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# Hedging in the Heston Model

I have simulated an underlying stock price, $$S_t$$ and a stochastic variance process, $$v_t$$ with the following stochastic differential equations from the Heston Universe: $$dS_t = \mu S_tdt + \sqrt{v_t}dW_{1,t}$$ $$dv_t = \kappa(\theta-v_t)dt + \sigma\sqrt{v_t}dW_{2,t}$$ Further, I have prices call options in the above Heston model with: $$C=S_t e^{−qt}P_1−Ke^{−rt}P_2$$ where $$P_1$$ and $$P_2$$ are in-the-money probability as definded in the original paper [1].

To do a perfect hedge of the call option in the simulated Heston world, I need a proportion of the stock and some proportion of another asset whose value depends on the variance. I want to be able to implement it in my simulations.

My question is now: how do I calculate these proportions?

I have calculated $$\Delta_C = e^{−qt}P_1$$ and $$Vega=Se^{−qt} \frac{∂P_1}{∂v_0}2\sqrt{v_0}−Ke^{−rt}\frac{∂P_2}{∂v_0}2\sqrt{v_0}$$, and I'm thinking that these are the quantaties to to buy of the underlying asset and of a another asset whose value depends on the variance to hedge the option.

However, I'm not sure that this is correct. Is this all that is needed to hedge an option in the Heston Universe?

[1] Heston, Steven L. (1993). "A Closed-Form Solution for Options with Stochastic Volatility with Applications to Bond and Currency Options". The Review of Financial Studies. 6 (2): 327–343. doi: 10.1093/rfs/6.2.327. JSTOR 2962057.

• Consider a portfolio which holds the call option, some proportion of the stock and some proportion of another asset whose value depends on the variance. This way, you make the market complete and can hedge the volatility risk. Then apply Itô’s Lemma to get the changes of this portfolio. Then find the trading strategy (number of the two assets you need to trade) to eliminate the stochastic terms. You’re then left with a deterministic portfolio change. – Kevin Oct 7 '20 at 14:15
• Which risks do you want to hedge? What is the end goal of this exercise? – Frido Rolloos Oct 7 '20 at 15:14
• @KeSchn this is exactly what I need. Do you know any sources that go through the math and the logic in debt? – Modvinden Oct 8 '20 at 7:50
• @Modvinden you should do your homework on your own. What have you tried thus far? Do you know Itô’s Lemma for two processes (and time)? Take the simple delta hedge from the Black-Scholes model and try to generalise each step. – Kevin Oct 8 '20 at 8:09
• @KeSchn What I have managed so far is the following. The price of a call option is $C=S_t e^{-qt} P_1 - Ke^{-rt} P_2$ where $P_1$ and $P_1$ in-the-money probability as definded in the original paper. Then we need $\Delta_C= e^{-qt}P_1$ of the underlying asset and $\Vega = S e^{-qt} \frac{\partial P_1}{\partial v_0} 2 \sqrt{v_0}-K e^{-rt} \frac{\partial P_2}{\partial v_0} 2 \sqrt{v_0}$ of the variance swap. Is this correct? – Modvinden Oct 8 '20 at 11:31

Let's denote the option you need to hedge by $$C_1$$, which I am assuming you have sold (if you bought it then just turn the signs around). Under Heston you will need to hedge both its delta and its vega.

You can use the underlying $$S$$ to hedge the delta, but not to hedge vega. The most straightforward way to hedge the vega of $$C_1$$ is to buy another option in the market, call it $$C_2$$.

Let $$\nu_1$$ be the vega of $$C_1$$ and $$\nu_2$$ the vega of $$C_2$$ then the number $$n$$ of options $$C_2$$ you need to buy to hedge vega is

$$$$n = \nu_1 / \nu_2$$$$

However, you know that $$C_2$$ also has delta, call it $$\Delta_2$$, and $$C_1$$ has delta, call it $$\Delta_1$$. To neutralize these delta's you need to buy a certain amount $$m$$ of $$S$$, which is easily solved to be

$$$$m = - \Delta_1 + n\Delta_2$$$$

That's all there is to it, except of course that you need to calculate these greeks under the Heston model. Fortunately for Heston there is a closed form solution for the price and you can then differentiate the price wrt to $$S$$ and $$\sigma$$ to find delta and vega.

• Thank you so much for the great response. So I can calculate $\Delta_1$ and $v_1$ by differentiating $C_1=S_t e^{−qt}P_1−Ke^{−rt}P_2$ wrt $S_t$ and $\sigma$. I understand this. However, I still need to find $\Delta_2$ and $v_2$ by differentiating $C_2$ wrt the underlying and its $sigma$. $\Delta_2$ and $v_2$ will then depend on the way I specify $C_2$ to be? I have read that $C_2$ is often a variance swap. How is $C_2$ often specified - I mean what expression/equation do people use for $C_2$? – Modvinden Oct 9 '20 at 11:34
• Within Heston (and actually that's the way the Heston PDE is derived) $C_2$ is another vanilla option. Of course you can take another instrument sensitive to volatility, but at this stage it will only make your life more complicated. For example, what is the sensitivity of the variance swap to the instantaneous volatility? That's not an easy question. My advice: stick to taking another option for $C_2$. Once you understand the nitty gritty then it's OK to talk to investment banks trying to sell you varswaps :-D – Frido Rolloos Oct 9 '20 at 11:58
• So $C_2$ is a vanilla option on a new underlying asset $S_{t,2}$? I want to implement this into my simulation, so then I need to specify dynamics of this option as well? And should this option also be generated with Heston? Thank you for the great answers!! – Modvinden Oct 9 '20 at 12:10
• Not another underlying, the same underlying, different strike and/or maturity. – Frido Rolloos Oct 9 '20 at 12:11
• There is no rule in theory. But you may want to choose a maturity and strike such that also the vanna of the total portfolio (target option + hedge instruments) is minimized. This way you reduce transaction costs. Vanna is the sensitivity of delta to vol or equivalently sensitivity of vega to underlying asset movements. Minimizing vanna will reduce the number of times you need to rebalance your hedge assuming you have some mismatch bandwidths. But for purely educational purposes you can choose any option. – Frido Rolloos Oct 9 '20 at 12:19


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# Difference between revisions of "2017 AMC 8 Problems/Problem 24"

## Problem 24

Mrs. Sanders has three grandchildren, who call her regularly. One calls her every three days, one calls her every four days, and one calls her every five days. All three called her on December 31, 2016. On how many days during the next year did she not receive a phone call from any of her grandchildren?

$\textbf{(A) }78\qquad\textbf{(B) }80\qquad\textbf{(C) }144\qquad\textbf{(D) }146\qquad\textbf{(E) }152$

## Solution 1

In $360$ days, there are $$360 \cdot \frac 23 \cdot \frac 34 \cdot \frac 45 = 144$$ days without calls. Note that in the last five days of the year, day $361$ and $362$ also do not have any calls, as they are not multiples of $3$, $4$, or $5$. Thus our answer is $144+2 = \boxed{\textbf{(D)}\ 146}$.

$$ Alternatively, there are $365\cdot \frac45 \cdot \frac 34 \cdot \frac23=\boxed{146}$ days without calls. Multiplying the fractions in this order prevents partial days, as $365$ is a multiple of $5$, $365\cdot \frac45$ is a multiple of $4$ and $365\cdot \frac45 \cdot \frac 34$ is a multiple of $3$.

## Solution 2

We use Principle of Inclusion Exclusion. There are $365$ days in the year, and we subtract the days that she gets at least $1$ phone call, which is $$\left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor$$.

To this result we add the number of days where she gets at least $2$ phone calls in a day because we double subtracted these days. This number is $$\left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor$$.

We now subtract the number of days where she gets three phone calls, which is $\left \lfloor \frac{365}{60} \right \rfloor$. Therefore, our answer is

$365 - \left( \left \lfloor \frac{365}{3} \right \rfloor + \left \lfloor \frac{365}{4} \right \rfloor + \left \lfloor \frac{365}{5} \right \rfloor \right) + \left( \left \lfloor \frac{365}{12} \right \rfloor + \left \lfloor \frac{365}{15} \right \rfloor + \left \lfloor \frac{365}{20} \right \rfloor \right) - \left \lfloor \frac{365}{60} \right \rfloor$

$= 365 - 285+72 - 6 = \boxed{\textbf{(D) } 146}$.

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I know, for various reason, some users preprocess their latex file using Perl or sed, say.

I'm considering doing this to, so I would like to seek your guidance, to smooth my entry in this area.

My use case is a simple macro expansion preprocessing

If $\ep$ is small...   -->  If $\epsilon$ is small...

So that a compilation session becomes

doc.tex  --preprocessor-->  post.tex  --pdftex--> post.pdf

And the following issues arise

1. There is generally a mismatch between line numbers in doc.tex and post.tex. So that pdftex error messages points to a line in post.tex. But, really, one wants the line number in doc.tex.

2. The same for synctex. Backward and foward search are with respect to post.tex. But, one really wants them to be with respect to doc.tex.

3. Sometimes, one wants to go back and forth from doc.tex and post.tex. The preprocessor knows the corresponds between lines of the two documents. But then, how to put this knowledge in a workable way. I mean, text editors does not have out-of-the-box support for these backward-forward trips.

So, do you know some software, workflow or technics which will flatten (1)-(3)?

