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# A Possible Decision Theory for Many Worlds Living

post by Evan Ward · 2019-05-04T21:20:42.127Z · score: 0 (8 votes) · LW · GW · 9 comments

## Contents

  Background
My Proposal
If one CoA is twice as choice-worthy as another, then I argue that we should commit to doing that CoA with 2:1 odds or 66% of the time based on radioactive particle decay.
Why?
What this Theory Isn't
Is it Incrementally Useful?
Crucial Considerations
Is RECMDT Safer if Applied Only with Particular Mindsets?
Converting Radioactive Decay to Random Bit Strings
Converting Random Bit Strings to Choices
What Does Application Look Like?
Can We Really Affect the Distribution of Other Worlds through Our Actions?
What if Many Worlds Isn't True?
None

Hey LessWrong! I may have gone in over my head as I am not well-versed in decision theory literature, but I tentatively believe I have a new decision theory for decision-making in a MWI universe. Let me know what you think!

--------------------------------------------

Originally posted at: https://www.evanward.org/a-decision-theory-for-many-worlds-living/

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Here, I describe a decision theory that I believe applies to Many-Worlds living that combines principles of quantum mechanical randomness, evolutionary theory, and choice-worthiness. Until someone comes up with a better term for it, I will refer to it as Random Evolutionary Choice-worthy Many-worlds Decisions Theory, or RECMDT.

### Background

If the Many World's Interpretation (MWI) of quantum mechanics is true, does that have any ethical implications? Should we behave any differently in order to maximize ethical outcomes? This is an extremely important question that I'm not aware has been satisfactorily answered. If MWI is true and if we can affect the distribution of worlds through our actions, it means that our actions have super-exponentially more impact on ethically relevant phenomena. I take ethically relevant phenomena to be certain fundamental physics operations responsible for the suffering and well-being associated with the minds of conscious creatures.

### My Proposal

We ought to make decisions probabilistically based on sources of entropy which correspond with the splitting of worlds (e.g. particle decay) and the comparative choice-worthiness of different courses of action (CoA). By choice-worthiness, I mean a combination of the subjective degree of normative uncertainty and expected utility of a CoA. I will go into determining choice-worthiness in another post.

If one CoA is twice as choice-worthy as another, then I argue that we should commit to doing that CoA with 2:1 odds or 66% of the time based on radioactive particle decay.

### Why?

Under a single unfolding of history, the traditional view is that we should choose whichever CoA available to us which has the highest choice-worthiness. When presented with a binary decision, the thought is that we should choose the most choice-worthy option given the sum of evidence every single time. However, the fact that a decision is subjectively choice-worthy does not mean it is guaranteed to actually be the right decision—it could actually move us towards worse possible worlds. If we think we are living in a single unfolding of history but are actually living under MWI, then a significant subset of the 3↑↑↑3+ (but a finite number) of existing worlds end up converging on similar futures, which are by no means destined to be good.

However, if we are living in a reality of constantly splitting worlds, I assert that it is in everyone's best interest to increase the variance of outcomes in order to more quickly move towards either a utopia or extinction. This essentially increases evolutionary selection pressure that child worlds experience so that they either more quickly become devoid of conscious life or more quickly converge on worlds that are utopian.

As a rough analogy, imagine having a planet covered with trillions of identical, simple microbes. You want them to evolve towards intelligent life that experiences much more well-being. You could leave these trillions of microbes alone and allow them to slowly incur gene edits so that some of their descendants drift towards more intelligent/evolved creatures. However, if you had the option, why not just increase the rate of the gene edits, by say, UV exposure? This will surely push up the timeline for intelligence and well-being and allow a greater magnitude of well-being to take place. Each world under MWI is like a microbe, and we might as well increase the variance, and thus, evolutionary selection pressure in order to help utopias happen as soon and as abundantly as possible.

### What this Theory Isn't

A key component of this decision heuristic is not maximizing chaos and treating different CoAs equally, but choosing CoAs relative to their choice-worthiness. For example, in a utopian world with, somehow, 99% of the proper CoAs figured out, only in 1 out of 100 child worlds must a less choice worthy course of action be taken. In other words, once we get confident in particular CoA, we can take that action the majority of the time. After all, the goal isn't for 1 world to end up hyper-utopian, but to maximize utility over all worlds.

If we wanted just a single world to end up hyper utopian, then we want to act in as many possible ways based on the results of true sources of entropy. It would be ideal to come up with any and flip a (quantum) coin and go off its results like Two-Face. Again, the goal is to maximize utility over all worlds, so we only want to explore paths in proportion to the odds that we think a particular path is optimal.

### Is it Incrementally Useful?

A key component of most useful decision theories is that they are useful insofar as they are followed. As long as MWI is true, each time RECMDT is deliberately adhered to, it is supposed to increase the variance of child worlds. Following this rule just once, depending on the likelihood of worlds becoming utopian relative to the probability of them being full of suffering, likely ensures many future utopias will exist.

### Crucial Considerations

While RECMDT should increase the variance and selection pressure on any child worlds of worlds that implement it, we do not know enough about the likelihood and magnitude of suffering at an astronomical level to guarantee that the worlds that remain full of life will overwhelmingly tend to be net-positive in subjective well-being. It could be possible that worlds with net-suffering are very stable and do not tend to approach extinction. The merit of RECMDT may largely rest on the landscape of energy-efficiency of suffering as opposed to well-being. If suffering is very energy inefficient compared to well-being, then that is good evidence in favor of this theory. I will write more about the implications of the energy-efficiency of suffering soon.

### Is RECMDT Safer if Applied Only with Particular Mindsets?

One way to hedge against astronomically bad outcomes may be to only employ RECMDT when one fully understands and is committed to ensuring that survivability remains dependent on well-being. This works because following this decision theory essentially increases the variance of child worlds like using birdshot instead of a slug. If one employs this heuristic only while having a firm belief and commitment to a strong heuristic to reduce the probability of net-suffering worlds, then it seems that yourself in child worlds will also have this belief and be prepared to act on it. You can also only employ RECMDT while you believe in your ability to take massive-action on behalf of your belief that survivability should remain dependant on well-being. Whenever you feel unable to carry out this value, you should perhaps not act to increase the variance of child worlds because you will not be prepared to deal with the worst-case scenarios in those child worlds.

Evidence against applying RECMDT only when one holds certain values strongly, however, is all the Nth-order effects of our actions. For decisions that have extremely localized effects where one's beliefs dominate the ultimate outcome, the plausible value of RECMDT over not applying it is rather small.

For decision with many Nth order effects, such as deciding which job to take (which, for example, has many unpredictable effects on the economy), it seems that one cannot control for the majority of the effects of one's actions after an initial decision is made. The ultimate effects likely rest on features of our universe (e.g. the nature of human market economies in our local group of many-worlds) that one's particular belief has little influence over. In other words, for many decisions, one can affect the world once, but they cannot control the Nth order effects through acting a second time. Thus, while certain mindsets are useful to hold dearly regardless of whether one employs RECMDT, it seems that it is not generally useful for one to not employ RECMDT if they are not holding any particular mindsets.

### Converting Radioactive Decay to Random Bit Strings

In order to implement this decision theory, agents much require access to a true source of entropy—pseudo-random number generators will NOT work. There are a variety of ways to implement this, such as by having an array of Geiger counters surrounding a radioactive isotope and looking at which groups of sensors get triggered first in order to yield a decision. However, I suspect one of the cheapest and most reliably random sensors would be built to implement the following algorithm from HotBits:

Since the time of any given decay is random, then the interval between two consecutive decays is also random. What we do, then, is measure a pair of these intervals, and emit a zero or one bit based on the relative length of the two intervals. If we measure the same interval for the two decays, we discard the measurement and try again, to avoid the risk of inducing bias due to the resolution of our clock.
John Walker
from HotBits

### Converting Random Bit Strings to Choices

We have a means above to generate truly random bit strings that should differ between child worlds. The next question is how do we convert these bit strings to choices regarding which CoA we will execute? This depends on the number of CoAs we were considering and the specific ratios that we arrived at for comparative choice-worthiness. We simply need to determine the least common multiple of all the individual odds of each CoA, and acquire a bit string that is long enough that its representation as a binary number is higher than the least common multiple. From there, we can use a simple preconceived encoding scheme to have the base 2 number encoded in the bit string select for a particular course of action.

For example, in a scenario where one CoA is 4x as choice-worthy as another, we need a random number that represents the digits 0 to 4 equally. Drawing the number 4 can mean we must do the less-choice worthy CoA, and drawing 0-3 can mean we do the more choice-worth CoA. We need at least 3 random bits in order to do this. Since 2^3 is 8 and there is no way to divide the states 5, 6, 7 equally to the states 0, 1, 2, 3, and 4, we cannot use this bit string if it is over 4, and must acquire another one until we acquire a number under 4. Once we select a bitstring with a number below our least-common-multiple, we can use the value of the bit string to select our course of action.

The above selection method prevents us from having to make any rounding errors, and it shouldn't take that many bits to implement as any given bit string of the proper length always has over a 50% chance of working out. Other encoding schemes introduce rounding errors, which only detract from the uncertainty of our choice-worthiness calculations.

### What Does Application Look Like?

I think everyone with solid choice-worthy calibrating ability should have access to truly random bits to choose courses of action from.

Importantly, the time of the production of these random bits is relevant. A one-year-old random bitstring captured from radioactivity is just as random as one captured 5 seconds ago, but employing the latter is key for ensuring the maximum number of recent sister universes make different decisions.

