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# HEALPix¶

 Classification Mixed Available forms Forward and inverse, spherical and elliptical projection Defined area Global Implemented by Alex Raichev and Michael Speth Options No special options for this projection

The HEALPix projection is area preserving and can be used with a spherical and ellipsoidal model. It was initially developed for mapping cosmic background microwave radiation. The image below is the graphical representation of the mapping and consists of eight isomorphic triangular interrupted map graticules. The north and south contains four in which straight meridians converge polewards to a point and unequally spaced horizontal parallels. HEALPix provides a mapping in which points of equal latitude and equally spaced longitude are mapped to points of equal latitude and equally spaced longitude with the module of the polar interruptions.

## Usage¶

To run a forward HEALPix projection on a unit sphere model, use the following command:

proj +proj=healpix +lon_0=0 +a=1 -E <<EOF
0 0
EOF
# output
0 0 0.00 0.00


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# gstat / Linear model of coregionalization

2 messages
 Hello Nick, Once you have your lmc fitted, call the object $psill.For example, if your lmc object is called g.fit with variables v , z call: g.fit$model$v$psill[1] to get coefficient of v for 1st structure of the nested model.g.fit$model$z$psill[1] to get coefficient of z for 1st structure of the nested model. g.fit$model$v$psill[2] to get coefficient of v for 2nd structure of the nested model.g.fit$model$z$psill[2] to get coefficient of z for 2nd structure of the nested model. For cross-variogram, please see the list of g.fit$model to see the name of the object storing v*z cross-variogram model coefficients. I hope this helps. There are other ways of doing this, anyway this one should work.Regards,Manuel  Manuel Luís Castro Ribeiro CERENA- Centro de Recursos Naturais e AmbienteInstituto Superior TécnicoAv. Rovisco Pais  1049-001 Lisboa PortugalPhone:  +351 21 841 74 41Mobile: +351 92 442 80 07 Web: www.cerena.pt   From: R-sig-Geo <[hidden email]> On Behalf Of Nick HammSent: quarta-feira, 29 de janeiro de 2020 04:22To: r-sig-geo <[hidden email]>; [hidden email]Subject: [R-sig-Geo] gstat / Linear model of coregionalization Dear allIs anybody using gstat for computing cross-variograms and fitting cross variogram models?  I have implemented this in a similar manner to the example code given in the gstat vignette (Section 10). I would like to extract the coefficients (i.g., the  b_{ij}^l of the linear model of coregionalization as given in Equation 4.37 of Goovaearts (1997) book (page 110).   \gamma_{ij} (h) = \sum_{l=0}^L b_{ij}^l g_l(h) \forall i, j     Is this possible? kind regards Nick  _______________________________________________ R-sig-Geo mailing list [hidden email] https://stat.ethz.ch/mailman/listinfo/r-sig-geo

format: plain


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Browse Questions

If the variance of a data is 121, then the standard deviation of the data is _______

$\begin{array}{1 1}(A)\; 55 \\(B)\;11 \\(C)\;97 \\(D)\;27 \end{array}$

$SD =\sqrt {\sigma^2}$
$\qquad= \sqrt {121}$
$\qquad= 11$
Hence B is the correct answer.


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Submit Essay

Once you submit your essay, you can no longer edit it.

# Riemann H. proved true, if settled by 2100

### Question

previous Metaculus questions:

The Riemann hypothesis is a conjecture stating that the nontrivial zeroes of the Riemann zeta function $\zeta(s)$ all have real part equal to $1/2$. The Riemann zeta function also has zeroes at the strictly negative even integers $-2, -4, -6, \ldots$, which are by definition its trivial zeroes. The hypothesis would therefore be correct if all zeroes of the Riemann zeta function other than these trivial zeroes had real part equal to $1/2$.

The conjecture is significant because it implies the tightest possible error bounds on a wide range of estimates in analytic number theory, starting from the tight asymptotic $\pi(x) = \textrm{Li}(x) + O(\sqrt{x} \log x)$ for the prime counting function $\pi(x)$. It has now become standard practice to prove theorems of analytic number theory conditional on the Riemann hypothesis or some of its closely related generalizations.

Will the Riemann Hypothesis be proved true by 2100?

This question will resolve positively if the Millennium Prize for the proof of the Riemann hypothesis is awarded before the resolve date of this question. It will resolve negatively if the Millennium Prize is awarded, according to rule 5.c of the Millennium Prize Rules, for the disproof of the Riemann hypothesis. It will resolve ambiguously if the Prize is not awarded for either achievement until the resolve date of the question, or if the Prize is awarded for a proof that the Riemann hypothesis is undecidable in ZFC set theory.

### Prediction

Note: this question resolved before its original close time. All of your predictions came after the resolution, so you did not gain (or lose) any points for it.

Note: this question resolved before its original close time. You earned points up until the question resolution, but not afterwards.

Current points depend on your prediction, the community's prediction, and the result. Your total earned points are averaged over the lifetime of the question, so predict early to get as many points as possible! See the FAQ.


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1)It takes 23 hours 56 minutes and 4 seconds for the earth to make one revolution (mean sidereal day). What is the angular speed of the earth?
2)Assume the earth is spherical. Relative to someone on the rotation axis, what is the linear speed of an object on the surface if the radius vector from the center of the earth to the object makes an angle of 3)59.0� with the axis of rotation. The radius of the earth is 6.37�103 km.
What is the acceleration of the object on the surface of the earth in the previous problem?


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## Intermediate Algebra: Connecting Concepts through Application

$6(x-4)(x-7)$
To factor by trial and error, first check if you can pull out the greatest common factor of the terms. Then determine the two numbers that, when multiplied, equal the third term, and, when added, equal the coefficient of the middle term. $6x^2-66x+168$ $6(x^2-11x+28)$ $6(x-4)(x-7)$


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# Tag Info

7

Lake breezes(similar to sea breezes) are fundamentally a feature of mesoscale meteorology and the peer reviewed reference Small Lake Daytime Breezes: Some Observational and Conceptual Evaluations details both the observational studies of lake breezes and the conceptual understanding behind the formation of the lake breeze. Since OP's question is How big ...

7

Your screenshot lists the raob as station 72357 on 2013-06-01 00Z. Station 72357 is OUN and the sounding for that day time is available at the SPC sounding archive. The SPC sounding analysis shows 3351 J kg$^{-1}$ CAPE and -76 J kg$^{-1}$ CIN for surface parcels. My speculation for your anomolous non-physical values of CAPE and CIN is that the algorithm ...

5

WRF model development is done in such a way that users can run the model independently before you start adding more complex options to ingest observations. There is even an "ideal" mode that new users can take advantage of to learn how the system works (not for simulating real Earth situations). In "real" mode, there are typically two types of simulations ...

3

For soil mosture you can initialize the model with a dataset that provides that information. If you are doing historical retrospective you are likely going to initialize from a reanalysis dataset and if you are doing a forecasting case then you are likely initializing from GFS, RAP, HRRR or some other model. For the case of initialization from GFS, the ...

2

I fly too, so I have first hand experience with movement of air below 13000 feet (Cessna 172). I have my simple version: in an otherwise calm weather a thermal has three stages which looked from profile looks like a tree. Base level near earth is the roots. Rising column is trunk. And cooling off and dissipating at high altitude is the canopy. The part that ...

2

Start with the original equation. Let's first write the Hydrostatic equation: $$\frac{\partial p}{\partial z}=-\rho g$$ So let's prove that $$-g=c_p\theta\frac{\partial \pi}{\partial z}$$ If we use the product rule, we observe $$-g=c_p(\frac{\partial \theta \pi}{\partial z}-\pi\frac{\partial \theta}{\partial z})$$ Since $\pi=\frac{T}{\theta}$, we can ...

2

So wind is caused by differences in atmospheric pressure which in turn is a result of temperature induced density differences. 1. Required temperature diference: A quick search brought up this paper: Miller et al. 2003 SEA BREEZE: STRUCTURE, FORECASTING, AND IMPACTS https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/2003RG000124 They calculate that ...

2

High pressures and low pressures at any height do not exist in isolation. Had to remind myself a lot of this (so anyone feel free to correct), but the first thing worth highlighting is that the Bermuda High is a warm core high pressure and so extends a great depth through the atmosphere. This image is a fairly typical cross section of the atmosphere at 35 ...

1

After looking back into this I found my error to lie in the long equation $\dfrac{\partial \pi}{\partial z}=\left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}\dfrac{\partial}{\partial z}\exp\left(\frac{R}{c_p}\log(p)\right)=\left(\frac{1}{p_0}\right)^{\frac{R}{c_p}}* \frac{T}{\theta}*{\frac{R}{c_p}}*\frac{1}{p}*\dfrac{\partial p}{\partial z}$. I was wrong, ...

1

The storm and basics: Here is an animation I got from this NOAA Tweet of the lightning in Harvey hours before landfall: https://mashable.com/2017/08/31/harvey-weather-satellite-maps-lightning/, where I found the imagery, offers some basic thoughts, including the key concept that: As with most hurricanes, you'll notice that the most active area of lightning ...

1

Extratropical cyclones are relatively well understood. They often follow the Norwegian Cyclone model, which was developed in the 1910's and 1920's. Because of their spatial extent, extratropical cyclones are often less intense than their tropical counterparts. Since they are so much larger and live longer over land, it is easy to get data about an ...

1

You're definitely on the right path with wanting to dig into raw data yourself. RAP has always been a pretty solid source of useful basic data. But in terms of making sense of the data, it indeed will take some practice. You can feel like you're not getting it at all, but suddenly pieces will start to click, much like riding a bike. Basically, you need to ...

Only top voted, non community-wiki answers of a minimum length are eligible


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# 68 centimeters in decimeters

## Conversion

68 centimeters is equivalent to 6.8 decimeters.[1]

## Conversion formula How to convert 68 centimeters to decimeters?

We know (by definition) that: $1\mathrm{cm}=0.1\mathrm{dm}$

We can set up a proportion to solve for the number of decimeters.

$1  cm 68  cm = 0.1  dm x  dm$

Now, we cross multiply to solve for our unknown $x$:

$x\mathrm{dm}=\frac{68\mathrm{cm}}{1\mathrm{cm}}*0.1\mathrm{dm}\to x\mathrm{dm}=6.800000000000001\mathrm{dm}$

Conclusion: $68  cm = 6.800000000000001  dm$

## Conversion in the opposite direction

The inverse of the conversion factor is that 1 decimeter is equal to 0.147058823529412 times 68 centimeters.

It can also be expressed as: 68 centimeters is equal to $\frac{1}{\mathrm{0.147058823529412}}$ decimeters.

## Approximation

An approximate numerical result would be: sixty-eight centimeters is about zero decimeters, or alternatively, a decimeter is about zero point one five times sixty-eight centimeters.

## Footnotes

[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).

Results may contain small errors due to the use of floating point arithmetic.


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Are there any other rational preference relations without utility function representations, besides Lexicographic?

It seems like lexicographic isn't that "special". Like yes it is special in that supposing it has a utility function gives you a bijection from the rationals to the reals, but I mean unique in some sense. I know that it satisfies a bunch of nice properties, but it still feels like there should be plenty of other (rational) preference relations that don't have utility representations.

Yes, there are many. Here are some examples :

Consider the weak preference relation $$\succsim$$ defined over $$\mathbb{R}^2$$ as:

• Example 1

$$(x_1, y_1) \succsim (x_2, y_2)$$

if and only if

either ($$x_1+y_1 > x_2 + y_2$$) or ($$x_1+y_1 = x_2 + y_2$$ and $$x_1 \geq x_2$$)

Claim : $$\succsim$$ cannot be represented by a utility function.

Proof : Suppose by contradiction that there existed a utility function $$u$$ representing these preferences. For each $$a > 0$$, we have $$(a, 0) \succ (0, a)$$, and therefore, $$u(a, 0) > u(0, a)$$. We can therefore assign to $$a$$ a non-degenerate interval of values satisfying the above inequality $$I(a) = [u(0, a), u(a, 0)]$$. For any $$a > b > 0$$, all commodity bundles generating utilities in the interval $$I(a)$$ are strictly preferred to those in the disjoint interval $$I(b)$$ and should therefore be assigned a greater utility level. Then in each of these intervals we can pick a distinct rational number in increasing order to represent preferences. Since $$a \in \mathbb{R_{++}}$$, there are uncountably many such intervals, but set of rational numbers are countable. This results in a contradiction.

