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# Better prediction of probabilities through auctions

I bet you think this is another post about Hypothesis, don’t you?

Nope! This is about something completely different. This is about prediction markets and proper scoring rules.

Which aren’t actually something I know that much about, so it’s very likely everything in this post is either obvious or a bad idea. This should just be considered a “here’s some stuff I thought of while reading about this and I couldn’t find anyone who wrote this down before so I’ll write it down now” post.

A scoring rule is a way of rewarding predictions. You have some possibilities A1, …, An which are disjoint and cover everything. So e.g. you could have A1 = “there will be a labour majority in the UK 2015 elections”, A2 = “there will be a conservative majority”, A3 = “none of the above”. You want people to assign probabilities to these events. They give you a probability distribution and depending on which outcome occurs you give them a reward (or penalise them for negative rules, which a lot of them are).

A scoring rule is proper if predicting the correct probabilities is an optimal solution and strictly proper if it’s the only optimal solution. For example a scoring rule that gives you a score of log(r_i) where r_k is the probability you assigned to outcome k and i is the outcome that occurs is a strictly proper scoring rule. A scoring rule that assigns you a score of 0 regardless of what happens is proper but not strictly proper (You can’t do better than 0, but anything you want to predict at all doesn’t do worse than that either).

When thinking about these things I noticed a particularly simple sort of proper scoring rule. It’s not strictly proper, but under a certain natural extension of the definitions can be made into a strictly proper extended scoring rule.

The rule is this:

Let $$q_1, \ldots, q_n$$ be any sequence of numbers in $$[0, 1]$$. They do not have to sum to $$1$$. Let $$r_1, \ldots, r_n$$ be your prediction. Your total cost is $$C = \sum\limits_{i, q_i < r_i} q_i$$. Let $$i$$ be the winning outcome. Then you score $$1 – C$$ if $$r_i > q_i$$ and $$-C$$ otherwise.

You might notice that you are basically conducting a simultaneous Vickrey auction on each coordinate.

Theorem: Under this scoring rule, a distribution $$r$$ obtains the optimal score if and only if $$r_i < q_i$$ whenever $$p_i < q_i$$ and $$q_i < r_i$$ whenever $$p_i < r_i$$.

Proof:

In order to simplify this proof we will show that this is true without the constraint that $$\sum r_i = 1$$. Showing that these are the optimal solutions in the general case also implies that they are the optimal solutions in the specific case.

Let $$t_i = 1$$ if $$r_i > q_i$$ else $$t_i = 0$$. Then your expected score is $$\sum t_i (p_i – r_i)$$, because you pay a cost of $$r_i$$ for each case where $$t_i = 1$$ and a cost of $$0$$ for each case where $$t_i = 0$$, and you get a benefit of $$1$$ with probability $$p_i$$ if $$t_i$$ = 1 and 0 otherwise.

This means that the multiplier for $$t_i$$ is negative if $$p_i < r_i$$, positive $$p_i > r_i$$ and $$0$$ when $$p_i = r_i$$. So the only sensible choices to achieve optimality are to set $$t_i = 0$$ when the coefficient is negative and $$1$$ when it is positive. It doesn’t matter what you set it to when it’s 0 because the expected value is the same either way.

QED

Note that this uses very little about your chosen distribution: All you’re actually doing is picking a set of values and saying “I think your current choice of estimates undervalues these and I’m prepared to stake money on it”. This is why it’s not a proper scoring rule: Any distribution which correctly identifies which values are underconfident will do.

There are two things you can do with this. One of them turns it into a proper scoring rule of sorts, the other just uses that feature as is. I’ll now explain both:

#### Randomizing for fun and expected profit

The solution which makes it a proper scoring rule is as follows: Let $$Q$$ be any random vector taking values in $$[0, 1]^n$$ with a smooth and strictly positive distribution function.

Now play exactly the above scoring rule, but you make your prediction and then $$Q$$ is drawn and you play the scoring rule against the drawn value.

Lets see that the only optimal choice of $$r_i$$ is $$p_i$$:

Consider again the profit you make from component $$i$$. Call this profit $$T_i$$. Then $$E(T_i|Q_i = q_i) = p_i – q_i$$ if $$r_i > q_i$$ or 0 if $$r_i \leq q_i)$$.

So $$E(T_i) = (p_i – E(q_i | q_i < r_i) ) P(q_i < r_i) = p_i P(q_i < r_i) – \int\limits_0^{r_i} x f_i(x) dx = \int\limits_0^{r_i} (p_i – x) f_i(x)$$

Where $$f_i$$ is the marginal distribution of $$Q_i$$. Now, as assumed, $$f_i$$ is strictly positive. That means that this integrand is strictly increasing for as long as $$p_i \leq x$$ and strictly decreasing once $$x > p_i$$, so the optimal value has to be at $$r_i = p_i$$.

QED

We can also derandomize this given any explicit choice of $$Q$$: You make your prediction and then when the results come in we give you what would have been the expected value of your choice.

Lets consider how that would work for the simple case where each $$q_i$$ is chosen uniformly at random. Splitting it up once more into benefit and cost, your expected reward is the probability that you would have won the winning index $$i$$. That is, it’s $$r_i$$.

Meanwhile your total cost is $$\sum E(Q_i | Q_i \leq r_i) P(Q_i \leq r_i = \sum \frac{1}{2} r_i^2$$. So your reward is $$r_i – \frac{1}{2} \sum r_i^2$$. And this is just our old friend the quadratic scoring rule! So it’s possible that by picking other interesting probability distributions for $$q$$ we could produce other interesting proper scoring rules. I don’t know. I haven’t really investigated.

#### Prediction markets

As Robin Hanson famously observed, you can turn any proper scoring rule into a prediction market. I don’t think what I’m about to describe is actually an instance of his scheme, but I don’t feel like I’ve sufficiently digested said scheme to be sure that it’s not. It’s certainly very close even if it’s not exactly the same thing.