Thus, people need access to recently created bit strings. These could be from a portable, personal Gieger counter, but it could also be from a centralized Gieger counter in say, the middle of the United States. The location does not matter as much as the recency of bit production. Importantly, however, bit strings should not ever be reused as this is not as random as using new bit strings as whatever made you decide to reuse them is non-random information.

### Can We Really Affect the Distribution of Other Worlds through Our Actions?

One may think that since everything is quantum mechanics including our brains, can we really affect the distribution of child worlds from our intentions and decisions? This raises the classic problem of free will and our place in a deterministic universe. I think the simplest question to ask is: do our choices have an effect on ethically-relevant phenomena? If the answer is no, then why should we care about decision theory in general? I think it's useful to think of the answer as yes.

### What if Many Worlds Isn't True?

If MWI isn't true, then RECMDT optimizes for worlds that will not exist at the potential cost to our own. This may seem to be incredibly dangerous and costly. However, as long as people make accurate choice-worthiness comparisons between different CoAs, then I will actually argue that adhering to RECMDT is not that risky. After all, choice-worthiness is distinct from expected-utility.

It would be a waste to have people, in a binary choice of actions with one having 9x more expected-utility than the other, choose the action with less expected-utility even 10% of the time. However, it seems best, even in a single unfolding of history, that where we are morally uncertain, we should actually cycle through actions based on our moral uncertainty via relative choice-worthiness.

By always acting to maximize choice-worthiness, we risk not capturing any value at all through our actions. While I agree that we should maximize expected-utility in both one shot and iterative scenarios alike and be risk neutral assuming we adequately defined our utility function, I think that given the fundamental uncertainty at play in a normative uncertainty assessment, it is risk neutral to probabilistically decide to implement different CoAs relative to their comparative choice-worthiness. Importantly, this is only the ideal method if the CoAs are mutually exclusive--if they are not, one might as well optimize for both moral frameworks.

Hence, while I think RECMDT is true, I also think that even if MWI is proven false, a decision theory exists which combines randomness and relative choice-worthiness. Perhaps we can call this Random Choice-worthy Decision Theory, or RCDT.

---------------------------------------------------------

Thanks for reading. Let me know what you think of this!

comment by Donald Hobson (donald-hobson) · 2019-05-04T11:37:05.143Z · score: 7 (4 votes) · LW · GW

I think that your reasoning here is substantially confused. FDT can handle reasoning about many versions of yourself, some of which might be duplicated, just fine. If your utility function is such that where . (and you don't intrinsically value looking at quantum randomness generators) then you won't make any decisions based on one.

If you would prefer the universe to be in than a logical bet between and . (Ie you get if the 3^^^3 th digit of is even, else ) Then flipping a quantum coin makes sense.

I don't think that randomized behavior is best described as a new decision theory, as opposed to an existing decision theory with odd preferences. I don't think we actually should randomize.

I also think that quantum randomness has a Lot of power over reality. There is already a very wide spread of worlds. So your attempts to spread it wider won't help.

comment by Alexei · 2019-05-05T01:14:20.206Z · score: 3 (2 votes) · LW · GW

If you would prefer the universe to be in ... If I was to make Evan's argument, that's the point I'd try to make.

My own intuition supporting Evan's line of argument comes from the investing world: it's much better to run a lot of uncorrelated positive EV strategies than a few really good ones, since the former reduces your volatility and drawdown, even while at the expense of EV measured in USD.

comment by Evan Ward · 2019-05-04T21:42:56.309Z · score: 1 (1 votes) · LW · GW

I'm sorry but I am not familiar with your notation. I am just interested in the idea: when an agent Amir is fundamentally uncertain about the ethical systems that he evaluates his actions by, is it better if all of his immediate child worlds make the same decision? Or should he hedge against his moral uncertainty, ensure his immediate child worlds choose courses of action that optimize for irreconcilable moral frameworks, and increase the probability that in a subset of his child worlds, his actions realize value?

It seems that in a growing market (worlds splitting at an exponential rate), it pays in the long term to diversify your portfolio (optimize locally for irreconcilable moral frameworks).

I agree that QM already creates a wide spread of worlds, but I don't think that means it's safe to put all of one's eggs in one basket when one has doubt that their moral system is fundamentally wrong.

comment by Donald Hobson (donald-hobson) · 2019-05-05T08:59:22.092Z · score: 1 (1 votes) · LW · GW

If you think that there is 51% chance that A is the correct morality, and 49% chance that B is, with no more information available, which is best.

Optimize A only.

Flip a quantum coin, Optimize A in one universe, B in another.

Optimize for a mixture of A and B within the same Universe. (Act like you had utility U=0.51A+0.49B) (I would do this one.)

If A and B are local objects (eg paperclips, staples) then flipping a quantum coin makes sense if you have a concave utility per object in both of them. If your utility is Then if you are the only potential source of staples or paperclips in the entire quantum multiverse, then the quantum coin or classical mix approaches are equally good. (Assuming that the resource to paperclip conversion rate is uniform. )

However, the assumption that the multiverse contains no other paperclips is probably false. Such an AI will run simulations to see which is rarer in the multiverse, and then make only that.

The talk about avoiding risk rather than expected utility maximization, and how your utility function is nonlinear, suggests this is a hackish attempt to avoid bad outcomes more strongly.

While this isn't a bad attempt at decision theory, I wouldn't want to turn on an ASI that was programmed with it. You are getting into the mathematically well specified, novel failure modes. Keep up the good work.

comment by Evan Ward · 2019-06-09T19:37:26.718Z · score: 1 (1 votes) · LW · GW

I really appreciate this comment and my idea definitely might come down trying to avoid risk rather than maximize expected utility. However, I still think there is something net positive about diversification. I write a better version of my post here: https://www.evanward.org/an-entropic-decision-procedure-for-many-worlds-living/ and if you could spare the time, I would love your feedback.

comment by Alexei · 2019-05-05T01:12:28.866Z · score: 3 (2 votes) · LW · GW

I'm actually very glad you wrote this up, because I have had a similar thought for a while now. And my intuition is roughly similar to yours. I wouldn't use terms like "decision theory," though, since around here that has very specific mathematical connotations. And while I do think my intuition on this topic is probably incorrect, it's not yet completely clear to me how.

comment by Evan Ward · 2019-06-09T19:32:33.446Z · score: 3 (2 votes) · LW · GW

I am glad you appreciated this! I'm sorry I didn't respond sooner. I think you are write about the term "decision theory" and have opted for "decision procedure" in my new, refined version of the idea at https://www.evanward.org/an-entropic-decision-procedure-for-many-worlds-living/

comment by Pattern · 2019-05-04T21:40:39.358Z · score: 1 (1 votes) · LW · GW

Intuitively, one does not want to take actions a and b with probabilities of 2/3 and 1/3, whenever the EU of a is twice that of b. Rather, it might be useful to not act entirely as utility estimates based on the uncertainty present - but if you are absolutely certain U(a) = 2*U(b), then it seems obvious one should take action a, if they are mutually exclusive. (If there is a 1/2 chance that U(a) = 1, and U(b) = 2, and a 1/2 chance that U(a) = 1, and U(b) = 1/2, then EU(a) = 1, and EU(b) = 1.5.)

comment by Evan Ward · 2019-06-09T19:34:40.715Z · score: 1 (1 votes) · LW · GW

I think you are right, but my idea applies more when one is uncertain about their expected utility estimates. I write a better version if my idea here https://www.evanward.org/an-entropic-decision-procedure-for-many-worlds-living/ and would love your feedback


text:
Confessions of a researchaholic

May 21, 2016

Another benefit of Latex

Filed under: Real — liyiwei @ 2:44 pm

The difficulty to express some emoticons will force the writing to appear more professional.

o$( \cap \_ \cap )$o
$\odot \omega \odot$
@\_@
=$\omega$=
$^-\bigtriangledown^-$

May 20, 2016

System surveys

Filed under: Real — liyiwei @ 11:41 am
Tags:

Few people have the time and patience to answer surveys, even when they do they might not provide genuine opinions (consciously or not), and the sampling is inherently biased.

A well designed system collects data and analyzes users on the go; just look at Google, Facebook, and Amazon.

May 19, 2016

encumbrance

Filed under: Real — liyiwei @ 5:31 pm

A new word I learned in the context of spending grant money.
🙂

For those who still email “newsletters”

Filed under: Real — liyiwei @ 3:17 pm

You are out of your fucking minds.

May 12, 2016

Assassin’s Creed

Filed under: Real — liyiwei @ 11:17 am

The premise of this game is about bringing back one’s ancestral memory over centuries, which means I should not play it at all.
🙂

May 11, 2016

Filed under: Real — liyiwei @ 9:47 am
Tags:

Hi Jun, Qi, and mamba:

You know what kinds of students I am looking for.

They trust your words more than mine.

And the fewer students I have, the more time I can spend on each of you.
🙂

Thanks!

May 10, 2016

Confidence is reflected at one’s worst

Filed under: Real — liyiwei @ 2:24 pm

(If you cannot see the content above, twitter is blocked.)

Managing emails

Filed under: Real — liyiwei @ 11:29 am

Sometimes when I received too many emails, I just deleted them all, and counted on that important ones will be resent again.
(I watch all svn/git commits and Moodle boards, so my research and teaching were never affected.)

When people asked why I did not reply their emails, I apologized with the above reason, so that they knew it is not personal.