• Example 2

$$(x_1, y_1) \succsim (x_2, y_2)$$

if and only if

either ($$\min(x_1,y_1) > \min(x_2,y_2)$$) or ($$\min(x_1,y_1) = \min(x_2,y_2)$$ and $$x_1+y_1 \geq x_2 + y_2$$)

Claim : $$\succsim$$ cannot be represented by a utility function.

Proof : Suppose by contradiction that there existed a utility function $$u$$ representing these preferences. For each $$a > 0$$, we have $$(a+1, a) \succ (a, a)$$, and therefore, $$u(a+1, a) > u(a, a)$$. We can therefore assign to $$a$$ a non-degenerate interval of values satisfying the above inequality $$I(a) = [u(a, a), u(a+1, a)]$$. For any $$a > b > 0$$, all commodity bundles generating utilities in the interval $$I(a)$$ are strictly preferred to those in the disjoint interval $$I(b)$$ and should therefore be assigned a greater utility level. Then in each of these interval we can pick a distinct rational number in increasing order to represent preferences. Since $$a \in \mathbb{R_{++}}$$, there are uncountably many such intervals, but set of rational numbers are countable. This results in a contradiction.

• @Giskard, I've updated example 1 with the proof that there is no utility representation. – Amit Oct 23 '19 at 21:46
• arent these just variations of lexiographic preference relations? – EconJohn Oct 24 '19 at 4:37
• Yes, these can be thought of as variations, but they are not called Lexicographic. – Amit Oct 24 '19 at 4:49


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2,129 Pages

Bigtymerxg4

(Difference between revisions) | User talk:Bigtymerxg4

Me in general

I'm a nerd. True and straight to the point. I love video games, anime, and the musics of such. I quote them very often in my life. In a past life, I was a writer. I'm odd and yet, I'm friendly. If I'm not at work, I'm in front of my computer. I like meeting new people, and I also like talking to others. If you see me on your talk page too often, oh well. Mines dead. So GG.

I'm currently obsessing over Paper Mario, Metroid Prime (again), FFVII Music, getting glad 3 and a bambi, and getting my life in order. I have MSN.

I've been banned again, so I'm using this time to try WoW out. Don't start asking me what Realm or whatever I'm in, it's nothing but a trial account.

• bigtymerxg4@hotmail.com - I was using MSN when most people on here didn't even know what the intarwebs were. Feel free to add me, if you want, but make sure to tell me who you are.

$Dee is awesome, X defines how awesome he is. If Y is -sleep, What does -sleep mean? Show your work.$

Characters

Smakaveli Lewcifer

Description

250px Arguably the worst named monk ever.

He hates his name and he makes it well known. He is rarely seen healing others, instead opting to smite for Cathedral of Flame runs with friends. When he does decide to heal, it's only in RA or Fort Aspenwood. There he can openly mock baed players and vent the frustration of his terribad name on them.

Stryke and Stryke Two Dethlok

Description

250px Two assassins, one family. One specializes in rolling PvE, while the latter spends his days in AB, picking off weak monks. While Stryke is methodical and planning, Stryke Two prefers to quickly gut a target and laugh at the resulting carnage. They don't get along well, hence them playing completely different from one another. It's even worse that Stryke Two is the younger of the two.

Wilde Deth

Description

250px Wilde is a, well wild one. His chiseled good looks, combined with brave and powerful bow spikes using his favorite elite, Glass Arrows, makes him into an easily approachable ladies man. However, he has one fear. Being that ranger with [x] weapon from [x] profession in his hands. Outside of that, he wins. He's a JQ favorite. Melandru's Shot + stupid kiters = massive lul.

Captain Dethlok

Description

250px Imbagonnnnnnnnnnnnn. Loud and commanding, she demands her presence be noted. She always seems to be in a bad mood. Probably from only being used for imbagon duties, as she's well equipped to PvP. Her favorite elite is Cruel Spear, but she's always being forced into using Focused Anger. She's hoping to prove one day that shes more than just a loud mouth behind a pretty face.

Prince Dethlok

Description

250px The nuker. Mainly used for nuking in AB/CM or crying/nuking in PvE farms. He has elite Flameforged now, but I CBA to upload new pictures.

Sir Dethlok

Description

250px He's the main character on the account. He was made shortly after purchasing Nightfall and Factions. Cloaks and scythes make him feel badass, so he had to work hard to be the main. He only uses green scythes, unlike everyone else who has their weapons created. Less work for him. He IS a conjurefag, and he WILL use Avatar of Lyssa with daggers if need-be. He only does this because he rolls PvE and PvP with little more effort than raising his cloak to step over the corpse of the last person to cross him. He also says 'ym' a lot.

King Dethlok

Description

250px The second main. He loves hitting stuff, lemme tell ya. His favorite two elites are Earth Shaker and Eviscerate. As he's a meta-whore, whenever it comes up, he HAS to have the build. Swords still suck. He also thinks Power Attack is imba.

Endless Hysteria/Dee Strongfist

Description

I usually name my PvP characters Endless Hysteria or Dee Strongfist. If you see me in-game, PM me or something. I need to redo my FL. =)

I do believe this helps. My current guild is The Butter Battlers [BB] Offensive Guild Name [ban]. We're Kurzick Luxon.
 This user is a member of the Kurzick alliance.
 This user is a Warrior by nature.
 This user is a Dervish by nature.
 This user is an Assassin by nature.
 This user plays for the territory of North America.
 PvP This user is primarily a PvP player.

imo

Listen to stuff I post there, cause I win.

The true standout here is my favorite: Fading Entity. Give it a listen.

Beggin by Frankie Valli which is another awesome song.

Outside of Guild Wars, I'm either working, or playing some other type of video game. I'm 21 22, and a father to a one year old child. He wins like his dad.

 The Protomen - Vengeance

Best. Song. Ever. Living Inside the Shell by Steve Conte/Shanti Snyder from Ghost in the Shell S.A.C. 2nd GiG.

Listen.

 Steve Conte/Shanti Snyder - Living Inside the Shell

Roaming inbetween the worlds of sleep and awake
Seems so far away from where I've been and untrue but unafraid
Intrusting -- my soul -- I know I must be taken to see the world that is
Not so far from now

Imaginations come and sweep the shores of my mind
Letting it be, visions pass, and emotions arise--
Letting them go, and beyond are doors I've never seen, opening one by one

(Wake up and show the light, wake up the time is right)
I hear a voice, hear a voice calling out to me
Look inside, see the light now ever holding you
All the truth is all you need to make of your reality, its right here

Finding out a galaxy of planets and stars within me
Listening to each of them singing the same silent melody
I've never seen such beauty in possibility -- no speck of doubt or fear

(Wake up and show the light, wake up the time is right)
I hear a voice, hear a voice calling out to me
I see inside, see the light now ever holding me
All the truth, all I need to make of this reality it's -- beauty within the shell

(Wake up and show the light, wake up the time is right)
I hear a voice, hear a voice calling out to me
I see inside, see the light now ever holding me
All the truth, all I need to make of this reality, it's inside
Right here within this shell

The sandglass starts for another time's beginning from within
Cotton fields, mama's arms are gently unfolding me into the new...

(Wake up and show the light, wake up the time is right)
Here from behind my sight, my thoughts my mind
Show from the light
The time is right
And from the depth within show the balance
Of outer and inner harmony
Mind and heart, soul and spirit undivided
Here's where the true strength and beauty lies
We'll see this before us with our own eyes
We'll see, with our own eyes... LOVE

(Wake up and show the light, wake up the time is right)
I hear a voice, hear a voice calling out to me
I see inside, see the light now ever holding me
All the truth, all I need to make of this reality it's -- beauty within the shell

Godbox's Random TB Creation of the Whenever I Make A New Build

What's up with that ugly space between the skillbars? D: Scavenger's Focus with QZ allows for constant maintaining, and it's not a stance! Free attackspamming. Use half-recharge Mantra of Recharge to be able to run boonmonks. Very strong spike. CLICK!

Also

Epic Quotes

explain why not all build that get vetted are written up. And also explain why i should waste time writing something up when people suggest changing a lot of it so i have to re-write everything? — LukeJohnson  19:44, 27 March 2009 (UTC)

Because its the rules dumbass. If you don't like them, go away. Rawrawr Dinosaur 19:46, 27 March 2009 (UTC)
Community content is available under CC-BY-NC-SA 2.5 unless otherwise noted.


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# Towards Optimal and Efficient Best Arm Identification in Linear Bandits

We give a new algorithm for best arm identification in linearly parameterised bandits in the fixed confidence setting. The algorithm generalises the well-known LUCB algorithm of Kalyanakrishnan et al. (2012) by playing an arm which minimises a suitable notion of geometric overlap of the statistical confidence set for the unknown parameter, and is fully adaptive and computationally efficient as compared to several state-of-the methods. We theoretically analyse the sample complexity of the algorithm for problems with two and three arms, showing optimality in many cases. Numerical results indicate favourable performance over other algorithms with which we compare.

## Authors

• 3 publications
• 26 publications
• 4 publications
• ### Best Arm Identification in Generalized Linear Bandits

Motivated by drug design, we consider the best-arm identification proble...
05/20/2019 ∙ by Abbas Kazerouni, et al. ∙ 0

• ### Gamification of Pure Exploration for Linear Bandits

We investigate an active pure-exploration setting, that includes best-ar...
07/02/2020 ∙ by Rémy Degenne, et al. ∙ 0

• ### Top-m identification for linear bandits

Motivated by an application to drug repurposing, we propose the first al...
03/18/2021 ∙ by Clémence Réda, et al. ∙ 0

• ### Explicit Best Arm Identification in Linear Bandits Using No-Regret Learners

We study the problem of best arm identification in linearly parameterise...
06/13/2020 ∙ by Mohammadi Zaki, et al. ∙ 0

• ### Gradient Ascent for Active Exploration in Bandit Problems

We present a new algorithm based on an gradient ascent for a general Act...
05/20/2019 ∙ by Pierre Ménard, et al. ∙ 0

• ### Best Arm Identification in Spectral Bandits

We study best-arm identification with fixed confidence in bandit models ...
05/20/2020 ∙ by Tomáš Kocák, et al. ∙ 0

• ### Probabilistic Sequential Shrinking: A Best Arm Identification Algorithm for Stochastic Bandits with Corruptions

We consider a best arm identification (BAI) problem for stochastic bandi...
10/15/2020 ∙ by Zixin Zhong, et al. ∙ 0

##### This week in AI

Get the week's most popular data science and artificial intelligence research sent straight to your inbox every Saturday.

## 1 Introduction

This paper is concerned with the problem of optimising an unknown linear function over a finite domain when given the ability to sequentially test and observe noisy function values at domain points of our choice. In the language of online learning this is the problem of best arm identification in linearly parameterised bandits; in classical statistics it is essentially the problem of adaptive, sequential composite hypothesis testing where each hypothesis corresponds to one domain point being optimal. From the point of view of causal inference, it can be interpreted as the problem of learning the best (i.e., most rewarding) intervention, from among a set of parameterised interventions available at hand with respect to an observed variable (Lattimore et al., 2016, Sen et al., 2016)

, with the key difference in this work being that the causal response of the variable to an intervention is modelled as being linear in the intervention value. The linear structure endows the model with complex but exploitable structure, in that it makes possible inference about the utility (function value) of an intervention (bandit arm) by using observations from other, correlated interventions, akin to what happens in standard (batch) prediction with linear regression.

In the linear bandit setting, each arm or action is associated with a fixed known feature vector

and the expected reward obtained by choosing to pull arm with feature vector is where

is a fixed but unknown vector. We specifically consider the probably-approximately-correct (PAC) objective of the learner (agent) declaring a guess for the identity of the optimal arm, after it has made an internally determined (and potentially random) number of sequential plays of arms, which is required to be correct with at least a given probability

– the fixed confidence best arm identification goal (Even-Dar et al., 2006). In this regard, our focus is on both the statistical and computational efficiency of adaptive arm-sampling strategies, i.e., designing strategies (a) whose number of plays is as close to the quantifiable information-theoretic limit on sample complexity across all strategies, and (b) which can determine in a computationally lightweight manner the next arm to play at each adaptive round.