You can use the original non-random form of the scoring rule to create what’s basically the world’s simplest prediction market:

You have n disjoint events as above that you want to predict, and you want to solicit the wisdom of the crowds in predicting it.

You do so by placing an initial non-zero bid on each option, $$b_i$$. The market is now open for bidding.

A bid consists of simply picking one of the outcomes. When bidding on outcome $$i$$ you pay the current market prediction, which is $$\frac{b_i}{\sum b_j}$$ and increment $$b_i$$ by one. If this is the outcome that occurs then you will get a reward of $$1$$ for your bid. People may bid as many times as they want on as many options as they want.

This scheme should converge to something corresponding to the true beliefs of the market. When the market believes that the current prediction is under confident they will buy shares in it, raising its value. When they believe it is overconfident they must also believe that some other value is underconfident (because the probabilities sum to 1 so if $$q_i > p_i$$ for some $$i$$ then $$q_j < p_j$$ for some $$j$$), so they will bid on that one, causing all the other probabilities to fall.

Convergence in theory is quite slow, basically oscillating like $$\frac{1}{n}$$ in the number of bids on either side of the true probabilities, but I think in practice that still quickly takes it within the realms of being finer precision than peoples’ predictions are going to be anyway.

I’ve no idea if this is a good prediction mechanism in general, but it’s appealingly simple and might be worth consideration in some contexts for that alone.

This entry was posted in Uncategorized on by .


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# More symmetry and Another Product

We’ve seen that $$\zeta(s)$$ satisfies the functional equation

$\zeta(1-s)=2^{1-s}\pi^{-s}\cos(\pi s/2)\Pi(s-1)\zeta(s).$

(Well, it still needs to be proved, but let’s just assume it’s correct for now.) The goal of this post is an even more symmetrical form that will yield the function $$\xi(s)$$ which we can develop into an incredibly useful product expression.

On our wish list for $$\xi(s)$$ we find three items:

1. It’s an entire function, i.e., a function that’s holomorphic everywhere in $$\mathbb{C}$$ without any poles.
2. It has zeros for all non-trivial zeros of the $$\zeta$$-function, and no others.
3. It’s perfectly symmetrical along the critical line, i.e., it satisfies $$\xi(1-s)=\xi(s)$$.

The first two requirements are fairly easy to satisfy. To make $$\zeta(s)$$ holomorphic, we just need to cancel out its single simple pole at $$s=1$$ by multiplying $$s-1$$. Keeping in mind the desired symmetry, we also multiply a factor $$s$$. To get rid of the trivial zeros $$s=-2,-4,-6,\ldots$$, plus the one we just introduced at $$s=0$$, we need to multiply by a function that has (simple) poles at exactly these points, and no zeros. From the functional equation, we know that the factorial function $$\Pi(s)$$ is a pretty good candidate, which has no zeros, but poles for $$s=-1,-2,-3,\ldots$$. So $$\Pi(s/2)$$ has poles at $$s=-2,-4,-6,\ldots$$, and $$\Pi(s/2-1)$$ has poles at $$s=0,-2,-4,\ldots$$, perfectly covering all the zeros we needed eliminated without introducing new ones. This leads us to

$\xi(s)=\zeta(s)s(s-1)\Pi(s/2-1)\hat{\xi}(s),$

where $$\hat{\xi}(s)$$ is just a placeholder for other factors that may be necessary achieve requirement 3, the symmetry around the critical line. We are free to choose anything for $$\hat{\xi}(s)$$ as long as it does not interfere with the other two requirements, i.e., it cannot have zeros or poles. (Spoiler alert: Pretty much the only entire functions satisfying this are constant and exponential functions, and that’s exactly what we will need.)

I won’t try to create any further tension as we’re almost there (and I suck at it anyway). Indeed, Riemann found the integral expression

$\pi^{-s/2}\Pi(s/2-1)\zeta(s)=-\frac{1}{s(1-s)}+\int_1^\infty\left(x^{s/2-1}+x^{-s-1/2}\right)\psi(x),\mathrm{d}x,$

where

$\psi(x)=\sum_{n=1}^\infty e^{-\pi n^2 x}$

is the exponential sum. I won’t go into the details, but it is now obvious that the right-hand side is unchanged when you substitute $$s$$ by $$1-s$$. In other words, if we define ((The factor 1/2 has just been introduced to make certain calculations more convenient.))

$\xi(s)=\frac{1}{2}\pi^{-s/2}s(s-1)\Pi(s/2-1)\zeta(s)$

we have found a function that satisfies all three requirements. Eureka!

So, why did we jump through all these hoops? I promised a nice product representation for $$\xi(s)$$, and we’re now in the position to get cracking. More specifically, it’s high time the zeros entered the scene!

Representing a function through a product over their zeros is a pretty common idea. Take, for instance, a polynomial function $$p(s)$$ of degree, say, $$d$$ with zeros $$\varrho_1,\ldots,\varrho_d$$. (Of course we are all aware of the fundamental theorem of algebra which ensures that every polynomial of degree $$d$$ has exactly $$d$$ zeros over the complex numbers, right?) Now, consider the expression

$\hat{p}(s)=p(0)\prod_{n=1}^d\left(1-\frac{s}{\varrho_n}\right).$

For $$s=0$$, all factors are $$1$$, so $$\hat{p}(0)=p(0)$$. Further, if $$s=\varrho_i$$, i.e., one of the zeros, then the corresponding factor is $$0$$, and hence the whole product, as is the polynomial at this point, so $$\hat{p}(\varrho_i)=p(\varrho_i)$$. Now, we found an expression that corresponds to our polynomial in $$d+1$$ points (the $$d$$ zeros and the origin). Since the polynomial has $$d+1$$ degrees of freedom (the $$d+1$$ coefficients of the polynomial), it is completely determined by these points, and hence must correspond to our expression: $$\hat{p}(s)=p(s)$$ for all $$s$$. (In all I’ve just said I implicitly assumed the zeros to be simple, and $$p(0)\neq0$$, but these are just technical obstacles.)