I look around, but I could only find example of Perl or sed preprocessing scripts.

I never saw the question of the workflow discuss somewhere.

EDIT

The line mismatch can originate from big expansion

A function is continuous if
\multiline_continuity_definition_I_often_use
Hence ...

where the definition is expanded to

\begin{description}

\item[Standard]
$\forall \epsilon ...$

\item[Nonstandard]
$x \approx y ...$

\end{description}

Edit 2

The point of this question is to discuss preprocessing technics.

My examples are very simplistic, but I hope they catch the main challenge of a preprocessing session.

-
What causes the line mismatch? Are you hard-wrapping your lines so that \ep --> \epsilon causes 'hard' changes to lineation? Soft-wrapping might be better in your case, unless it is something else causing the mismatch. – jon Mar 29 '13 at 20:38
Even if the editor is hard-wrapping, the described preprocessor shouldn't move the line breaks. – T. Verron Mar 29 '13 at 20:41
Given that TeX is mostly a macro processing engine with some typesetting capabilities added, My use case is a simple macro expansion preprocessing must be the strangest use case for wanting to invoke an external processor. Why on earth can't you use \def\eps{\epsilon} ???? – David Carlisle Mar 29 '13 at 21:10
@DavidCarlisle It was not easy to write this question. I tried to catch the main points of a preprocessing session. It's true that my epsilon example is not so good. – Nicolas Essis-Breton Mar 29 '13 at 21:16
@NicolasEssis-Breton, perhaps but it seems that all three of your issues go away if you use TeX to do your re-processing so I think you'd need to give an example of something that can't so easily be done in TeX (there are examples: pulling fragments from a database, translating between different languages, ....) – David Carlisle Mar 29 '13 at 21:20

There is no full solution for this problem, as far as I know, because it will vary depending on your needs.

But here are two hints which should rule out your three problems for most cases:

## Use LaTeX as much as possible

Your first example could easily be solved by

\newcommand{\ep}{\epsilon}

\newcommand{\multiline_continuity_definition_I_often_use}{% You might prefer a shorter name :)
\begin{description}

\item[Standard]
$\forall \epsilon ...$

\item[Nonstandard]
$x \approx y ...$

\end{description}
}

The key point is to avoid shadok logic, that is "Don't waste time developing a tool you don't really need". Even if the tool looks beautiful.

## Try to separate what is in the document and what isn't

This is for "real" examples where you need a preprocessor. Usually, that will be because they involve dynamic data (if you need to do always the exact same thing, latex should be enough) and difficult computations/processing.

What you can do then is separate this dynamic part from your document. For example, \input a file which will only contain the part which will need preprocessing. Then preprocess the document (following these \input, or simply filtering the auxiliary files based on their extension, for example), and compile.

Latex can output the errors in a file-line-error style, and then the line-numbering won't be messed up. If you are using a multifile-aware editor (like, for example, emacs + AUCTeX + refTeX), this can be made almost transparently.

About synctex, it shouldn't cause any issue either. And for the back-and-forth between the base document and the processed one... Well, since you only change small auxiliary files, there should be no problem.

A good example of this workflow is the sagetex package. It allows users to make some computations with the symbolic computation software sage, and typeset their results directly in the .pdf document.

It uses some post-processing to make it even more transparent to the user, but the idea is the same:

1. The user inputs the computations in the .tex file (using a few commands)
2. A first run of pdflatex extracts all these sage bits in a document.sage file.
3. A run of sage on that file creates a document.sageout containing the results.
4. A second run of pdflatex inserts these results in the final document.

It is a fully-developed package, seriously optimised for user-friendliness, but you don't need such a carefully-designed solution. For example, you can create the extra files by hand (with the help of your editor, it can be as easy as typing \input{filename} and hitting a few keys).

And once again, this isn't a full solution, you will probably need some work to find out what really suits your needs... Take it as hints towards your workflow, nothing more.

-
+1 yes sagetex is a good example (rewriting sage in tex would be inconvenient:-) – David Carlisle Mar 29 '13 at 22:08


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# Vectors problem

i have just joined ur sit e
can u help me out
if we have the initial point of two vectors and equations of these vectors , how can we find the intersection point of these vectors

HallsofIvy
Homework Helper
?? Vectors are "free floating"- only specific representations have initial points. And even those may not intersect- they may not reach far enough.
If you mean you have two lines, one passing through $(x_0,y_0,z_0)$ in the same direction as vector Ai+ Bj+ Ck, the other passing through $(x_1, y_1, z_1)$, in the same direction as vector Ui+ Vj+ Wk, then the first line can be written in parametric equations $x= x_0+ At$, $y= y_0+ Bt$, $z= z_0+ Ct$ and the second line can be written in parametric equations $x= x_1+ Ut$, $y= y_1+ Vt$, $z= z_1+ Wt$. Set "x=x", "y= y", "z= z", and solve the resulting three equations for s and t.
Of course, in general, three equations won't give just two unknown values- in three dimensions most lines are "skew" and don't intersect. If you are working in 2-dimensions, (x, y), just ignore the z equations and you have two equations to solve for s and t.


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## Electronic Journal of Probability

### Harnack inequalities for SDEs driven by time-changed fractional Brownian motions

#### Abstract

We establish Harnack inequalities for stochastic differential equations (SDEs) driven by a time-changed fractional Brownian motion with Hurst parameter $H\in (0,1/2)$. The Harnack inequality is dimension-free if the SDE has a drift which satisfies a one-sided Lipschitz condition; otherwise we still get Harnack-type estimates, but the constants will, in general, depend on the space dimension. Our proof is based on a coupling argument and a regularization argument for the time-change.

#### Article information

Source
Electron. J. Probab., Volume 22 (2017), paper no. 71, 23 pp.

Dates
Accepted: 11 July 2017
First available in Project Euclid: 13 September 2017

https://projecteuclid.org/euclid.ejp/1505268102

Digital Object Identifier
doi:10.1214/17-EJP82

Mathematical Reviews number (MathSciNet)
MR3698740

Zentralblatt MATH identifier
06797881

#### Citation

Deng, Chang-Song; Schilling, René L. Harnack inequalities for SDEs driven by time-changed fractional Brownian motions. Electron. J. Probab. 22 (2017), paper no. 71, 23 pp. doi:10.1214/17-EJP82. https://projecteuclid.org/euclid.ejp/1505268102

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• [6] Deng, C.-S.: Harnack inequalities for SDEs driven by subordinate Brownian motions. J. Math. Anal. Appl. 417, (2014), 970–978.
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• [20] Wang, F.-Y.: Harnack Inequalities for Stochastic Partial Differential Equations. Springer, New York 2013.
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• [22] Wang, L. and Zhang, X.: Harnack inequalities for SDEs driven by cylindrical $\alpha$-stable processes. Potential Anal. 42, (2015), 657–669.
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# zbMATH — the first resource for mathematics

Generic hardness of inversion on ring and its relation to self-bilinear map. (English) Zbl 1455.94198
Summary: In this paper, we study the generic hardness of the inversion problem on a ring, which is a problem to compute the inverse of a given prime $$c$$ by just using additions, subtractions and multiplications on the ring. If the characteristic of an underlying ring is public and coprime to $$c$$, then it is easy to compute the inverse of $$c$$ by using the extended Euclidean algorithm. On the other hand, if the characteristic is hidden, it seems difficult to compute it. For discussing the generic hardness of the inversion problem, we first extend existing generic ring models to capture a ring of an unknown characteristic. Then we prove that there is no generic algorithm to solve the inversion problem in our model when the underlying ring is isomorphic to $$\mathbb{Z}_p$$ for a randomly chosen prime $$p$$ assuming the hardness of factorization of an unbalanced modulus. We also study a relation between the inversion problem on a ring and a self-bilinear map. Namely, we give a construction of a self-bilinear map based on a ring on which the inversion problem is hard, and prove that natural complexity assumptions including the multilinear computational Diffie-Hellman (MCDH) assumption hold w.r.t. the resulting sef-bilinear map.