Broadly, there are two different approaches towards solving such pure exploration problems: (i) uniform sampling-elimination based and (ii) adaptive sampling-UCB based. The algorithm of Tao et al. (2018) is based on the former approach while those of Soare et al. (2014) and Xu et al. (2017) are based on the latter, adaptive-sampling idea, but both have sample complexity guarantees that do not depend finely on the problem instance and linear structure, and thus are worst-case optimal at best.

The -static algorithm of Soare et al. (2014) is a static algorithm which fixes the schedule of arm plays before collecting any observations. Hence, it is not able to adapt towards pulling arms which may be more “informative" for identifying the best arm for the given instance, and can consequently only be worst-case optimal. The LinGapE algorithm Xu et al. (2017) is a fully adaptive algorithm which performs well experimentally. However it requires to solve optimization problems at start, one for every pair of arms ( denotes the number of arms), which is computationally inefficient for large values of . Finally, the -ElimTil- algorithm of Tao et al. (2018) is an elimination based algorithm, and though its sample complexity scales only linearly with the dimension , it requires to sample arms in each round which is already far from optimal even for the case of the canonical (unstructured) multi-armed bandit (MAB) having the standard basis vectors as the arms. This can again only be worst-case optimal. A summary of the sample complexity bounds of the above algorithms appears in Table 1.

A very recent departure from this worst-case sample complexity dependence is the work of Fiez et al. (2019), that shows a provably instance-optimal best arm identification algorithm for linear (and more generally transductive) bandits. However, implementation of this algorithm requires computing an arguably costly rounding procedure to determine (in phases) a schedule of arms to play, as well as solving a minimax optimization problem which may be computationally very expensive111In fact, these aspects of their algorithm have prevented us from successfully implementing and testing it..

Our contributions and organization. In contrast with existing work, we aim to take a qualitatively different route towards the design of linear best arm identification, by drawing inspiration from the upper confidence bound principle, which is known to give sample-optimal performance for canonical MAB (Kalyanakrishnan et al., 2012, Jamieson et al., 2014). In this conceptually simple and elegant approach, in each decision round the learner constructs a statistically plausible confidence set for the underlying bandit instance (the weight vector in our linear setting) based on past observations, and then plays the arm that best appears to reduce the uncertainty about the optimal linear arm.

We generalise the Lower Upper Confidence Bound (LUCB) algorithm of Kalyanakrishnan et al. (2012) to the linear bandit setting. To achieve this, we introduce a new geometric “maximum overlap” principle as a basis for the learner to identify which arm is most informative to play at any given round. This results in a fully data-dependent arm selection strategy which we call Generalized-LUCB (GLUCB) (Section 3). We then proceed to rigorously analyse the sample complexity of GLUCB for certain specialized (yet instructive) cases in Section 4, and finally compare its empirical performance with other state-of-the-art methods in Section 5.

As a comment on the execution times as compared to the other algorithms proposed for this problem, our proposed algorithm GLUCB, improves significantly on the time complexity over LinGapE (Xu et al., 2017) and -ElimTil- (Tao et al., 2018) as it does not require solving any offline optimization problems.

## 2 Problem Statement and Notation

We study the problem of best arm identification in linear multi-arm bandits (LMABs) with the arm set , where is finite but possibly large. We will interchangeably use and the set , whenever the context is clear. Each arm is a vector in . The quantity will, henceforth, be called the ambient dimension. At every round the agent chooses an arm , and receives a reward , where is assumed to be a fixed but unknown vector and is zero-mean noise assumed to be conditionally -subGaussian, i.e., , Let The goal of the agent is, given an error probability , to identify with probability , by pulling as few arms as possible (in literature, this is known as the “fixed- regime Kaufmann et al. (2016)). Henceforth, we will call this the LMAB (linear multi armed bandit) problem. When restricted to the case where and is the standard ordered basis the problem reduces to the Standard LMAB (SMAB) problem studied, for instance, in Kalyanakrishnan et al. (2012).

In the rest of the paper, we will assume that and that the agent has information of some upper bound on say, . Let be a positive definite matrix, then we denote by , the matrix norm induced by . Let for any , be the gap between the largest expected reward and the expected reward for arm . Denote by , the smallest reward gap.

## 3 The GLUCB Algorithm

This section is organized as follows. We begin with a description of the ingredients required to construct GLUCB, including “MaxOverlap.” Thereafter, we formally describe the GLUCB algorithm. Finally, we show how GLUCB is a generalization of LUCB.

To begin with, note that any algorithm for the best arm identification problem requires the following ingredients:

1. a stopping rule: which decides when the agent must stop sampling arms, and is a function of past observations, arms chosen and rewards only,

2. a sampling rule: which determines, based on the arms played and rewards observed hitherto, which arm to pull next (clearly, this rule is invoked only if the stopping rule decides not to stop); and

3. a recommendation rule: which, when the stopping rule decides to stop, chooses the index of the arm that is to be reported as the best.

Each of these steps will now be developed in detail and combined to give the full GLUCB algorithm. Towards this we first introduce some technical desiderata.

Let be a sequence of arms played until time by any adaptive strategy (i.e., a strategy which chooses to play an arm depending on the past arm pulls and their corresponding observations) and let be the received rewards. The (regularized)

least squares estimate

of at time is given by

 θt := V−1tbt, where Vt := λI+t∑s=1xsxTs, and bt:=t∑s=1xsys.

By standard results on least squares confidence sets for adaptive sampling (Abbasi-Yadkori et al., 2011b), it can be shown that with high probability, lies in the confidence ellipsoid333recall that noise is assumed to be -sub Gaussian

 Et(θt,Vt):={θ∈Rd∣∣∣∥θt−θ∥Vt≤βt}, where βt:=R√dlogtδ.

Notice that the ellipsoid is time-indexed, since, as more arms chosen, the estimate changes and so does the ellipsoid. In the sequel, we sometimes denote by in the interest of space.

We also define a set , where and are vectors in . Next, for any , define , as the cone of parameters such that, if , then is the optimal arm. Clearly partition the entire parameter space , modulo the degenerate regions where more than one arm is optimal. Furthermore, let be the index of the arm that currently appears to be the best.

### 3.1 Ingredients of the GLUCB algorithm

Following the intuition in (Soare et al., 2014, Sec. 3), we observe that a good choice for a stopping rule could be to stop the algorithm when the confidence ellipsoid is completely contained within one of the cones . Therefore, we wish to design an algorithm which minimizes the overlap of the current confidence ellipsoid with every cone which currently seems to be suboptimal, i.e., all the cones other than the current home cone of . That way, the algorithm can quickly insert completely into one of the cones, and because contains with high probability, so does since, now, This also means that upon stopping, will be the arm recommended.

###### Definition 1 (MaxOverlap).

The MaxOverlap of set on set is defined to be the maximum distance of set from the boundary of another set .

Here denotes the closure of the set and its topological boundary (Rudin (1964)). Hence, at time , our algorithm GLUCB is defined as sampling the arm

 at+1:=Kargmina=1MaxOverlap(E(θt,Vt+xaxTa);Rcht).

The following result shows that the MaxOverlap-based arm sampling rule reduces to a concrete prescription for the linear bandit setting. (Due to space constraints, proof details are omitted and can be found in the Appendix.)

###### Proposition 1.

At time step , define the arm

 lt:=argmaxi≠ht,i∈[K]MaxOverlap(E(θt,Vt);HalfSpace(i,ht)).

Then,

We are now ready to describe the ingredients of GLUCB. At every time step define the “Advantage” of arm as

Stopping rule: The algorithm stops when the “Advantage” defined above becomes non-positive for every arm other than the current best arm
Sampling rule: Play the arm which minimizes the current max Advantage: In case of a tie, the agent selects an arm uniformly randomly.
Recommendation rule: Once the algorithm stops, the current best arm is recommended as the guess for the best arm.

Note that the GLUCB algorithm (Algorithm 1) reduces to the well-known LUCB algorithm of Kalyanakrishnan et al. (2012) for the SMAB problem. If we consider the case when , and the arms being the standard basis , we see that the arm in algorithm 1 corresponds to which is what LUCB would suggest. Indeed, when , . However, with we obtain, . Also, it is easy to check the stopping criterion also reduces to that of LUCB. Hence, Algorithm 1, when applied to the unstructured case, plays the current best and the closest arm simultaneously every time till the algorithm stops.

We now provide some preliminary theoretical results regarding the sample complexity performance of GLUCB.

## 4 Analysis of GLUCB

The following result proves the correctness of G-LUCB.

###### Theorem 1.

Let be any arbitrary sampling strategy. Algorithm 1 returns the optimal arm upon stopping with probability at least

We will now analyze the sample complexity of GLUCB when and . For this we first present the following useful result on the convexity of a certain norm-based function on the probability simplex.

###### Lemma 1.

For any , and , where and , the function is convex in

### 4.1 Analysis of GLUCB for Linear MAB with K=2 arms

Let for which the arm set is . Let and with . For this simple case, it is clear that the set We aim to analyze the sample complexity of GLUCB by tracking the possible sample paths of (which turns out to be tractable in this setting). Let be the number of times arm has been pulled till time . Then, playing GLUCB guarantees that,

###### Theorem 2.

In any round ,

The proof of the result relies on the following observations.

###### Lemma 2.

Whenever

• , and

• there is a tie.

The lemma tells us that for any , if then there is a tie. Next, we show that whenever an arm (say arm 1 w.l.o.g) is played for times while arm 2 for times we will be forced to play arm 2.

###### Lemma 3.

Let Then,

Infact, for the two arm case, the sample complexity of GLUCB is optimal. The following ‘potential function’-based result formally establishes this fact.
Let us consider two algorithms and , where is GLUCB and is any other algorithm. Let for We can now state

###### Theorem 3.

, .

On the other hand, a detailed analysis of the information-theoretic lower bound on sample complexity, e.g., Kaufmann et al. (2016) or Fiez et al. (2019), yields the following result where is the optimal vector of arm frequencies in the min-max optimisation problem of the lower bound (termed in Fiez et al. (2019)).

###### Lemma 4.

[Lower bound for ]. For , with any two arms and such ,

By combining theorem 3 and lemma 4 We have the following corollary.

###### Corollary 1.

The (expected) sample complexity of GLUCB for the 2-arm setting is at most , where , where is the probability simplex over 2 arms and represents the optimal arm.

###### Note 1.

The quantity is the usual information theoretic lower bound on best arm identification sample complexity (Fiez et al., 2019), with the sample complexity of GLUCB being only away from it; the extra factor arises because of weaker concentration bounds for adaptive strategies.

### 4.2 Linear MAB with K=3 arms

This section deals with a representative example of the linear bandit. Let , and the arm set Let . This setup is particularly interesting when is close to 0. An algorithm which is optimal for standard MAB will quickly discard arm 2, and would continue to sample arms 1 and 3 until stopping. However, this is not the optimal strategy, since pulling arm 2 gives valuable information about . We will see that this is what GLUCB does.

As compared to the algorithms designed for best arm identification standard MAB, GLUCB identifies the structure (if any) present in the arms and tries to exploit it. For the particular case just described, Arm 3 is always dominated by the other two arms, i.e., we will see in the key Lemma 5 that in order to minimize the uncertainity in any direction , the reduction obtained by pulling Arm 3 is always dominated by that for some other arm.

###### Theorem 4.

If GLUCB run on the above problem instance and is its stopping time, then,

 P{(τ≤4βt2Δ2minsin2(ω)+4β2tΔminsin(ω)+max{4β2t(sin(ω)Δmin+1),4β2tμ23(sin(ω)Δmincos(ω)+1−sin(ω))})∧{hτ∗==a∗}}≥1−δ.

The proof of this result relies on the following key lemma.

###### Lemma 5.

If , then Arm 3 is never played.

Next, the following lemma shows an upper-bound on the time taken by GLUCB to discard arm 2 as or . We show this by bounding the number of samples required by GLUCB such that .

###### Lemma 6.

With probability , for all , where , Arm 3

Finally we bound the number of samples needed by GLUCB to stop once the set has frozen.

###### Lemma 7.

The number of samples needed for G-LUCB to stop once in steady state is upper bounded by

###### Note 2.

The term inside the max in the theorem statement, is small and can be absorbed into the leading terms. Hence, the sample complexity of GLUCB can be written as .

###### Note 3.

A crucial observation here is that the geometry of the problem enters the sample complexity (in terms of ), which. since is small, reduces the sample complexity compared to that of a standard MAB algorithm running on the instance.