This appeared to be working fine so far, but follow my strategy at your own risk.
🙂

May 9, 2016

Recursive interviews

Filed under: Real — liyiwei @ 6:20 pm
Tags: , ,

When you were interviewing the candidates I was also interviewing you.

You opened yourself like a book.
🙂

April 30, 2016

Strength in numbers

Filed under: Real — liyiwei @ 12:12 pm
Tags:

In his PhD application, a student indicated that he is collaborating with a famous professor on a recent SIGGRAPH project.

I asked that professor about the student. He told me that the project has 20 students (most are not co-authors), and the particular student is probably not even a secondary helper.

In comparison, most of my projects so far involve 0 or 1 student, and none over 2 students.
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## 机器学习代写|流形学习代写manifold data learning代考|Topological Spaces

Topological spaces were introduced by Maurice Fréchet (1906) (in the form of metric spaces), and the idea was developed and extended over the next few decades. Amongst those who contributed significantly to the subject was Felix Hausdorff, who in 1914 coined the phrase “topological space” using Johann Benedict Listing’s German word Topologie introduced in $1847 .$

A topological space $\mathcal{X}$ is a nonempty collection of subsets of $\mathcal{X}$ which contains the empty set, the space itself, and arbitrary unions and finite intersections of those sets. A topological space is often denoted by $(\mathcal{X}, \mathcal{T})$, where $\mathcal{T}$ represents the topology associated with $\mathcal{X}$. The elements of $\mathcal{T}$ are called the open sets of $\mathcal{X}$, and a set is closed if its complement is open. Topological spaces can also be characterized through the concept of neighborhood. If $\mathbf{x}$ is a point in a topological space $\mathcal{X}$, its neighborhood is a set that contains an open set that contains $\mathbf{x}$.
Let $\mathcal{X}$ and $\mathcal{Y}$ be two topological spaces, and let $U \subset \mathcal{X}$ and $V \subset \mathcal{Y}$ be open subsets. Consider the family of all cartesian products of the form $U \times V$. The topology formed from these products of open subsets is called the product topology for $\mathcal{X} \times \mathcal{Y}$. If $W \subset \mathcal{X} \times \mathcal{Y}$, then $W$ is open relative to the product topology iff for each point $(x, y) \in \mathcal{X} \times \mathcal{Y}$ there are open neighborhoods, $U$ of $x$ and $V$ of $y$, such that $U \times V \subset W$. For example, the usual topology for $d$-dimensional Euclidean space $\Re^{d}$ consists of all open sets of points in $\Re^{d}$, and this topology is equivalent to the product topology for the product of $d$ copies of $\Re$.

One of the core elements of manifold learning involves the idea of “embedding” one topological space inside another. Loosely speaking, the space $\mathcal{X}$ is said to be embedded in the space $\mathcal{Y}$ if the topological properties of $\mathcal{Y}$ when restricted to $\mathcal{X}$ are identical to the topological properties of $\mathcal{X}$. To be more specific, we state the following definitions. A function $g: \mathcal{X} \rightarrow \mathcal{Y}$ is said to be continuous if the inverse image of an open set in $\mathcal{Y}$ is an open set in $\mathcal{X}$. If $g$ is a bijective (i.e., one-to-one and onto) function such that $g$ and its inverse $g^{-1}$ are continuous, then $g$ is said to be a homeomorphism. Two topological spaces $\mathcal{X}$ and $\mathcal{Y}$ are said to be homeomorphic (or topologically equivalent) if there exists a homeomorphism from one space onto the other. A topological space $\mathcal{X}$ is said to be embedded in a topological space $\mathcal{Y}$ if $\mathcal{X}$ is homeomorphic to a subspace of $\mathcal{Y}$.

## 机器学习代写|流形学习代写manifold data learning代考|Riemannian Manifolds

In the entire theory of topological manifolds, there is no mention of the use of calculus. However, in a prototypical application of a “manifold,” calculus enters in the form of a “smooth” (or differentiable) manifold $\mathcal{M}$, also known as a Riemannian manifold; it is usually defined in differential geometry as a submanifold of some ambient (or surrounding) Euclidean space, where the concepts of length, curvature, and angle are preserved, and where smoothness relates to differentiability. The word manifold (in German, Mannigfaltigkeit) was coined in an “intuitive” way and without any precise definition by Georg Friedrich Bernhard Riemann (1826-1866) in his 1851 doctoral dissertation (Riemann, 1851; Dieudonné, 2009); in 1854, Riemann introduced in his famous Habilitations lecture the idea of a topological manifold on which one could carry out differential and integral calculus.

A topological manifold $\mathcal{M}$ is called a smooth (or differentiable) manifold if $\mathcal{M}$ is continuously differentiable to any order. All smooth manifolds are topological manifolds, but the reverse is not necessarily true. (Note: Authors often differ on the precise definition of a “smooth” manifold.)

We now define the analogue of a homeomorphism for a differentiable manifold. Consider two open sets, $U \in \Re^{r}$ and $V \in \Re^{s}$, and let $g: U \rightarrow V$ so that for $\mathbf{x} \in U$ and $\mathbf{y} \in V, g(\mathbf{x})=$ y. If the function $g$ has finite first-order partial derivatives, $\partial y_{j} / \partial x_{i}$, for all $i=1,2, \ldots, r$, and all $j=1,2, \ldots, s$, then $g$ is said to be a smooth (or differentiable) mapping on $U$. We also say that $g$ is a $\mathcal{C}^{1}$-function on $U$ if all the first-order partial derivatives are continuous. More generally, if $g$ has continuous higher-order partial derivatives, $\partial^{k_{1}+\cdots+k_{r}} y_{j} / \partial x_{1}^{k_{1}} \cdots \partial x_{r}^{k_{r}}$, for all $j=1,2, \ldots, s$ and all nonnegative integers $k_{1}, k_{2}, \ldots, k_{r}$ such that $k_{1}+k_{2}+\cdots+k_{r} \leq r$, then we say that $g$ is a $\mathcal{C}^{\top}$-function, $r=1,2, \ldots$. If $g$ is a $\mathcal{C}^{r}$-function for all $r \geq 1$, then we say that $g$ is a $\mathcal{C}^{\infty}$-function.

If $g$ is a homeomorphism from an open set $U$ to an open set $V$, then it is said to be a $\mathcal{C}^{r}$-diffeomorphism if $g$ and its inverse $g^{-1}$ are both $\mathcal{C}^{r}$-functions. A $\mathcal{C}^{\infty}$-diffeomorphism is simply referred to as a diffeomorphism. We say that $U$ and $V$ are diffeomorphic if there exists a diffeomorphism between them. These definitions extend in a straightforward way to manifolds. For example, if $\mathcal{X}$ and $\mathcal{Y}$ are both smooth manifolds, the function $g: \mathcal{X} \rightarrow \mathcal{Y}$ is a diffeomorphism if it is a homeomorphism from $\mathcal{X}$ to $\mathcal{Y}$ and both $g$ and $g^{-1}$ are smooth. Furthermore, $\mathcal{X}$ and $\mathcal{Y}$ are diffeomorphic if there exists a diffeomorphism between them, in which case, $\mathcal{X}$ and $\mathcal{Y}$ are essentially indistinguishable from each other.

## 机器学习代写|流形学习代写manifold data learning代考|Topological Spaces

Maurice Fréchet (1906) 引入了拓扑空间（以度量空间的形式），这个想法在接下来的几十年中得到发展和扩 展。对这个主题做出重大贡献的人中有 Felix Hausdorff，他在 1914 年使用 Johann Benedict Listing 的德语单词 Topologie 创造了“拓扑空间”一词。1847.

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## 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。


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ISL Ingénierie LJ Hydro Consultancy Experience and strong · PDF file 2018. 6. 21. · LJ Hydro Consultancy . have chosen to join forces to deliver high performance projet // EXAMPLE

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• LJ Hydro Consultancy Access Road No. 7 House No. 110; Unit 10, Ban Sailom, Chanthabury District, Vientiane, Lao PDR. PO Box: 4818 Phone: +856 21 265 064, 65, 66; Fax: +856 21 214 049 Web site : www.ljhc.asia

ISL Ingénierie - Head office 75 boulevard Mac Donald 75019 Paris FRANCE Phone: +33 1 55 26 99 99 Fax: +33 1 40 34 63 36 Web site : www.isl.fr

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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

ISL Ingénierie &

LJ Hydro Consultancy

Experience and strong partnership to achieve high quality project in Asia

• Some examples of projects realized with our partnership

// EXAMPLE 3 : Nam Ngiep - 1 HPP

Technical assistance for the construction works of a 167 m high RCC gravity dam, with an installed capacity of 240 MW.

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Contacts : ISL Ingénierie Jérôme Partiot, head of business development Mob: +33 6 77 90 98 80, [email protected]

LJ Hydro Consultancy Julian Chin, Commercial Manager Mob: + 856 20 7888 1222, [email protected]

ISL Ingénierie & LJ Hydro Consultancy have chosen to join forces to deliver

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Technical assistance for the construction works of a 177 m high RCC arch dam, with an installed capacity of 650 MW.

ISL Ingénierie is a French engineering company, specialised in dam engineering, hydropower and water resources development, including environmental impact assessment. ISL Ingénierie is entirely independent from any financial or industrial group. ISL’s worldwide experience counts numerous references on hydroelectric potential studies and hydroelectric developments, either on technical, environmental or regulatory aspects. It provides a complete project team which brings a rigorous multi-dimensional expertise.