Why shouldn’t the same hold for infinitely many zeros? Holomorphic function are really just polynomials of infinite degree, so we would expect an expression like

$\xi(s)=\xi(0)\prod_\varrho\left(1-\frac{s}{\varrho}\right),$

where the product runs over all the zeros of $$\xi(s)$$ which, by our construction, are exactly the non-trivial zeros of $$\zeta(s)$$. Of course, this argumentation is all very vague, and for the above expression to be correct, we at the very least need convergence of the product. As an additional obstacle, the product converges only conditionally, i.e., the convergence depends on the ordering of the zeros. It turns out the “right” ordering is to pair every zero $$\varrho$$ with the mirrored $$1-\varrho$$. Recalling what we learnt earlier, it suffices then to prove that $$\sum_{\Im\varrho>0}1/|\varrho(1-\varrho)|$$ converges.

Riemann states in his paper (though he makes no attempt to prove it, so neither shall I feel under any obligation to do so) that the number of zeros of $$\zeta(s)$$ (or $$\xi(s)$$, if you prefer) in the critical strip with imaginary part between $$0$$ and $$T$$ is approximately

$\frac{T}{2\pi}\log\frac{T}{2\pi}-\frac{T}{2\pi}.$

This implies, very roughly, that the $$n$$th zero has modulus about $$n/\log n$$, so the above sum behaves similar to $$\sum(\log n)^2/n^2$$ which indeed converges, and hence the product does.

There’s a lot of hand-waving and unproved arguments in this article, some of which we may return to, some which we just leave as it is. But one conclusion will hopefully stick: We have the functions $$\zeta(s)$$ and $$\xi(s)$$ (which are essentially the same for all we are concerned about), and represented them as two products, one over the prime numbers, and one over the zeros in the critical strip. This intrinsically links these two together! We will go on to use the zeros to calculate primes, both approximately and exactly.

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1. Engineering
2. Computer Science
3. 1why are network logs important to the security of our...

# Question: 1why are network logs important to the security of our...

###### Question details

1)Why are network logs important to the security of our data?

2)Why is it important to audit passwords and software licensing?


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# 1 Discriminant Analysis

Discriminant analysis is a classification problem, where two or more groups or clusters or populations are known a priori and one or more new observations are classified into one of the known populations based on the measured characteristics.

Discriminant analysis works by creating one or more linear combinations of predictors, creating a new latent variable for each function. These functions are called discriminant functions. The first function created maximizes the differences between groups on that function. The second function maximizes differences on that function, but also must not be correlated with the previous function. This continues with subsequent functions with the requirement that the new function not be correlated with any of the previous functions.

A couple of things need to know for discriminant analysis

• We have data from known groups.

• We want to determine one or more functions based on explanatory variables to classify those groups.

• The classification is based on group means.

• Our goal is to be able to classify new data into one of the known groups based on the classification defined by discriminant function(s).

## 1.1 Linear Discriminant Analysis (LDA)

Linear Discriminant Analysis(LDA) is a classification method originally developed in 1936 by R.A.Fisher. It is simple, mathematically robust and often produces models whose accuracy is as good as more complex methods.

We can think of discriminant analysis as drawing lines in space to (in some sense) optimally separate groups. In the linear case with two groups, we would have a discriminant function $z = a_1x_1 + a_2x_2 + ... + a_px_p$ The vector $$a = (a_1, a_2, ..., a_p)'$$ maximizes the ratio of the between-groups variance of $$z$$ to its within- groups variance. LDA carries the following assumptions:

• The data from each group have multivariate normal distributions.

• The covariances of each group are the same.

For two groups and assuming both groups are equally probable (e.g. have equal prior probabilities), we can compute $$z_i$$ from the $$i_{th}$$ observation and compare it with $z_c = \frac{\bar{z}_1 + \bar{z}_2}{2}$

If group 1 has smaller mean $$z$$ value, a new observation is classified as being in group 1 if {z_i < z_c} and in group 2 otherwise. In the case of more than 2 groups, there would be multiple (linear in the case of LDA) discriminant functions partitioning the space.

## 1.2 Quadratic Discriminant Analysis (QDA)

In quadratic discriminant analysis, we still assume multivariate normality but within-group covariances are no longer assumed to be equal and the discriminant functions are no longer linear, so QDA carries the following assumptions:

• The data from each group have multivariate normal distributions.

• The covariances of each group are not all the same.

# 2 k-Nearest Neighbors

The KNN or k-nearest neighbors algorithm is one of the simplest machine learning algorithms and is an example of instance-based learning, where new data are classified based on stored, labeled instances. More specifically, the distance between the stored data and the new instance is calculated by means of some kind of a similarity measure. This similarity measure is typically expressed by a distance measure such as the Euclidean distance, cosine similarity or the Manhattan distance. In other words, the similarity to the data that was already in the system is calculated for any new data point that you input into the system. Then, you use this similarity value to perform predictive modeling. Predictive modeling is either classification, assigning a label or a class to the new instance, or regression, assigning a value to the new instance. Whether you classify or assign a value to the new instance depends of course on your how you compose your model with KNN.

The k-nearest neighbor algorithm adds to this basic algorithm that after the distance of the new point to all stored data points has been calculated, the distance values are sorted and the k-nearest neighbors are determined. The labels of these neighbors are gathered and a majority vote or weighted vote is used for classification or regression purposes. In other words, the higher the score for a certain data point that was already stored, the more likely that the new instance will receive the same classification as that of the neighbor. In the case of regression, the value that will be assigned to the new data point is the mean of its k nearest neighbors.