##### MSC:
 94A60 Cryptography 11T71 Algebraic coding theory; cryptography (number-theoretic aspects)
Full Text:
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Math., 126, 3, 649-673 (1987) · Zbl 0629.10006 [40] Yamakawa, T.; Hanaoka, G.; Kunihiro, N., Generalized hardness assumption for self-bilinear map with auxiliary information, (ACISP (2016)), 269-284 · Zbl 1346.94135 [41] Boneh, D.; Zhandry, M., Multiparty key exchange, efficient traitor tracing, and more from indistinguishability obfuscation, (CRYPTO (2014)) · Zbl 1310.94130 [42] Khurana, D.; Rao, V.; Sahai, A., Multi-party key exchange for unbounded parties from indistinguishability obfuscation, (ASIACRYPT 2015 I (2015)), 52-75 · Zbl 1337.94044 [43] Goldreich, O.; Levin, L. A., A hard-core predicate for all one-way functions, (STOC (1989)), 25-32 [44] Gorbunov, S.; Vaikuntanathan, V.; Wichs, D., Leveled fully homomorphic signatures from standard lattices, (STOC 2015 (2015)), 469-477 · Zbl 1321.94062
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## Found 132 Documents (Results 1–100)

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### Inference on breakdown frontiers. (English)Zbl 1454.62110

MSC:  62G05 62G10
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### Rank-based classifiers for extremely high-dimensional gene expression data. (English)Zbl 1416.62608

MSC:  62P10 62H30 68T05
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MSC:  68P15
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### Invariance of rank for generalized Jordan products of quadratic matrix. (Chinese. English summary)Zbl 1399.15002

MSC:  15A03 15A24
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### Coherence invariant maps on tensor products. (English)Zbl 1366.15001

MSC:  15A03 15A04 15A69
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### Efficient and robust path openings using the scale-invariant rank operator. (English)Zbl 1386.68209

MSC:  68U10 94A08
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### A modified neural network for solving general singular convex optimization with bounded variables. (English)Zbl 1327.90204

Gao, David (ed.) et al., Advances in global optimization. Selected papers based on the presentations at the 3rd world congress on global optimization in engineering and science, WCGO, Anhui, China, July 8–12, 2013. Cham: Springer (ISBN 978-3-319-08376-6/hbk; 978-3-319-08377-3/ebook). Springer Proceedings in Mathematics & Statistics 95, 363-374 (2015).
MSC:  90C25 68T05
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### Birth and death master equation for the evolution of complex networks. (English)Zbl 1395.60090

MSC:  60J28 92D25 60J80
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### On abelian groups with right-invariant isometries. (English. Russian original)Zbl 1303.20057

Sib. Math. J. 55, No. 3, 574-577 (2014); translation from Sib. Mat. Zh. 55, No. 3, 701-705 (2014).
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### Rank-one solutions for homogeneous linear matrix equations over the positive semidefinite cone. (English)Zbl 1293.15013

MSC:  15A24 15B48
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### On rank invariance of generalized block Pick matrices of matrix-valued Carathéodory functions. (Chinese. English summary)Zbl 1255.47018

MSC:  47A57 30E05

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### Graph reductions, binary rank, and pivots in gene assembly. (English)Zbl 1237.05181

MSC:  05C76 05C25 92D10
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### Tests and estimates of shape based on spatial signs and ranks. (English)Zbl 1359.62150

MSC:  62G10 62M30
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MSC:  62J05
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### The invariance for the product of three matrices on weighted generalized inverse. (Chinese. English summary)Zbl 1183.15006

MSC:  15A09 15A03

### Limit theorems for permutations of empirical processes with applications to change point analysis. (English)Zbl 1129.60033

MSC:  60F17 62G09
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### Extremal ranks of some symmetric matrix expressions with applications. (English)Zbl 1123.15001

MSC:  15A03 15A09 15A24
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### A simple derivation of Prelec’s probability weighting function. (English)Zbl 1194.91082

MSC:  91B14 91B16
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### Invariance properties of a triple matrix product involving generalized inverses. (English)Zbl 1105.15005

MSC:  15A09 15A03
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### Robust statistical methods with R. (English)Zbl 1097.62020

Boca Raton, FL: Chapman and Hall/CRC (ISBN 1-58488-454-1/hbk). xi, 197 p. (2006).
MSC:  62F35 62-01 65C60 62G35 62-04

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### Depth and multivariate generalization of the Wilcoxon-Mann-Whitney test. (English)Zbl 1151.62328

MSC:  62H15 65C60
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### Proving program invariance and termination by parametric abstraction, Lagrangian relaxation and semidefinite programming. (English)Zbl 1111.68503

Cousot, Radhia (ed.), Verfication, model checking, and abstract interpretation. 6th international conference, VMCAI 2005, Paris, France, January 17–19, 2005. Proceedings. Berlin: Springer (ISBN 3-540-24297-X/pbk). Lecture Notes in Computer Science 3385, 1-24 (2005).
MSC:  68Q60 68N30 90C22
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### Rawlsian maximin, Dutch books, and non-additive expected utility. (English)Zbl 1094.91017

MSC:  91B08 91B14 91B16
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### Weakly dependent stochastic sequences and their applications. Vol. XV: Recent topics on limit theorems for functionals of processes. With contents and index of this series. (English)Zbl 1085.60005

Tokyo: Sanseido (ISBN 4-385-36236-X/hbk). vi, 421 p. (2005).

### On multivariate signed rank tests. (English)Zbl 1055.62067

MSC:  62H15 62G10
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### Robust adaptive signal processing methods for heterogeneous radar clutter scenarios. (English)Zbl 1153.94329

MSC:  94A13 93C40 93E10
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### Multivariate nonparametric tests. (English)Zbl 1100.62567

MSC:  62G10 62H15
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MSC:  15A09
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### Rank invariance and automorphisms of generalized Kac-Moody superalgebras. (English)Zbl 1101.17014

Ho, Chat Yin (ed.) et al., Finite groups 2003. Proceedings of the Gainesville conference on finite groups, Gainesville, FL, USA, March 6-12, 2003. In honour of John Thompson to his 70th birthday. Berlin: Walter de Gruyter (ISBN 3-11-017447-2/hbk). 291-305 (2004).
MSC:  17B67 17B69

### An affine-invariant generalization of the Wilcoxon signed-rank test for the bivariate location problem. (English)Zbl 1064.62054

MSC:  62G10 62H15
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### On SO($$n$$)-invariant rank 1 convex functions. (English)Zbl 1077.49021

MSC:  49K20 74G65
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### Rank 1 convex hull of SO(n)-invariant functions. (English)Zbl 1049.49014

Miehe, Christian (ed.), IUTAM symposium on computational mechanics of solid materials at large strains. Proceedings of the IUTAM symposium, Stuttgart, Germany, August 20–24, 2001. Dordrecht: Kluwer Academic Publishers (ISBN 1-4020-1170-9/hbk). Solid Mech. Appl. 108, 21-32 (2003).

### The maximal and minimal ranks of $$A - BXC$$ with applications. (English)Zbl 1036.15004

MSC:  15A03 15A09 15A24
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### Semiconvexity of invariant functions of rectangular matrices. (English)Zbl 1053.49009

MSC:  49J45 74B20 74G65
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### On rank invariance of Schwarz-Pick-Potapov block matrices of matrix-valued Carathéodory functions. (English)Zbl 1018.15001

Böttcher, Albrecht (ed.) et al., Toeplitz matrices and singular integral equations. The Bernd Silbermann anniversary volume. Basel: Birkhäuser. Oper. Theory, Adv. Appl. 135, 161-181 (2002).
MSC:  15A03 15A54 15B57

### Upper and lower bounds for ranks of matrix expressions using generalized inverses. (English)Zbl 1016.15003

Reviewer: V.P.Kostov (Nice)
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### On Schwarz-Pick-Potapov block matrices of matrix-valued functions which belong to the extended Potapov class. (English)Zbl 1033.47014

MSC:  47A56 30E05
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### Invariance of numerical character of matrix products and their statistical applications. (English)Zbl 0991.15006

MSC:  15A09 62M05 62J05

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MSC:  62P15
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### Applications of weighted approximations via quantile inequalities. (English)Zbl 0929.62057

Szyszkowicz, Barbara (ed.), Asymptotic methods in probability and statistics. A volume in honour of Miklós Csörgő. ICAMPS ’97, an international conference at Carleton Univ., Ottawa, Ontario, Canada, July 1997. Amsterdam: North-Holland/ Elsevier. 151-167 (1998).

### Conformal fourth-rank gravity, non-vanishing cosmological constant and anisotropy. (English)Zbl 0908.53062

MSC:  53Z05 83D05
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### A coefficient of agreement adjusted for bias in paired ordered categorical data. (English)Zbl 0884.62123

MSC:  62P10 62H17 62G99
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### Binary vectors partially determined by linear equation systems. (English)Zbl 0877.15004

MSC:  15A06 15A03 15B36
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### Limit theorems for rank statistics. (English)Zbl 0933.62039

MSC:  62G20 62E20 60F17
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### A multivariate Chernoff-Savage theorem with applications to rank statistics from multivariate populations. (English)Zbl 0874.62059

Brunner, E. (ed.) et al., Research developments in probability and statistics. Festschrift in honor of Madan L. Puri on the occasion of his 65th birthday. Utrecht: VSP. 125-140 (1996).

### Regression rank scores estimation in ANOCOVA. (English)Zbl 0898.62087

MSC:  62J05 62G05 62G20
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### Power robustification of approximately linear tests. (English)Zbl 0843.62053

MSC:  62G10 62F15 62F03
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### On rank one convex functions which are homogeneous of degree one. (English)Zbl 0911.49012

Bandle, C. (ed.) et al., Calculus of variations, applications and computations. Proceedings of the 2nd European conference on elliptic parabolic problems, Mont-à-Mousson, June 1994. Harlow: Longman Scientific & Technical. Pitman Res. Notes Math. Ser. 326, 84-99 (1995).
MSC:  49J45 74B20 49J10 90C26 74S30 74P10

### Daniels’s association measures under right censoring. (English)Zbl 0822.62035

MSC:  62G10 62H20 62P10
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### Subelliptic estimates of polynomial differential operators and applications to rigidity of abelian actions. (English)Zbl 0836.57026

Reviewer: B.Helffer (Orsay)
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### Empirical-type processes in weighted metrics and change-point problems. (English)Zbl 0836.60033

MSC:  60F25 60F17

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### Weakly dependent stochastic sequences and their applications. Vol. III: Order statistics based on weakly dependent data. (English)Zbl 0942.62055

Tokyo: Sanseido. vii, 341 p. (1993).