LJ Hydro Consultancy is a hydro- power consulting company based in Vientiane, Laos. LJHC provides the complete suite of hydropower services at all levels of development. It promotes efficient open communications with stakeholders and establishes expert’s teams to provide optimization services, emergency maintenance and troubleshooting for projects of all installed capacity. Its team is made of highly professional people who approach work with integrity and always strive to add value by understanding their clients’ needs and complexity of the overall project.

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Technical assistance for the construction works of a run- of-river power plant, with an installed capacity of 2x15 MW inlcuding a clay core earthfill dam.

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# Oriented cover

## 1 Introduction

$\displaystyle w(\widetilde{X}) = w(X)\circ p_*,$
which corresponds to the intuition that the cover $\widetilde{X}$$\widetilde{X}$ is orientable if all loops in $\widetilde{X}$$\widetilde{X}$ project to orientable loops in $X$$X$.

Lemma 1.1. The space $\widetilde{X}$$\widetilde{X}$ is orientable if and only if the orientation character $w(X)$$w(X)$ factors through $\pi$$\pi$.

Proof. Consider the diagram

$\displaystyle \xymatrix{ \pi_1(\widetilde{X}) \ar[dr]^-{w(\widetilde{X})} \ar[d] ^-{p_*}& \\ \pi_1(X) \ar[d]^-{q}\ar[r]^{w(X)} & \Zz_2 \\ \pi \ar@{-->}[ur]^-{w} & }$

If there exists a $w$$w$ such that $w(X) = w\circ q$$w(X) = w\circ q$, then $w(\widetilde{X}) = w\circ q\circ p_* = 0$$w(\widetilde{X}) = w\circ q\circ p_* = 0$. Conversely if $w(\widetilde{X}) = 0$$w(\widetilde{X}) = 0$ then the map

$\displaystyle \begin{array}{rcl} w:\pi & \to & \Zz_2 \\ {[\alpha]} & \mapsto & w(X)(\alpha)\end{array}$

for any representative $\alpha\in \pi_1(X)$$\alpha\in \pi_1(X)$ of $[\alpha]\in \pi$$[\alpha]\in \pi$ is well defined and factors $w(X)$$w(X)$.

$\square$$\square$

In light of this we make the following definition.

## 2 Definition

An orientable cover $(\widetilde{X},\pi,w)$$(\widetilde{X},\pi,w)$ of a (connected) space $X$$X$ with an orientation character $w(X)\in H^1(X;\Zz_2) = \Hom(\pi_1(X),\Zz_2)$$w(X)\in H^1(X;\Zz_2) = \Hom(\pi_1(X),\Zz_2)$ is a regular covering of $X$$X$ with group of covering translations $\pi$$\pi$, together with an orientation character $w:\pi \to \Zz_2$$w:\pi \to \Zz_2$ such that

$\displaystyle \xymatrix{ w(X): \pi_1(X) \ar[r] & \pi \ar[r]^-{w} & \Zz_2. }$
An oriented cover is an orientable cover together with a choice of lift $\widetilde{b}$$\widetilde{b}$ of a basepoint $b\in X$$b\in X$ to $\widetilde{X}$$\widetilde{X}$.

## 3 Lifts correspond to orientations

Let $(\widetilde{M},\pi,w)$$(\widetilde{M},\pi,w)$ be an oriented cover of a connected manifold $M^m$$M^m$. A choice of lift $\widetilde{b}\in\widetilde{M}$$\widetilde{b}\in\widetilde{M}$ corresponds to a choice of $w$$w$-twisted fundamental class $[\widetilde{M}]\in H_m(M;\Z^w)$$[\widetilde{M}]\in H_m(M;\Z^w)$. Given a lift $\widetilde{b}$$\widetilde{b}$ a fundamental class $[\widetilde{M}]\in H_m(M;\Z^w)$$[\widetilde{M}]\in H_m(M;\Z^w)$ is uniquely determined by setting its restriction to $H_n(\widetilde{M},\widetilde{M}-\{\widetilde{b}\}; \Z)$$H_n(\widetilde{M},\widetilde{M}-\{\widetilde{b}\}; \Z)$ to be $1\in \pi$$1\in \pi$ and extending equivariantly. Conversely, given a fundamental class $[\widetilde{M}]\in H_m(M;\Z^w)$$[\widetilde{M}]\in H_m(M;\Z^w)$ define $\widetilde{b}$$\widetilde{b}$ to be the lift such that the restriction of $[\widetilde{M}]\in H_m(M;\Z^w)$$[\widetilde{M}]\in H_m(M;\Z^w)$ to $H_n(\widetilde{M},\widetilde{M}-\{\widetilde{b}\};\Z)$$H_n(\widetilde{M},\widetilde{M}-\{\widetilde{b}\};\Z)$ is $1\in \pi$$1\in \pi$.

## 4 Examples

The two most important examples of oriented covers are the universal cover $(\widetilde{X},\pi_1(X),w(X))$$(\widetilde{X},\pi_1(X),w(X))$ and the orientation double cover $(X^w,\Zz_2,\id_{\Zz_2})$$(X^w,\Zz_2,\id_{\Zz_2})$. These correspond to the two extreme cases of factoring the orientation character via $\pi_1(X)$$\pi_1(X)$ and $\Zz_2$$\Zz_2$ respectively. Every oriented cover is a regular cover of $X^w$$X^w$ and has $\widetilde{X}$$\widetilde{X}$ as a regular cover - this is a consequence of the fact that if $H_1, H_2$$H_1, H_2$ are normal subgroups of $G$$G$ with $H_1$$H_1$ a subgroup of $H_2$$H_2$ then $H_1$$H_1$ is a normal subgroup of $H_2$$H_2$. Thus for $(\overline{X},\pi,w)$$(\overline{X},\pi,w)$ any oriented cover of $X$$X$ the following diagram commutes:

$\displaystyle \xymatrix{ && \pi_1(\widetilde{X})=\{1\} \ar[dl] \ar[dd] && \\ & \pi_1(\overline{X})\ar[dr]^-{p_*} \ar[dl]&&&\\ \pi_1(X^w)\ar[rr] && \pi_1(X) \ar[dd]^{\id} \ar[dr]^-{q} \ar[rr]^{w(X)} && \Z_2\\ &&&\pi \ar[ur]^-{w} & \\ && \pi_1(X)\ar[ur]&& }$

## 5 Orientations of the orientation double cover

Let $M^m$$M^m$ be a connected manifold and let $(M^w,\Zz_2,\id_{\Zz_2})$$(M^w,\Zz_2,\id_{\Zz_2})$ be the orientation double cover. Following [Ranicki2002, Proposition 4.48] there is a short exact sequence of $\Z[\Z_2]$$\Z[\Z_2]$-modules

$\displaystyle \xymatrix@R=1mm{0 \ar[r] &\Z^{-} \ar[r] & \Z[\Z_2] \ar[r] & \Z \ar[r] &0 \\ & 1 \ar@{|->}[r] & 1-T, && \\ &&a+bT \ar@{|->}[r] & a+b. &}$

Applying $-\otimes_{\Z[\Z_2]} S(M^w)$$-\otimes_{\Z[\Z_2]} S(M^w)$ we obtain another short exact sequence which induces the following exact sequence in homology:

$\displaystyle \xymatrix{\ldots \ar[r] & H_{n+1}(M) \ar[r] & H_n(M;\Z^w) \ar[r] & H_n(M^w) \ar[r]^-{p_*} & H_n(M) \ar[r] & \ldots}$
As $M$$M$ is $m$$m$-dimensional we have that $H_{m+1}(M)=0$$H_{m+1}(M)=0$ and so by the long exact sequence
$\displaystyle H_m(M;\Z^w)\cong \ker(p_*:H_m(M^w)\to H_m(M)).$

In other words, a $w$$w$-twisted fundamental class of a connected manifold $M$$M$ can be thought of as a fundamental class of the orientation double cover that projects to zero in $H_m(M)$$H_m(M)$. In the case that the manifold $M$$M$ is already orientatable, the orientation double cover $(M^w,\Zz_2,\id_{\Zz_2})$$(M^w,\Zz_2,\id_{\Zz_2})$ consists of two disjoint copies of $M$$M$ and a $w$$w$-twisted orientation corresponds to an orientation of $M^w$$M^w$ where the two copies of $M$$M$ are given opposite orientations. In the case that $M$$M$ is non-orientable $H_m(M;\Z)=0$$H_m(M;\Z)=0$ so a $w$$w$-twisted orientation is precisely an orientation of the cover $M^w$$M^w$.


text:
# Scott topology, but for graphs

Hi,

Would it be possible to define an order theoretic topology on graphs? I am thinking about the scott topology. There would be an associated continuous map on graphs.