The choice of K is essential in building the KNN model. In fact, k can be regarded as one of the most important factors of the model that can strongly influence the quality of predictions. One appropriate way to look at the number of nearest neighbors k is to think of it as a smoothing parameter. For any given problem, a small value of k will lead to a large variance in predictions. Alternatively, setting k to a large value may lead to a large model bias. Thus, k should be set to a value large enough to minimize the probability of misclassification and small enough (with respect to the number of cases in the example sample) so that the K nearest points are close enough to the query point. Thus, like any smoothing parameter, there is an optimal value for k that achieves the right trade off between the bias and the variance of the model. KNN can provide an estimate of K using an algorithm known as cross-validation.

############################################################################
# South African Heart Disease Data
############################################################################
# The Coronary Risk‐Factor Study data involve 462 males between the ages of 15 and 64 from
# a heart‐disease high‐risk region of the Western Cape, South Africa.
# The response is "chd", the presence (chd=1) or absence (chd=0) of coronary heart disease.
# There are 9 covariates:
# sbp:  systolic blood pressure cumulative tobacco (kg)
# tobacco:  cumulative tobacco (kg)
# ldl:  low densiity lipoprotein cholesterol
# famhist:  family history of heart disease (Present, Absent)
# typea:  type‐A behavior
# obesity
# alcohol:  current alcohol consumption
# age:  age at onset

attach(Heart)
dim(Heart)
## [1] 462  10
head(Heart)
##   sbp tobacco  ldl adiposity famhist typea obesity alcohol age chd
## 1 160   12.00 5.73     23.11 Present    49   25.30   97.20  52   1
## 2 144    0.01 4.41     28.61  Absent    55   28.87    2.06  63   1
## 3 118    0.08 3.48     32.28 Present    52   29.14    3.81  46   0
## 4 170    7.50 6.41     38.03 Present    51   31.99   24.26  58   1
## 5 134   13.60 3.50     27.78 Present    60   25.99   57.34  49   1
## 6 132    6.20 6.47     36.21 Present    62   30.77   14.14  45   0
Heart$chd = as.factor(Heart$chd)

Num_train = round(dim(Heart)[1]*0.8)
Index_train = sample(nrow(Heart), Num_train)
Heart_train = Heart[Index_train,]
Heart_test = Heart[-Index_train,]
HeartFull = glm(chd~., data=Heart_train, family=binomial)
summary(HeartFull)
##
## Call:
## glm(formula = chd ~ ., family = binomial, data = Heart_train)
##
## Deviance Residuals:
##     Min       1Q   Median       3Q      Max
## -1.8360  -0.8041  -0.4356   0.8536   2.5328
##
## Coefficients:
##                 Estimate Std. Error z value Pr(>|z|)
## (Intercept)    -6.627964   1.500123  -4.418 9.95e-06 ***
## sbp             0.012319   0.006516   1.891 0.058668 .
## tobacco         0.092421   0.030004   3.080 0.002068 **
## ldl             0.194224   0.066003   2.943 0.003254 **
## adiposity       0.030069   0.032798   0.917 0.359255
## famhistPresent  0.986346   0.256582   3.844 0.000121 ***
## typea           0.043229   0.013649   3.167 0.001540 **
## obesity        -0.083934   0.048969  -1.714 0.086523 .
## alcohol        -0.001928   0.004860  -0.397 0.691607
## age             0.035942   0.013504   2.662 0.007776 **
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
##     Null deviance: 478.49  on 369  degrees of freedom
## Residual deviance: 373.42  on 360  degrees of freedom
## AIC: 393.42
##
## Number of Fisher Scoring iterations: 5
HeartStep = step(HeartFull, scope=list(upper=~., lower=~1), k=2, trace=0)
summary(HeartStep)
##
## Call:
## glm(formula = chd ~ sbp + tobacco + ldl + famhist + typea + obesity +
##     age, family = binomial, data = Heart_train)
##
## Deviance Residuals:
##     Min       1Q   Median       3Q      Max
## -1.9079  -0.8098  -0.4351   0.8751   2.5622
##
## Coefficients:
##                 Estimate Std. Error z value Pr(>|z|)
## (Intercept)    -7.070551   1.419165  -4.982 6.29e-07 ***
## sbp             0.012583   0.006407   1.964 0.049544 *
## tobacco         0.090811   0.029558   3.072 0.002124 **
## ldl             0.209208   0.064446   3.246 0.001169 **
## famhistPresent  0.976077   0.255085   3.826 0.000130 ***
## typea           0.042194   0.013575   3.108 0.001881 **
## obesity        -0.051441   0.032901  -1.564 0.117932
## age             0.042571   0.011631   3.660 0.000252 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
##     Null deviance: 478.49  on 369  degrees of freedom
## Residual deviance: 374.37  on 362  degrees of freedom
## AIC: 390.37
##
## Number of Fisher Scoring iterations: 5
train_phat=HeartStep$fitted.values logistic_train_pred = (train_phat>0.5) table(Heart_train$chd, logistic_train_pred)
##    logistic_train_pred
##     FALSE TRUE
##   0   206   35
##   1    58   71
logistic_train_err = (63+30)/nrow(Heart_train)
logistic_train_err
## [1] 0.2513514
test_phat=predict(HeartStep,newdata = Heart_test)
logistic_test_pred = (test_phat>0.5)
table(Heart_test$chd, logistic_test_pred) ## logistic_test_pred ## FALSE TRUE ## 0 55 6 ## 1 21 10 logistic_test_err = (23+4)/nrow(Heart_test) logistic_test_err ## [1] 0.2934783 library(MASS) LDA = lda(chd ~. ,data=Heart_train) LDA_train_pred=predict(LDA, Heart_train[,-10])$class
table(Heart_train$chd, LDA_train_pred) ## LDA_train_pred ## 0 1 ## 0 209 32 ## 1 59 70 LDA_train_err = mean(LDA_train_pred != Heart_train$chd)
LDA_train_err
## [1] 0.2459459
LDA_test_pred=predict(LDA, Heart_test[,-10])$class table(Heart_test$chd, LDA_test_pred)
##    LDA_test_pred
##      0  1
##   0 50 11
##   1 16 15
LDA_test_err = mean(LDA_test_pred != Heart_test$chd) LDA_test_err ## [1] 0.2934783 QDA = qda(chd ~. ,data=Heart_train) QDA_train_pred=predict(QDA, Heart_train[,-10])$class
table(Heart_train$chd, QDA_train_pred) ## QDA_train_pred ## 0 1 ## 0 210 31 ## 1 52 77 QDA_train_err = mean(QDA_train_pred != Heart_train$chd)
QDA_train_err
## [1] 0.2243243
QDA_test_pred=predict(QDA, Heart_test[,-10])$class table(Heart_test$chd, QDA_test_pred)
##    QDA_test_pred
##      0  1
##   0 49 12
##   1 17 14
QDA_test_err = mean(QDA_test_pred != Heart_test$chd) QDA_test_err ## [1] 0.3152174 library(class) knn_test_pred <- knn(Heart_train[,-5], Heart_test[,-5], Heart_train$chd,k=20)
table(knn_test_pred, Heart_test$chd) ## ## knn_test_pred 0 1 ## 0 52 23 ## 1 9 8 knn_test_err = mean(knn_test_pred != Heart_test$chd)
knn_test_err
## [1] 0.3478261