### On rank statistics: An approach via metrics on the permutation group. (English)Zbl 0770.62036

Reviewer: B.K.Kale (Pune)
MSC:  62G10 62A01
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MSC:  60F17
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### Differential topology. Transl. from the Spanish. (English)Zbl 0760.57001

North-Holland Mathematics Studies. 173. Amsterdam etc.: North-Holland. xv, 603 p. (1992).
MSC:  57-01 57Rxx 58-01

### Rank tests for time series analysis. A survey. (English)Zbl 0768.62075

New directions in time series analysis. Part I, Proc. Workshop, Minneapolis/MN (USA) 1990, IMA Volumes in Mathematics and Its Applications 45, 111-153 (1992).
MSC:  62M10 62G10 62G20

### Weak convergence of conditional empirical processes and conditional quantile estimators for weakly dependent sequences. (English)Zbl 0772.62025

Information theory, statistical decision functions, random processes, Trans. 11th Prague Conf., Prague/Czech. 1990, Vol. B, 201-226 (1992).

### Invariance faible de la statistique de rang multidimensionnelle. Applications. (Weak invariance of the multidimensional rank statistic. Applications). (French)Zbl 0736.60031

MSC:  60F17 62M10

### A weak convergence result useful in robust autoregression. (English)Zbl 0751.60019

MSC:  60F05 60F17 62M10
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### Rank of quantized universal enveloping algebras and modular functions. (English)Zbl 0726.17011

MSC:  17B37 81R50 11F22
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MSC:  60F17
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### Maximal selektierte Rangstatistiken. (Maximally selected rank statistics). (German)Zbl 0705.62100

Dortmund: Univ., FB Statistik, Diss. 132 S. (1990).

### A non-central Banach valued limit theorem and its application. (English. Russian original)Zbl 0784.60010

J. Sov. Math. 61, No. 1, 1901-1905 (1992); translation from Zap. Nauchn. Semin. Leningr. Otd. Mat. Inst. Steklova 177, 114-119 (1989).
MSC:  60B12 60G15 60F17

### The non-central Banach-valued limit theorem and its applications. (Russian)Zbl 0711.60007

Zap. Nauchn. Semin. Leningr. Otd. Mat. Inst. Steklova 177, 114-119 (1989).
Reviewer: I.S.Borisov
MSC:  60B12 60F17 60G15
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MSC:  62N05
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### Inner product in the sense of Aczél. (English)Zbl 0688.39004

Reviewer: J.Aczél

### Rank order statistics for time series models. (English)Zbl 0682.62064

MSC:  62M10 60F17
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### An invariance principle for processes indexed by two parameters and some statistical applications. (English)Zbl 0675.60025

Reviewer: U.Rösler
MSC:  60F17 62E20

### Differential topology. (Topologîa diferencial.) (Spanish)Zbl 0633.57001

Madrid: Consejo Superior de Investigaciones Cientîficas. 453 p. (1988).
Reviewer: D.Andrica
MSC:  57-01 57Rxx 58-01

### An invariance principle for the changepoint problem. (English)Zbl 0657.62054

Reviewer: R.Zielinski

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## Monday, July 14, 2008 ... //

### LHC doomsday: humor

The LHC has been mostly cooled down. The temperatures in sectors 23, 34, 45, 56 are around 2 K while 67 has 2.5 K, 78 and 81 (next to LHCb from both sides) have around 18 K, and 12 has again around 2.5 K.

Recently there have been several funny stories in the media about the LHC doom:

As a bonus, the last story is almost 100% accurate. ;-)

See also a funny picture about the LHC, the CMS stop button, or click the "lhc" category in "other texts" below for dozens of LHC articles.


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Please use this identifier to cite or link to this item: http://hdl.handle.net/10525/604

 Title: A Periodic Lotka-Volterra System Authors: Tsvetkov, D. Keywords: Periodic Lotka-Volterra System Predator-Prey Issue Date: 1996 Publisher: Institute of Mathematics and Informatics Bulgarian Academy of Sciences Citation: Serdica Mathematical Journal, Vol. 22, No 2, (1996), 109p-116p Abstract: In this paper periodic time-dependent Lotka-Volterra systems are considered. It is shown that such a system has positive periodic solutions. It is done without constructive conditions over the period and the parameters. URI: http://hdl.handle.net/10525/604 ISSN: 1310-6600 Appears in Collections: Volume 22 Number 2