-
I suppose you could build the nerve of a graph and use that as a topology, but it won't be the same thing as the Scott topology in the case of the graph of some arbitrary partial order, for example. I guess the answer is that "yes, you can define many topologies", but the real question is which is the most useful one for your applications? Probably the closest notion to a continuous map for graphs would be a graph homomorphism, but you don't really need a topology to define that.. –  Mikola Jun 24 '11 at 19:15

In the theory of Borel equivalence relations under Borel reducibility (see this MO answer, also Greg Hjorth's excellent survey article), which is concerned with general questions surrounding the relative difficulty of isomorphism relations and other relations arising in mathematics, particularly of the difficulty of their classification problems---the theory ultimately arranges these relations into a complex hierarchy under Borel reducibility---the class of countable graphs $\Gamma=\langle\mathbb{N},E\rangle$ is considered by identifying the graph $\Gamma$ with its edge relation $E\in 2^{\mathbb{N}^2}$, and using the ordinary product topology, and is thereby realized as a standard Borel space, which enjoys various uniqueness properties. See for example page 2 of these notes by Simon Thomas and Scott Schneider. The basic open sets of this topology on the space of countable graphs are therefore determined by specifying finitely many edges and finitely many non-edges.
(Note, however, that for the purpose of the Borel reducibility theory, the principal focus is on the resulting $\sigma$-algebra of Borel sets, rather than on the topology of open sets.)
This comment is coming far too late, but I had a look at the Borel-reducibility notion and I think it is quite interesting. I might have been thinking about functors in terms of Borel reducibility. For instance, an equivalence relation in category theory is over morphisms like $fg=h$ A functor maps diagrams to diagrams, so $F: fg=h \rightarrow F(f)F(g)=F(h)$ I was trying to see functors as equivalence relations going into more or less "simple" (a term used to compare Borel Relations) equivalence relations. Joel's post is actually insightful as to my thinking. Thanks Joel. –  Ben Sprott Jul 13 '11 at 12:32


text:
# Do I really have to specify the width of my wrapfigure twice?

I'm trying to add a figure to my document. Because I want text to wrap around it, I use the wrapfigure environment:

\begin{wrapfigure}{r}{1cm}
\begin{center}
\includegraphics[width=1cm]{"My figure"}
\end{center}
\caption{A caption of my figure}
\end{wrapfigure}

As you can see, I want my figure to be shrunk down (the file is quite large) so I set the width to 1 cm. But now I must also tell the wrapfigure to be 1 cm wide. Why can't wrapfigure just automatically be as wide as the figure it contains?

-
With this you can put white space around the figure. –  Sigur May 17 at 0:57
@Sigur Yes, I realize that. However, I have many large documents with many figures, where I don't want any white space. Isn't there a way of saving myself the trouble of having to enter the width twice? It can be quite annoying when I'm trying out a few different sizes for a figure. –  Superbest May 17 at 1:01
You could define a custom environment using \newenvironment tweaked as you wish? –  cfr May 17 at 1:05

wrapfig actually doesn't care what it contains. As such, there is no connection between the image and the wrapfigure environment width. You can, however, set the width to \linewidth so that it fills the container it's in:

\documentclass{article}
\usepackage{wrapfig,graphicx}
\begin{document}

\begin{wrapfigure}{r}{3cm}
\includegraphics[width=\linewidth]{example-image}
\caption{A caption of my figure}
\end{wrapfigure}
Lorem ipsum dolor sit amet, consectetur adipiscing elit.
Maecenas luctus id tellus semper mattis. Integer quis neque
sem. Etiam semper, tortor in lacinia feugiat, nisi mi cursus
metus, et molestie felis nisl quis sapien. Fusce vel ligula
eget nunc gravida malesuada ac sit amet risus. Duis et urna
vestibulum, iaculis metus nec, placerat sem. Fusce suscipit
quis velit mattis fermentum. Curabitur odio urna, tempor ac
orci a, euismod elementum est. Mauris id mollis est, a consequat
purus. Mauris quis risus non ipsum accumsan elementum ut in
quam. Proin suscipit non augue dapibus tempus. Mauris
scelerisque, turpis et adipiscing interdum, neque nunc
tristique ipsum, a sollicitudin libero eros eget tellus. Donec
fringilla cursus diam, eget malesuada erat dapibus in. Aliquam
pellentesque vitae enim ut rutrum. In auctor nunc ac laoreet
molestie. Proin scelerisque convallis tortor, sit amet egestas
nulla accumsan ac. Donec urna justo, fringilla et vulputate eu,

\end{document}

Try also setting the default in the preamble:

\setkeys{Gin}{width = \linewidth}

-
Why not \usepackage{etoolbox} \AtBeginEnvironment{wrapfigure}{\setkeys{Gin}{width = \linewidth}} so inside wrapfigure there's no need to manually specify \linewidth as the width? –  Gonzalo Medina May 17 at 2:00

Or this, for example, defines a custom command which wraps your graphic in the necessary environments. It requires 3 arguments with a fourth optional.

\documentclass{article}
\usepackage{wrapfig,graphicx,kantlipsum}
\newcommand{\mywrapfigure}[4][r]{%
\begin{wrapfigure}{#1}{#2}
\centering
\includegraphics[width=#2]{#3}
\caption{#4}
\end{wrapfigure}}

\begin{document}
\mywrapfigure{25mm}{example-image-a}{A caption of my figure}
\kant[1-2]

\mywrapfigure[l]{30mm}{example-image-b}{A caption of my other figure}
\kant[3]
\end{document}

-
You are loosing two arguments for wrapfigure. –  Gonzalo Medina May 17 at 1:34
@GonzaloMedina It was only supposed to be an example of how to set up a custom command based on the example in the question. Usually when I write custom macros, I just define them to accommodate the arguments I need - I don't typically accommodate everything the commands/environments I use in the definition support. Perhaps this is bad policy but I prefer to think of it as a virtue of customisation! (It keeps things simpler and focuses on the particular features I need without requiring me to worry about the rest.) I never use wrapfig, though, so maybe I've missed an essential? –  cfr May 17 at 1:56
For that matter, I'm neglecting almost every feature \includegraphics offers... –  cfr May 17 at 1:57
+1 for the code, but I'll accept the other question since it is the most direct answer to my question as written. –  Superbest May 17 at 4:34


text:
# Mathematical and Computational Methods in Physics and Engineering (fall 2016)

## Daily outline for Mon, Sep 19, 2016 9:10 AM (1 year ago)

### Review

assigned today:

Due today:
• standard-complex functions:

I can manipulate complex functions.

1. Show how a complex function can produce a vector field.
2. Derive the Cauchy-Riemann equations.
3. Problem 13.4.25
• Quiz on problem 13.4.25. Gave $$u(x,y)=x^2-y^2$$ and asked for v.
• we talked about how plotting the equilines for v is easier than for u.
• They voted for what of the 6 complex problems are the hardest (really they ranked them). The winner seemed to be the $$\sqrt[n]{a+i b}$$ problem.
• We considered the Laplacian equation of harmonic functions and talked about the ramifications of them.
• I don't think this went well. I need to structure a better exercise for this. I thought about proving the mean value theorem for harmonic functions but didn't really think there was time.
• I asked them to prove that $$\sqrt[3]{a+i b}$$ has at most one real root.
• Some tackled it geometrically, but not all.
• we talked about how 1 has complex cube roots and how weird that was
• We also talked about how the trick (keep adding $$2\pi$$) has an end point. I asked if every believed that as soon as $$6\pi$$ gives you a repeat you can be convinced that there's no reason to go any further.
• I asked them to talk about what further resources they thought they might need.
• I found that it helped them focus by trying to figure out problems with tough changes I might make in the quiz.
• Ela asked about how to deal with it if I gave a map and asked for the complex function that produces it.
• I gave them two examples on the board to think about that: constant vectors up at 45 degrees and a upward field that got stronger as you get higher.

No new resources requested.

edit


text:
## Algebra 2 Common Core

Published by Prentice Hall

# Skills Handbook - Operations With Rational Expressions - Exercises - Page 984: 11

#### Answer

$$\frac{74x}{35}$$

#### Work Step by Step

A common denominator of $5$ and $7$ is $35$, so $$\frac{7x}{5}+\frac{5x}{7}=\bigg(\frac{7x}{5}\bigg)\bigg(\frac{7}{7}\bigg)+\bigg(\frac{5x}{7}\bigg)\bigg(\frac{5}{5}\bigg)=\frac{49x}{35}+\frac{25x}{35}=\frac{74x}{35}.$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
Kernel is Normal Subgroup of Domain

Jump to navigation Jump to search

Theorem

Let $\phi$ be a group homomorphism.

Then the kernel of $\phi$ is a normal subgroup of the domain of $\phi$:

$\map \ker \phi \lhd \Dom \phi$

Proof

Let $\phi: G_1 \to G_2$ be a group homomorphism, where the identities of $G_1$ and $G_2$ are $e_{G_1}$ and $e_{G_2}$ respectively.

$\map \ker \phi \le \Dom \phi$

Let $k \in \map \ker \phi, x \in G_1$.

Then:

 $\ds \map \phi {x k x^{-1} }$ $=$ $\ds \map \phi x \map \phi k \paren {\map \phi x}^{-1}$ Homomorphism to Group Preserves Inverses $\ds$ $=$ $\ds \map \phi x e_{G_2} \paren {\map \phi x}^{-1}$ Definition of Kernel of Group Homomorphism $\ds$ $=$ $\ds \map \phi x \paren {\map \phi x}^{-1}$ Definition of Identity Element $\ds$ $=$ $\ds e_{G_2}$ Definition of Inverse Element

So:

$x k x^{-1} \in \map \ker \phi$

This is true for all $k \in \map \ker \phi$ and $x \in G_1$.

From Subgroup is Normal iff Contains Conjugate Elements, it follows that $\map \ker \phi$ is a normal subgroup of $G_1$.