text:
# Properties

 Label 3920.2.a.bq Level $3920$ Weight $2$ Character orbit 3920.a Self dual yes Analytic conductor $31.301$ Analytic rank $1$ Dimension $2$ CM no Inner twists $1$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$3920 = 2^{4} \cdot 5 \cdot 7^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 3920.a (trivial)

## Newform invariants

 Self dual: yes Analytic conductor: $$31.3013575923$$ Analytic rank: $$1$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{2})$$ Defining polynomial: $$x^{2} - 2$$ Coefficient ring: $$\Z[a_1, a_2, a_3]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 35) Fricke sign: $$1$$ Sato-Tate group: $\mathrm{SU}(2)$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of $$\beta = \sqrt{2}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + ( -1 + \beta ) q^{3} + q^{5} -2 \beta q^{9} +O(q^{10})$$ $$q + ( -1 + \beta ) q^{3} + q^{5} -2 \beta q^{9} + ( -2 + 2 \beta ) q^{11} + ( -2 - 2 \beta ) q^{13} + ( -1 + \beta ) q^{15} + ( 2 + 2 \beta ) q^{17} -2 \beta q^{19} + ( 1 - \beta ) q^{23} + q^{25} + ( -1 - \beta ) q^{27} - q^{29} + 6 q^{31} + ( 6 - 4 \beta ) q^{33} -2 q^{39} + ( -5 - 2 \beta ) q^{41} + ( -5 + \beta ) q^{43} -2 \beta q^{45} -2 q^{47} + 2 q^{51} + ( -4 + 2 \beta ) q^{53} + ( -2 + 2 \beta ) q^{55} + ( -4 + 2 \beta ) q^{57} + ( 4 - 6 \beta ) q^{59} + ( -3 + 6 \beta ) q^{61} + ( -2 - 2 \beta ) q^{65} + ( -11 + \beta ) q^{67} + ( -3 + 2 \beta ) q^{69} + ( 4 - 6 \beta ) q^{71} + ( 2 - 2 \beta ) q^{73} + ( -1 + \beta ) q^{75} + ( -12 - 2 \beta ) q^{79} + ( -1 + 6 \beta ) q^{81} + ( -1 - 9 \beta ) q^{83} + ( 2 + 2 \beta ) q^{85} + ( 1 - \beta ) q^{87} + ( -3 - 4 \beta ) q^{89} + ( -6 + 6 \beta ) q^{93} -2 \beta q^{95} + ( 6 + 4 \beta ) q^{97} + ( -8 + 4 \beta ) q^{99} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2q - 2q^{3} + 2q^{5} + O(q^{10})$$ $$2q - 2q^{3} + 2q^{5} - 4q^{11} - 4q^{13} - 2q^{15} + 4q^{17} + 2q^{23} + 2q^{25} - 2q^{27} - 2q^{29} + 12q^{31} + 12q^{33} - 4q^{39} - 10q^{41} - 10q^{43} - 4q^{47} + 4q^{51} - 8q^{53} - 4q^{55} - 8q^{57} + 8q^{59} - 6q^{61} - 4q^{65} - 22q^{67} - 6q^{69} + 8q^{71} + 4q^{73} - 2q^{75} - 24q^{79} - 2q^{81} - 2q^{83} + 4q^{85} + 2q^{87} - 6q^{89} - 12q^{93} + 12q^{97} - 16q^{99} + O(q^{100})$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
1.1
 −1.41421 1.41421
0 −2.41421 0 1.00000 0 0 0 2.82843 0
1.2 0 0.414214 0 1.00000 0 0 0 −2.82843 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Atkin-Lehner signs

$$p$$ Sign
$$2$$ $$-1$$
$$5$$ $$-1$$
$$7$$ $$1$$

## Inner twists

This newform does not admit any (nontrivial) inner twists.

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 3920.2.a.bq 2
4.b odd 2 1 245.2.a.h 2
7.b odd 2 1 3920.2.a.bv 2
7.c even 3 2 560.2.q.k 4
12.b even 2 1 2205.2.a.n 2
20.d odd 2 1 1225.2.a.k 2
20.e even 4 2 1225.2.b.g 4
28.d even 2 1 245.2.a.g 2
28.f even 6 2 245.2.e.e 4
28.g odd 6 2 35.2.e.a 4
84.h odd 2 1 2205.2.a.q 2
84.n even 6 2 315.2.j.e 4
140.c even 2 1 1225.2.a.m 2
140.j odd 4 2 1225.2.b.h 4
140.p odd 6 2 175.2.e.c 4
140.w even 12 4 175.2.k.a 8