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# tempur pedic dream pillow

When we consider eigenvalues as functions of $$A$$, we use the notation $$\lambda_j(A)$$, $$j=1,\dots,n$$. The Cauchy residue formula gives an explicit formula for the contour integral along $$\gamma$$: $$\oint_\gamma f(z) dz = 2 i \pi \sum_{j=1}^m {\rm Res}(f,\lambda_j), \tag{1}$$ where $${\rm Res}(f,\lambda)$$ is called the residue of $$f$$ at $$\lambda$$ . These equations are key to obtaining the Cauchy residue formula. Using residue theorem to compute an integral. Many classical functions are holomorphic on $$\mathbb{C}$$ or portions thereof, such as the exponential, sines, cosines and their hyperbolic counterparts, rational functions, portions of the logarithm. Suppose C is a positively oriented, simple closed contour. We consider the function $$f(z) = \frac{e^{i\pi (2q-1) z}}{1+(2a \pi z)^2} \frac{\pi}{\sin (\pi z)}.$$ It is holomorphic on $$\mathbb{C}$$ except at all integers $$n \in \mathbb{Z}$$, where it has a simple pole with residue $$\displaystyle \frac{e^{i\pi (2q-1) n}}{1+(2a \pi n)^2} (-1)^n = \frac{e^{i\pi 2q n}}{1+(2a \pi n)^2}$$, at $$z = i/(2a\pi)$$ where it has a residue equal to $$\displaystyle \frac{e^{ – (2q-1)/(2a)}}{4ia\pi} \frac{\pi}{\sin (i/(2a))} = \ – \frac{e^{ – (2q-1)/(2a)}}{4a} \frac{1}{\sinh (1/(2a))}$$, and at $$z = -i/(2a\pi)$$ where it has a residue equal to $$\displaystyle \frac{e^{ (2q-1)/(2a)}}{4ia\pi} \frac{\pi}{\sin (i/(2a))} =\ – \frac{e^{ (2q-1)/(2a)}}{4a} \frac{1}{\sinh (1/(2a))}$$. Cauchy Residue Formula. The Cauchy residue trick: spectral analysis made “easy”. Perturbation Theory for Linear Operators, volume 132. For $$I = \mathbb{R}$$, then this can be done using Fourier transforms as: $$K(x,y) = \frac{1}{2\pi} \int_\mathbb{R} \frac{e^{i\omega(x-y)}}{\sum_{k=0}^s \alpha_k \omega^{2k}} d\omega.$$ This is exactly an integral of the form above, for which we can use the contour integration technique. We thus obtain an expression for projectors on the one-dimensional eigen-subspace associated with the eigenvalue $$\lambda_k$$. Residue theorem. Do not simply evaluate the real integral – you must use complex methods. See the detailed computation at the end of the post. [3] Tosio Kato. Proof. If a proof under general preconditions ais needed, it should be learned after studenrs get a good knowledge of topology. The contour $$\gamma$$ is defined as a differentiable function $$\gamma: [0,1] \to \mathbb{C}$$, and the integral is equal to $$\oint_\gamma f(z) dz = \int_0^1 \!\!f(\gamma(t)) \gamma'(t) dt = \int_0^1 \!\! Thus the gradient of $$\lambda_k$$ at a matrix $$A$$ where the $$k$$-th eigenvalue is simple is simply $$u_k u_k^\top$$, where $$u_k$$ is a corresponding eigenvector. If the function $$f$$ is holomorphic and has no poles at integer real values, and satisfies some basic boundedness conditions, then$$\sum_{n \in \mathbb{Z}} f(n) = \ – \!\!\! \sum_{ \lambda \in {\rm poles}(f)} {\rm Res}\big( f(z) \pi \frac{\cos \pi z}{\sin \pi z} ,\lambda\big).$$This is a simple consequence of the fact that the function $$z \mapsto \pi \frac{\cos \pi z}{\sin \pi z}$$ has all integers $$n \in \mathbb{Z}$$ as poles, with corresponding residue equal to $$1$$. Springer Science & Business Media, 2011. Unlimited random practice problems and answers with built-in Step-by-step solutions. We also consider a simple closed directed contour $$\gamma$$ in $$\mathbb{C}$$ that goes strictly around the $$m$$ values above. 0. §33 in Theory of Functions Parts I … Required fields are marked *. Series. [1] Gilbert W. Stewart and Sun Ji-Huang. We thus need a perturbation analysis or more generally some differentiability properties for eigenvalues or eigenvectors [1], or any spectral function [2]. So, now we give it for all derivatives f(n)(z) of f. This will include the formula for functions as a special case. Note that several eigenvalues may be summed up by selecting a contour englobing more than one eigenvalues. Find more Mathematics widgets in Wolfram|Alpha. Proposition 1.1. Your email address will not be published. The desired integral is then equal to $$2i\pi$$ times the sum of all residues of $$f$$ within the unit disk. The function $$F$$ can be represented as$$F(A) = \sum_{k=1}^n f(\lambda_k(A)) = \frac{1}{2i \pi} \oint_\gamma f(z) {\rm tr} \big[ (z I – A)^{-1} \big] dz,$$where the contour $$\gamma$$ encloses all eigenvalues (as shown below). The Cauchy residue theoremgeneralizes both the Cauchy integral theorem(because analytic functionshave no poles) and the Cauchy integral formula(because f(x)/(x-a)nfor analytic fhas exactly one pole at x=awith residue Res(f(x)/(x-a)n,a)=f(n)(a)/n! If you are already familiar with complex residues, you can skip the next section. [u(x(t),y(t)) +i v(x(t),y(t))] [ x'(t) + i y'(t)] dt,$$ where $$x(t) = {\rm Re}(\gamma(t))$$ and $$y(t) = {\rm Im}(\gamma(t))$$. (7.13) Note that we could have obtained the residue without partial fractioning by evaluating the coeﬃcient of 1/(z −p) at z = p: 1 1−pz z=p = 1 1−p2. Consistency of trace norm minimization. Matrix Perturbation Theory. Here is a very partial and non rigorous account (go to the experts for more rigor!). For holomorphic functions $$Q$$, we can compute the integral $$\displaystyle \int_0^{2\pi} \!\!\!  The dependence on \(z$$ of the form $$\displaystyle \frac{1}{z- \lambda_j}$$ leads to a nice application of Cauchy residue formula. Wolfram Web Resources. \sum_{ \lambda \in {\rm poles}(f)} {\rm Res}\big( f(z) \pi \frac{1}{\sin \pi z} ,\lambda\big).\) See [7, Section 11.2] for more details. Singular value decompositions are also often used, for a rectangular matrix $$W \in \mathbb{R}^{n \times d}$$. The Cauchy residue theorem can be used to compute integrals, by choosing the appropriate contour, looking for poles and computing the associated residues. Join the initiative for modernizing math education. $$Taking the trace, the cross-product terms $${\rm tr}(u_j u_\ell^\top) = u_\ell^\top u_j$$ disappear for $$j \neq \ell$$, and we get:$$ {\rm tr} \big[ z (z I – A – \Delta)^{-1} \big] – {\rm tr} \big[ z (z I – A)^{-1} \big]= \sum_{j=1}^n \frac{ z \cdot u_j^\top \Delta u_j}{(z-\lambda_j)^2} + o(\| \Delta \|_2). The Residue Theorem has Cauchy’s Integral formula also as special case. Question on evaluating $\int_{C}\frac{e^{iz}}{z(z-\pi)}dz$ without the residue theorem. Journal of Machine Learning Research, 9:1019-1048, 2008. Note that similar constructions can be used to take into account several poles. [5] Jan R. Magnus. Note that this extends to piecewise smooth contours $$\gamma$$. Cauchy's Residue Theorem contradiction? By expanding the expression on the basis of eigenvectors of $$A$$, we get $$z (z I- A – \Delta)^{-1} – z (z I- A)^{-1} = \sum_{j=1}^n \sum_{\ell=1}^n u_j u_\ell^\top \frac{ z \cdot u_j^\top \Delta u_\ell}{(z-\lambda_j)(z-\lambda_\ell)} + o(\| \Delta \|_2). The Cauchy Residue Theorem Before we develop integration theory for general functions, we observe the following useful fact. By expanding the product of complex numbers, it is thus equal to$$\int_0^1 [ u(x(t),y(t)) x'(t) \ – v(x(t),y(t))y'(t)] dt +i \int_0^1 [ v(x(t),y(t)) x'(t) +u (x(t),y(t))y'(t)] dt,$$which we can rewrite in compact form as (with $$dx = x'(t) dt$$ and $$dy = y'(t)dt$$):$$\oint_\gamma ( u \, dx\ – v \, dy ) + i \oint_\gamma ( v \, dx + u \, dy ).$$We can then use Green’s theorem because our functions are differentiable on the entire region $$\mathcal{D}$$ (the set “inside” the contour), to get$$\oint_\gamma ( u \, dx\ – v \, dy ) + i \oint_\gamma ( v \, dx + u \, dy ) =\ – \int\!\!\!\!\int_\mathcal{D} \! 1 Residue theorem problems Bernd Schroder¨ Louisiana Tech University, College of Engineering and Science The Residue Theorem. Hints help you try the next step on your own. If around $$\lambda$$, $$f(z)$$ has a series expansions in powers of $$(z − \lambda)$$, that is, $$\displaystyle f(z) = \sum_{k=-\infty}^{+\infty}a_k (z −\lambda)^k$$, then $${\rm Res}(f,\lambda)=a_{-1}$$. Econometric Theory, 1(2):179–191, 1985. Complex integration: Cauchy integral theorem and Cauchy integral formulas Deﬁnite integral of a complex-valued function of a real variable Consider a complex valued function f(t) of a real variable t: f(t) = u(t) + iv(t), which is assumed to be a piecewise continuous function deﬁned in the closed interval a ≤ t … sur les intégrales définies, prises entre des limites imaginaires, Polynomial magic III : Hermite polynomials, The many faces of integration by parts – II : Randomized smoothing and score functions, The many faces of integration by parts – I : Abel transformation. See an example below related to kernel methods. SIAM Journal on Matrix Analysis and Applications 23.2: 368-386, 2001. 0) = 1 2ˇi I. C. f(z) z z. Deﬁnition Let f ∈ Cω(D\{a}) and a ∈ D with simply connected D ⊂ C with boundary γ. 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# Integration by Parts

### Integration by Parts

If u and v be two functions of x, then the integral of product of these two functions is given by $$\int{uv}.dx=u\int{vdx-\int{\left[ \frac{du}{dx}\int{vdx} \right]}}dx$$.

In applying the above rule care has to be taken in the selection of the first function (u) and the second function (v) depending on which function can be integrated easily. Normally we use the following methods for making this choice:

If both functions are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable. Usually we use the following preference order for the first function.

(Inverse, Logarithmic, Algebraic, Trigonometric, Exponent)

This rule is called as ILATE which states that the inverse function should be assumed as the first function while performing integration. Hence, the functions are to be assumed in the order from left to right depending on the type of functions involved. Various other cases have been discussed in detail in the later sections.

Examples

1. Evaluate∫ xsec²x dx.

Solution: Let I = ∫ xsec²x dx.

= x (tanx) – ∫ tanx dx.

= x tanx – log|secx|+c

2. Evaluate∫logx/x² dx.

Solution: ∫logx/x² dx = (logx)(-1/x) + ∫1/x.1/x dx

=-1/x  logx – 1/x² + c

3. ∫(1ogx)² dx

Solution: ∫(1ogx)² dx = (1ogx)²x – ∫x.2logx.1/x dx

= (1ogx)² x – 2∫logx. dx

= (1ogx)² x – 2(xlogx-∫x.1/x dx)

= (logx)² – 2x.logx + x + c


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# Tetration Forum

Full Version: base holomorphic tetration
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actually I think that the regular tetration at the lower fixed point is not continuable to bases $>e^{1/e}$.

Actually I think the Shell-Thron region is the domain of holomorphy, every point on the boundary is singular. This became clear to me some days ago:

1. The regular iteration depends on the derivative of $\exp_b$ at the fixed point.
2. If $a$ is the fixed point then the derivation is $\log(a)$
3. The fixed point is given by $a = \exp(-W(-\log(b)))$, or $\log(a)=-W(-\log(b))$.
4. The problematic value of the derivation at the fixed point is $|\log(a)|=1$. In this case $\exp_b^{\circ t}$ is not analytic at $a$. I guess this implies that $b\mapsto\exp_b^{\circ t}(1)$ is not analytic at $b$.
5. $|\log(a)|=1$ is exactly at those $b$ in the image of the unit circle under of the inverse of $b\mapsto -W(-\log(b))$, i.e. under $z\mapsto\exp(z \exp(-z))$ which is just the boundary of the Shell-Thron region.
Interesting, that it is the natural boundary (it would remind me of the situation with the Jacobi theta functions in the nome $q$: the (unit, I think) circle is the natural boundary, delimited by a dense set of singularities.). This would suggest that the STR and the region outside are two totally different domains, and there may not be a "true" tetrational that solves them both, or, if there is, it would not agree with the regular iteration. Which makes me wonder. There's this graph:

http://math.eretrandre.org/tetrationforu...926#pid926

which shows the possible region of convergence of Robbins' tetration, overlapping the STR a little. Could one test there and calculate it out really really really far and try to see if it does indeed disagree with the value of the regular iteration as would be expected if this hypothesis was correct?

Then again, the hypothesis could also be incorrect. I'm a little dubious about point #4: how does $\exp^{t}_b(w)$ being non-analytic at w = a imply it non-analytic in *b* when w = *1*? Can you do a graph of the derivative $\frac{\partial}{\partial b} \exp^{t}_b(1)$ for the regular iteration, in the same way that you did the graph of the regular iteration you posted here earlier?
I also noticed something in that paper where you describe the $g$-coefficients for the regular iteration. You have the iterating formula

$g_n = \frac{1}{{f_1}^n - f_1} \left(f_n {g_1}^n - {g_1}{f_n} + \sum_{m=2}^{n-1} f_m {g^m}_n - g_m {f^m}_n\right)$.

where $g = reg$
But at n = 2, we have a sum from "2 to 1". How do you do that? Is it equal to the sum from 2 to 2, minus the value of the summand evaluated at 2 (like how the sum from 2 to 3 is that from 2 to 4 minus the value of the summand at 4)? Wouldn't that just be 0? If I assume so, then I get

$g_2 = \frac{\log(a)^2 \left(\log(a)^t\right)^2 - \log(a)^t \log(a)^2}{\log(a)^2 - \log(a)}$

which disagrees with what you gave in your earlier post here.

bo198214: closed one tex tag to make the post readible
(11/12/2009, 07:08 PM)mike3 Wrote: [ -> ]Then again, the hypothesis could also be incorrect. I'm a little dubious about point #4: how does $\exp^{t}_b(w)$ being non-analytic at w = a imply it non-analytic in *b* when w = *1*?

