined: Fri Sep 28, 2018 12:15 am

### Re: Result of Only Adding One of the Reactants

Yes, because by increasing the concentration of any one reactant, the denominator (i.e. the product of reactant concentrations) of the quotient Q will become greater than the denominator of K at equilibrium. Thus, Q < K, and reaction will proceed to balance this larger denominator out by creating more product.

Timothy_Yueh_4L
Posts: 57
Joined: Fri Sep 28, 2018 12:28 am

### Re: Result of Only Adding One of the Reactants

By adding more NH3 to the chemical reaction, you increase the concentration of the that particular chemical on the reactants side which gives an Q constant that is lower than the equilibrium constant, Kc. This will result with the forward reaction being favored as more product will be formed until the original equilibrium constant is attained.

Melody P 2B
Posts: 35
Joined: Fri Sep 29, 2017 7:06 am

### Re: Result of Only Adding One of the Reactants

Then, if you were to remove some reactant would the reaction shift left? Towards the reactants to compensate for the loss?

Edward Xie 2E
Posts: 63
Joined: Fri Sep 28, 2018 12:23 am

### Re: Result of Only Adding One of the Reactants

Then, if you were to remove some reactant would the reaction shift left? Towards the reactants to compensate for the loss?

Yep, since Q > K the reaction will shift towards the reactants.

Return to “Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions”

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# Sentiment analysis of automotive Tweets

The automotive industry is a multi-billion dollar company relying on positive business-client relations. With so much at stake, it is critical for these types of businesses to monitor and protect their online reputation. One way of obtaining social media data about companies is to monitor Twitter data and use the machine learning models to calculate the sentiment of the tweets. It has been shown in other work that in fact the sentiment of these tweets is correlated to the movement of the stock market.

In this blog post, we outline the methodology that was used to build a machine learning sentiment classification model, as well as the infrastructure to collect, process and store live twitter data. This was followed by some exploratory data analysis where we used topic modeling to filter irrelevant topics. Finally we used the CAP model to study the possible influence of the twitter sentiment signals calculated by the machine learning model on the return of the stocks and provide the results of the analysis.

In this work five automotive brands are examined: Tesla, Ford, Toyota, Mercedes and Porsche. The focus of this post is to outline the mathematical and statistical analysis methods, as well as to set up the computational infrastructure needed to undertake such a study of tweet sentiments and financial returns. In a follow up post, we will focus on the analysis of a large data set to robustly quantify and analyse the trends observed in this work.

# Sentiment Analysis Model

The natural language model that we used in this project was the Bag-of-words model (BOW). Using a large corpus of documents, all occurring words are analysed and ranked from most to least frequent. Then a function is used that maps a given sentence to a vector representing the occurrence of words from the vocabulary in the sentence.

More sophisticated language models exist on the market, however, as shown in Ref. 1, the simple BOW model is already capable of achieving about an 80% accuracy in sentiment identification problems, making it a good initial model for this pilot study.

Based on this BOW model, we generated feature vectors of the corpus in the training set that consisted of amazon reviews, yelp reviews, and the sentiment140 data set. These feature vectors and their corresponding labels were used to train two binary classification models, the logistic regression (LR) and naive Bayes classifiers (NB).

While more sophisticated sentiment classifier models exist, we chose to use these simpler approaches. This is because the LR and NB classifiers both provide probabilities in their categorizations that will be used as uncertainty estimates. It was indicated by Max Margenot that a sentiment classifier model based on logistic regression achieved about a 78% accuracy, while a more sophisticated LSTM neural network scored about 81.5% on the sentiment140 dataset and took significantly longer to train. This indicates that the NB and LR are sufficient for this task.

In Fig.1, we provide the training scores of the different sentiment classifier models that were tested. The training data was split into training and testing (80/20 %) split. The LR and NB models both scored well on accuracy and recall.

## Model Threshold

Once the two models were trained the final model used consisted of an ensemble of the LR, NB and pre-built textblob (denoted as TB) classifiers. The output of the ensemble model was the sentiment that corresponded to the classifier that had the greatest confidence in the result. Furthermore, we added the constraint that required the ensemble model to only make predictions if the confidence of the model was larger than the input threshold value $T$. The mathematical description of the model is $\text{Prediction} = \text{max}\lbrace \text{Pred}(\text{LR}),\text{Pred}(\text{NB}),\text{Pred}(\text{TB}) \rbrace,$

with the condition that

This threshold value was varied from 0 (predictions made with 0% confidence), up to the value 1 (100% prediction confidence).

In the above Figure the fraction of true/false positive predictions for a test data set is plotted as a function of the threshold. As the threshold increases, the number of tweets identified as positive increases while the false positives decrease as expected.

# Data collection and cleaning

With the model in place, the next step of the project was to collect twitter data based on keywords related to the car brands that were examined. The keywords used in this study were

• keywords: daimler; BMW; mercedes benz; toyota; ford; tesla; porsche.

We used the twython library which allowed us to listen to live twitter feed based on the above keywords.

An AWS EC2 server was used to collect all tweets into an SQLite database. Once the twitter data was collected the database was downloaded and the text was processed as follows:

• punctuation was removed,
• text was coverted to lower case,
• URLS were removed
• “_” was removed,
• “RT” was removed,
• ”@” sign was removed,
• hashtags were removed,
• numbers were removed,
• text was lemmatized.

Duplicates were also removed from the data set. For any days with missing data, such as weekends for the stock market returns, the median was used. The histogram of the tweet length before the preprocessing steps is given in the figure below:

Many of the tweets contained URLS causing a large number of tweets to reach the 140 character limit. After processing the tweets and removing duplicates the histogram looks more Gaussian:

Another issue in the data set was that for specific car brands, their tweets contained many topics not related to cars. To filter out irrelevant tweets, topic modelling was employed on the database. The topic modelling algorithms required an optimal value for the “number of topics” parameter in the prebuilt (latent dirichlet allocation or non-negative matrix factorization) functions used. The number of topics parameter ($k$) was determined by maximizing the average cohesion score for the top $m$ ranked words that describe a specific topic. The cohesion function acts on a string by calculating the average cosine similarity score (using a prebuilt word embedding model from spacy ‘en_core_web_lg’) for all pairs of words in that string.

The python function which computes this string cohesion score is given below:

This function was used to find irrelevant topics in the twitter database, for example some car tweets were related to people warning each other about a police chase.

# Preliminary Results

Finally after building the model, creating the AWS server to collect the tweets and cleaning the database, we conduct the statistical analysis. Using appropriate SQL queries, we processed the tweets of automotive brands on a day-by-day basis for the two-week period using different model thresholds. These day-by-day sentiments of the tweets are shown below for Tesla:

The $y$-axis represents the number of tweets related to Tesla, that were identified as being “positive” for a given day in the data set. There is a spike around August 8th for $T=0$. As the prediction certainty threshold value ($T$) is increased, the strengths of these spikes decreases. Furthermore, the baseline number of positively identified tweets decreases as we the certainty threshold of the model is increased.

## Combining the sentiment curves

In Fig. 5 there are specific peaks in the sentiment curves that do not vanish when the certainty threshold is increased. For example, on August 6th, there is a small consistent bump for all thresholds. Intuitively, this means that this signal is more important than others. These are the signals that we want to extract from the data. However, the curves for different $T$ values have vastly different scales. The scale for the $T=0$ curves have a baseline around 30%, while for $T$=0.8 the baseline is around 3%.