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
35.2.e.a 4 28.g odd 6 2
175.2.e.c 4 140.p odd 6 2
175.2.k.a 8 140.w even 12 4
245.2.a.g 2 28.d even 2 1
245.2.a.h 2 4.b odd 2 1
245.2.e.e 4 28.f even 6 2
315.2.j.e 4 84.n even 6 2
560.2.q.k 4 7.c even 3 2
1225.2.a.k 2 20.d odd 2 1
1225.2.a.m 2 140.c even 2 1
1225.2.b.g 4 20.e even 4 2
1225.2.b.h 4 140.j odd 4 2
2205.2.a.n 2 12.b even 2 1
2205.2.a.q 2 84.h odd 2 1
3920.2.a.bq 2 1.a even 1 1 trivial
3920.2.a.bv 2 7.b odd 2 1

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(\Gamma_0(3920))$$:

 $$T_{3}^{2} + 2 T_{3} - 1$$ $$T_{11}^{2} + 4 T_{11} - 4$$ $$T_{13}^{2} + 4 T_{13} - 4$$ $$T_{17}^{2} - 4 T_{17} - 4$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$-1 + 2 T + T^{2}$$
$5$ $$( -1 + T )^{2}$$
$7$ $$T^{2}$$
$11$ $$-4 + 4 T + T^{2}$$
$13$ $$-4 + 4 T + T^{2}$$
$17$ $$-4 - 4 T + T^{2}$$
$19$ $$-8 + T^{2}$$
$23$ $$-1 - 2 T + T^{2}$$
$29$ $$( 1 + T )^{2}$$
$31$ $$( -6 + T )^{2}$$
$37$ $$T^{2}$$
$41$ $$17 + 10 T + T^{2}$$
$43$ $$23 + 10 T + T^{2}$$
$47$ $$( 2 + T )^{2}$$
$53$ $$8 + 8 T + T^{2}$$
$59$ $$-56 - 8 T + T^{2}$$
$61$ $$-63 + 6 T + T^{2}$$
$67$ $$119 + 22 T + T^{2}$$
$71$ $$-56 - 8 T + T^{2}$$
$73$ $$-4 - 4 T + T^{2}$$
$79$ $$136 + 24 T + T^{2}$$
$83$ $$-161 + 2 T + T^{2}$$
$89$ $$-23 + 6 T + T^{2}$$
$97$ $$4 - 12 T + T^{2}$$


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# Understanding op amp solution

Discussion in 'Homework Help' started by Hitman6267, Jun 2, 2010.

1. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0

I also don't know V1 V2... are the voltages at what resistors.

2. ### ELECTRONERD AAC Fanatic!

May 26, 2009
1,146
16
Nodal analysis is typically used to analyze an op amp with a single voltage source on the input.

You need to remember the ideal attrributes of an op amp:

• Infinite Gain
• Infinite Input Impedance
• Zero Output Impedance
• V+ = V-
• I+ = I- = 0
Consider the resistor in series with Vs. Since the V+ input has infinite input impedance, the current flow theoretically can't go into your V+ input. By KCL, the current has to go somewhere, so it goes through the feedback network. Based on those hints, see if you can solve it now.

3. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
Could you tell me the effects of having the grounds the way we do ? Thats what is throwing me off. I have no idea what they do.

4. ### ELECTRONERD AAC Fanatic!

May 26, 2009
1,146
16
Take a look at your first attachement. You see how V+ is connected to GND? Therefore, at that point, it is 0V. Also keep in mind that V+ = V-, so V- = 0 as well. I'll show a quick example of solving for the current in the first resistor in series with Vs.

$I_R_1 = \frac{V_S - 0}{R_1}$

5. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
I get that, here's where I'm facing trouble.
After the current passes the first R in the feedback cable does it divide itself between the R linked to the ground and the other ?

V2= -Vs -2 isrc R , is relative the the ground linked to the Vs. The V0 they calculated is relative to which ground ?

6. ### ELECTRONERD AAC Fanatic!

May 26, 2009
1,146
16
Think of KCL, all your resistors are equal to 0.5kΩ. So after the first feedback resistor, the current splits equally into each of the next resistors.

Try taking the op amp out of your drawing and just simplify all the resistors. Remember that the first resistor in series with Vs is also in series with the first feedback resistor.

7. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
Are the first R in the feedback loop and the R connected to the ground in parallel ?

8. ### ELECTRONERD AAC Fanatic!

May 26, 2009
1,146
16
Didn't I say to take your op amp out of your drawing and solve for the resistor network?

Look at the attachement, this can be solved by very simple source transformations.

Or look at the Vo resistor, it's in parallel with one of the R's. You could simplify the feedback network that way also.

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9. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
First thank you for the effort of drawing it, second I can't do that unless I know if they're considered to be in parallel . In your drawing would the resistor that would result from the transformation of Vs and R1 be considered in parallel with R2 ?

10. ### ELECTRONERD AAC Fanatic!

May 26, 2009
1,146
16
It's almost like a puzzle, there's a certain way you have to solve it and you won't be able to solve it otherwise. In this case you'll have to work backwards, because R4 and R5 are in parallel with -8V across each of them. Or, do the source transformations yourself and you'll see which are in parallel or not.

Last edited: Jun 2, 2010
11. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
In your drawing between R4 and R5 there is a resistor missing. Did you remove it somehow ? If it still were there would R5 and R4 still be considered in parallel ?

12. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,637
1,299
All you need is to find equivalent resistance of this circuit :

Rz = 1.6R = 800Ω

Then we need to find the voltage on on this node

So if we find V1 then we know the current that is flow through Vs

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• ###### be3z tytułu.PNG
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13. ### Hitman6267 Thread Starter Member

Apr 6, 2010
82
0
That's what I don't know how to do.

14. ### The Electrician AAC Fanatic!

Oct 9, 2007
2,644
468
All the grounds are connected to the same point. Imagine that the circuit was sitting on a copper sheet; all the grounds could be connected to that sheet of copper. Ground is the reference for all the voltage measurements in your circuit.