Ya indeed, perhaps I get some stronger arguments later.
For illustration: I was considering the limit formula. Only if b is inside the STR the fixed point is attracting and hence reachable from 1 by repeated iteration.
If the base crossed the boundary one has to apply the limit formula for the inverse, and on the boundary the functions by approximating the fixed point from inside and from outside are different (no continuations of each other).

Quote: Can you do a graph of the derivative $\frac{\partial}{\partial b} \exp^{t}_b(1)$ for the regular iteration, in the same way that you did the graph of the regular iteration you posted here earlier?

On the real axis you will not see anything special about the derivative I guess.

(11/12/2009, 07:37 PM)mike3 Wrote: [ -> ]I also noticed something in that paper where you describe the $g$-coefficients for the regular iteration. You have the iterating formula

$g_n = \frac{1}{{f_1}^n - f_1} \left(f_n {g_1}^n - {g_1}{f_n} + \sum_{m=2}^{n-1} f_m {g^m}_n - g_m {f^m}_n\right)$.

where $g = reg$
But at n = 2, we have a sum from "2 to 1". How do you do that? Is it equal to the sum from 2 to 2, minus the value of the summand evaluated at 2 (like how the sum from 2 to 3 is that from 2 to 4 minus the value of the summand at 4)? Wouldn't that just be 0? If I assume so, then I get

yes, 0 is just the classical sense.

Quote:$g_2 = \frac{\log(a)^2 \left(\log(a)^t\right)^2 - \log(a)^t \log(a)^2}{\log(a)^2 - \log(a)}$

which disagrees with what you gave in your earlier post here.

No, it agrees. $f_2 = \log(a)/2$.
1. The limit formula will simply fail to converge when the fixpoint is no longer attracting, no? (unless you reverse it, but then you get a different function so this cannot be interpreted as continuation of the formula) If so, that just marks the region of convergence of that limit formula, it does not imply that it is a natural boundary, just as the failure of the convergence of a Taylor series beyond its radius of convergence does not imply that such radius is a natural boundary.

2. If $e^{1/e}$ is a singularity, as you seem to suspect, the derivative should explode at that point -- consider $\frac{d}{dx} \sqrt{x}$ as $x \rightarrow 0^{+}$. Though I just realized, it need not be the first derivative, but a higher one may do it. It can't be a smooth point. (Though you have to watch out (!) and not be misled by numerical or convergence error (is that a type of numerical error?) esp. considering slowdown of convergence as $e^{1/e}$ is approached, and also the rounding errors of numerical differentiation if you're differentiating it numerically.)

3. Wouldn't it be $\frac{\log(a)^2}{2}$? Or do you mean the series for f expanded about the fixed point?
Not sure if this will help anything, but here's a graph showing the root test for the first 33 Taylor coefficients in b for $^{1/2} b$ expanded about b = 1.42. I used the regular formula

$^{t} b = \lim_{n \rightarrow \infty} \log_b^n\left(F - \left(F - \exp^n_b(1)\right) \log(F)^t\right)$

(this is just a rearrangement of the formula mentioned in my original post to this thread)

with n = 8192, 2048 digits of precision (just to be safe) and the simple, straightforward numerical differentiation via the difference quotient (delta step 10^-20). Took about an hour(!) to compute due to the extreme precision... I could probably do some more if I could parallelize the job but Pari/GP doesn't have parallel computation facilities (no threads, no MPI, no nothing like that). I suppose I could recode the formula as a C or Fortran program for a parallel job, interfaced with an arbitrary-precision package like GMP or MPFR (I think Pari/GP uses something like this under the hood), and crunch even more coeffs, because crunching these coeffs is tediously slow, esp. if I have to add more precision (I'm pretty sure up to the 30th or 31st coeff is right, not sure about the accuracy of the last 2 but they should be, though only a recalc with more precision will tell.).

Coeffs:
Code:
a_0 = 1.246220033102832033637391168 a_1 = 0.4479211001478450057502788291 a_2 = -0.1944285662380565757140341278 a_3 = 0.1431678738611080328964523214 a_4 = -0.1449673997741212873243507213 a_5 = 0.1821593012238922403824879947 a_6 = -0.2634074264259034150890239199 a_7 = 0.4170984777615359795375604795 a_8 = -0.7022041478879047165932451401 a_9 = 1.235350783631857820858433698 a_10 = -2.247058971388908630541707138 a_11 = 4.196782696009382012696224987 a_12 = -8.009130240912803756451562486 a_13 = 15.56210572774662623695298001 a_14 = -30.70293022611311616071677783 a_15 = 61.37465434532134439666317750 a_16 = -124.0937577676786262630256434 a_17 = 253.4276217341761588591503345 a_18 = -522.1524296306182164337119282 a_19 = 1084.317735417423446029022288 a_20 = -2267.611476993890273952854904 a_21 = 4772.099072723291344681868206 a_22 = -10098.44557704069780712228148 a_23 = 21465.93884981489863800142138 a_24 = -45708.11493010257551050715076 a_25 = 96243.61297567708695783556454 a_26 = -185475.2857026224205407755534 a_27 = 186483.8146778817735118583707 a_28 = -451252.6193153016119357011782 a_29 = 73309968.03777933438145964027 a_30 = -1983477241.135794800710911594 a_31 = 19226263007.10400347804764423 a_32 = 346411096888.4104287944798334

Graph of root test (testing for convergence radius, as you mention):

[attachment=632]

The big question, of course, is does it dip below 0.42, the expected radius of convergence, and stay below it, out to infinity? If it does, then that would indicate a closer singularity. If it drops to 0.02466... then that would indicate a singularity at $b = e^{1/e}$. (Of course, the graph alone isn't a proof.)
(rmv double post)
(11/13/2009, 02:22 AM)mike3 Wrote: [ -> ]1. The limit formula will simply fail to converge when the fixpoint is no longer attracting, no? (unless you reverse it, but then you get a different function so this cannot be interpreted as continuation of the formula) If so, that just marks the region of convergence of that limit formula, it does not imply that it is a natural boundary, just as the failure of the convergence of a Taylor series beyond its radius of convergence does not imply that such radius is a natural boundary.

yes, yes, I know my argumentation is very weak. Its more a feeling. Lets see whether sometime I can make this more precise, or whether it just turns out to be misleading.

Quote:2. If $e^{1/e}$ is a singularity, as you seem to suspect, the derivative should explode at that point -- consider $\frac{d}{dx} \sqrt{x}$ as $x \rightarrow 0^{+}$. Though I just realized, it need not be the first derivative, but a higher one may do it. It can't be a smooth point.

I dont think so, if all derivatives from one approaching sector exists the function can still be non-analytic at that point. In this case one says f has an asymptotic powerseries development. This is particularly true for the non-integer iteration of $e^x-1$. You can derive a powerseries at 0 and you have an (or rather two) analytic function(s) whose derivatives are compatible with the powerseries development at 0 when approaching 0 from one direction. But the function itself is not analytic, because the two analytic functions defined in opposite sectors are not continuations of each other.

Quote:3. Wouldn't it be $\frac{\log(a)^2}{2}$? Or do you mean the series for f expanded about the fixed point?

Mike, I explained in my previous posts with what functions I work. The function $f$ in the powerseries formula (which is $g$ in my original post) is $\ln(a)(e^x-1)$ where $a$ is the lower fixed point.
This is a linear conjugation of $b^x$ which moves the lower fixed point to 0.
Hmm.

Oh yeah, about the thing with the "g": I was thinking the function was $b^x$ not $\log(a) (e^x - 1)$, although now that I read it again I see it now. I guess the inconsistent notation with the paper was throwing me off. So then would I be right in interpreting "$g$" in the posts as what is called "$f$" in the paper, and what is called "$g^{\circ t}$" in the posts as what is called "$g$" in the paper? (i.e. I'm trying to figure out how to map the notation from your posts to that of the paper so I can get the recursive-generating formula going)
(11/13/2009, 10:33 AM)mike3 Wrote: [ -> ]So then would I be right in interpreting "$g$" in the posts as what is called "$f$" in the paper, and what is called "$g^{\circ t}$" in the posts as what is called "$g$" in the paper?