Another problem is that not all of the curves are to be taken with the same weight. The values with $T=0$ carry less weight high $T$ values. But how do we determine this weight for a given $T$?

So we are faced with two problems:

• Is it possible to remove the scale associated with the different thresholds?
• Once we have similarly scaled data, how can we combine the different curves together to get unique signal?

This first problem can be solved by normalizing the data. We chose to do this with the Z-transformation for each curve from the different thresholds, this will result in the set of curves

for the positive and negative sentiment time series signals, where $x_t$ represents the day and $T$ the threshold.

Now we need to know how to combine them. The solution is inspired by Bayesian statistics. The positive and negative sentiment curves for different thresholds are combined into a final true positive and false positive curve, denoted $P(x_t)$ and $\tilde{P}(x_t)$, respectively, are written as

Where $W_{TP}(T)$, $W_{FP}(T)$ are the true-positive and false-positive weights at a given $T$. They represent how seriously a given sentiment curve should be taken into account for a specific $T$. Their exact expressions are

Here, the symbol $\hat{y}$ represents the sentiment prediction of the model and $y$ is the true sentiment. Therefore, $P(\hat{y}=1|y=1,T)$ indicates the probability of a true positive, $P(y=1|T)$ is the prior probability of positive tweets, and $P(T)$ is the prior distribution for the threshold. (This last prior is taken to be a uniform distribution from 0 to 1)

Similar expressions hold for the true negative and false negative sentiment signals. For the incoming automotive tweets, the true distributions of positive and negative sentiments are not known. However, we will assume that the distributions obtained from the testing set will be the same (or similar) to the live automotive tweets. If the model has been well trained then we would expect our model to generalize and this assumption should be approximately correct. In the figure below the weight $W(T)$ as a function of the threshold are plotted in dimensionless units.

Fig. 6 shows that as the threshold value $T$ increases, the positive sentiment time series prediction, $\text{Pos}(x_t,T)$, for that threshold is weighted more heavily.

In contrast, in Fig. 7 as the threshold value $T$ increases, the negative sentiment time series prediction, $\text{Neg}(x_t,T)$, for those thresholds is weighted much less.

This indicates that our ensemble model is not good at making confident predictions about negative sentiment tweets, but it does seem to do well at understanding which tweets are positive. Therefore, when combining predictions for different thresholds it considers high confidence positive sentiment signals with more weight, while ignoring the high confidence negative sentiment signals.

Using the Bayesian weighting average formula found earlier, we calculate the normalized positive and negative sentiment signals. The one for Tesla during the two-week period is shown below

Figure 8 indicates a signal spike for August 3,2018 and August 6, 2018 indicating a large number of positive tweets in the data set for Tesla during those two days. In the next section, we will analyse this data within a financial framework to examine correlations between these sentiment signal time series and the return of the stocks for vehicle brands.

## Analysing stock returns and sentiment signals

Now that the data is normalized and the curves at different thresholds have been combined together into a single sentiment signal time series, we want to analyse the correlation between this time-series and the stock return time series. However, the next problem that we encounter is:

• how can we relate the sentiment signals to the return of the stock?

There is a financial model, known as the CAPM(note we use 0 for the risk free return rate) that relates the return of an asset to the return of the market through a linear model. The base CAP model states that the return of a stock $r$ should be proportional to the return of the market $r_{M}$ by

Here the proportionality constant to the market is $\beta_M$, $\alpha$ is a small offset and the parameter $x_t$ represents the day. This base hypothesis can be extended to include additional coefficients $\beta_P$ and $\beta_N$ that allow the inclusion of the effects of the positive and negative sentiment signals from Figs. 8 and 9 through a generalized linear multifactor model

A Bayesian linear regression analysis was conducted to determine the distributions of the linear fit coefficients. In the Bayesian framework, we need to determine the posterior distribution of a parameter $\beta$ given the times series data, denoted $D$. This posterior distribution is denoted $P(\beta|D)$.

This Bayesian fitting was carried out using the emcee Python package, that employs a parallelized Markov chain Monte Carlo algorithm to compute the posterior distributions of the parameters of the fit from a likelyhood function. Mathematically speaking we are carrying out the integrals given by

where the likelyhood function of the data given the model is

here $\sigma$ is the width of the distribution and $\chi^2$ is the least square residuals between the linear model and data.

The priors of the parameters $P(\beta_{N})$, $P(\beta_{M})$, $P(\beta_{P})$, $P(\alpha)$ are uniformly distributed in the range (-100,100). The prior $P(\sigma)$ is a scale invariant Jeffreys prior. We calculated the posterior distributions for all 5 car brands. The distribution for the coefficient $\beta_P$ for Tesla is shown below,

Fig. 10 gives the median value of $\beta_P$ of the model as well as the 68% and 96% confidence intervals. These confidence intervals demonstrate the power of using the Bayesian linear regression to quantify uncertainty. We observe that while the median value of the fit is negative and excludes zero at the 68% confidence level, there isn’t enough statistical power to exclude 0 at the 96% confidence level. In a future post we will revisit this analysis with a larger data set.

Based on the determined median values of the $\beta$ parameters in Fig. 11, we summarize the correlations that were not excluded at the 68% confidence level for the 5 investigated car brands.

Green indicates a positive trend, red a negative trend, and orange means the correlation was excluded at the 68% confidence region.

The main findings of this table are:

• For all examined vehicle brands, the median value of $\beta_M$, representing the sensitivity of the market return to the stock return was always positive and seemed to be the strongest effect of the linear model.

• The stock returns for Tesla are positively correlated with negative twitter sentiments

• The stock returns for Toyota and Porsche are positively correlated with positive twitter sentiments

• The stock returns for Tesla and Ford are negatively correlated with positive twitter sentiments

• The stock returns for Mercedes are negatively correlated with negative twitter sentiments

In future work, it would be useful to investigate the underlying reasons for the correlations. It is counter intuitive that positive tweet sentiments are anti-correlated with the return of the automotive stocks for some companies.

## Posteriors of the correlations

The results of the last section give the general “sign/direction” of the correlations, but not the numerical strengths. The next task is to compute the posterior distributions of these correlations. The Pearson’s coefficient between time series $x$ and $y$ is

where $N$ is the number of data points (or days). Here $\bar{x}$ and $\bar{y}$ are the mean values of the time series. The correlation coefficients that investigate are

The posterior distribution of $\rho(P,r)$ for Tesla is shown in Fig. 12 below

Fig 12. shows that there is a minimum correlation coefficient that is possible for the data set, at around -0.7 for Tesla, and most of the distribution is centered around negative values as indicated by the $\beta_P$ value in Fig. 10.

We summarize the median values of all correlations in Fig. 11.