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,637
1,299
We can try solve this circuit without finding equivalent resistance.
We draw this

Is = I1
I3 = I1 + I2
I5 = I3 + I4
V1 = I2*R2
V2 = I3*R3 + I2*R2
Vout = I5*R5 + I4*R4

Vs/R = (0 - V1) / R1 ----> V1 = -R1/R * Vs

I2 = V1/R2 = Vs/R2*R1/R

I3 = I1 + I2 = Vs/R + Vs/R2*R1/R = Vs/R *( (R1+R2) )/R2) =
Vs/R*(1+R1/R2)

V2 =
V2 = I3*R3 + I2*R2 = Vs*(R1/R + R3/R + R3/R*R1/R2)

I4 =
(Vs*(R1/R + R3/R + R3/R*R1/R2))/R4 = Vs/R4 * (R1/R + R3/R + R3/R*R1/R2) = Vs/R4 * [ (R3*(R2+R1))/(R*R2) + R1/R ]

I4*R4 = (R1/R + ((R1 + R2)*R3)/(R*R2)) *Vs

I5 = I3 + I4 and so on
We finally find the voltage gain:
Au = - (R1*R5*( R2 + R3 + R4 ) + R1*R4*( R2 + R3 ) + R2*R5*( R3 + R4 ) + R2*R3*R4 ) / (R*R2*R4)

It would be much faster if we substitute the nambers in the first place.

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LACK 3D Print Enclosure & Custom Thermometer

I have been getting a lot of use out of my Prusa i3 Mk3S! It’s a great machine that makes great FDM models and needs little maintenance compared to the model I had to work with at my job at the Oregon State University Makerspace. I bed level there and nozzle swap there and this printer has had no issues. However, the printer is in open air, meaning if I were to print in a temperature sensitive material such as ABS, ASA, or TPU then I would need to control the temperature of the ambient air to a greater degree then I currently am. This is usually done by placing the 3D printer in an enclosure so that the ambient air stays warm and makes your sensitive plastics less prone to the warping and curling that occurs as they cool.

Given I had gotten a roll of ASA due to its admirable properties of UV resistance and greater tensile strength, I figured it was time to build myself an enclosure.

I decided to replicate a classic design in the 3D printing world for the Prusa i3. The Ikea Lack Table 3D print enclosure. I sourced my parts and information on how to build this enclosure from this article on PrusaPrinters Blog.

These Lack tables are cheap at $9 a piece and stack on top of each other perfectly. So, I was able to stack three so the printer would be at chest height and use the middle table as a bench for spare filament rolls, putty scrapers, and maintenance manuals. I printed the parts included in the article from Orange PLA, scored some pieces of acrylic to the size of the walls of the enclosure and installed a fire alarm in the enclosure. I do recommend a fire alarm, even though I trust the safety measures of the Prusa, putting a hot metal bed and nozzle in an enclosure made of that excellent wood particle and glue Ikea furniture didn’t feel right without additional precautions. With everything together I figured I had to leave my mark on it to make the project my own in some way. One missing component was a thermometer to verify that the enclosure was keeping a higher temperature than the surroundings. I found and older LM335 Linear Temperature Sensor and build an interface around to measure the analog value from the sensor and display it on a beautiful yellow seven segment display. Determining how to go from analog voltage to digital number for temperature on the display took a few values from the datasheet of the LM335 and the ATmega328P. The sensor has a sensitivity of 10mV/K and needs the be calibrated with a potentiometer to the optimal voltage value for room temperature. Also take note the sensor accuracy of +/-1C and a total range of -40C to 125C, for use as a print enclosure temp sensor that is just fine. The noise we should expect from this sensor is also important to our characterization. From figure 6 in the datasheet we can find the noise spectral density of this component. The noise spectral density comes in units of$\frac{V}{sqrt{Hz}}\$ and as we can see on this plot and by the units used in the calculation of noise, the noise will changed based on the bandwidth of the circuit. The lower the bandwidth the more noise, seems backwards, right? Regardless the circuit bandwidth here is determined by the sample rate of the ADC. The noise units are in the range of nV and given 10mV per degree, most of the noise will come from the quantization noise of the ATmega168PB’s integrated ADC.

The integrated ADC on the ATmega168 is a 10-bit ADC that when paired with a 5V reference, gives 5mV of resolution. $$LSB = \frac{Vref}{2^n}$$.

Once the sensor conditioning and interpretation was settled, the code to support the LTC-4627 seven segment display was written. The display updates each segment based on an ISR timer that prevents the display from flickering.

The enclosure was built up in Fusion 360 and got a solid barrier terminal connector that matched the ones on the Prusa i3 power supply. Once it was working and installed in the enclosure I was pleased with my work and got to test printing.

The firmware, EAGLE schematics, and enclosure files are all in the GitHub repo for this project.


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# How do we calculate the irradience of an EM wave?

I've been told that the Irradience = time average of the electric field squared. However, I've also been told that it's equal to the total Amplitude of the electric field squared. I'm kinda confused as to which interpretation it is.

There are 2 calculations for the irradience which have given me 2 different values.

The first is done via an algebraic method:

$$E_\text{tot} = A\cos(\omega t + \phi_1) + A\cos(\omega t + \phi_2)$$ $$=2A\cos(\frac{\phi_1 - \phi_2}{2})\cos(\omega t + \frac{\phi_1 + \phi_2}{2})$$ $$= 2A \cos(k\frac{r_2-r_1}{2})\cos(\omega t - k\frac{r_1 + r_2}{2})$$ If we were to take the time average of the electric field squared in this case, we would end up with Irradience being equal to: $$=\frac{1}{2}A^2\cos^2([k\{r_1 - r_2\}]/2)$$

This is clearly not just the total Amplitude squared.

However, if we were to now preform these calculations via complex notation:

$$E_\text{tot} = A_1e^{i(\omega t + \phi_1)} + A_2e^{i(\omega t + \phi_2)} = (A_1e^{i\phi_1} + A_1e^{i\phi_1})e^{i\omega t}$$

Now, if we were to find the irradience, it would end up being equal to the total amplitude squared.