We will now see that this is indeed the optimal strategy.

#### 4.2.1 Lower-bound for the three-arm case

By (Fiez et al., 2019, Theorem 1), the expected sample complexity of any PAC best arm identification algorithm for LMAB is lower bounded as:

 E[τ]log12.4δ≥minw∈P3maxa′≠1∥xa′−x1∥2W−1Δ2a′≥minw∈P3∥x3−x1∥2W−1Δ2min,

where . By solving the above optimization problem, we have

###### Theorem 5.

For , and , the expected sample complexity is lower bounded as

## 5 Experiments

In this section, we compare the performance of GLUCB with XY-static Soare et al. (2014), LUCB Kalyanakrishnan et al. (2012), LinGapE Xu et al. (2017) and X-ElimTilp with Tao et al. (2018) through experiments in three synthetic settings and simulations based on real data. For LinGapE we implement the version of the algorithm which has been analyzed in their paper. For implementation of X-ElimTil0 with the setting as mentioned in Tao et al. (2018).

### 5.1 Experiments based on synthetic data

Throughout we assume noise independent. The results reported are averaged over 100 trials under each setting. We report the average number of samples to stop in each case, for each algorithm. The empirical probability of error in each case was found to be 0.

1. Dataset 1: This is the setting introduced by Soare et al. (2014) for linear bandits. We set up the linear bandit problem with arms, where features are the canonical bases and an additional arm with so that the first arm is the best arm, with the th arm being the most ambigous arm. We test by varying . With , this setup resembles the case we analyzed in 4.

2. Dataset 2: In this dataset, feature vectors are sampled uniformly at random the surface of the unit sphere centered at the origin. We pick the two closest arms, say and , and then set for . This makes as the best arm. We test the algorithm for .

3. Dataset 3: This setup is important as this shows the efficiency of GLUCB in the case when there may be many arms which competing for the second best arm. For a given value of , the armset contains feature vectors from where where . was fixed to be . We conduct the experiment by varying .

### 5.2 Experiments based on real data

We conduct an experiment on Yahoo! Webscope dataset R6A which consists of features of 36-dimensions accompanied with binary outcomes. We change the situation as is done in Xu et al. (2017) so that it can be adopted for best arm identification setting. We construct the 36-dimensional feature set by the random sampling from the dataset, and the reward is generated as with probability and with probability , where is the regularized least squared estimator fitted for the original dataset. We choose the vectors such that . For the detailed procedure, we refer the reader to the paper of Xu et al. (2017).

## 6 Conclusion and future work

We have generalised the LUCB best arm identification algorithm to bandits with linear structure via a new MaxOverlap rule to reason under uncertainty. The resulting GLUCB algorithm is computationally very attractive as compared to many state-of-the-art algorithms for linear bandits. In particular, it does not require solving optimisation problems which are inefficient when is large. Viewed from another perspective, the algorithm leverages the fact that a strategy which tries to greedily maximize the gap between the current best and second best arms is optimal for the BAI problem. We show that for the special case of two arms GLUCB is better than any causal algorithm for BAI problem. We also show orderwise optimality in the case of three arms.

In light of the analysis presented and the performance of GLUCB in our experiments, we conjecture that our algorithm is optimal for any set of arms. Proving this forms part of our future work. We conduct several experiments based on synthetically designed environments and real world dataset and show the superior performance of GLUCB over other algorithms. Furthermore, we believe that the factor of in the sample complexity results is due to the general concentration bound for adaptive sequences and can be improved, which also remains as a future work. More generally, it is interesting to ask if the general max-overlap principle works for other, non-linear bandit reward structures as well.

## References

• Abbasi-Yadkori et al. (2011a) Yasin Abbasi-Yadkori, David Pal, and Csaba Szepesvari. Improved Algorithms for Linear Stochastic Bandits. In Proc. NIPS, pages 2312–2320, 2011a.
• Abbasi-Yadkori et al. (2011b) Yasin Abbasi-Yadkori, Dávid Pál, and Csaba Szepesvári. Improved algorithms for linear stochastic bandits. In J. Shawe-Taylor, R. S. Zemel, P. L. Bartlett, F. Pereira, and K. Q. Weinberger, editors, Advances in Neural Information Processing Systems 24, pages 2312–2320. Curran Associates, Inc., 2011b.
• Even-Dar et al. (2006) Eyal Even-Dar, Shie Mannor, and Yishay Mansour.

Action elimination and stopping conditions for the multi-armed bandit and reinforcement learning problems.

J. Mach. Learn. Res., 7:1079–1105, December 2006. ISSN 1532-4435.
• Fiez et al. (2019) Tanner Fiez, Lalit Jain, Kevin Jamieson, and Lillian Ratliff. Sequential experimental design for transductive linear bandits. arXiv preprint arXiv:1906.08399, 2019.
• Garivier and Kaufmann (Jun. 2016) Aurélien Garivier and Emilie Kaufmann. Optimal best arm identification with fixed confidence. In Conference On Learning Theory, pages 998–1027, Jun. 2016.
• Horn and Johnson (2012) Roger A. Horn and Charles R. Johnson. Matrix Analysis. Cambridge University Press, New York, NY, USA, 2nd edition, 2012. ISBN 0521548233, 9780521548236.
• Jamieson et al. (2014) Kevin Jamieson, Matthew Malloy, Robert Nowak, and Sébastien Bubeck. lil´ucb: An optimal exploration algorithm for multi-armed bandits. In Conference on Learning Theory, pages 423–439, 2014.
• Kalyanakrishnan et al. (2012) Shivaram Kalyanakrishnan, Ambuj Tewari, Peter Auer, and Peter Stone. Pac subset selection in stochastic multi-armed bandits. In ICML, 2012.
• Kaufmann et al. (2016) Emilie Kaufmann, Olivier Cappé, and Aurélien Garivier. On the complexity of best-arm identification in multi-armed bandit models. J. Mach. Learn. Res., 17(1):1–42, January 2016. ISSN 1532-4435.
• Lattimore et al. (2016) Finnian Lattimore, Tor Lattimore, and Mark D Reid. Causal bandits: Learning good interventions via causal inference. In Advances in Neural Information Processing Systems, pages 1181–1189, 2016.
• Rudin (1964) Walter Rudin. Principles of mathematical analysis, volume 3. McGraw-hill New York, 1964.
• Sen et al. (2016) Rajat Sen, Karthikeyan Shanmugam, Murat Kocaoglu, Alexandros G Dimakis, and Sanjay Shakkottai. Contextual bandits with latent confounders: An nmf approach. arXiv preprint arXiv:1606.00119, 2016.
• Soare et al. (2014) Marta Soare, Alessandro Lazaric, and Rémi Munos. Best-arm identification in linear bandits. CoRR, abs/1409.6110, 2014.
• Tao et al. (2018) Chao Tao, Saúl Blanco, and Yuan Zhou. Best arm identification in linear bandits with linear dimension dependency. In Jennifer Dy and Andreas Krause, editors,

Proceedings of the 35th International Conference on Machine Learning

, volume 80 of Proceedings of Machine Learning Research, pages 4877–4886, Stockholmsmässan, Stockholm Sweden, 10–15 Jul 2018. PMLR.
• Xu et al. (2017) Liyuan Xu, Junya Honda, and Masashi Sugiyama. Fully adaptive algorithm for pure exploration in linear bandits. In

International Conference on Artificial Intelligence and Statistics

, 2017.

## 7 Appendix

### 7.1 Proof of Prop. 3.1

As mentioned in Sec. 3, at time , our algorithm samples arm

 at+1 :=Kargmina=1MaxOverlap(E(θt,Vt+xaxTa);Rcht) =Kargmina=1maxθ∈E(θt,Vt+xaxTa),θ∈Rchtmin~θ∈∂(Rcht)∥∥θ−~θ∥∥2 =Kargmina=1maxθ∈E(θt,Vt+xaxTa),θ∈⋃i≠ht{θ′:θ′Txi>θ′Txht}min~θ∈⋃i≠ht{θ′:θ′Txi=θ′Txht}∥∥θ−~θ∥∥2 =Kargmina=1maxi≠htmaxθ∈E(θt,Vt+xaxTa),θTxi>θTxhtmin~θ:~θTxi=~θTxht∥∥θ−~θ∥∥2

Where we have used the fact that To get a closed form solution for the above, we solve the optimization problem : explicitly to obtain:

 at+1=Kargmina=1maxi≠htmaxθ∈E(θt,Vt+xaxTa)(θT(xi−xht))+

The agent must stop choosing arms if this value is zero, i.e., when no longer intersects any of the suboptimal cones. Due to the over arms the above strategy is inefficient to implement. A slight modification of the above which is easier to implement, is as follows. recalling the definition of HalfSpace, define the cone () which has the maximum overlap with the current cone.

 lt :=argmaxi≠ht,i∈[K]MaxOverlap(E(θt,Vt);HalfSpace(i,ht)) ≡argmaxi≠htmaxθ∈E(θt,Vt),θTxi>θTxhtmin~θ:~θTxi=~θTxht∥∥θ−~θ∥∥2 ≡argmaxi≠htmaxθ∈E(θt,Vt)θT(xi−xht) =argmaxi≠ht(θTt(xi−xht)+βt∥xi−xht∥V−1t),

which is straightforward to implement on a computer. Then play an arm according to the following rule. If play:

 at+1:= argmina∈[K]MaxOverlap(E(θt,Vt+xaxTa);HalfSpace(lt,ht)) ≡ = Kargmina=1maxθ∈E(θt,Vt+xaxTa)θT(xlt−xht) = Kargmina=1[θTt(xlt−xht)+βt∥xlt−xht∥(Vt+xaxTa)−1] = Kargmina=1∥xlt−xht∥(Vt+xaxTa)−1 ≡ Kargmaxa=1\absxTaV−1t(xlt−xht)√1+∥xa∥2V−1t.

The last step uses the Matrix Inversion Lemma Horn and Johnson (2012).

### 7.2 Lower Bound on Sample Complexity

We begin by restating the result of [Garivier and Kaufmann (Jun. 2016)] in the special case of linear bandits. Let be a given set of arms in . Let be any vector in . For any arbitrary vector , we define . Define the set .

###### Lemma 8 (General change-of-measure based lower bound of Garivier and Kaufmann (Jun. 2016)).

Let For any strategy and any linear bandit with the unknown parameter vector

. Let the noise be normal with a variance parameter of

. Then the expected sample complexity of any strategy

 Eθ[τδ]≥T∗(θ)kl(δ,1−δ)

where, for and

 T∗(θ)−1:=supw––∈PKinfξ∈Alt(θ)K∑a=1waKL(N(θTxa,1/2),N(ξTxa,1/2)),

where

is defined as the set of all probability distributions on

and is the KL-divergence between any two probability distributions and , and

is normal distribution with mean

and variance .

###### Theorem (Lower Bound).

Let For any strategy and any linear bandit problem with the unknown parameter vector ,

where the expectation is under and , where are the arms in

###### Proof.

Recall the definition from Lemma 8,

where is the KL divergence between any two distributions. Hence, we have,

 T∗(θ)−1= supw––∈PKinfξ∈Alt(θ)K∑a=1wa(θTxa−ξTxa)2 = supw––∈PKmina′≠1infξ:ξTx′a≥ξTx1+ϵK∑a=1wa(θTxa−ξTxa)2

for some We will first consider the inner part of the expression above (a convex program), which can be re-written as

 infξ:ξTx′a≥ξTx1+ϵ(θ−ξ)TW(θ−ξ) (1)

where is defined as in the theorem. Writing the Lagrangian, we get

 L(ξ,λ)=(θ−ξ)TW(θ−ξ)−λ(ξTx′a−ξTx1−ϵ).

Setting , we get

 ΔξL(ξ,λ): −2Wθ+2Wξ−λ(xa′−x1)=0 ⇒ξ =θ+λ2W−1(xa′−x1).

Substituting this value of into the Lagrangian, we obtain:

 L(λ)=θ


text:
nVidia driver causes flickering viewport background

I noticed a flickering background whenever I click a button or orbit in blender 2.8 or 2.79

It only happens when I use my NVIDIA 940mx but not when I switch to the integrated Intel 620.

I can't record the effect with a screen recorder, so it might be related to the NVIDIA driver interacting with the display brightness/gamma/contrast.