# Conclusions and further work

In this article, we focused more on the mathematical details and the methodology constructed with python that allowed us to carry out the end-to-end data science project. The main limitation of our work was the small data set that was used to compute the correlations. We found correlations at the 68% confidence level, but they may yet be excluded at the 96% confidence region. However, this proof-of-principle project produced interesting results and provides us the tools for future investigations. In the next upcoming blog post, we will use a large data set, with several months worth of data to study the model proposed here. In that upcoming post we will examine in detail the following effects:

• How do the sentiment correlations change as a function of time?
• What are the results of using a sliding window approach for the correlations?
• Does changing the normalization method change the correlations? For example using a min-max scaler as opposed to the Z-transformed data?
• What are the results when we include/remove outliers in the data set?


text:
ホーム > メンバー > 中村健太郎

## プレプリント

Zariski density of crystalline representations for any $p$-adic field
Preprint 2011, arXiv:1104.1760v1 [math.NT]

Abstract. The aim of this article is to prove Zariski density of crystalline representations in the rigid analytic space associated to the universal deformation ring of a $d$-dimensional mod $p$ representation of $\operatorname{Gal}(\bar{K}/K)$ for any $d$ and for any $p$-adic field $K$.
This is a generalization of the results of Colmez, Kisin ($d=2$, $K=\mathbb{Q}_p$), of the author ($d=2$, any $K$), of Chenevier (any $d$, $K=\mathbb{Q}_p$). A key ingredient for the proof is to construct a $p$-adic family of trianguline representations. In this article, we construct (an approximation of) this family by generalizing Kisin's theory of finite slope subspace $X_{fs}$ for any $d$ and for any $K$.

Deformations of trianguline $B$-pairs and Zariski density of two dimensional crystalline representations
Preprint 2010, arXiv:1006.4891v1 [math.NT]

Abstract.The aim of this article is to study deformation theory of trianguline $B$-pairs for any $p$-adic field. For benign $B$-pairs, a special good class of trianguline $B$-pairs, we prove a main theorem concerning tangent spaces of these deformation spaces. These are generalizations of Bellaiche-Chenevier's and Chenevier's works in the $\mathbb{Q}_p$ case, where they used $(\phi,\Gamma)$-modules over the Robba ring instead of using $B$-pairs. As an application of this theory, in the final chapter, we prove a theorem concerning Zariski density of two dimensional crystalline representations for any $p$-adic field, which is a generalization of Colmez and Kisin's results in the $\mathbb{Q}_p$ case.

## 出版論文

2 次元クリスタリン表現のZariski 稠密性
RIMS Kôkyûroku Bessatsu B25: Algebraic Number Theory and Related Topics 2009, eds. T. Ichikawa, M. Kida, T. Yamazaki, June (2011), 161-184.
Classification of two dimensional split trianguline representations of $p$-adic fields
Compositio Mathematica, Volume 145 Part 4 (2009), pp. 865-914.
Available by subscription at Cambridge Journals Online.

Abstract. The aim of this article is to classify two-dimensional split trianguline representations of $p$-adic fields. This is a generalization of a result of Colmez who classified two-dimensional split trianguline representations of $\operatorname{Gal}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$ for $p\neq 2$ by using $(\varphi,\Gamma)$-modules over a Robba ring. In this article, for any prime $p$ and for any $p$-adic field K, we classify two-dimensional split trianguline representations of $\operatorname{Gal}(\overline{K}/K)$ using $B$-pairs as defined by Berger.

## 口頭発表

### 2011年度

• Zariski density of crystalline representations for any $p$-adic field, Boston-Keio Summer Workshop、September 17, 2011
• Zariski density of crystalline representations for any $p$-adic field
• Zariski density of crystalline representations for any $p$-adic field
代数セミナー、慶應義塾大学、April 21, 2011

### 2010年度

• Zariski density of crystalline representations
数論幾何セミナー、北海道大学、March 7, 2011
• Zariski density of crystalline representations
九州代数数的整数論2011、九州大学、February 17, 2011
• $GL_2(\mathbb{Q}_p)$の $p$進局所ラングランズ対応について
城崎新人セミナー、城崎、February 15, 2011
• Zariski density of crystalline representations
第9回仙台広島整数論集会、東北大学、July 22, 2010
• Construction of $p$-adic families of trianguline representations
玉原数論幾何研究集会、玉原、May 24, 2010

### 2009年度

• Trianguline表現と二次元クリスタリン表現のザリスキー稠密性について
数論幾何セミナー、名古屋大学、January 12, 2010
• 二次元クリスタリン表現の Zariski 稠密性
RIMS研究集会「代数的整数論とその周辺」2009, 東京大学, December 8, 2009
• Trianguline representations and Zariski density of two dimensional cristalline representations
Keio-Yonsei Cooperative Research Conference on Mathematical Sciences 2009, Yonsei University, November 19, 2009
• Zariski density of two dimensional trianguline representations of $p$-adic fields
Workshop: $p$-進特殊関数と数論幾何, October 29, 2009
• Zariski density of two dimensional cristalline representations
代数セミナー、九州大学、October 22, 2009
• Trianguline表現の分類とZariski稠密性について
日本数学会2009年度秋季総合分科会、代数分科会・一般講演、大阪大学、September 25, 2009
• $p$-進表現入門 Ⅰ、Ⅱ
第17回（2009年度）整数論サマースクール「l-進ガロア表現とガロア表現の整数論」、アピカルイン京都、August 17-21, 2009
• Zariski density of two dimensional trianguline representations of $p$-adic fields
$p$-adic Automorphic Forms and Arithmetic Geometry、気仙沼大島、July 28-31, 2009
• $p$-進体の二次元三角表現のZariski稠密性
第８回広島整数論集会、広島大学、July 24, 2009
• Zariski density of two dimensional trianguline representations of $p$-adic fields
代数セミナー、慶應義塾大学、June 18, 2009
• Zariski density of two dimensional trianguline representations of $p$-adic fields
大阪大学整数論 ＆ 保型形式セミナー、大阪大学、June 5, 2009

### 2008年度

• Classification of two dimensional split trianguline representations of $p$-adic fields
Industrious Number Theory、KIAS ソウル、March 11, 2009


text:
J. Korean Ceram. Soc. > Volume 47(4); 2010 > Article
 Journal of the Korean Ceramic Society 2010;47(4): 325. doi: https://doi.org/10.4191/kcers.2010.47.4.325
 Thermal Expansion and Dielectric Properties of CaO-ZnO-B2O3-SiO2 Glass-Added Al2O3 Composites for LTCC Applications Tae-Hun Byeon, Hyo-Sung Park, Hyun-Ho Shin, Sang-Ok Yoon, Chang-Yong Oh1 Department of Advanced Ceramic Materials Engineering, Gangneung-Wonju National University1Temen Technology, Inc., Gangneung Science Industry Park ABSTRACT Varying quantities of a high-thermal-expansion glass, 50CaO-20ZnO-$20B_2O_3-10SiO_2$ (CZBS), were added to alumina and sintered at $875^{circ}C$ for 2 h for low temperature co-firing ceramic (LTCC) applications. As the amount of glass addition increased from 40 wt% to 70 wt%, the apparent density of the sintered product increased from 88.8% to 91.5%, which was also qualitatively confirmed by microstructural observation. When the glass addition was very high, e.g., 70 wt%, an apparent formation of secondary phases such as $CaZn_2AlZnSiAlO_7$, $Ca_2Al(AlSi)O_7$, $Ca_2Al_2SiO_7$, $Ca_2ZnSi_2O_7$ and ZnO, was observed. Both the dielectric constant and the coefficient of thermal expansion increased with the glass addition, which was qualitatively consistent with the analytical models, while the experimental values were lower than the predicted ones due to the presence of pores and secondary phases. Key words: Calcium zinc borosilicate glass, Alumina, Coefficient of thermal expansion, Dielectric constant, LTCC
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text:
USIP COMPONENT PACKAGE

# uSiP Component Package

Article by:

## Overview

 Name uSiP (Plastic System In Package or Micro System In Package, TI) Synonyms SIL (TI) Variants SIL-10C (10 pin uSiP) Similar To Mounting SMD Pin Count 8-33 Pitch Solderability Thermal Resistance Depends on variant and exact component, see the Thermal Resistance section below. Dimensions 3D Models Common Uses Small point-of-load power modules.