Why is this the case?

If at a position you have two waves of amplitude $A$ which are in phase with one another the irradiance is proportional to $(A+A)^2 = 4A^2$ not $A^2+A^2 = 2 A^2$.
$E^2 = A^2+A^2+2AA\cos \phi = 4A^2 \cos^2 \frac \phi 2$


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# Ordering of Large Cardinal Axioms

One of the unexplained phenomena about large cardinal axioms is that they tend to be in a linear order (in terms of consistency strength). So it brings to mind other natural questions about this "order".

So my question is - For any two large cardinal axioms $A1$, $A2$ such that $ZFC + A1$ proves $ZFC + A2$ is consistent, can we define a large cardinal axiom $A3$ such that $ZFC + A1$ proves $ZFC + A3$ is consistent, and $ZFC + A3$ proves $ZFC + A2$ is consistent?

Meaning, is the order known\thought to be densely ordered (assuming the ordering itself is not some random phenomena)? Alternatively, can we build an example (even a highly contrived one) of $A1$, $A2$, where any axiom we add to $ZFC$ is either stronger than $ZFC + A1$, weaker than $ZFC + A2$, or equiconsistent to one of them (meaning the ordering is not dense)?

P.S - I know there is no agreed upon definition of "large cardinal axiom", but if we discuss their linear order we should be able to ask questions like this as well.

• We expect the consistency strength degrees of large cardinals to be pre-well-ordered. Essentially, because we ultimately expect to identify these degrees with certain "mice", and these should be orderable via comparison processes. – Andrés E. Caicedo Mar 15 '16 at 15:20
• A good place to read about why this leads to a (pre-)well-ordering is MR1720574 (2000k:03101). Löwe, Benedikt(1-CA); Steel, John R.(1-CA) An introduction to core model theory. (English summary) Sets and proofs (Leeds, 1997), 103–157, London Math. Soc. Lecture Note Ser., 258, Cambridge Univ. Press, Cambridge, 1999. Particularly see Section 2.5 on the mouse order. (Note the link above is to a postscript file.) – Andrés E. Caicedo Mar 15 '16 at 15:20
• @Andrés: Are these mice nice? And if there is just an iterable mouse, is it nouse? And if it's not, does that mean that the iterable mouse is a terrible mouse? – Asaf Karagila Mar 15 '16 at 15:23

If by large cardinal axioms you mean axioms which directly allude to cardinals, then the answer is negative.

Namely, if a large cardinal axiom is something like "There exists a cardinal $\kappa$ with such and such properties" and these properties imply certain amount of inaccessibility, in which case we rule out things like "$\sf ZFC$ is consistent" or "$0^\#$ exists", then the answer is negative.

To see that, simply note that there will always be a smallest large cardinal. If you want it to be such that $V_\kappa\models\sf ZFC$, then the least worldly cardinal is your smallest large cardinal. Of course this is debatable, and we don't have a mathematical definition of what is a large cardinal axiom. But much like pornography, "we know it when we see it".

So if we accept that worldly cardinals are the smallest large cardinals, then "There are two worldly cardinals" proves the consistency of "There is a worldly cardinal", but there are no intermediate notions to be found.

Of course you might argue that $\sf ZFC$ with "the theory $\sf ZFC+\exists \kappa$ worldly is consistent" proves the consistency of the existence of worldly cardinals, but now this becomes a question of whether or not this is a large cardinal axiom. Some people might argue that it is, others might argue that it's not.

For what it's worth, statement about "this theory is consistent" are arithmetic, so if we only assume $\sf PA$ with "this theory is consistent" then we can prove the consistency of said theory. But this is really not what we mean when we talk about large cardinals, is it now?

• Of course, I am strictly treating the case where the two extensions of ZFC have different consistency strength. One could argue that we cannot find anything intermediate between "There is a weakly compact cardinal" and "There is a successor of a regular cardinal with the tree property", because the two theories have been shown to be equiconsistent. – Asaf Karagila Mar 15 '16 at 15:26
• Interesting. Though the example with worldly cardinals makes me wonder what would happen if I'd amend the question to demand a large cardinal axiom that states the existence of an unbounded class of a certain type of cardinals (instead of a single cardinal). – Alon Navon Mar 15 '16 at 20:59
• That would usually be the the same. A worldly limit of inaccessible cardinals is the smallest large cardinal above "proper class of inaccessible cardinals" in terms of consistency strength. – Asaf Karagila Mar 15 '16 at 21:32


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2 "fixed" diagram

A math paper should follow all the usual rules of grammar, so in particular there should be subjects and verbs and the sort of punctuation you'd expect to find in a piece of nontechnical writing. I would prefer to write the following:

The formula for a circle is $$x^2+y^2=r^2.$$

If I had to use the wording in the original question, I would write

This is the formula for a circle: $$x^2+y^2=r^2.$$

Occasionally, the aesthetics of the page make punctuation look awkward. For example, one might write:

Therefore, the following diagram commutes: $$M×N -> M⊗RN \ xymatrix{M\times N\ar[r]\ar[dr] & M\otimes_R N\ar[d] | \ & | \ | v v A }$$

with no punctuation after the diagram. There isn't any sensible location for a period at the end of a sentence, so I'd leave it out.

A math paper should follow all the usual rules of grammar, so in particular there should be subjects and verbs and the sort of punctuation you'd expect to find in a piece of nontechnical writing. I would prefer to write the following:

The formula for a circle is $$x^2+y^2=r^2.$$

If I had to use the wording in the original question, I would write

This is the formula for a circle: $$x^2+y^2=r^2.$$

Occasionally, the aesthetics of the page make punctuation look awkward. For example, one might write:

Therefore, the following diagram commutes: $$\xymatrix{M\times N\ar[r]\ar[dr] & M\otimes_R N\ar[d] \ & A}$$

with no punctuation after the diagram. There isn't any sensible location for a period at the end of a sentence, so I'd leave it out.