Maybe someone has experienced this problem and knows how to fix it? Also, is there any disadvantage of using the integrated graphics? Especially on 2.8 and eevee?

• try File > User Preferences > System, and under the Window Draw Method select Triple Buffer. – rob Dec 3 '18 at 12:12
• thx, window draw method was not the solution. it's independent from this setting. just appearing with nvidia gfx enabled. intel is working fine. – Simon Sais Dec 3 '18 at 14:56
• In the OpenGL settings of the Nvidia controls, try disabling Sync to Vblank – cegaton Mar 13 at 15:53
• I have the same problem with my GTX1050 (MSI PS63 laptop.) I've tried toggle Window Fullscreen and it was help me. It's working. Does anyone to know another solution? How we can fix it without full screen mode. – Antony Sep 1 at 18:27

$$`$$Putting the software in fullscreen may tell the NVIDIA card that it can take full control of it, so those bugs don't happen.


text:
help-gnu-emacs
[Top][All Lists]

## Re: problem with toolbar commands

 From: coskun Subject: Re: problem with toolbar commands Date: Thu, 14 Feb 2008 11:37:58 -0800 (PST) User-agent: G2/1.0

```On Feb 14, 12:53 am, Ralf Angeli <address@hidden> wrote:
> * coskun (2008-02-14) writes:
> > Emacs+AucTeX and MiKTeX work fine separately, but when I try to use
> > toolbar commands LaTeX, view, pdfLaTeX, etc. I get the following
> > message.
>
> > Running `LaTeX' on `test' with ``latex \nonstopmode\input{test.tex}''
>
> > However, compilation never gets completed, no output is generated.
>
> Does the latex command work if you use it in a Windows prompt?
>
> --
> Ralf

Yes it does. It works fine if I latex the file in a Windows prompt.
All other commands such as yap, etc. also work in a prompt. However,
they don't work if I call these commands from the menu bar in Emacs
22. Emacs freezes every time I try it. (AuCTeX is also working without
problems.)

I suspected the problem is about path names and Vista. I added c:
\Program Files\Emacs\bin and c:\Program Files\MiKTeX 2.7\miktex\bin to
the value of Path variable, but it didn't help. Any ideas?

Best,
Coskun

```


text:
Timezone: »

On Feature Collapse and Deep Kernel Learning for Single Forward Pass Uncertainty
Joost van Amersfoort · Lewis Smith · Andrew Jesson · Oscar Key · Yarin Gal

Inducing point Gaussian process approximations are often considered a gold standard in uncertainty estimation since they retain many of the properties of the exact GP and scale to large datasets. A major drawback is that they have difficulty scaling to high dimensional inputs. Deep Kernel Learning (DKL) promises a solution: a deep feature extractor transforms the inputs over which an inducing point Gaussian process is defined. However, DKL has been shown to provide unreliable uncertainty estimates in practice. We study why, and show that with no constraints, the DKL objective pushes `far-away'' data points to be mapped to the same features as those of training-set points. With this insight we propose to constrain DKL's feature extractor to approximately preserve distances through a bi-Lipschitz constraint, resulting in a feature space favorable to DKL. We obtain a model, DUE, which demonstrates uncertainty quality outperforming previous DKL and other single forward pass uncertainty methods, while maintaining the speed and accuracy of standard neural networks.

#### Author Information

##### Lewis Smith (University of Oxford)

Lewis Smith is a DPhil student supervised by Yarin Gal. His main interests are in the reliability and robustness of machine learning algorithms, Bayesian methods, and the utilisation of structure (such as invariances in the data). He is also a member of the [AIMS CDT](www.aims.robots.ox.ac.uk). Before joining OATML, he recieved his masters degree in physics from the University of Manchester.


text:
Posted on Categories:Stochastic Porcesses, 数学代写, 随机过程

# 数学代写|随机过程Stochastic Porcesses代考|STAT507 Stationarity

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## 数学代写|随机过程Stochastic Porcesses代考|Stationarity

Definition 2.2.1. We say that the stochastic process ${X(t), t \in T}$ is stationary, or strict-sense stationary (SSS), if its distribution function of order $n$ is invariant under any change of origin:
$$F\left(x_1, \ldots, x_n ; t_1, \ldots, t_n\right)=F\left(x_1, \ldots, x_n ; t_1+s, \ldots, t_n+s\right)$$
for all $s, n$, and $t_1, \ldots, t_n$.
Remark. The value of $s$ in the preceding definition must be chosen so that $t_k+s \in T$, for $k=1, \ldots, n$. So, if $T=[0, \infty)$, for instance, then $t_k+s$ must be nonnegative for all $k$.

In practice, it is difficult to show that a given stochastic process is stationary in the strict sense (except in the case of Gaussian processes, as will be seen in Section 2.4). Consequently, we often satisfy ourselves with a weaker version of the notion of stationarity, by considering only the cases where $n=$ 1 and $n=2$ in Definition 2.2.1.
If ${X(t), t \in T}$ is a (continuous) SSS process, then we may write that
$$f(x ; t)=f(x ; t+s) \quad \forall s, t$$
and
$$f\left(x_1, x_2 ; t_1, t_2\right)=f\left(x_1, x_2 ; t_1+s, t_2+s\right) \quad \forall s, t_1, t_2$$

## 数学代写|随机过程Stochastic Porcesses代考|Ergodicity

In statistics, to estimate an unknown parameter of a distribution function, for example, the parameter $\lambda$ of an r.v. $X$ having a $\operatorname{Poi}(\lambda)$ distribution, we draw a random sample of $X$. That is, we take $n$ observations, $X_1, \ldots, X_n$, of $X$ and we assume that the $X_k$ ‘s have the same distribution function as $X$ and are independent. Next, we write that the estimator $\hat{\lambda}$ of $\lambda$ (which is the mean of the distribution) is the arithmetic mean of the observations. Similarly, to estimate the mean $m_X(t)$ of a stochastic process ${X(t), t \in T}$ at time $t$, we must first take observations $X\left(t, s_k\right)$ of the process. Next, we define
$$\widehat{m_X(t)}=\frac{1}{n} \sum_{k=1}^n X\left(t, s_k\right)$$
Thus, we estimate the mean $m_X(t)$ of the s.p. by the mean of a random sample taken at time $t$. Of course, the more observations of the process at time $t$ we have, the more precise the estimator $m_X(t)$ should be. Suppose, however, that we only have a single observation, $X\left(t, s_1\right)$, of $X(t)$. Since we cannot estimate $m_X(t)$ in a reasonable way from a single observation, we would like to use the values of the process for the other values of $t$ to estimate $m_X(t)$. For this to be possible, it is necessary (but not sufficient) that the mean $m_X(t)$ be independent of $t$.

Definition 2.3.1. The temporal mean of the s.p. ${X(t), t \in \mathbb{R}}$ is defined by
$$\langle X(t)\rangle_S=\frac{1}{2 S} \int_{-S}^S X(t, s) d t$$

## 数学代写|随机过程随机过程代考|平稳性

$$F\left(x_1, \ldots, x_n ; t_1, \ldots, t_n\right)=F\left(x_1, \ldots, x_n ; t_1+s, \ldots, t_n+s\right)$$

。在前面的定义中，必须选择$s$的值，以便$t_k+s \in T$表示$k=1, \ldots, n$。因此，例如，如果$T=[0, \infty)$，那么$t_k+s$对于所有$k$必须是非负的 在实践中，很难证明给定的随机过程在严格意义上是平稳的(高斯过程除外，如第2.4节所述)。因此，我们通常只考虑定义2.2.1中$n=$ 1和$n=2$的情况，从而满足于平稳性概念的较弱版本。如果${X(t), t \in T}$是一个(连续的)SSS进程，那么我们可以写
$$f(x ; t)=f(x ; t+s) \quad \forall s, t$$

$$f\left(x_1, x_2 ; t_1, t_2\right)=f\left(x_1, x_2 ; t_1+s, t_2+s\right) \quad \forall s, t_1, t_2$$

## 数学代写|随机过程Stochastic processes代考|Ergodicity

.

$$\widehat{m_X(t)}=\frac{1}{n} \sum_{k=1}^n X\left(t, s_k\right)$$

$$\langle X(t)\rangle_S=\frac{1}{2 S} \int_{-S}^S X(t, s) d t$$

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。


text:
1 Introduction

It is a challenging task to model the emerging high-dimensional clinical data with survival outcomes. For its simplicity and efficiency, penalized Cox models are significantly useful for accomplishing such tasks.

hdnom streamlines the workflow of high-dimensional Cox model building, nomogram plotting, model validation, calibration, and comparison.

2 Build Survival Models

To build a penalized Cox model with good predictive performance, some parameter tuning is usually needed. For example, the elastic-net model requires to tune the $$\ell_1$$-$$\ell_2$$ penalty trade-off parameter $$\alpha$$, and the regularization parameter $$\lambda$$.

To free the users from the tedious and error-prone parameter tuning process, hdnom provides several functions for automatic parameter tuning and model selection, including the following model types:

Function name Model type Auto-tuned hyperparameters
fit_lasso() Lasso $$\lambda$$
fit_alasso() Adaptive lasso $$\lambda$$
fit_enet() Elastic-net $$\lambda$$, $$\alpha$$
fit_aenet() Adaptive elastic-net $$\lambda$$, $$\alpha$$
fit_mcp() MCP $$\gamma$$, $$\lambda$$
fit_mnet() Mnet (MCP + $$\ell_2$$) $$\gamma$$, $$\lambda$$, $$\alpha$$
fit_scad() SCAD $$\gamma$$, $$\lambda$$
fit_snet() Snet (SCAD + $$\ell_2$$) $$\gamma$$, $$\lambda$$, $$\alpha$$
fit_flasso() Fused lasso $$\lambda_1$$, $$\lambda_2$$

In the next, we will use the imputed SMART study data to demonstrate a complete process of model building, nomogram plotting, model validation, calibration, and comparison with hdnom.

Load the packages and the smart dataset:

library("hdnom")
## Registered S3 methods overwritten by 'ggplot2':
##   method         from
##   [.quosures     rlang
##   c.quosures     rlang
##   print.quosures rlang
data("smart")
x <- as.matrix(smart[, -c(1, 2)])
time <- smart$TEVENT event <- smart$EVENT
y <- survival::Surv(time, event)

The dataset contains 3873 observations with corresponding survival outcome (time, event). 27 clinical variables (x) are available as the predictors. See ?smart for a detailed explanation of the variables.

Fit a penalized Cox model by adaptive elastic-net regularization with fit_aenet() and enable the parallel parameter tuning:

suppressMessages(library("doParallel"))
registerDoParallel(detectCores())

fit <- fit_aenet(x, y, nfolds = 10, rule = "lambda.1se", seed = c(5, 7), parallel = TRUE)
names(fit)
## [1] "model"       "alpha"       "lambda"      "model_init"  "alpha_init"
## [6] "lambda_init" "pen_factor"  "type"        "seed"        "call"

Adaptive elastic-net includes two estimation steps. The random seed used for parameter tuning, the selected best $$\alpha$$, the selected best $$\lambda$$, the model fitted for each estimation step, and the penalty factor for the model coefficients in the second estimation step are all stored in the model object fit.

3 Nomogram Visualization

Before plotting the nomogram, we need to extract some necessary information about the model: the model object and the selected hyperparameters:

model <- fit$model alpha <- fit$alpha
lambda <- fit$lambda adapen <- fit$pen_factor

Let’s generate a nomogram object with as_nomogram() and plot it:

nom <- as_nomogram(
fit, x, time, event,
pred.at = 365 * 2,
funlabel = "2-Year Overall Survival Probability"
)

plot(nom)

According to the nomogram, the adaptive elastic-net model selected 6 variables from the original set of 27 variables, effectively reduced the model complexity.

Information about the nomogram itself, such as the point-linear predictor unit mapping and total points-survival probability mapping, can be viewed by printing the nom object directly.

4 Model Validation

It is a common practice to utilize resampling-based methods to validate the predictive performance of a penalized Cox model. Bootstrap, $$k$$-fold cross-validation, and repeated $$k$$-fold cross-validation are the most employed methods for such purpose.

hdnom supports both internal model validation and external model validation. Internal validation takes the dataset used to build the model and evaluates the predictive performance on the data internally with the above resampling-based methods, while external validation evaluates the model’s predictive performance on a dataset which is independent to the dataset used in model building.