The uSiP package is used for the TPSM265R1 family of point-of-load power modules (this component incorporates the SMPS buck controller, MOSFET, inductor and some of the capacitors, hence being a complete power modules).

## Thermal Resistance

The thermal resistance of a uSiC package depends on the variant and also on the components soldered onto the sub-assembly. For that reason, thermal resistances are not given per package, but per component in the component’s datasheet. For example, a graph of $$\theta_{JA}$$ versus copper area on the PCB is shown below from the TPSM265R1 datasheet, which uses the SIL-10C uSiP package:


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# Complex Variables (math.CV)

• In this paper we study an analogue of the classical Riemann-Hilbert problem stated for the classes of difference and $q$-difference systems. The Birkhoff's existence theorem was generalized in this paper.
• We study some spectral properties of Volterra-type integral operators $V_g$ and $I_g$ with holomorphic symbol $g$ on the Fock--Sobolev spaces $\mathcal{F}_{\psi_m}^p$. We showed that $V_g$ is bounded on $\mathcal{F}_{\psi_m}^p$ if and only if $g$ is a complex polynomial of degree not exceeding two, while compactness of $V_g$ is described by degree of $g$ being not bigger than one. We also identified all those positive numbers $p$ for which the operator $V_g$ belongs to the Schatten $\mathcal{S}_p$ classes. Finally, we characterize the spectrum of $V_g$ in terms of a closed disk of radius twice the coefficient of the highest degree term in a polynomial expansion of $g$.
• We construct a Brody hyperbolic Horikawa surface that is a double cover of $\mathbb{P}^2$ branched along a smooth curve of degree $10$. We also construct Brody hyperbolic double covers of Hirzebruch surfaces with branch loci of the lowest possible bidegree.
• In this paper we construct a complete injective holomorphic immersion $\mathbb{C}\to\mathbb{C}^2$ whose image is dense in $\mathbb{C}^2$. The analogous result is obtained for any closed complex submanifold $X\subset \mathbb{C}^n$ for $n>1$ in place of $\mathbb{C}\subset\mathbb{C}^2$. We also show that, if $X$ intersects the unit ball $\mathbb{B}^n$ of $\mathbb{C}^n$ and $K$ is a connected compact subset of $X\cap\mathbb{B}^n$, then there is a Runge domain $\Omega\subset X$ containing $K$ which admits a complete holomorphic embedding $\Omega\to\mathbb{B}^n$ whose image is dense in $\mathbb{B}^n$.
• We study when there exists a dense holomorphic curve in a space of holomorphic maps from a Stein space. We first show that for any bounded convex domain $\Omega\Subset\mathbb{C}^n$ and any connected complex manifold $Y$, the space $\mathcal{O}(\Omega,Y)$ contains a dense holomorphic disc. Our second result states that $Y$ is an Oka manifold if and only if for any Stein space $X$ there exists a dense entire curve in every path component of $\mathcal{O}(X,Y)$. In the second half of this paper, we apply the above results to the theory of universal functions. It is proved that for any bounded convex domain $\Omega\Subset\mathbb{C}^n$, any fixed-point-free automorphism of $\Omega$ and any connected complex manifold $Y$, there exists a universal map $\Omega\to Y$. We also characterize Oka manifolds by the existence of universal maps.
• The basin of attraction of a uniformly attracting sequence of holomorphic automorphisms that agree to a certain order in the common fixed point, is biholomorphic to $\mathbb{C}^n$. We also give sufficient estimates how large this order has to be.
• Kähler-Einstein metrics for polarized families of Calabi-Yau manifolds define a natural hermitian metric on the relative canonical bundle. The fact that the curvature form is equal to the pull-back of the Weil-Petersson form up to a numerical constant is being used for the construction of a Kähler form on the total space of a given family, whose restriction to the fibers is Ricci flat.


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## A not so intuive amortization problem

A loan at ${j_1} = 9\%$ is being repaid by monthly payments of $750 each. The total principal repaid in the 12 monthly installments of the 8th year is$400. What is the total interest paid in the 12 installments of the 10th year?

The answer is of course \$8,524.76
The answer makes use of the fact that the entries in the principal repaid column (especially in a no rounding scenario) are in the ratio of 1 + i.
An interesting question of course is whether it's possible to solve for the amount of the loan and loan term from the given information.
Apparently, it is.


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+0

# Band performances

0
62
1
+79

A band has a portfolio of 15 songs. 5 are acoustic. For a concert the band will play 10distint songs, 4 will be acoustic. The 4 acoustic will be played consecutively during the 2nd half, but withou ending in anacoustic song. How many different sequences of the songs are possible?

Dec 18, 2018

#1
+3580
+1

$$\text{looking at the 2nd half of the show first}\\ \text{the songs will be AAAA!A}\\ \text{there are }4!=24 \text{ ways to arrange the 4 acoustic songs, and}\\ \text{10 ways to select the !A song at the end}\\ \text{this is then 240 ways to arrange the 2nd half of the show}\\ \text{the first half of the show is then 5 non-acoustic songs chosen from 9}\\ \text{since we care about order there are }\dbinom{9}{5}5! = 15120\\ \text{ arrangements for the first half} \\ \text{so in total we have}\\ 240 \times 15120 = 3628800 \text{ possible concert arrangements}$$

.
Dec 18, 2018


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# Time Series Forecast / Transfer Function

I'm trying to interpret the forecast values from an ARIMAX function, and I'm confused about what's happening in the actual forecasted values as I change the values for the predictor during the forecast period.

The ARIMAX model shows one of the predictors (Spend) has the following (significant) transfer function coefficients

Numerator (lag 0)= .029 Denominator (lag 2) = .038

I held the forecast values of the predictor "Spend" constant and the compared different level - i.e. 6 periods at 100,000, then 6 periods at 200,000

One would expect that doubling the value of "Spend" would dramatically increase the predicted forecast - holding everything else constant. But it does not.

The effect was two periods of positive forecast values, followed by 4 periods of negative forecast values. I suspect this may have something to do with the denominator being larger than the numerator in the transfer function. (that's a guess). However, I also cannot explain why the forecast is positive for two periods and then goes negative, and stays negative. Why wouldn't it oscillate given the 2-period lag in the denominator?

(I've been trying to make good use of this post: Transfer function in forecasting models - interpretation)

Your help is appreciated, as always!

• Any chance you can post the data? Lag 2 in denominator could be spurious. – forecaster Oct 13 '18 at 2:49
• yes, I could use some assistance in figuring out how to post a .csv file; if there's a way, I could also email it. dropbox.com/s/mobnwmmgyft9bhs/sample_data.csv?dl=0 – SD_Data_Scientist Oct 13 '18 at 3:04
• posted my public dropbox link. see if that works. – SD_Data_Scientist Oct 13 '18 at 3:12
• Yes it does, I’ll try to work something up tomorrow – forecaster Oct 13 '18 at 3:24
• Very much appreciate your expertise! – SD_Data_Scientist Oct 13 '18 at 3:26

Identifying transfer functions in Arima is more of an art than science. Automatic procedures aren't always right. I have used SPSS in the past, but currently don't have access to it. So I'll try to do in SAS and R you could easily do it in SPSS.