$\blacksquare$


text:
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# Use the Midpoint Rule with the given value of $n$ to approximate the integral. Round the answer to four decimal places.$\displaystyle \int^1_0 \sqrt{x^3 + 1}\, dx$, $n = 5$

## 1.1097

Integrals

Integration

### Discussion

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##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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### Video Transcript

Yeah, yeah. So five dot to problem number one. So we're asked to use Riemann sums to estimate this? Uh integral. So our estimate is going to be using five rectangles. So in the interval From 0 to 1, If you divide that into five, everyone is going to be .2. So point to 0.4 0.6 .8. And then one. Now we're also asked to use mid points. So what are the midpoint is going to be? The mid points are going to be .1 0.3 0.5 0.7 and 0.9. So if I were to estimate this with rectangles, I'm going to use the function value at the midpoint. The width of each rectangle is still going to be .2. So my estimate Ah for this integral. So the integral from 0 to 1 Cube root, x cubed plus one dx is going to be estimated by the area, which is the sum of these five rectangles. So what is the width of each rectangle? So the width of each rectangle Is going to be what one? So I got the width is one divided by five. So the width of each rectangle is 1/5. Yeah And then the height of each rectangle is going to be evaluating at the midpoint. So you're going to get .1 Q plus one. And let me just do it sort of this way plus the square root of 0.3 cubed plus one Plus the square root of .5 cubed Plus one plus the square root of .7 cubed plus one Plus the square root of .9 cubed plus one. So all of that turns out to be our estimate. Now there are easier ways of doing this with a calculator than doing all of that out. Let's try that. So let's just create my function is the square root. So the square root no. Of X cubed plus one. And I'm going to store that my function F of X. And so now what I need to do is just create the sequence of all of those mid points. Or I could just on it just sum them all up. I could just say, hey, the area is going to be, What is it? 1/5, so 1 5th Times F of .1. Yeah. Plus F point too. Excuse me, F A 0.1. And then fo 0.3 Plus F 5.5. Yeah, half of mhm. Seven F 0.9. So 1/5 Uh f of one F 2 F three f of five. F seven F. Of nine. So 12345 is the case there. And so what you see there is 1.109 6 7. So this area estimate Is about 1.109. Let me make sure right, that right? 1.109 67 Okay, so that is the estimate of this integral.

Florida State University

#### Topics

Integrals

Integration

##### Catherine R.

Missouri State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Michael J.

Idaho State University

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## G = C3×C4⋊Q8order 96 = 25·3

### Direct product of C3 and C4⋊Q8

direct product, metabelian, nilpotent (class 2), monomial, 2-elementary

Series: Derived Chief Lower central Upper central

 Derived series C1 — C22 — C3×C4⋊Q8
 Chief series C1 — C2 — C22 — C2×C6 — C2×C12 — C6×Q8 — C3×C4⋊Q8
 Lower central C1 — C22 — C3×C4⋊Q8
 Upper central C1 — C2×C6 — C3×C4⋊Q8

Generators and relations for C3×C4⋊Q8
G = < a,b,c,d | a3=b4=c4=1, d2=c2, ab=ba, ac=ca, ad=da, bc=cb, dbd-1=b-1, dcd-1=c-1 >

Subgroups: 84 in 68 conjugacy classes, 52 normal (12 characteristic)
C1, C2, C2 [×2], C3, C4 [×6], C4 [×4], C22, C6, C6 [×2], C2×C4, C2×C4 [×6], Q8 [×4], C12 [×6], C12 [×4], C2×C6, C42, C4⋊C4 [×4], C2×Q8 [×2], C2×C12, C2×C12 [×6], C3×Q8 [×4], C4⋊Q8, C4×C12, C3×C4⋊C4 [×4], C6×Q8 [×2], C3×C4⋊Q8
Quotients: C1, C2 [×7], C3, C22 [×7], C6 [×7], D4 [×2], Q8 [×4], C23, C2×C6 [×7], C2×D4, C2×Q8 [×2], C3×D4 [×2], C3×Q8 [×4], C22×C6, C4⋊Q8, C6×D4, C6×Q8 [×2], C3×C4⋊Q8

Smallest permutation representation of C3×C4⋊Q8
Regular action on 96 points
Generators in S96
(1 26 14)(2 27 15)(3 28 16)(4 25 13)(5 34 42)(6 35 43)(7 36 44)(8 33 41)(9 63 46)(10 64 47)(11 61 48)(12 62 45)(17 22 30)(18 23 31)(19 24 32)(20 21 29)(37 94 60)(38 95 57)(39 96 58)(40 93 59)(49 84 70)(50 81 71)(51 82 72)(52 83 69)(53 87 91)(54 88 92)(55 85 89)(56 86 90)(65 76 80)(66 73 77)(67 74 78)(68 75 79)
(1 2 3 4)(5 6 7 8)(9 10 11 12)(13 14 15 16)(17 18 19 20)(21 22 23 24)(25 26 27 28)(29 30 31 32)(33 34 35 36)(37 38 39 40)(41 42 43 44)(45 46 47 48)(49 50 51 52)(53 54 55 56)(57 58 59 60)(61 62 63 64)(65 66 67 68)(69 70 71 72)(73 74 75 76)(77 78 79 80)(81 82 83 84)(85 86 87 88)(89 90 91 92)(93 94 95 96)
(1 32 12 34)(2 29 9 35)(3 30 10 36)(4 31 11 33)(5 14 24 45)(6 15 21 46)(7 16 22 47)(8 13 23 48)(17 64 44 28)(18 61 41 25)(19 62 42 26)(20 63 43 27)(37 82 77 89)(38 83 78 90)(39 84 79 91)(40 81 80 92)(49 75 87 58)(50 76 88 59)(51 73 85 60)(52 74 86 57)(53 96 70 68)(54 93 71 65)(55 94 72 66)(56 95 69 67)
(1 57 12 74)(2 60 9 73)(3 59 10 76)(4 58 11 75)(5 69 24 56)(6 72 21 55)(7 71 22 54)(8 70 23 53)(13 96 48 68)(14 95 45 67)(15 94 46 66)(16 93 47 65)(17 92 44 81)(18 91 41 84)(19 90 42 83)(20 89 43 82)(25 39 61 79)(26 38 62 78)(27 37 63 77)(28 40 64 80)(29 85 35 51)(30 88 36 50)(31 87 33 49)(32 86 34 52)

G:=sub<Sym(96)| (1,26,14)(2,27,15)(3,28,16)(4,25,13)(5,34,42)(6,35,43)(7,36,44)(8,33,41)(9,63,46)(10,64,47)(11,61,48)(12,62,45)(17,22,30)(18,23,31)(19,24,32)(20,21,29)(37,94,60)(38,95,57)(39,96,58)(40,93,59)(49,84,70)(50,81,71)(51,82,72)(52,83,69)(53,87,91)(54,88,92)(55,85,89)(56,86,90)(65,76,80)(66,73,77)(67,74,78)(68,75,79), (1,2,3,4)(5,6,7,8)(9,10,11,12)(13,14,15,16)(17,18,19,20)(21,22,23,24)(25,26,27,28)(29,30,31,32)(33,34,35,36)(37,38,39,40)(41,42,43,44)(45,46,47,48)(49,50,51,52)(53,54,55,56)(57,58,59,60)(61,62,63,64)(65,66,67,68)(69,70,71,72)(73,74,75,76)(77,78,79,80)(81,82,83,84)(85,86,87,88)(89,90,91,92)(93,94,95,96), (1,32,12,34)(2,29,9,35)(3,30,10,36)(4,31,11,33)(5,14,24,45)(6,15,21,46)(7,16,22,47)(8,13,23,48)(17,64,44,28)(18,61,41,25)(19,62,42,26)(20,63,43,27)(37,82,77,89)(38,83,78,90)(39,84,79,91)(40,81,80,92)(49,75,87,58)(50,76,88,59)(51,73,85,60)(52,74,86,57)(53,96,70,68)(54,93,71,65)(55,94,72,66)(56,95,69,67), (1,57,12,74)(2,60,9,73)(3,59,10,76)(4,58,11,75)(5,69,24,56)(6,72,21,55)(7,71,22,54)(8,70,23,53)(13,96,48,68)(14,95,45,67)(15,94,46,66)(16,93,47,65)(17,92,44,81)(18,91,41,84)(19,90,42,83)(20,89,43,82)(25,39,61,79)(26,38,62,78)(27,37,63,77)(28,40,64,80)(29,85,35,51)(30,88,36,50)(31,87,33,49)(32,86,34,52)>;