Exactly.
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## On lifting.(English)Zbl 0538.20014

Lie group representations II, Proc. Spec. Year, Univ. Md., College Park 1982-83, Lect. Notes Math. 1041, 209-249 (1984).
[For the entire collection see Zbl 0521.00012.]
Let $$F$$ be a global field, $$C_ F$$ be the adèle class group of $$F$$, $$\varepsilon: C_ F\to C^*$$ be a character of order $$n$$ and let $$L\supset F$$ be the corresponding cyclic extension, $${\mathcal G}=\text{Gal}(L/F).$$ Let $$A$$ be the adèle ring of $$F$$, $$G_ A=\text{GL}_ n(A)$$, $${\mathcal F}_ a(G)$$ be the set of equivalent classes of irreducible automorphic representations of $$G_ A$$ and $${\mathcal F}_ a^{\varepsilon}(G)=\{\pi \in {\mathcal F}_ a(G);\;\pi \otimes \varepsilon \cong \pi \}$$ where one considers $$\varepsilon$$ as a character of $$G_ A$$. For any place $${\mathfrak p}$$ of $$F$$ we denote by $$F_{{\mathfrak p}}$$ the completion of $$F$$ at $${\mathfrak p}$$ and by $$\varepsilon_{{\mathfrak p}}: F^*_{{\mathfrak p}}\to C^*$$ the restriction of $$\varepsilon$$ to $$F^*_{{\mathfrak p}}$$. Let $${\mathfrak p}_ 1$$, $${\mathfrak p}_ 2$$ be two non-archimedean places such that $$\varepsilon_{{\mathfrak p}_ 1}$$, $$\varepsilon_{{\mathfrak p}_ 2}$$ are primitive characters of order $$n$$,
$$\tilde {\mathcal F}_ a(G)=\{\pi \in {\mathcal F}^{\varepsilon}_ a(G)$$; local components of $$\pi$$ at $${\mathfrak p}_ 1$$ and $${\mathfrak p}_ 2$$ are cuspidal} and
$${\hat C}^ 0_ L=\{\chi \in \hat C_ L=\operatorname{Hom}(C_ L,C^*)$$; $$\chi^{\sigma}\neq \chi_{{\mathfrak p}_ 1}$$, $$\chi^{\sigma}\neq \chi_{{\mathfrak p}_ 2}$$ for any $$\sigma\in G-\{e\}$$, where $$\chi_{{\mathfrak p}_ 1},\chi_{{\mathfrak p}_ 2}$$ are local components of $$\chi$$ at $${\mathfrak p}_ 1$$ and $${\mathfrak p}_ 2\}$$.
The author first proves that there exists a natural one-to-one correspondence between $$\tilde {\mathcal F}^{\varepsilon}_ a(G)$$ and $${\mathcal G}$$-orbits on $${\hat C}^ 0_ L$$. Using Arthur’s results with this theorem, one can show that there exists a “natural” one-to-one correspondence between $${\mathcal F}^{\varepsilon}_ a(G)$$ and $${\mathcal G}$$-orbits on $${\hat C}_ L$$. Also, the author proves the following local variant of the theorem. Let $$F$$ be a local non-archimedean field, $$\varepsilon:F^*\to C^*$$ be a character of order $$n$$. $$G=\text{GL}_ n(F)$$. We denote by the same letter $$\varepsilon$$ the character of $$G$$ given by $$\varepsilon(g)=\varepsilon(\det g),$$ $$g\in G$$. Let $${\mathcal F}(G)$$ be the set of equivalent classes of smooth irreducible representations of $$G$$ and $${\mathcal F}^{\varepsilon}(G)=\{\pi \in {\mathcal F}(G);\;\pi \otimes \varepsilon \simeq \pi \}$$. Let $$L\supset F$$ be the cyclic extension which corresponds to $$\varepsilon$$ by the local class field theory and $$G=\text{Gal}(L/F)\,(\cong \mathbb Z/n\mathbb Z)$$. Then there exists a natural one-to-one correspondence between $${\mathcal F}^{\varepsilon}(G)$$ and $${\mathcal G}$$-orbits on $$\operatorname{Hom}(L^*,G^*)$$.
Reviewer: E. Abe

### MSC:

 20G05 Representation theory for linear algebraic groups 20G35 Linear algebraic groups over adèles and other rings and schemes 11R39 Langlands-Weil conjectures, nonabelian class field theory 20G25 Linear algebraic groups over local fields and their integers 20G30 Linear algebraic groups over global fields and their integers 11S37 Langlands-Weil conjectures, nonabelian class field theory 22E50 Representations of Lie and linear algebraic groups over local fields 11F70 Representation-theoretic methods; automorphic representations over local and global fields

Zbl 0521.00012


text:
Contents

Definition

Definition

A commutative ring $R$ is w-contractible if every faithfully flat pro-étale morphism $Spec A \to Spec R$ has a section.

Properties

Proposition

For every commutative ring $R$, there is a w-contractible $A$, def. , equipped with a faithfully flat pro-étale morphism $Spec A \to Spec R$.

Proposition

For $R$ w-contractible, the profinite set $\pi_0(Spec R)$ is an extremally disconnected profinite set.

part of (Bhatt-Scholze 13, theorem 1.8)

References

Last revised on November 21, 2013 at 09:38:30. See the history of this page for a list of all contributions to it.


text:
## anonymous 5 years ago Can somebody explain how to simplify this expression? Negative 2y to the negative 1 power, all inside parentheses, with a negative 2 exponent outside the parentheses.

1. amistre64

using math notation for staters

2. amistre64

$(-2y^{-1})^{-2}$

3. amistre64

this?

4. amistre64

$-2^{-2} y^2$ $-\frac{y^2}{4}$ maybe?

Is it still a minus (if it was inside the paren)?

6. anonymous

Yes, you set it up right, amistre.

I suspect the answer may be $y ^{2}\over 4$

Maybe amisstrte64 will come back and verify his answer.

9. anonymous

I think it is 4y^2

10. anonymous

...because it looks like this is the expression... (-2y^-1)^-2 Multiplying exponents (-2y)^2 Expanding brackets 4y^2

I went this direction, but I do see your logic. I may have violated order of operations:$(-2y ^{-1})^{-2}$ to:$1\over (-2y ^{-1})^{2}$then to:$1\over 4 y ^{-2}$ finally$y ^{2}\over 4$

12. anonymous

$(-2y^{-1})^{-2}=(-2)^{-2}y^2=\frac{y^2}{(-2)^2}=\frac{y^2}{4}$\]

Help me out here satellite73

14. anonymous

if that was the problem to beginwith

15. anonymous

we get $\frac{y^2}{4}$ yes?

16. anonymous

hello radar! here to celebrate amistre's 1000 medal

that is what i got, amistre got a -y^2/4

18. anonymous

hope it is soon cause i got to run

before you run look at gianfranco solution above, what is wrong with that approach beside getting a different answer???

20. anonymous

well if the question is as he wrote it, you have to raise (-2) to the power of -2, not the same as $-2^{-2}$

21. anonymous

you should get a 4 in the denominator

22. anonymous

gianfraco was acting as if it was $((-2y)^{-1})^{-2}$

23. anonymous

but that is now how i read the problem. i read it as only the y being raised to the power of -1. if there were no parentheses that is what it means

24. anonymous

i read it the way you did

I can see that that would make a difference.

26. anonymous

of course. like the difference between $(2\times 3)^2$ and $2\times 3^2$

Understand. Thanks for clearing up a few things.

28. anonymous

O never mind everybody..I think I've figured it out.


text:
Moving arrows and entities through a sentence

Currently, the two arrows are pointing to jumped and over in the sentence. I was wondering how I can move these two arrows from The(at the beginning of the sentence) to dog.

EDIT : I am trying to reproduce slides 12-25 from this talk.

My only option is to include images with arrows pointing to different words.

\begin{document}
\frame{
\frametitle{Learning Algorithm(Contd.)}

\begin{figure}
\includegraphics[width=.9\textwidth]{Fig1.PNG}
\label{fig:Fig1}
\end{figure}

}

\frame{
\frametitle{Learning Algorithm(Contd.)}

\begin{figure}
\includegraphics[width=.9\textwidth]{Fig2.PNG}
\label{fig:Fig2}
\end{figure}

}
\end{document}

• Can you provide a compilable MWE showing what you've done so far? – Richard May 28 '16 at 4:54
• Rather than post image files of the output of your code, please post the code itself. – Mico May 28 '16 at 6:35
• We won't know why you can't move the arrows unless we understand how you're placing the arrows at present. For that, we need to see the code, not an image file. Please note that I haven't claimed you should "resort [to] tikz". – Mico May 28 '16 at 7:00
• From your code I can conclude that you use beamer as document class and that you so far use images (drawn in unknown program), which you include in frame as image. What is your question actually? How to make animation in which you in desired sequence uncover/hide your images? Or how to draw new images? Or how to draw/wrote this images with LaTeX? – Zarko May 28 '16 at 10:02
• If you do not know how to insert the arrows with latex, you could at least provide a MWE that prints the text about the fox and the dog. – user36296 May 28 '16 at 10:51

I suggest you use a custom array environment to stack the word/arrow/explanation triplets. Observe the use of the \color macro to set colors.

\documentclass{beamer}

\usepackage{mathtools} % for "\mathclap" macro
\newcommand\stacktab[2]{%
$\begin{array}[t]{@{}c@{}} \text{#1} \\ \uparrow \\ \mathclap{#2} \end{array}$}
\newcommand\uIN{\upsilon^{}_{\!IN}}
\newcommand\uOUT{\upsilon^{}_{\!OUT}}

\begin{document}
\begin{frame}
The quick brown fox \stacktab{jumped}{\uOUT}
\stacktab{\color{blue}over}{\uIN} the lazy dog.