There are two forms of identifying Transfer functions (see here for more details):

1. Prewhitening and looking cross correlation functions (see for example Box, Jenkins and Reinsel).
2. Lagged regression of independent variable, with AR(1) error term and plot the effect of coefficients and see if you have a pattern that can determine transfer function. (See Pankratz)

Procedure #1 is awfully complex, no one uses it in practice when you have more than 2 independent variables.

Going by #2, I lagged your spend and intent variable 4 and 5 times respectively and ran an arima model in R using auto.arima (in SPSS you could do the same by just running an AR(1) instead of auto.arima).

Spend on Profit:

Intent on Profit:

You do see a up and down pattern in Spend and you could potentially see a delay effect on intent. Without knowing exactly what these variable means I have no way to provide additional insights on why these variable behave the way they do. Does this makes sense?

IF you can provide additional context I can model this using SAS and provide you transfer function modeling and interpretation. Here is the R code for replication.

input <- read.csv("transfer.csv")

input.ts <- ts(input,frequency = 1)

input.lagged <- cbind(input.ts,spend1,spend2,spend3,spend4,intent1,intent2,intent3,intent4,intent5)

new_lagged_data <- na.remove(input.lagged)

model <- auto.arima(y = new_lagged_data[,1],xreg = new_lagged_data[,-1])

barplot(model$coef[c(1,3,4,5,6)]) barplot(model$coef[c(2,7,8,9,10,11)])

• Thank you for your analysis. Intent is a latent measure obtained from a monthly assessment of customers attitudes. i.e. purchase intent. Spend is monthly spend on marketing. I was able to work out some of the math, and determined that the denominator as a kind of exponential decay for each time period. (I'm sure that everyone already knew that). – SD_Data_Scientist Oct 14 '18 at 18:00
• My concern was that when the marketing spend is held constant, the forecasted sales actually turns negative after the first two periods, which seems roughly consistent with the plots you provided above. I'm not entirely certain, but it seems that is the result of the "decay" coefficient being larger than the numerator (which I believe is interpreted as the auto-regressive part of variable) – SD_Data_Scientist Oct 14 '18 at 18:08
• How do real forecasters (hahaha) perform "what-if" scenarios for significant independent variables? because of SPSS limitations, I'm forced to essentially re-run the model in SPSS each time with different what-if values for each independent variable in the forecast period.... that seems awfully slow. – SD_Data_Scientist Oct 14 '18 at 18:12
• I would be interested to see what SAS produces and how it compares with SPSS output. Let me know if you have everything you need from me. – SD_Data_Scientist Oct 14 '18 at 22:23
• My mistake - for Spend, there is no denominator in the transfer function. This clears up a lot. Don't do complex time series late at night... – SD_Data_Scientist Oct 14 '18 at 23:11

A useful model for the 28 observations is here and also here as a familiar regression equation .

When identifying via pre-whitening it is often necessary to include any needed differencing operators. Some software provides recognize that the transfer function itself does not necessarily include the differencing operators and they provide the option to include differencing operators or not. For example if y and x are both upwards trending series , it is entirely possible for y=3x might be a useful model even though both y and x ( by themselves) are non-stationary.

In this case unwanted/unneeded differencing and unwarranted dynamic structure (INTENT) injects unneeded structure causing your consternation.

There are two seasonal pulses and 1 level/step shift at or around 12/2017. I have have also attached the litmus test for model adequacy ... the plot of the model errors and also the acf of the model errors. . It would be interesting ( and probably revealing) to see the plot of the residuals and the residual acf of the SPSS model.

I would suggest that any complete/thorough analysis would include this presentation.

In terms of why neither SAS or SPSS include the automatic identification of deterministic structure https://pdfs.semanticscholar.org/09c4/ba8dd3cc88289caf18d71e8985bdd11ad21c.pdf when predictor series are employed, you might ask them as the last time I looked they were not available.

The actual/fit/forecast graph is here ## ## and the statistical summary of the model is here


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# How can I transform $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^k\sin(\pi rn)}$ into a modular form?

Let

$$f_k(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^k\sin(\pi zn)}$$

be a family of holomorphic functions on the upper-half plane $$\mathbb{H}=\{a+bi|b>0\}$$ for each odd natural number $$k$$. These functions clearly satisfy the periodicity condition $$f_k(z+2)=f_k(z)$$, but more surprisingly they also satisfy the formula

$$$$z^{\frac{1-k}{2}}f_k(z)+z^{\frac{k-1}{2}}f_k\left(\frac{1}{z}\right)=\sum_{j=0}^{k+1}a_{k,j}z^{\frac{k+1}{2}-j}\tag{1}$$$$

where $$a_{k,j}$$ are given by

$$$$a_{k,j}:=\frac{(-1)^{\frac{k+1}{2}}(2\pi)^k}{(k+1)!}{k+1\choose j}B_j\left(\frac{1}{2}\right)B_{k+1-j}\left(\frac{1}{2}\right)\tag{2}$$$$

where $$B_j(\cdot)$$ are the Bernoulli polynomials. I discovered and proved this formula a year ago and didn't think much of it, but recently I have begun thinking about there recurrences in the context of modular/cusp forms and $$L$$-functions. More specifically, these two formulas seem quite similar to those of a cusp form of weight $$1-k$$ over the congruence subgroup $$\Gamma(2)$$. The match isn't perfect because of the extraneous polynomials that pop up on the LHS of (2), but it still makes one wonder.

Another piece of evidence that the functions $$f_k(\cdot)$$ are connected to modular forms is that they naturally induce $$L$$-functions in the same way that modular forms do. In this context I mean that if a modular form $$g$$ of weight $$k$$ has $$q$$-expantion $$g(z)=\sum_{n}a_n n^{\frac{k-1}{2}}q^n$$ then it induces the $$L$$-function $$L(g,s)=\sum_{n}a_n n^{-s}$$. The $$L$$-function attached to $$f_k(z)$$ is

\begin{align} L(f_k,s)&=2i\frac{1-2^{1-\frac{k}{2}-s}}{1-2^{-\frac{k}{2}-s}}\zeta\left(s-\frac{k}{2}\right)\zeta\left(s+\frac{k}{2}\right)\\ &=2i\frac{1-2^{1-\frac{k}{2}-s}}{1-2^{-\frac{k}{2}-s}}\prod_{p}\frac{1}{1-\left(p^{\frac{k}{2}}+p^{-\frac{k}{2}}\right)p^{-s}+p^{-2s}} \tag{3}\end{align}

Making $$L(f_k,s)$$ a degree 2 $$L$$-function with a critical line $$\Re(s)=\frac{1}{2}$$. $$L(f_k,s)$$ is not of the Selberg class of $$L$$-functions since there is no function $$\gamma_k(s)$$ such that $$\Lambda_k(s):=\gamma_k(s)f_k(s)$$ satisfies $$\Lambda_k(1-s)=\Lambda_k(s)$$.