G:=Group( (1,26,14)(2,27,15)(3,28,16)(4,25,13)(5,34,42)(6,35,43)(7,36,44)(8,33,41)(9,63,46)(10,64,47)(11,61,48)(12,62,45)(17,22,30)(18,23,31)(19,24,32)(20,21,29)(37,94,60)(38,95,57)(39,96,58)(40,93,59)(49,84,70)(50,81,71)(51,82,72)(52,83,69)(53,87,91)(54,88,92)(55,85,89)(56,86,90)(65,76,80)(66,73,77)(67,74,78)(68,75,79), (1,2,3,4)(5,6,7,8)(9,10,11,12)(13,14,15,16)(17,18,19,20)(21,22,23,24)(25,26,27,28)(29,30,31,32)(33,34,35,36)(37,38,39,40)(41,42,43,44)(45,46,47,48)(49,50,51,52)(53,54,55,56)(57,58,59,60)(61,62,63,64)(65,66,67,68)(69,70,71,72)(73,74,75,76)(77,78,79,80)(81,82,83,84)(85,86,87,88)(89,90,91,92)(93,94,95,96), (1,32,12,34)(2,29,9,35)(3,30,10,36)(4,31,11,33)(5,14,24,45)(6,15,21,46)(7,16,22,47)(8,13,23,48)(17,64,44,28)(18,61,41,25)(19,62,42,26)(20,63,43,27)(37,82,77,89)(38,83,78,90)(39,84,79,91)(40,81,80,92)(49,75,87,58)(50,76,88,59)(51,73,85,60)(52,74,86,57)(53,96,70,68)(54,93,71,65)(55,94,72,66)(56,95,69,67), (1,57,12,74)(2,60,9,73)(3,59,10,76)(4,58,11,75)(5,69,24,56)(6,72,21,55)(7,71,22,54)(8,70,23,53)(13,96,48,68)(14,95,45,67)(15,94,46,66)(16,93,47,65)(17,92,44,81)(18,91,41,84)(19,90,42,83)(20,89,43,82)(25,39,61,79)(26,38,62,78)(27,37,63,77)(28,40,64,80)(29,85,35,51)(30,88,36,50)(31,87,33,49)(32,86,34,52) );

G=PermutationGroup([(1,26,14),(2,27,15),(3,28,16),(4,25,13),(5,34,42),(6,35,43),(7,36,44),(8,33,41),(9,63,46),(10,64,47),(11,61,48),(12,62,45),(17,22,30),(18,23,31),(19,24,32),(20,21,29),(37,94,60),(38,95,57),(39,96,58),(40,93,59),(49,84,70),(50,81,71),(51,82,72),(52,83,69),(53,87,91),(54,88,92),(55,85,89),(56,86,90),(65,76,80),(66,73,77),(67,74,78),(68,75,79)], [(1,2,3,4),(5,6,7,8),(9,10,11,12),(13,14,15,16),(17,18,19,20),(21,22,23,24),(25,26,27,28),(29,30,31,32),(33,34,35,36),(37,38,39,40),(41,42,43,44),(45,46,47,48),(49,50,51,52),(53,54,55,56),(57,58,59,60),(61,62,63,64),(65,66,67,68),(69,70,71,72),(73,74,75,76),(77,78,79,80),(81,82,83,84),(85,86,87,88),(89,90,91,92),(93,94,95,96)], [(1,32,12,34),(2,29,9,35),(3,30,10,36),(4,31,11,33),(5,14,24,45),(6,15,21,46),(7,16,22,47),(8,13,23,48),(17,64,44,28),(18,61,41,25),(19,62,42,26),(20,63,43,27),(37,82,77,89),(38,83,78,90),(39,84,79,91),(40,81,80,92),(49,75,87,58),(50,76,88,59),(51,73,85,60),(52,74,86,57),(53,96,70,68),(54,93,71,65),(55,94,72,66),(56,95,69,67)], [(1,57,12,74),(2,60,9,73),(3,59,10,76),(4,58,11,75),(5,69,24,56),(6,72,21,55),(7,71,22,54),(8,70,23,53),(13,96,48,68),(14,95,45,67),(15,94,46,66),(16,93,47,65),(17,92,44,81),(18,91,41,84),(19,90,42,83),(20,89,43,82),(25,39,61,79),(26,38,62,78),(27,37,63,77),(28,40,64,80),(29,85,35,51),(30,88,36,50),(31,87,33,49),(32,86,34,52)])

42 conjugacy classes

 class 1 2A 2B 2C 3A 3B 4A ··· 4F 4G 4H 4I 4J 6A ··· 6F 12A ··· 12L 12M ··· 12T order 1 2 2 2 3 3 4 ··· 4 4 4 4 4 6 ··· 6 12 ··· 12 12 ··· 12 size 1 1 1 1 1 1 2 ··· 2 4 4 4 4 1 ··· 1 2 ··· 2 4 ··· 4

42 irreducible representations

 dim 1 1 1 1 1 1 1 1 2 2 2 2 type + + + + + - image C1 C2 C2 C2 C3 C6 C6 C6 D4 Q8 C3×D4 C3×Q8 kernel C3×C4⋊Q8 C4×C12 C3×C4⋊C4 C6×Q8 C4⋊Q8 C42 C4⋊C4 C2×Q8 C12 C12 C4 C4 # reps 1 1 4 2 2 2 8 4 2 4 4 8

Matrix representation of C3×C4⋊Q8 in GL5(𝔽13)

 9 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1
,
 1 0 0 0 0 0 12 2 0 0 0 12 1 0 0 0 0 0 12 0 0 0 0 0 12
,
 1 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 12 0 0 0 1 0
,
 1 0 0 0 0 0 12 2 0 0 0 0 1 0 0 0 0 0 0 5 0 0 0 5 0

G:=sub<GL(5,GF(13))| [9,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1],[1,0,0,0,0,0,12,12,0,0,0,2,1,0,0,0,0,0,12,0,0,0,0,0,12],[1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,12,0],[1,0,0,0,0,0,12,0,0,0,0,2,1,0,0,0,0,0,0,5,0,0,0,5,0] >;

C3×C4⋊Q8 in GAP, Magma, Sage, TeX

C_3\times C_4\rtimes Q_8
% in TeX

G:=Group("C3xC4:Q8");
// GroupNames label

G:=SmallGroup(96,175);
// by ID

G=gap.SmallGroup(96,175);
# by ID

G:=PCGroup([6,-2,-2,-2,-3,-2,-2,144,313,151,938,230]);
// Polycyclic

G:=Group<a,b,c,d|a^3=b^4=c^4=1,d^2=c^2,a*b=b*a,a*c=c*a,a*d=d*a,b*c=c*b,d*b*d^-1=b^-1,d*c*d^-1=c^-1>;
// generators/relations

׿
×
𝔽


text:
# Question #6856d

Mar 3, 2017

The volume of ${\text{Ca(OH)}}_{2}$ solution will be 8.33 mL.

#### Explanation:

The equation for the reaction is

$\text{2HCl + Ca(OH)"_2 → "CaCl"_2 + 2"H"_2"O}$.

1. Calculate the moles of $\text{HCl}$

$\text{Moles of HCl" = "0.025 00" color(red)(cancel(color(black)("L HCl"))) ×( "0.500 mol HCl")/(1 color(red)(cancel(color(black)("L HCl")))) = "0. 012 50 mol HCl}$

2. Calculate the moles of ${\text{Ca(OH)}}_{2}$

$\text{Moles of Ca(OH)"_2 = "0.012 50" color(red)(cancel(color(black)("mol HCl"))) × ("1 mol Ca(OH)"_2)/(2 color(red)(cancel(color(black)("mol HCl")))) = "0.006 250 mol Ca(OH)"_2}$

3. Calculate the molarity of the ${\text{Ca(OH)}}_{2}$

$\text{Volume of Ca(OH)"_2 = "0.006 250" color(red)(cancel(color(black)("mol Ca(OH)"_2))) × ("1 L Ca(OH)"_2)/(0.750 color(red)(cancel(color(black)("mol Ca(OH)"_2)))) = "0.008 33 L" = "8.33 mL}$


text:
# automaton.synchronize¶

Create a new transducer, equivalent to the first one, but where the transitions advance along all the tapes at the same rate, for as long as possible. The transitions before the final states may have the empty word on one of the tapes, to allow for words of different lengths on the input and output.

Preconditions:

• The input automaton is a transducer.
• Input.has_bounded_lag

Postconditions:

• Result.is_synchronized

## Examples¶

In [1]:
import vcsn
ctx = vcsn.context("lat<law_char, law_char>, b")
ctx

Out[1]:
$\{\ldots\}^* \times \{\ldots\}^*\to\mathbb{B}$

The following automaton is not synchronized (the first transition is already not synchronized):

In [2]:
a = ctx.expression(r"(abc|\e)(d|v)*(\e|wxyz)").standard()
a

Out[2]:

The lag is bounded, because every cycle (here, the loop) has a lag of 0.

In [3]:
a.has_bounded_lag()

Out[3]:
True

Apart from pure spontaneous transitions, the only transitions with $\varepsilon$ in them are right before the final state.

In [4]:
s = a.synchronize()
s

Out[4]:
In [5]:
s.is_synchronized()

Out[5]:
True

This is a (for the most part) letter-to-letter transducer equivalent to the input.

In [6]:
s.proper().letterize().minimize().strip()

Out[6]:


text:
# How do I allow user to edit title of Folder in Document Library?

How can I allow a user to edit the Title of folder separately from the folder's name? I've done it by accident, messing with the content types.

I found that I could add a custom content type that inherits from folder, but then there is the drop-down that lets you pick which type of folder. I'd like to keep it simple. Also, I need to programmatically deploy such a change.

• Have you checked the Permission on the folder? – user2536 Jan 24 '12 at 22:33

## 1 Answer

If you enable management of content types on your list folder should show up as an option (Assuming WSS3 like in your tag - this no longer works in 2010) From there you can add fields to your folder or in the case of title, make it required or optional instead of hidden.

All of these settings should be accessible through the API as well.

• SPList.ContentTypesEnabled should allow you to enable management of content types.
• SPList.ContentTypes["Folder"].FieldLinks["Title"].Hidden should allow you to flip hidden to false

I have not tested this but theoretically it should work. Let me know how it goes.