4.1 Internal Validation

validate() allows us to assess the model performance internally by time-dependent AUC (Area Under the ROC Curve) with the above three resampling methods.

Here, we validate the performance of the adaptive elastic-net model with bootstrap resampling, at every half year from the first year to the fifth year:

val_int <- validate(
x, time, event,
model.type = "aenet",
alpha = alpha, lambda = lambda, pen.factor = adapen,
method = "bootstrap", boot.times = 10,
tauc.type = "UNO", tauc.time = seq(1, 5, 0.5) * 365,
seed = 42, trace = FALSE
)

print(val_int)
## High-Dimensional Cox Model Validation Object
## Random seed: 42
## Validation method: bootstrap
## Bootstrap samples: 10
## Model type: aenet
## glmnet model alpha: 0.15
## glmnet model lambda: 0.4322461
## glmnet model penalty factor: specified
## Time-dependent AUC type: UNO
## Evaluation time points for tAUC: 365 547.5 730 912.5 1095 1277.5 1460 1642.5 1825
summary(val_int)
## Time-Dependent AUC Summary at Evaluation Time Points
##                365     547.5       730     912.5      1095    1277.5
## Mean     0.6736888 0.6980639 0.6883915 0.6877188 0.7171367 0.7329403
## Min      0.6702928 0.6943209 0.6852430 0.6838311 0.7104863 0.7274906
## 0.25 Qt. 0.6726013 0.6964793 0.6873324 0.6851874 0.7140472 0.7293532
## Median   0.6740403 0.6983876 0.6879825 0.6884841 0.7163633 0.7321747
## 0.75 Qt. 0.6749226 0.6996930 0.6896045 0.6896149 0.7210478 0.7371324
## Max      0.6760820 0.7029066 0.6916833 0.6916965 0.7236975 0.7389807
##               1460    1642.5      1825
## Mean     0.6801672 0.6841585 0.6935327
## Min      0.6713776 0.6770946 0.6800381
## 0.25 Qt. 0.6776301 0.6821142 0.6924683
## Median   0.6799323 0.6831381 0.6956346
## 0.75 Qt. 0.6825063 0.6864491 0.6966764
## Max      0.6920046 0.6939752 0.6997915

The mean, median, 25%, and 75% quantiles of time-dependent AUC at each time point across all bootstrap predictions are listed above. The median and the mean can be considered as the bias-corrected estimation of the model performance.

It is also possible to plot the model validation result:

plot(val_int)
##                365     547.5       730     912.5      1095    1277.5
## Mean     0.6736888 0.6980639 0.6883915 0.6877188 0.7171367 0.7329403
## Min      0.6702928 0.6943209 0.6852430 0.6838311 0.7104863 0.7274906
## 0.25 Qt. 0.6726013 0.6964793 0.6873324 0.6851874 0.7140472 0.7293532
## Median   0.6740403 0.6983876 0.6879825 0.6884841 0.7163633 0.7321747
## 0.75 Qt. 0.6749226 0.6996930 0.6896045 0.6896149 0.7210478 0.7371324
## Max      0.6760820 0.7029066 0.6916833 0.6916965 0.7236975 0.7389807
##               1460    1642.5      1825
## Mean     0.6801672 0.6841585 0.6935327
## Min      0.6713776 0.6770946 0.6800381
## 0.25 Qt. 0.6776301 0.6821142 0.6924683
## Median   0.6799323 0.6831381 0.6956346
## 0.75 Qt. 0.6825063 0.6864491 0.6966764
## Max      0.6920046 0.6939752 0.6997915


text:
output.to from Sideway
Draft for Information Only

# Content

` Iteration of Complex Function Iteration of Function Iteration of Quadratic Polynomials The Set of Iteration The Basin of Attraction to ∞ Involution or conjugation Julia Sets  Computer Algorithm finding the Filled-In Julia Sets  Prettier Pictures The Mandelbrot Set  Computer Algorithm to Plot the Mandelbrot Set  A Prettier Picture  Properties of the Mandelbrot Set  Misiurewicz Points  Big Open Conjecture`

source/reference:

# Iteration of Complex Function

## Iteration of Function

fⁿ is called the nth iterate of f

```Let f(z)=z+1
f²(z)=f(f(z))=f(z+1)=(z+1)+1=z+2
f³(z)=f(f²(z))=f(z+2)=(z+2)+1=z+3
…
fⁿ(z)=f(fⁿ⁻¹(z))=f(z+(n-1))=(z+(n-1))+1=z+n```
```Let f(z)=3z
f²(z)=f(f(z))=f(3z)=3(3z)=3²z
f³(z)=f(f²(z))=f(3²z)=3(3²z)=3³z
…
fⁿ(z)=f(fⁿ⁻¹(z))=f(3ⁿ⁻¹z)=3(3ⁿ⁻¹z)=3ⁿz```

The general quadratic polynomial is of the form p(z)=az²+bz+d, where constants a, b, d∊ℂ. However, for each triple of constants (a,b,d), there is exactly one constant c such that polynomial f(z)=z²+c behave the same as p(z)=az²+bz+d under iteration.

```Given p(z)=az²+bz+d, where constants a, b, and d∊ℂ.
Define functons φ(z)=az+b
2-(b
2)²
.
⇒f(φ(z))=(φ(z))²+c=(az+b
2-(b
2)²
2=a(az²+bz+d)+b
2
⇒f(φ(z))=a(p(z))+b
2=φ(p(z))
⇒φ⁻¹(f(φ(z)))=p(z)
∴p(z)=φ⁻¹(f(φ(z))) for all z
rewrite as p=φ⁻¹∘f∘φ where ∘ means composed with
⇒p∘p=(φ⁻¹∘f∘φ)∘(φ⁻¹∘f∘φ)=φ⁻¹∘f∘φ∘φ⁻¹∘f∘φ=φ⁻¹∘f∘f∘φ
⇒p²=φ⁻¹∘f²∘φ
⇒p³=φ⁻¹∘f³∘φ
…
⇒pⁿ=φ⁻¹∘fⁿ∘φ```

It thus suffices to study the iteration of quadratic polynomials of the form f(z)=z²+c.

## The Set of Iteration

The Julia set, named after the French mathematician Gaston Julia (1893-1978), of f(z)=z²+c is the set of all z∊ℂ for which the behavior of the iterates is "chaotic" in a neighborhood. the Fatou set, named after the French mathematician {oerre Fatpi (1878-1929), is the set of all z∊ℂ for which the iterates behave "normally" in a neighborhood.

The iterates of f behave normally near z if nearby point remain nearby under iteration. The iterates of behave chaotically at z if in any small neighborhood of z the behavior of the iterates depends sensitively on the initial point.

For example,

```f(z)=z²+c. Let c=0, f(z)=z², then fⁿ(z)=z2ⁿ
Let z=re𝑖θ, then fⁿ(z)=r2ⁿe𝑖2ⁿθ
If |z|<1, then |fⁿ(z)|=|z|2ⁿ→0 as n→∞, so fⁿ(z)→0 as n→∞.
If |z|>1, then |fⁿ(z)|=|z|2ⁿ→∞ as n→∞, so fⁿ(z)→∞ as n→∞.
If |z|=1, then z=re𝑖θ=e𝑖θ, so fⁿ(z)=e𝑖2ⁿθ, thus |fⁿ(z)|=1 for all n
In any little disk around a point z with |z|=1,
there are points w with |w|>1, for which fⁿ(w)→∞,
and other points w with |w|<1, for which fⁿ(w)→0.
The unit circle {z:|z|=1} is thus
the locus of chaotic behavior, whereas {z:|z|>1} with iterates attracted to ∞
and {z:|z|<1} with iterates attracted to 0 form the locus of normal behavior.```

Julia set J(f)={z:|z|=1}, the unit circle. and Fatou set F(f)={z:|z|>1}∪{z:|z|<1}

## The Basin of Attraction to ∞

Let f(z)=z²+c. Let A(∞)={z:fⁿ(z)→∞} "basin of attraction to ∞"

By theorem. The set A(∞) is open, connected and unbounded. It is contained in the Fatou set of f. The Julia set of f coincides with the boundary of A(∞) , which is a closed and bounded subset of ℂ.

Where

• The Julia set J(f) is a closed and bounded set.
• The Fatou set F(f) is open and unbounded and contains A(∞).
• J(f)F(f)

## Involution or conjugation

Let f(z)=z²-2. f²(z)=f(z²-2)=(z²-2)²-2. But it is hard to calculate and understand the iterates, fⁿ(z).

Conjugate f with

```φ(w)=w+1
w,φ:{w:|w|>1}→ℂ\[-2,2].```

f maps [-2,2] to[-2,2] and ℂ\[-2,2] to ℂ\[-2,2]. Thus look at φ⁻¹∘f∘φ.

```Let f(z)=z²-2. φ(w)=w+1
w,
⇒f(φ(w))=(φ(w))²-2=(w+1
w)²-2=w²+2w
w+1
w²-2=w²+1
w²=φ(w²)
⇒φ⁻¹(f(φ(w)))=w²
Let g(w)=w²
⇒φ⁻¹(f(φ(w)))=g(w)
⇒ f(φ(w))=φ(g(w))
Let φ(w)=z⇒w=φ⁻¹(z)
⇒ f(z)=φ(g(φ⁻¹(z)))```

Thus, on ℂ\[-2,2], the function f(z)=z²-2 behaves like g(w)=w² behaves on the exterior of the closed unit disk. Since the iterates gⁿ(w) tend to ∞ for |w|>1. therefore fⁿ(z)→∞ as n→∞ for all z∊ℂ\[-2,2]. Thus A(∞)=ℂ\[-2,2], and thus J(f)=[-2,2], the closed interval from -2 to 2 on the real axis. Both the Julia sets of  f(z)=z² and f(z)=z²-2 are the only smooth Julia set examples amongst all f(z)=z²+c.

## Julia Sets

For f(z)=z²+c, where c∊ℂ,the Julia set of f is the boundary of A(∞), where A(∞) is the "basin of attraction to infinity", i.e. A(∞)={z∊ℂ:fⁿ(z)→∞ as n→∞}. Let K(f) be the filled-in Julia set of f for which z∊ℂ and {fⁿ(z)} stays bounded, i.e. K(f)={z∊ℂ : {fⁿ(z)} is bounded}. For examples

```f(z)=z² (c=0): Then K(f)={z:|z|≤1} and J(f)={z:|z|=1}
f(z)=z²-2 (c=-2): Then K(f)=[-2,2] and J(f)=[-2,2]```

For f(z)=z²-1 (c=-1)

```Let f(z)=z²-1
z=0: f(0)=-1, f²(0)=f(-1)=0, f³(0)=f(0)=-1, f⁴(0)=0, …. The orbit is thus 0, -1, 0, -1, 0, -1, ….
This is called a periodic orbit, and 0 is a periodic point of period 2. clearly, 0∊K(f).
z=1: 1, 0, -1, 0, -1, 0, -1, …. So 1 isnot itself a periodic point (the orbit never returen to 1),
but it's the next best thing;it runs into a periodic orbit. It thus is called 1 a pre-periodic point.
Again, 1∊K(f).
z=1+√5
2:(1+√5
2)²
-1=1+2√5+5
4-1=2+2√5
4=1+√5
2=z.
The point 1+√5
2is a fixed point of f and thus belongs to K(f) as well.```