I suspect that the fact that the $$L$$ function generated is "almost" of the Selberg class corresponds to the fact that $$f_k(z)$$ is "almost" a cusp form. My overall goal with connecting $$f_k(z)$$ to modular forms would be to learn something non-trivial about the cosecant function $$\frac{1}{\sin(\pi zr)}$$ or to use the fact that the vector space of cusp forms $$S_k(\Gamma(2))$$ is finite dimensional and has a simple basis to find a new way to compute

$$\sigma_k^{(o)}(n):=\sum_{\substack{d|n \\ d\equiv 1\mod{2}}}d^k$$

since the $$q$$-expantion of $$f_k(z)$$ is

$$f_k(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\sigma_k^{(o)}(n)}{n^k}q^n$$

• Have a look at Armin Straub's work, e.g., "Special values of trigonometric Dirichlet series and Eichler integrals" (arxiv.org/abs/1407.5119 - in particular page 5) – Marcus Dec 31 '20 at 9:07
• @Marcus This is so funny! When I started researching and proved the main identity, the first thing I did was prove results about $f_k(\rho)$ being a rational times $\pi^k$ when $\rho$ is a quadratic irrational. It is such a small world! – Milo Moses Dec 31 '20 at 17:40
• @Marcus The main theory in the linked paper seems to be of relation functions similar to $f_k(z)$ back to themselves using the theory of differential equations, and so if I am not mistaken my result does not follow from theirs and theirs does not follow from mine. Is that correct? – Milo Moses Dec 31 '20 at 17:48
• I haven’t checked that, but it’s in the same vein. For a link to Eisenstein series, see arminstraub.com/downloads/pub/ramanujanzeta.pdf – Marcus Jan 1 at 9:05
• Also very similar, though not exactly the same: Have a look at Ramanujan's Notebooks - Part II (by Berndt), Chapter 14 on Infinite Series, Formula 1.14 (p. 245 on top) – Marcus Jan 8 at 20:20

The generalized Eisenstein series

$$$$\hat{E}_{k,\chi,\psi}(z):=\sum_{n=1}^{\infty}\left(\sum_{d|n}\psi(d)\chi(n/d)d^{k-1}\right)q^n\tag{1}$$$$

Is a eigenform over the space $$M_k(RL,\chi\psi)$$ for characters $$\psi$$ and $$\chi$$ with conductors $$L$$ and $$R$$ respectively, whenever $$L>1$$ and $$\psi(-1)\chi(-1)=(-1)^k$$. More generally, for any $$t>0$$ the function $$\hat{E}_{k,\chi,\psi}\left(z^t\right)$$ is a modular form over $$M_k(RLt,\chi\psi)$$ according to this online textbook, Theorem 5.8.

We now set $$\psi=1$$ to be the trivial character and $$\chi_=\chi_2$$ to be the unique character modulo $$2$$. This means that $$L=2>1$$ and $$\psi(-1)\chi(-1)=1$$ so the conditions of (1) are satisfied if $$k$$ is even, making $$\hat{E}_{k,\chi_2,1}(z)$$ a modular (eigen) form. Now, if $$k'$$ is any odd integer we can let $$k=1-k'$$ be an even integer, so

$$$$\hat{E}_{1-k',\chi_2,1}(z)=\sum_{n=1}^{\infty}\left(\sum_{d|n}\chi_2(n/d)d^{-k'}\right)q^n$$$$

is a modular form of degree $$1-k'$$ over $$\Gamma_1(2)$$. Since $$\sum_{d|n}\chi_2(n/d)d^{-k'}=\frac{\sigma^{(o)}_{k}(n)}{n^k}$$, this means that

$$$$\hat{E}_{1-k,\chi_2,1}(z)=\sum_{n=1}^{\infty}\frac{\sigma^{(o)}_{k}(n)}{n^k}q^n\tag{2}$$$$

Relating this to $$f_{k}(z)$$ is now a straightforward task. Namely, we see that

\begin{align*} f_k(2z+1)&=2i\sum_{n=1}^{\infty}\frac{(-1)^{n+1}\sigma_k^{(o)}(n)}{n^k}\exp\left(\frac{2\pi in(2z+1)}{2}\right)\\ &=-2i\sum_{n=1}^{\infty}\frac{\sigma_k^{(o)}(n)}{n^k}\exp\left(2\pi inz\right)\\ &=-2i\hat{E}_{1-k,\chi_2,1}(z) \end{align*}

Thus, the transformation $$f_k(2z+1)i/2$$ turns $$f_k(z)$$ into a eigenform. Working out that $$f_k(2z+1)i/2$$ is an eigenform just from the formulae given in the problem statement seems hard, but also I am not aware of any general formula for Eisenstein series which yields the formulae given in the problem statement.

I find this stuff absolutely fascinating, and I'll be looking more into this and undoubtedly coming up with lots more fun stuff, so if anyone actually ends up reading this and cares about I'm writting then I will add more answers/edit this answer as time goes on to make the solution more complete. Thank you for your time.


text:
## Saturday, August 29, 2015

### Double Pendulum Concluded (Part 5): Discretization

 Path traced out in state space by a double pendulum
In this final part, we discretize the equations for the double pendulum, covered in Parts 1, 2, 3, and 4. As frequently noted, for this problem, we are in favor of symplectic methods, because qualitatively, the algorithms have very good long-time energy behavior, and you don't get nonsense like the pendulum speeding up uncontrollably. And in order to do this, we rephrased in terms of Hamiltonian mechanics in Part 4, from which we recall (setting $\mathbf{q} = (\theta_1,\theta_2)$):
$\dot{\mathbf{q}} = I^{-1} \mathbf{p}$$\dot{\mathbf{p}} = \frac{1}{2} \dot{\boldsymbol{\theta}}^t \frac{\partial I}{\partial \mathbf{q}}\dot{\boldsymbol{\theta}} - \frac{\partial V}{\partial\mathbf q}$
which, for this system, is
$\dot{p}_1 = \frac{1}{2}\dot{\mathbf{q}}^t \begin{pmatrix} 0 & m_2\ell_1\ell_2 \sin(\theta_2-\theta_1) \\ m_2\ell_1\ell_2 \sin(\theta_2-\theta_1) &0\end{pmatrix} \dot{\mathbf{q}} + (m_1 + m_2) g\ell_1 \sin \theta_1$$\dot{p}_2 = \frac{1}{2}\dot{\mathbf{q}}^t \begin{pmatrix} 0 & -m_2\ell_1\ell_2 \sin(\theta_2-\theta_1) \\ -m_2\ell_1\ell_2 \sin(\theta_2-\theta_1) &0\end{pmatrix} \dot{\mathbf{q}} + m_2 g\ell_2 \sin \theta_2,$
with the symmetric moment of inertia
$I = \begin{pmatrix} (m_1 +m_2) \ell_1^2 & m_2\ell_1\ell_2\cos(\theta_2-\theta_1) \\ m_2 \ell_1 \ell_2 \cos(\theta_2 -\theta_1) & m_2 \ell_2^2\end{pmatrix}.$
We use a one-step, first order method to simulate (with time step $h$). The usual thing to do is to consider the derivatives $\dot{\mathbf{q}}$ and $\dot{\mathbf{p}}$ as difference quotients, and solving for the later time step (denoted with the superscript $n+1$) in terms of the earlier (superscript $n$):
$\frac{\mathbf{q}^{n+1} - \mathbf{q}^n}{h} = I^{-1} \mathbf{p}$and$\frac{\mathbf{p}^{n+1} - \mathbf{p}^n}{h} = \frac{1}{2} \mathbf{p}^t I^{-1} \frac{\partial I}{\partial \mathbf{q}} I^{-1} \mathbf{p} - \frac{\partial V}{\partial \mathbf{q}}.$
Now of course, the subtlety is how to evaluate the quantities on the right-hand side; since now we don't assume that we have direct access to $\dot{\mathbf q}$, we have re-expressed it in terms of $\mathbf{p}$ in the second equation. Also, since $\partial I/\partial{\mathbf{q}}$ is some higher-order tensor, the notation becomes very confusing, very fast. If we write
$B = \frac{\partial I}{\partial \theta_1}$ then we find $\partial I/\partial \theta_2 = -B$, since, as we saw, $I$ only depends on $\theta_2 -\theta_1$. Therefore, we rewrite it as
$\frac{\mathbf{p}^{n+1} - \mathbf{p}^n}{h} = \frac{1}{2}\begin{pmatrix} \mathbf{p}^t B \mathbf{p} \\ -\mathbf{p}^t B \mathbf{p} \end{pmatrix}- \frac{\partial V}{\partial \mathbf{q}}.$
The Euler-B method uses $\mathbf{p}^{n+1}$ and $\mathbf{q}^n$ on on the right-hand side (we choose it instead of Euler-A, which would use $\mathbf{q}^{n+1}$ and $\mathbf{p}^n$, because the moment-of-inertia tensor $I$ depends nonlinearly on $\mathbf{q}$ and makes the resultant equations harder to solve—we'll see where that comes in, in a little bit).