\bigskip
\stacktab{\color{magenta}The}{\uOUT} quick brown fox jumped
\stacktab{\color{red}over}{\uIN} the lazy dog.

\bigskip
The quick brown fox jumped over the \stacktab{lazy}{\uIN}
\stacktab{\color{purple}dog}{\ \uOUT}. % shift "\UOUT" slightly to the right

\end{frame}
\end{document}


text:
# Understanding the significance of BLAT score calculations for read fragment alignment

I'm attempting to reconstruct the read fragments that form a hit to a target, excluding/skipping gaps. BLAST's outfmt 6, and alternatives such as Diamond's M8 format only give the number of gap opens on the alignment. One can't just use the start and end positions of the alignment as indices to slice the read DNA sequence, as this will include the nucleotides that were gapped by the alignment.

PSL appeared to be the only format I could find that gave the actual start positions (and lengths) of each of the pieces that formed a hit (that is, each gap forces a new block).

The regular BLAST output is human readable and is a nightmare to parse. The XML format mentioned has a structure for each HSP, but an HSP can still include gaps:

  [...]
<Hit_hsps>
<Hsp>
<Hsp_num>1</Hsp_num>
<Hsp_bit-score>51.6</Hsp_bit-score>
<Hsp_score>122</Hsp_score>
<Hsp_evalue>9.8e-06</Hsp_evalue>
<Hsp_query-from>2</Hsp_query-from>
<Hsp_query-to>98</Hsp_query-to>
<Hsp_hit-from>35</Hsp_hit-from>
<Hsp_hit-to>65</Hsp_hit-to>
<Hsp_query-frame>2</Hsp_query-frame>
<Hsp_hit-frame>0</Hsp_hit-frame>
<Hsp_identity>24</Hsp_identity>
<Hsp_positive>27</Hsp_positive>
<Hsp_gaps>1</Hsp_gaps>
<Hsp_align-len>32</Hsp_align-len>
<Hsp_qseq>ITAIGAGLQGPAGCEVIDAGGLLVMPGGIDTH</Hsp_qseq>
<Hsp_hseq>IAAVGTGLE-PAGAEIIDAGGLLVMPGGIDVH</Hsp_hseq>
<Hsp_midline>I A+G GL+ PAG E+IDAGGLLVMPGGID H</Hsp_midline>
</Hsp>
</Hit_hsps>
</Hit>

I'm trying to be lazy, I don't want to have to parse the Hsp_qseq and Hsp_hseq to determine exactly where my gaps are. I just want co-ordinates of where my read hits a protein. The PSL format (below) tells me this (tStarts):

match   mis-    rep.    N's     Q gap   Q gap   T gap   T gap   strand  Q               Q       Q       Q       T               T       T       T       block   blockSizes      qStarts  tStarts
match   match           count   bases   count   bases           name            size    start   end     name            size    start   end     count
---------------------------------------------------------------------------------------------------------------------------------------------------------------
[...]
21      4       0       0       1       8       1       24      ++      <PROTEIN>   662     321     354     <READ>      101     2       101     2       8,17,   321,337,        2,50,

As an alternative, I've executed the BLAT aligner as a means to find alignments of translated DNA reads against a small protein database. I selected BLAT as it provided the easiest means to access information on the actual alignment blocks (with the smallest amount of parsing). That is, the PSL has a comma delimited field that enumerates each of the starting positions of a query to a target.

However, BLAT outputs a significantly greater number of alignments than alternative tools. This is likely due to its identity percentage threshold defaulting to 25% when emulating a blastx style query. I can cope with this (apologies to my disks for the I/O), but for filtering purposes, the output PSL format appears to leave me nothing but raw match counts to work with.

The PSL gives me what I want, except now I don't have information on its significance. I know that the BLAT FAQ describes a Perl script for adding "web-based identity and score calculations". I can actually get identity percentages and scores from this script, but I cannot tell what the score actually represents. Is it comparable to bit-score? How can I recover something like an e-value from a PSL output file?

• Just out of curiosity: can you elaborate on how BLAT 'provided the easiest means to access information on the actual alignment blocks (with the smallest amount of parsing)'? What about blast HSPs? – holmrenser May 18 '17 at 9:31
• @holmrenser Don't you only get the HSPs by parsing the XML output? This is some laze on my part, as the HSPs in my XML described the hits from the perspective of the translated DNA, but I want to know where the alignments on the original sequence are without too much of my own error-prone interference. BLAT just gives the offset. It might be that I've overlooked something in the HSP data though... – Sam Nicholls May 18 '17 at 9:37
• No! Not at all! The default output format for blast is a simple text file and there are flags to get one HSP per line (no sequence, only the scores etc). You can get XML as well but that's just a way of making your life harder IMO. Please edit your question and clarify what the original (scientific) question you are asking is. Are you looking for exact matches? Distant homologs? Finally, it would be better if you could include an example of your PSL file if that's really what you want parsed. – terdon May 18 '17 at 10:49
• @terdon Despite my dislike of XML parsing, it seems a more structured way to get out the components of a hit compared to trying to infer structure from a sequence of lines. For what it's worth BLAT doesn't have flags for outputting an HSP per line. – Sam Nicholls May 18 '17 at 11:08
• Fair enough (and sorry, didn't mean to appear pushy). I was just wondering if using another tool (like BLAST which would give you the e-values in a parseable table format) wouldn't be a simpler approach. I realize that "redo the analysis with this tool instead" is not a perfect solution though :). – terdon May 18 '17 at 12:09

It would appear that BLAT and exonerate (as suggested by @terdon) are the only pieces of software that provide enough information for "HSPFragments" to be parsed from their outputs, at least according to BioPython's documentation, anyway:

Most search programs only have HSPs with one HSPFragment in them, making these two objects inseparable. However, there are programs (e.g. BLAT and Exonerate) which may have more than one HSPFragment objects in any given HSP. If you are not using these programs, you can safely consider HSP and HSPFragment as a single union.

It would seem the options are sticking with BLAT and generating scores afterwards with the script described in their FAQ, or using exonerate which provides a raw score as part of its default output. What these scores actually mean I will have to investigate, but for now I think the case is closed!

I would suggest using blastx and parsing the output XML. As an example, have a look at this: http://www.perlmonks.org/?node_id=1006367.

The relevant part of the hit would be the following:

<Hit>
<Hit_num>1</Hit_num>
<Hit_id>gnl|BL_ORD_ID|1515029</Hit_id>
<Hit_def>43989.cce_0262 (Cyanothece ATCC 51142)</Hit_def>
<Hit_accession>1515029</Hit_accession>
<Hit_len>65</Hit_len>
<Hit_hsps>
<Hsp>
<Hsp_num>1</Hsp_num>
<Hsp_bit-score>40.0466</Hsp_bit-score>
<Hsp_score>92</Hsp_score>
<Hsp_evalue>0.00664016</Hsp_evalue>
<Hsp_query-from>155</Hsp_query-from>
<Hsp_query-to>253</Hsp_query-to>
<Hsp_hit-from>12</Hsp_hit-from>
<Hsp_hit-to>44</Hsp_hit-to>
<Hsp_query-frame>-1</Hsp_query-frame>
<Hsp_hit-frame>0</Hsp_hit-frame>
<Hsp_identity>17</Hsp_identity>
<Hsp_positive>27</Hsp_positive>
<Hsp_gaps>0</Hsp_gaps>
<Hsp_align-len>33</Hsp_align-len>
<Hsp_qseq>LRGAICSMEHIEEALGKLKDWARKLIELLLGPR</Hsp_qseq>
<Hsp_hseq>ITGAVCLMDYLEKVLEKLRELAQKLIETLLGPQ</Hsp_hseq>
<Hsp_midline>+ GA+C M+++E+ L KL++ A+KLIE LLGP+</Hsp_midline>
</Hsp>
</Hit_hsps>
</Hit>

You should be able to get the position information from the following tags: <Hsp_query-from>,<Hsp_query-to>,<Hsp_hit-from>,<Hsp_hit-to>.

To translate back to you original DNA sequence I can imagine you need the <Hsp_query-frame> tag.

And then of course you get the evalue for free: <Hsp_evalue>

• An HSP is still allowed to contain gaps, though. So I'd still have to parse one of the qseq or hseq :( – Sam Nicholls May 18 '17 at 12:06
• Correct me if I'm wrong, but the tags I mentioned (Hsp_query-from, Hsp_query-to, Hsp_hit-from, Hsp_hit-to) are coordinates on the original sequence. – holmrenser May 18 '17 at 13:26
• You are correct (I believe)! But the HSP can contain gaps, so I can't just take those co-ordinates as given to extract the DNA from the read between those positions (as I don't want the gapped nucleotides). – Sam Nicholls May 18 '17 at 13:32
• The fact that the read aligns to that position, and that the coordinates are on the original sequence means you actually can. Whether or not you want to do a gapped alignment is a different matter I would think. – holmrenser May 18 '17 at 13:35
• Not for what I need - long story short is I'm trying to "flatten" the reads into a pseudo-consensus. I need to know exactly where (what actual position) the gaps are, so I can handle them appropriately. I can't just use the starts and ends of the read as given, because taking that subsequence will include one (or more) nucleotides that should be removed to accommodate the gap(s). – Sam Nicholls May 18 '17 at 13:38