Other z

```z=-2: -2, 3, 8, 63,  …, →∞. Thus -2∊A(∞).
z=: -2, 3, 8, 63,  …, →∞. Thus -2∊A(∞).```

First, if z lies on the real axis and is smaller than -(1+√5
2
) or larger than (1+√5
2
), then z∊A(∞).

More generally, if z∊ℂ and |z|>1+√5
2
, then z∊A(∞).

And more generally, if z∊ℂ and |fⁿ(z)|>1+√5
2
for some n, then z∊A(∞).

A similar condition holds for general quadratic polynomials f(z)=z²+c,

By theorem, Let f(z)=z²+c, and let R= 1+√(1+4|c|)
2
. Let z₀∊ℂ. If for some n>0, there are |fⁿ(z)|→∞as n →∞, i.e. z₀∊A(∞), so z₀∉K(f).

### Computer Algorithm finding the Filled-In Julia Sets

Computer algorithm:

1. Given c, let R= 1+√(1+4|c|)
2
.
2. Choose a window W={x+𝑖y: a≤x≤b, c≤y≤d} to display. If you want to see the entire Julia set, you'd want BR(0)⊂W. But at some point it also may be interesting to zoom into the Julia set!
3. The computer can't keep checking all iterates fⁿ(z), at some point it'll have to stop. So pick a largest number, maxiter, up to which it will check. The larger this number, the more accurate your picture will, get, but the slower the calculation will be.
4. For each pixel on your screen, choose a point z in your window W, corresponding to that pixel.
5. Calculate the iterates of z: f(z), f(f(z)), f(f(f(z))), …. If one of these iterates satisfies that |fⁿ(z)|>R, color the initial pixel white.
6. If you reach the maximum number of iterations, maxiter, without having left BR(0), there's a good chance that z belongs to K(f). Color this pixel black.

### Prettier Pictures

The pictures of the filled-in Julia sets produced so far are only black and white. Fro colorful pictures, choose colors c₀, c₁, c₂, …, cmaxiter for A(∞) as well as a color for the filled-in Julia set. If |z|>R, color the corresponding pixel with color c₀. Otherwise, if |f(z)|>R, color the corresponding pixel with color c₁. Otherwise, if |f(f(z))|>R, color the corresponding pixel with color c₂. …. Otherwise, i.e.  if |fⁿ(z)|>R for all n≤maxiter, color the corresponding pixel with corresponding color cₙ for the filled-in Julia set. The larger constant maxiter, the more detail of the pictures, but the longer the calculation will take. For small windows (i.e. zooms into the Julia set), a higher precision is required to avoid round-off errors.

## The Mandelbrot Set

The Julia set of f(z)=z²+c is either "in one piece" or "totally dusty"

By definition. The Mandelbrot set M is the set of all parameters c∊ℂ for which the Julia set J(f) of f(z)=z²+c is connected in one piece. That is M={c∊ℂ: J(z²+c) is connected}

Note: The Mandelbrot set is a subset of the parameter space (the space of all possible c-values), whereas Julia sets are sets of z-values.

Examples

```0∊M since J(z²)={z"|z|=1}
-2∊M since J(z²-2)=[-2,2]
¼∊M according to picture of J(z²+¼)
1∉M according to picture of J(z²+1)```

By theorem. Let f(z)=z²+c. Then J(f) is connected if and only if z=0 does not belong to A(∞), that is if and only if the orbit {fⁿ(0)} remains bounded under iteration.

In fact, it is possible to show the following:

By theorem. A complex number c belongs to M if and only if |fⁿ(0)|≤2 for all n≥1 (where f(z)=z²+c).

### Computer Algorithm to Plot the Mandelbrot Set

Computer algorithm

1. Choose a window W={cx+𝑖cy:cxmin≤cx≤cxmax , cymin≤cy≤cymax } to display. If you want to see the entire Mandelbrot set, you'd want something like W={c: -2≤cx≤0.75 , -1.5≤cy≤1.5 }
2. As before, pick a largest number of iterations, maxiter. The larger this number, the more accurate the picture will get, but the slower the calculation will be.
3. For each pixel on the screen, choose a parameter c in the window W, corresponding to that pixel. We'll look at the corresponding polynomial f(z)=z²+c.
4. Calculate the iterates of 0 under this polynomial: f(0)=c, f(f0))=c²+c, f(f(f(0)))=(c²+c)²+c, ….  If one of these iterates satisfies that |fⁿ(0)|>2, color the initial pixel white.
5. If you reach the maximum number of iterations, maxiter, without having left B₂(0), there's a good chance that the parameter c belongs to the Mandelbrot set M. Color this pixel black.

### A Prettier Picture

Again, you can use different colors for those parameters c∊ℂ for which 0 escapes to infinity under iteration, depending on how quickly the escape happens"

• I  |f(0)|=|c|>2, color the corresponding pixel in color zero.
• Otherwise, if |f(f0))|=|c²+c|>2, color the corresponding pixel in color one.
• Otherwise, if |f(f(f0)))|=|(c²+c)²+c|>2, color the corresponding pixel in color two.
• ….
• If |fⁿ(0)|≤2 for all n≤maxiter, color the pixel corresponding to c black (or whatever other color you choose for your M).

Zooming into the Mandelbrot set and coloring parameters by escape time yields beautiful pictures.

### Properties of the Mandelbrot Set

Properties of the Mandelbrot Set

• M is a connected set (Douadym Hubbard, 1982)
• M is contained in the disk of radius 2, centered at zero.
• The boundary of M is very intricate- this is where you'll find the most beatutiful zooms.
• Moreover, for c-values near the boundary of M, their Julia sets have many different patterns. Here are some examples:
The boundary of the main cardioid is given by c=½e𝑖θ-¼e2𝑖θ,0≤θ<2π.
Writing θ=2πα, 0≤α<1. we can distinguish whether α is a rational or an irrational number.
• Rational α:
• Let c=½e2π𝑖α-¼e4π𝑖α where ,0≤α<1 is a rational number. Then α is of the form p/q.
• The parameter c is an attachment point of another "bud" to the Mandelbrot set, and the Julia set for parameter values within the bud.
• Example: α=½. Then c=½eπ𝑖-¼e2π𝑖=-½-¼=-0.75. A picture for J(z²-0.75)
• Irrational α:
• Let c=½e2π𝑖α-¼e4π𝑖α where ,0≤α<1 is a irrational number. Thus there are no values p and q such that α is of the form p/q.
• Julia sets for such values look more intricate and come in several "flavors"!
• For  example: α= 1+√5
2
. Then c≡-0.390540870218401…-0.586787907346969…i. For  f(z)=z²+c, the interior of K(f) has a so-called "Siegel disk", in which iteration looks like a rotation by angle α.

### Misiurewicz Points

Many points in the boundary of M are co-called Misiurewicz points:

• A point c∊ℂ is called a Misiurewicz point if the orbit of 0 under f(z)=z²+c is pre-periodic, but not periodic.
• Example: c=𝑖: Then f(z)=z²+𝑖 and the orbit of 0 under f is 0, 𝑖, -1+𝑖, -𝑖, -1+𝑖, -𝑖, ….
• Clearly, Misiurewicz points c belong to M since the orbit of 0 under f(z)=z²+c is bounded.
• Properties:
• Let c be a Misiurewicz point. Then J(f)=K(f), i.e. K(f) has no interior.
• Misiurewicz points are dense in ∂M.
• The Mandelbrot set is self-similar under magnification near Misiurewicz points. Note: The Mandelbrot is "quasi-self-similar" everywhere: small, slightly different versions of itself can be found at arbitrary small scales.

### Big Open Conjecture

Here is one of the big outstanding conjectures in the field of complex dynamics:

By Conjecture. The Mandelbrot set is locally connected, that is, for every c∊M and every open set V with c∊V, there exists an open set U such that c∊U⊂V and U∩M is connected.

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# Establishing impact force from deceleration data

I am trying to establish an impact force. The impact came from a spear (wooden with a stone tip) hitting a soft (meat) target. The acceleration was measured by a GCDC X400-2 accelerometer.

Due to the spear impacting a non-consistent fluid/solid target, the deceleration was not regular. This is the reason I do not know how to apply the F=ma formula that is clearly the one need here.

I need to establish the amount of force with which the spear impacted the target and the amount of energy that was dissipated throughout the impacting into the target.

I was thinking about using the energy in momentum formula, but this goes into relativism, which is beyond me.

Below is the acceleration data for the impact. It is inverted, due to accelerometer mounting position:

Time m/s^2 31.721039 -5.568076923 31.726013 -12.8724359 31.731018 -23.52961538 31.736023 -22.93089744 31.741028 -15.62653846 31.746002 -2.394871795 31.751007 4.67 31.756012 -3.472564103 31.761017 -9.878846154 31.766022 -14.54884615 31.770996 -21.97294872 31.776001 -21.61371795 31.781006 -12.33358974 31.786011 -7.005 31.791016 -9.998589744 31.79599 -8.920897436 31.800995 0.838205128 31.806 18.56025641 31.811005 81.30589744 31.81601 173.3887179 31.820984 153.9902564 31.825989 49.87320513 31.830994 7.364230769 31.835999 -38.19820513 31.840973 17.06346154 31.845978 138.0044872 31.850983 75.13910256 31.855988 -12.21384615 31.860993 -3.771923077 31.865967 123.455641 31.870972 229.1892308 31.875977 149.44 31.880982 26.58307692 31.885956 -15.32717949 31.890961 0 31.895966 -17.00358974 31.900971 -20.65576923 31.905976 0.478974359 31.91095 -3.233076923 31.915955 -0.538846154 31.92096 -5.268717949 31.925782 -12.51320513 31.930786 -10.29794872 31.935791 -4.610128205 31.940949 2.45474359 31.945954 -3.652179487 31.950959 -10.59730769  The average velocity of this spear in flight was 13.89 m/s. This was calculated as the average velocity from the time of release to the time of impact. This throwing distance was 5m and the flight time was 0.36s. There was no deceleration noticed during this flight time.

For reference, this is for a PhD in Archaeology and hence I am not the best at physics in the world!! If there is any more information that you need, please contact me and I will be happy to provide it (along with larger data samples if that would help. I cropped the data to keep this post smaller.)

Cheers!

• What archaeological question are you trying to answer? Impact force might not be the best measure of what you are trying to quantify. The amount of energy which was dissipated is probably a better measure; this is the kinetic energy of the spear, $K=\frac12 mv^2$ where $v=13.89m/s$. – sammy gerbil Sep 21 '17 at 19:03


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# Why NP-Complete reduction is not reversible?

I have read the question asked here Is polynomial reduction reversible and the logic actually makes sense to me. In other words, if A is polynomially reducible to B, it means that A <= B in terms of hardness.

However, when I thought more about the reduction process, I am confused. As far as I can understand, the NP-Complete reduction process is as follows:

(1) Find some NP-Complete problem (i.e. A), convert its inputs and output to the inputs and output of the problem we want to prove (i.e. B)

(2) Show that if A is YES, then B is YES

(3) Show that if B is YES, then A is YES.

Then I also read several classic reduction proofs, and here is my confusing part. In all the reductions I read, step (1) seems to be a "one-to-one mapping" between the inputs of A to the inputs B. For example, minimum set cover (MSC) to minimum vertex cover (MVC), edges in MVC is elements in MSC and nodes in MVC is sets in MSC. Since it's a "one-to-one mapping" of inputs A to inputs B, if we can convert the inputs of A to the inputs of B, then we can always convert the inputs of B to the inputs of A. In this case, isn't step (1) a reversible process? I am not sure if "one-to-one mapping" is the correct terminology, but the main idea I want to say is "the specific way" of mapping inputs of A to the inputs of B. Is there any reduction example that maps the input from A to the inputs of B, but the mapping is not reversible, i.e. can't map the inputs of B back to A?

What's more, no matter if we convert A to B or convert B to A, I think step (2) and step (3) above are still the same. Therefore, I am really confused about which part makes the NP-Complete reduction not reversible?

Here is a concrete example. Let $$A$$ be the problem in which all instances are NO instances, and let $$B$$ be the problem with a unique YES instance $$1$$. The function $$f(x) = 0$$ is a reduction from $$A$$ to $$B$$, but there is no reduction from $$B$$ to $$A$$, since there is nothing we can map $$1$$ into.


text:
## Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B

These Solutions are part of Selina Concise Mathematics Class 10 ICSE Solutions. Here we have given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B.

Other Exercises

Question 1.
(i) In the given figure 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.

(ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4 cm. Find CD.

(iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.

Solution:

Question 2.
In the figure given below, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find:
(i) AB.
(ii) the length of tangent PT. (2014)

Solution:
PT is tangent and PDC is secant out to the circle
∴ PT² = PC x PD
PT² = (5 + 7.8) x 5 = 12.8 x 5
PT² = 64 ⇒ PT = 8 cm
In ΔOTP
PT² + OT² = OP²
8²+x² = (x + 4)²
⇒ 64 +x² = x² + 16 + 8x
64- 16 = 8x
⇒ 48 = 8x
x = $$\frac { 48 }{ 8 }$$ = 6 cm
AB = 2 x 6 = 12 cm

Question 3.
In the following figure, PQ is the tangent to the circle at A, DB is the diameter and O is the centre of the circle. If ∠ ADB = 30° and ∠ CBD = 60°, calculate
(i) ∠ QAB
(iii) ∠ CDB.