Once we solve for $\mathbf{p}^{n+1}$, then it is clear how to solve for $\mathbf{q}^{n+1}$ (noting we evaluate $I$ at $\mathbf{q}^n$):
$\mathbf{q}^{n+1} = \mathbf{q}^n + h I^{-1} \mathbf{p}^{n+1}.$
Now $\mathbf{p}^{n+1}$ is harder:
$\mathbf{p}^{n+1} = \mathbf{p} ^n + h \left( \frac{1}{2} \begin{pmatrix} (\mathbf{p}^{n+1})^t B \mathbf{p}^{n+1} \\-(\mathbf{p}^{n+1})^t B \mathbf{p}^{n+1}\end{pmatrix}- \frac{\partial V}{\partial \mathbf{q}}\right)$
where, again, the $B$ and $\partial V/\partial{\mathbf{q}}$ are evaluated at $\mathbf{q}^n$. The trouble is, of course, that $\mathbf{p}^{n+1}$ appears on both sides, and it's a nonlinear function of $\mathbf{p}^{n+1}$ (though only quadratic nonlinearity, but nonlinearity, nonetheless). How do we solve this? By...

#### Using Newton's Method.

To clear the clutter, we let $\mathbf{x}$ be the unknown, namely $\mathbf{p}^{n+1}$, and simply write $\mathbf{p} = \mathbf{p}^n$ as known given data. Thus, bringing all of that junk to one side of the equation, we want to solve $F(\mathbf{x}) = 0$ where $F$ is:
$F(\mathbf{x}) = \mathbf{x} - \mathbf{p} - h\left( \frac{1}{2} \mathbf{x}^t B \mathbf{x}\begin{pmatrix} 1 \\ - 1\end{pmatrix}- \frac{\partial V}{\partial \mathbf{q}}\right).$

Newton's method is to consider the iteration $\mathbf{x}_{m+1} = \mathbf{x}_m - F'(\mathbf{x}_m)^{-1}F(\mathbf{x}_m)$. Now
$F'(\mathbf{x}) = \mathrm{Id} + h \begin{pmatrix} -\mathbf{x}^t B \\ \mathbf{x}^t B \end{pmatrix}$
where $\mathrm{Id}$ is the identity matrix. This, for small enough $h$, is certainly invertible. In fact, the Newton iterations tend to be stable simply because the presence of an identity matrix with small $h$ makes the problem well-conditioned. And as a starting point for Newton's method, we can use $\mathbf{x}_0 = \mathbf{p} = \mathbf{p}^n$, the value of $\mathbf{p}$ at the current timestep; it should be reasonably close, because it is the smooth solution to differential equation defined by a smooth vector field. Finally, we should stop the Newton iterations when the residual $F(\mathbf{x}_m)$ falls within a small tolerance, say, $10^{-7}$; we then simply accept $\mathbf{x}_m$ as the new value $\mathbf{p}^{n+1}$. Here's how to say that in MATLAB:

   A = [ mm*r1^2, m2*r1*r2*cos(q(1)-q(2));
        m2*r1*r2*cos(q(1)-q(2)), m2*r2^2]; % matrix I
   C = [0 , -m2*r1*r2*sin(q(1)-q(2)) ;
         -m2*r1*r2*sin(q(1)-q(2)), 0];     % matrix dI/dq
   B = -A\(C/A);                           % derivative of inverse matrix
                                           % -I^-1 dI/dq I^-1
   dVdq = [mm*g*r1*sin(q(1)); m2*g*r2*sin(q(2))];

   x = p;          % initialize Newton iteration
   F = [1.0; 1.0]; % set initially to something that'll give a big norm
   iter = 0;
   while (norm(F) > 1.0e-7 && iter <= itmax)
      F = x - p + h *(1/2* [x'*B*x; -x'*B*x] + dVdq); % current F
      DF = eye(2) + h*([x'*B; -x'*B]);                % Jacobian
      x = x - DF\F;                                   % update iteration
      iter = iter+1;
   end
   p = x;            % update p
   q = q + h* (A\p); % update q, using updated p

Of course, we should save up the values of $q$ as a history if we want to see where it's been in state space. Here's a video of just that: we save the values of $q$ and plot them as a trail following the current point. But in this time we show the angles as being on the abstract phase space, a torus, rather than as the actual angular displacements of a pendulum (as shown in Part 1):

The picture that opened this post is simply the final frame of the movie, showing everywhere in state space where the pendulum has been. It is this abstract representation of a physical system that leads to intriguing visualizations! The idea is, hopefully, in some other physical system, the state space of the variables might more clearly point to interesting features of the system. We shall frequently keep this philosophy in mind in our posts!


text:
# Math Help - Polar coordinate simple problem

1. ## Polar coordinate simple problem

I am trying to convert dxdy in cartesian into polar plane.
and have to prove: dxdy = rd $\theta$dr

Here is my conversion proof:

x = rcos $\theta$
y = rsin $\theta$

dx = -rsin $\theta$d $\theta$ + cos $\theta$dr
dy = rcos $\theta$d $\theta$ + sin $\theta$dr

ignoring 2nd order differentials, and keeping only drd $\theta$,
we have:
dxdy = -rsin $\theta$d $\theta$sin $\theta$dr
+ rcos $\theta$drcos $\theta$d $\theta$
= rd $\theta$dr(-sin $^2 \theta$ + cos $^2 \theta$)

I need to equate (-sin $^2 \theta$ + cos $^2 \theta$) = 1,
but i can't because of the -ve sign. Where did I go wrong?