• To finish the code::> lib = GetLib(web, libname); lib.ContentTypes["Folder"].FieldLinks["Title"].Hidden=false; lib.ContentTypes["Folder"].Update(); – oglester Jan 25 '12 at 14:57
• Yeah that update call would be important. Good catch there. Glad I could help. – SpartanDonut Jan 25 '12 at 15:01


text:
# The fatigue stress concentration factor is defined as

The ratio of endurance limit with stress concentration to the endurance limit without stress concentration

The ratio of endurance limit without stress concentration to the endurance limit with stress concentration

The product of the endurance limits with and without stress concentration

All of the above

Please do not use chat terms. Example: avoid using "grt" instead of "great".


text:
# 6.3 Equations of Line of Best Fit

A set of two variables are related non linearly can be converted to a linear equation.  The line of best fit can be written in the form

Y = mX + c

where
X and Y are in terms of x and/or y
m is the gradient,
c is the Y-intercept

Recall: To find equation of straight line
(1)  Equation of a straight line if gradient (m) and one points $\left({x}_{1},{y}_{1}\right)$ are given:
$Y − y 1 = m ( X − x 1 )$
(2)  Equation of a straight line if gradient (m) and y-intercept (c) are given:
$Y = m X + c$


text:
other querymodes : Identifierquery Coordinatequery Criteriaquery Referencequery Basicquery Scriptsubmission TAP Outputoptions Help

2006ApJS..165..512K - Astrophys. J., Suppl. Ser., 165, 512-550 (2006/August-0)

Morphological analysis of H I features. II. Wavelet-based multifractal formalism.

KHALIL A., JONCAS G., NEKKA F., KESTENER P. and ARNEODO A.

Abstract (from CDS):

The 2D wavelet transform modulus maxima method is used to obtain quantitative information on the fractal/multifractal nature and anisotropic structure of Galactic atomic hydrogen (H I) from the Canadian Galactic Plane Survey. Five mosaics were analyzed in the second quadrant of the Milky Way, corresponding to the Local, Perseus, and Outer spiral arms, as well as two interarm'' regions located between these three spiral arms. A monofractal signature is found for all five mosaics. An anisotropic signature is detected: the rms roughness fluctuations of the mosaics in the latitude direction differ from those in the longitude direction. This anisotropy is scale-independent for the interarm regions while it is scale-dependent for the spiral arms. The longitudinal matter distribution of H I structure is similar for all five mosaics while the latitudinal distribution is smoother in the spiral arms. These results hold for all physical length scales studied, from ∼2 pc in the Local arm to ∼44 pc in the Outer arm. Several hypotheses are investigated to provide a physical explanation: the scale-height gradient, the density wave, star formation activity, photolevitation of dusty clouds, random motion of H I clouds, corrugation, and turbulence. The slopes of the power spectra for an increasing number of velocity channels were compared for 11 sections of the Local arm column density mosaic. All slopes are identical within the uncertainties (-3.0) and we do not detect for the Galactic plane the change in the power law index predicted by Lazarian and Pogosyan.


text:
# Fakultät für Physik (inkl. GAUSS)

## Browse by

A measurement of the $W$ boson polarisation in top quark decays is presented. The top anti-top pairs were produced at a centre of mass energy of $\sqrt{s}=8$ \tev \ using proton-proton collisions at the LHC. The data sample ...


text:
10 ml of a mixture of CH_4, C_2H_4 and CO_2 were exploded with excess of air . After explosion , there was contra

### Question Asked by a Student from EXXAMM.com Team

Q 1210856710.     10 ml of a mixture of CH_4, C_2H_4 and CO_2 were exploded with excess of air . After explosion , there was contraction on cooling of 17 ml and after treatment with KOH, there was further reduction of 14ml . What is the composition of the mixture?
A

a = 4.5, b = 4  c=1.5

B

a = 4, b = 4.5  c=1.5

C

a = 5, b = 4  c=1

D

None of these

#### HINT

(Provided By a Student and Checked/Corrected by EXXAMM.com Team)

#### Access free resources including

• 100% free video lectures with detailed notes and examples
• Previous Year Papers
• Mock Tests
• Practices question categorized in topics and 4 levels with detailed solutions
• Syllabus & Pattern Analysis


text:
1. ## matrix inversion

Find the values of k for which A is invertible.

How would i go on about solving this question??

thanks

2. Originally Posted by b00yeah05

Find the values of k for which A is invertible.

How would i go on about solving this question??

thanks
a matrix is invertible if and only if it's determinant is non-zero. so the easiest way, perhaps, is to find the expression that gives the determinant of the matrix and then pick appropriate values for k

3. Originally Posted by b00yeah05

Find the values of k for which A is invertible.

How would i go on about solving this question??

thanks
k= all real numbers EXCEPT for the one(s) that make the determinant equal to zero.

To get the determinant, make life easy for yourself by first doing the row op R1 --> R1 - R3. Then expand along the top row.

By inspection you should see that k = 2 will make the determinant equal to zero since you'll have a row of identical entries.

4. well...yes i see that if its a 2 the det = 0 which means that there is no inverse.
so does that mean....k = all real numbers except 2? is that the answer???

5. Originally Posted by b00yeah05
well...yes i see that if its a 2 the det = 0 which means that there is no inverse.
so does that mean....k = all real numbers except 2? is that the answer???
i didn't check, but yes, if k = 2 makes the determinant zero, that is the answer

6. if i was to go on to find the inverse of the matrix..would i just substitute any value into k apart from 2? and find the inverse?

7. Originally Posted by b00yeah05
if i was to go on to find the inverse of the matrix..would i just substitute any value into k apart from 2? and find the inverse?
it's not that kind of a question. you are to find k for which the matrix has an inverse, if you were to find the inverse as it is now, you'd have k's in it. don't worry about that, just know that you can find the inverse as long as k is anything but 2

8. yes, thanks i understand that but the question leading on to that is solving linear systems. of :

(i) x - y - 2z = 7
2x - 2y - 2z = 2
x - y + 2z = 13

and

(ii) x - y = 5
2x = 3
x - y +2z = 1

i have to find the solutions to that. if i find the inverse of the matrix above cant i just use matrices to solve these linear euqation as it is x = A-1*b

9. Originally Posted by b00yeah05
yes, thanks i understand that but the question leading on to that is solving linear systems. of :

(i) x - y - 2z = 7
2x - 2y - 2z = 2
x - y + 2z = 13

and

(ii) x - y = 5
2x = 3
x - y +2z = 1

i have to find the solutions to that. if i find the inverse of the matrix above cant i just use matrices to solve these linear euqation as it is x = A-1*b
A is not invertible in the first question, since k = 2. so don't even bother with that one (no solution)

for the second, k = 0, so just use the method you mentioned (if you must, there are actually many ways to solve a system), to find the solution

10. sorry to be a nusense but why is k=0 for the second one?

11. Originally Posted by b00yeah05
sorry to be a nusense but why is k=0 for the second one?
do you know how do find the matrix A? you get it from the coefficients of the variables in order

for the first question: $A = \begin{bmatrix} 1 & -1 & -2 \\ 2 & -2 & -2 \\ 1 & -1 & 2 \end{bmatrix} \implies k = -2$

ah, my bad, k = -2 not 2

for the second: $A = \begin{bmatrix} 1 & -1 & 0 \\ 2 & 0 & 0 \\ 1 & -1 & 2 \end{bmatrix} \implies k = 0$

12. Originally Posted by mr fantastic
k= all real numbers EXCEPT for the one(s) that make the determinant equal to zero.

To get the determinant, make life easy for yourself by first doing the row op R1 --> R1 - R3. Then expand along the top row.

By inspection you should see that k = 2 will make the determinant equal to zero since you'll have a row of identical entries.
I didn't mean that "By inspection ...." bit to imply k = 2 was the only solution that made the determinant equal to zero. I put it in to point out that at least one of the values k cannot be is obvious.

Originally Posted by b00yeah05
well...yes i see that if its a 2 the det = 0 which means that there is no inverse.
so does that mean....k = all real numbers except 2? is that the answer???
Did you work out the determinant and confirm that k = 2 is the only solution that solves det(A) = 0?? If you had done so instead of cutting corners (and I even suggested how to make life easier - to the point of almost trivial - in getting the determinant) you would have found that there is another value of k.

I don't want to sound rude - please take the following on board as a wake up call:

The questions you've been asking suggest to me that you haven't made much of an effort to get on top of this material. And to not even bother working out the determinant to get the other value of k - that suggests something to. This service is not here to do your assignment work - it's expected you'll do work too. Especially when you're given clear guidance on what to do.

13. yes i worked out the det(A) when k=2 it is 0. which means that at k=2, the matrix has no inverse. but i do not know how to find the value of k for which the matrix is invertible...i sort of understand what jhevon is saying but i dont know how to do it with the co-efficients

14. Originally Posted by b00yeah05
yes i worked out the det(A) when k=2 it is 0. which means that at k=2, the matrix has no inverse. but i do not know how to find the value of k for which the matrix is invertible...
the matrix is invertible for ANY value of k that is not 2. any real number, of course.

i sort of understand what jhevon is saying but i dont know how to do it with the co-efficients
look at the matrices i gave you and match them up to the matrix you posted in your first post. do you realize the matrices are identical except for the k's? that's all i did. i realize i can get one of my matrices by replacing k with -2 in the original, and i can get the other by replacing k with 0 in the original.

15. yes i realise that they are identical..but what i meant was why did you use -2 and 0. isnt it possible for you to use any value..for eg....5 or 10 since k= all real numbers except 2.

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