Solution:
(i) PAQ is a tangent and AB is the chord, ∠ QAB = ∠ ADB (Angles in the alternate segment)
= 30°
(ii) OA = OD (Radii of the same circle)
∴ ∠ OAD = ∠ ODA = 30°
But OA ⊥ PQ
∴ ∠ PAD = ∠ OAP – ∠OAD = 90° – 30° = 60°
(iii) BD is diameter
∴ ∠ BCD = 90° (Angle in semi circle)
Now in ∆ BCD,
∠ CDB + ∠ CBD + ∠ BCD = 180°
⇒ ∠ CDB + 60° + 90° = 180°
⇒ ∠ CDB = 180°- (60° + 90°) = 180° – 150° = 30°

Question 4.
If PQ is a tangent to the circle at R; calculate:
(i) ∠ PRS
(ii) ∠ ROT.

Given O is the centre of the circle and angle TRQ = 30°.
Solution:

PQ is tangent and OR is the radius
∴ OR ⊥ PQ
∴ ∠ORT = 90° =
∠TRQ = 90° – 30° = 60°
But in ∆ OTR, OT = OR (Radii of the circle)
∴ ∠ OTR = 60° or ∠STR = 60°
But ∠PRS = ∠STR (Angles in the alternate segment) = 60°
In ∆ ORT, ∠ OTR = 60°, ∠ TOR = 60°
∴ ∠ROT= 180°-(60°+ 60°)= 180°-120° = 60°

Question 5.
AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersect AB produced in D. Show that BC = BD.
Solution:

Given: In a circle, O is the centre,
AB is the diameter, a chord AC such that ∠ BAC = 30°
and a tangent from C, meets AB in D on producing. BC is joined.
To Prove: BC = BD
Construction: Join OC
Proof : ∠ BCD = ∠ BAC =30°(Angle in alternate segment)
Arc BC subtends ∠ DOC at the centre of the circle and ∠ BAC at the remaining part of the circle.
∴ ∠ BOC = 2 ∠ BAC = 2 x 30° = 60°
Now in ∆ OCD,
∠ BOC or ∠ DOC = 60° (Proved)
∠ OCD = 90° (∵ OC ⊥ CD)
∴ ∠ DOC + ∠ ODC = 90°
⇒ 60° + ∠ ODC = 90°
∴ ∠ ODC = 90°- 60° = 30°
Now in ∆BCD,
∵∠ ODC or ∠ BDC = ∠ BCD (Each = 30°)
∴ BC = BD Q.E.D.

Question 6.
Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that ∆PQR is isosceles.
Solution:

In a circumcircle of ∆PQR, a tangent TPS is drawn through P which is parallel to QR
To prove : ∆PQR is an isosceles triangle.
Proof:
∵ TS $$\parallel$$ QR
∠TPQ = ∠PQR (Alternate angles) ….(i)
∵ TS is tangent and PQ is the chord of the circle
∴ ∠TPQ = ∠RP (Angles in the alternate segment) ….(ii)
From (i) and (ii),
∠PQR = ∠QRP
∴ PQ = PR (Opposite sides of equal angles)
∴ ∆PQR is an isosceles triangle
Hence proved

Question 7.
Two circles with centres O and O’ are drawn to intersect each other at points A and B. Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O’ at A. Prove that OA bisects angle BAC.

Solution:

Given: Two circles with centre O and O’ intersect each other at A and B, O lies on the circumference, of the other circle. CD is a tangent at A to the second circle. AB, OA are joined.
To Prove: OA bisects ∠ BAC.
Construction: Join OB, O’A, O’B and OO’
Proof: CD is the tangent and AO is the chord
∠ OAC = ∠ OBA …(i)
(Angles in alt. segment)
In ∆ OAB, OA = OB (Radii of the same circle)
∴ ∠ OAB = ∠ OBA ….(ii)
From (i) and (ii),
∠ OAC = ∠ OAB
∴ OA is the bisector of ∠ BAC Q.E.D.

Question 8.
Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠ CPA = ∠DPB.

Solution:
Given:Two circles touch each other internally at P. A chord AB of the bigger circle intersects the smaller circle at C and D. AP, BP, CP and DP are joined.
To Prove: ∠ CPA = ∠ DPB
Construction: Draw a tangent TS at P to the circles given.
Proof:
∵ TPS is the tangent, PD is the chord.
∴ ∠ PAB = ∠ BPS …(i) ( Angles in alt. segment)
Similarly we can prove that
∠ PCD = ∠ DPS …(ii)
Subtracting (i) from (ii), we gel
∠ PCD – ∠ PAB = ∠ DPS – ∠ BPS
But in ∆ PAC,
Ext. ∠ PCD = ∠ PAB + ∠ CPA
∴ ∠ PAB + ∠ CPA – ∠ PAB = ∠ DPS – ∠ BPS
∠ CPA = ∠ DPB           Q.E.D.

Question 9.
In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
Solution:

Given: ABCD is a cyclic quadrilateral and diagonal AC bisects ∠ BCD. AT A. a tangent TAS is drawn. BD is joined.
To Prove: TS || BD.
Proof: ∠ ADB = ∠ ACB …….(i) (Angles in the same segment)
Similarly ∠ ABD = ∠ ACD ……..(ii)
But ∠ ACB = ∠ ACD (AC is the bisector of ∠ BCD)
∴ ∠ ADB = ∠ ABD |From (i) and (ii)]
TAS is a tangent and AB is chord
∴ ∠ BAS = ∠ ADB (Angles in all segment)
But ∠ ADB = ∠ ABD (Proved)
∴ ∠ BAS = ∠ ABD
But these are alternate angles.
∴ TS || BD.      Q.E.D.

Question 10.
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠ BCG = 108° and O is the centre of the circle,
Find:
(i) angle BCT
(ii) angle DOC

Solution:

Join OC, OD and AC.
(i) ∠ BCG + ∠ BCD = 180° (Linear pair)
⇒ 108° + ∠ BCD = 180°
(∵∠ BCG = 108° given)
∴∠ BCD = 180° – 108° = 72°
BC = CD (given)
∴ ∠ DCP = ∠ BCT
But ∠ BCT + ∠ BCD + ∠ DCP = 180°
∴∠ BCT + ∠ BCT + 72° = 180°
(∵∠ DCP = ∠ BCT)
2 ∠ BCT = 180° – 72° = 108°
∴∠ BCT = $$\frac { { 108 }^{ \circ } }{ 2 }$$ = 54°
(ii) PCT is the tangent and CA is chord
∴ ∠ CAD = ∠ BCT = 54°
But arc DC subtends ∠ DOC at the centre and
∠ CAD at the remaining part of the circle
∴ ∠ DOC = 2 ∠ CAD = 2 x 54° = 108°.

Question 11.
Two circles intersect each other at points A and B. A straight line PAQ cuts the circles at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T arc concyclic.
Solution:
Given: Two circles intersect each other at point A and B. PAQ is a line which intersects circles at P, A and Q. At P and Q, tangents are drawn to the circles which meet at T.
To Prove: P, B, Q, T are concyclic.

Construction: Join AB, BP and BQ.
Proof: TP is the tangent and PA. a chord
∴ ∠ TPA = ∠ ABP
(angles in alt. segment)
Similarly we can prove that
∠ TQA = ∠ ABQ …,(ii)
Adding (i) and (ii), we get
∠ TPA + ∠ TQA = ∠ ABP + ∠ ABQ
But in ∆ PTQ,
∠ TPA + ∠ TQA + ∠ PTQ = 180°
⇒ ∠ TPA + ∠ TQA = 180° – ∠ PTQ
⇒ ∠ PBQ = 180°- ∠ PTQ
⇒ ∠ PBQ + ∠PTQ = 180°
But there are the opposite angles of the quadrilateral
∴ Quad. PBQT is a cyclic
Hence P, B. Q and T are concyclic     Q.E.D.

Question 12.
In the figure; PA is a tangent to the circle. PBC is secant and AD bisects angle BAC.
Show that triangle PAD is an isosceles triangle. Also shaw that:

Solution:
Given: In a circle PA is the tangent, PBC is the secant and AD is the bisector of ∠BAC which meets the secant at D.
To Prove:
(i) ∆ PAD is an isosceles triangle.
(ii) ∠CAD = $$\frac { 1 }{ 2 }$$ [(∠PBA – ∠PAB)]
Proof:
(i) PA is the tangent and AB is chord.
∠PAB = ∠C ….(i)
(Angles in the alt. segment)
AD is the bisector is ∠BAC
∴ ∠1 = ∠2 ….(ii)
Ext. ∠ADP = ∠C + ∠1
= ∠PAB + ∠2 = ∠PAD
∴ ∆ PAD is an isosceles triangle.
(ii) In A ABC,
Ext. ∠PBA = ∠C + ∠BAC
∴∠BAC = ∠PBA – ∠C
⇒ ∠1 + ∠2 = ∠PBA – ∠PAB [from (i)]
⇒ 2 ∠1 = ∠PBA – ∠PAB
⇒ ∠1 = – [∠PBA – ∠PAB]
⇒ ∠CAD = – [∠PBA – ∠PAB] Q.E.D.

Question 13.
Two circles intersect each other at points A and B. Their common tangent touches the circles v at points P and Q as shown in the figure . Show that the angles PAQ and PBQ arc supplementary. [2000]

Solution:
Given: Two circles intersect each other at A and
B. A common tangent touches the circles at P and
Q. PA. PB, QA and QB are joined.
To Prove: ∠ PAQ + ∠ PBQ = 180°
or ∠ PAQ and ∠ PBQ are supplementary.
Construction: Join AB.
Proof: PQ is the tangent and AB is the chord
∴ ∠ QPA = ∠ PBA (alternate segment) ,…(i)
Similarly we can prove that
∠ PQA = ∠ QBA ,…(ii)
Adding (i) and (ii), we get
∠ QPA + ∠ PQA = ∠ PBA + ∠ QBA
But ∠ QPA + ∠ PQA = 180° – ∠ PAQ ,…(iii) (In ∆ PAQ)
and ∠ PBA + ∠ QBA = ∠ PBQ ,…(iv)
from (iii) and (iv)
∠ PBQ = 180° – ∠ PAQ
⇒ ∠ PBQ + ∠ PAQ = 180°
= ∠ PAQ + ∠ PBQ = 180°
Hence ∠ PAQ and ∠ PBQ arc supplementary Q.E.D.

Question 14.
In the figure, chords AE and BC intersect each other at point D.
(i) If ∠ CDE = 90°.
AB = 5 cm, BD = 4 cm and CD 9 cm;
Find DE.
(ii) If AD = BD, show that AE = BC.

Solution:

Question 15.
Circles with centres P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles arc congruent; show that CE = BD.

Solution:

Given: Two circles with centre P and Q intersect each other at A and B. CBD is a line segment and EBM is tangent to the circle with centre Q, at B. Radii of the cirlces are equal.
To Prove: CE = BD
Proof: EBM is the tangent and BD is the chord
∴ ∠ DBM = ∠ BAD (Anglesi in alt. segment)
But ∠ DBM = ∠ CBE (Vertically opposite angles)
∴ ∠ BAD = ∠ CBE
∵ In the same circle or congruent circles, if angles are equal, then chords opposite to them are also equal.
∴ CE = BD Q.E.D.

Question 16.
In the following figure O is the centre of ti.e circle and AB is a tangent to it at point B. ∠BDC = 65°, Find ∠BAO

Solution:
Given: ∠BDC = 65° and AB is tangent to circle with centre O.
⇒ OB ⊥ AB
In ∆BDC
∠DBC + ∠BDC + ∠B CD = 180°
90° + 65° + ∠BCD =180°
⇒ ∠BCD = 25°
∵ OE = OC = radius
⇒ ∠OEC = ∠OCE
⇒ ∠OEC = 25°
Also, ∠BOE = ∠OEC + ∠OCE
[Exterior angle = sum of opposite interior angles in a ∆]
⇒ ∠BOE = 25°+ 25°
⇒∠BOE = 50°
⇒ ∠BOA = 50°
In ∆AOB
∠AOB + ∠BAO + ∠OBA = 180°
50° + ∠BAO + 90° = 180°
⇒ ∠BAO = 40°

Hope given Selina Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords Ex 18B  are helpful to complete your math homework.

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