2. The Jacobian is r, so $dxdy=r\cdot drd\theta$

3. Here's a geometric/visual proof:

Consider the graph of a polar rectangle in the cartesian plane with area $dxdy$ (a polar rectangle has height in radius and length in angle). Consider the polar rectangle as subtracting a "sliver" of a smaller circle (let's say with radius $r_{2}$ from a "sliver" of a larger circle (radius $r_{2}$). Both have the same angle width of $\theta_{2}-\theta_{1}$; subtracting the smaller angle from the larger angle. So the area of the polar rectangle is $dxdy=\frac{1}{2}(\theta_{2}-\theta_{1})r_{2}^{2}-\frac{1}{2}(\theta_{2}-\theta_{1})r_{1}^{2}$ Well, $\theta_{2}-\theta_{1}=d\theta$; it's the change in $\theta$. So let's rewrite this as:

$dxdy=\frac{1}{2}d\theta(r_{2}^{2}-r_{1}^{2})=\frac{1}{2}d\theta(r_{2}+r_{1})(r_{2}-r_{1})$

Again, we can make substitutions. $\frac{1}{2}(r_{2}+r_{1})$ can really be thought of as the average of radii, and as this polar rectangle gets infinitely smaller and smaller, the average will simply equal $r$. And as with $d\theta$, $r_{2}-r_{1}=dr$, or the change in $r$. Making all the substitutions, we finally arrive at:

$dxdy=rdrd\theta$

4. Hi Pinkk, that was pretty neat. Thanks for your rather rigorous geometrical proof. But I am still interested in my analytic approach, and I need to know how to get rid of the -ve sign.

And I do not want to use the Jacobi method for the time being.

Thanks again.

5. Originally Posted by chopet
I am trying to convert dxdy in cartesian into polar plane.
and have to prove: dxdy = rd $\theta$dr

Here is my conversion proof:

x = rcos $\theta$
y = rsin $\theta$

dx = -rsin $\theta$d $\theta$ + cos $\theta$dr
dy = rcos $\theta$d $\theta$ + sin $\theta$dr

ignoring 2nd order differentials, and keeping only drd $\theta$,
we have:
dxdy = -rsin $\theta$d $\theta$sin $\theta$dr
+ rcos $\theta$drcos $\theta$d $\theta$
= rd $\theta$dr(-sin $^2 \theta$ + cos $^2 \theta$)

I need to equate (-sin $^2 \theta$ + cos $^2 \theta$) = 1,
but i can't because of the -ve sign. Where did I go wrong?

Hello Chopet!

What you are effectively trying to do is derive the jacobian from first principles.

The reason your approach is going wrong is that you are forgetting that the $\mathrm{d}x$'s, $\mathrm{d}y$'s, $\mathrm{d}r'$s and $\mathrm{d}\theta$'s all have associated directions i.e. you should be treating them as vectors.

Now you should know that the area of a parallelogram formed by two vectors is given by the magnitude of the cross product of those two vectors. So:

$
\mathrm{d}x \, \mathrm{d}y = |\mathbf{dx} \times \mathbf{dy}|$

$
= |(-r\sin \theta \, \boldsymbol{\mathrm{d}\theta} + \cos \theta \mathbf{dr}) \times (r\cos \theta \, \boldsymbol{\mathrm{d}\theta} + \sin \theta \mathbf{dr})|
$

$= |-r \sin^2 \theta \, \boldsymbol{\mathrm{d} \theta} \times \mathbf{dr}
+r \cos^2 \theta \, \mathbf{dr} \times \boldsymbol{\mathrm{d}\theta}|$

$= | r (\sin^2 \theta +\cos^2 \theta) \, \mathbf{dr} \times \boldsymbol{\mathrm{d}\theta}|$
$= r \, \mathrm{d}r \,\mathrm{d} \theta$

Now you'll notice about the above that the terms of order $\mathrm{d}r^2$ and $\mathrm{d} \theta^2$ drop out because their corresponding vector cross products are zero rather than because you arbitrarily decide they're smaller.

Finally you should also note that had you written your two original equations in matrix form then taking the determinant of that matrix would be equivalent to the vector cross product that I've just shown and infact is just our old friend the Jacobian!


text:
# [NTG-context] How to break laws in physics?

Hans Hagen pragma at wxs.nl
Mon Jan 9 19:39:52 CET 2006

Mojca Miklavec wrote:

>\definetypeface [iwona] [mm] [math] [iwona] [default]
>\setupbodyfont[iwona]
>
>${\partial\rho\over\partial t}+\vec{\nabla}\vec{\jmath}=0$
>
>(If I fail on the exam, I'll try to explain to professor that I found
>a few mistakes in the Maxwell equations and lost the whole time trying
>to fix them before I was able to learn further :)
>
>
ok, so  it's may fault that you failed that examn ... if so ... at
eurotex/bachotex i'll buy you a beer

>
>
>I've found three "kind-of-bugs" in font itself (\jmath and \imath were
>missing, \partial is mirrored, vectors placed strangely - the latter
>may not be connected with the font itself).
>
>
can you communicate this with J&J ...

>So I tried to redefine \jmath to use dotlessj from italic version of
>font and found another "bug" in ConTeXt. dotlessj is commented out in
>enco-ec.tex. Well, I understand that it may not be present in every
>font, but it's worse: it's already defined in the default encoding, so
>\dotlessj results in double quotes, which is worse than no char at
>all. enco-ec has some other faults present, I tried to fix a few of
>them ([t|s]cedilla should be [t|s]commaaccent, \Eth and \eth were
>commented out, while they are OK, one glyph is present twice, I
>redefined it to map to the first slot, since usually the second one is
>empty, upper/lower casing may still need some revision).
>
>Is there any chance that any numerical definitions of characters
>disappear from enco-def slowly? If anyone is using any other encoding
>than the standard one, the characters map to something completely
>weird then.
>
>
actually all chars in enco-def that use funny numbers should be present
in a (math) font vector; better a wrong something that nothing (we need
a fallback)

the alternative is to have a def which is mostly mapping to nothing and
then make the tex encoding on top of that (quite an effort and it may
temporary introduce bugs)

>Some fixes for enco-def:
>
>% DELETE THIS ONE: IT'S WORSE THAN NOTHING
>\definecharacter eth              {\dstroke} % wrong but better that nothing
>% and replace by:
>\definecharacter eth              {\unknownchar}
>
>
ok, gone are the dstrokes (only you use them -)

>\definecharacter scommaaccent     {\buildtextbottomcomma s}
>\definecharacter Scommaaccent     {\buildtextbottomcomma S}
>\definecharacter tcommaaccent     {\buildtextbottomcomma t}
>\definecharacter Tcommaaccent     {\buildtextbottomcomma T}
>
>\definecharacter Ohorndotbelow       {\buildtextbottomdot\Ohorn}
>\definecharacter ohorndotbelow       {\buildtextbottomdot\ohorn}
>\definecharacter Uhorndotbelow       {\buildtextbottomdot\Uhorn}
>\definecharacter uhorndotbelow       {\buildtextbottomdot\uhorn}
>
>

>enco-ini:
>
>Any chances to add something like:
>\definecharacter bottomcomma {,} % comma, lowered a bit
>\def\buildtextbottomcomma{\bottomaccent{.25ex}{0}{5}{\textbottomcomma}}
>
>I have no idea what \textbottomcomma should be, but probably just a
>comma, lowered a bit. The current definition with ' is a bit strange.
>
>
hm, so you suppose that i have an idea ...

\definetypeface[xxx][rm][serif][modern][default][encoding=qx]

{\xxx \scommaaccent}

\definecharacter textbottomcomma  {\hbox{\lower.35ex\hbox{\tx,}}}

\definecharacter scommaaccent     {\buildtextbottomcomma s}

\def\buildtextbottomcomma{\bottomaccent{.15ex}{0}{5}{\tx,}}

\scommaaccent \textbottomcomma

\stoptext

Hans