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http://mathhelpforum.com/calculus/138149-calculating-normal-acceleration.html	# Thread: 1. ## Calculating Normal Acceleration Hi guys, I have a 3D curve that is defined by a parameterization in the form of x(t), y(t), and z(t). In the context of the problem the curve is a waterslide and I need to calculate the normal acceleration at various points along the curve. I can calculate the velocity at any point by a simple conservation of energy from the top of the slide to the point of interest, but what do I need to do to calculate normal acceleration? 2. You have x(t), y(t), z(t). Velocity vector v = (dx/dt,dy/dt,dz/dt). Acceleration vector a=dv/dt = $(\frac{d^2x}{dt^2},\frac{d^2y}{dt^2},\frac{d^2z}{d t^2})$. v direction tangential to the curve. Find vector $\tau$ = v/\|v\| \|n\|=1 tangential to the curve. Tangential acceleration $a_t=a\tau$ Normal acceleration ${a_n}^2=a^2-{a_t}^2$. 3. Thank you for helping, but I must admit that I do not totally get it yet. Perhaps someone could help me with a simple example and then I will be able to apply it to my more complicated problem. Imagine that my curve is defined by the simple parametrization: x(t)=sin(t) y(t)=cos(t) z(t)=.01t I am interested in the point where t=90 degrees, which represents one specific point on the curve. Based on the conservation of energy and the height at that point, I know that the rider will be traveling at 10m/s. How do I calculate his normal acceleration? 4. $r(t) = ( \sin t, \cos t, 0.01t )$ $r'(t) = \frac{dr(t)}{dt} = ( \cos t, -\sin t, 0.01 )$ $a(t) = r''(t) = \frac{d^2r(t)}{dt^2} = ( -\sin t, -\cos t, 0 )$ $T = \frac{r'(t)}{\|r'(t)\|}$ - unit vector tangential to the curve $N = \frac{T'}{\|T'\|}$ - unit vector normal to the curve $a(t) = Ta_T + Na_N$ where $a_T = T\cdot a(t) = \frac{\|r'(t) \cdot a(t)\|}{\|r'(t)\|}$ $\|r'(t)\| = \sqrt{1.01}$. $r'(t) \cdot a(t) = ( \cos t, -\sin t, 0.01 )( -\sin t, -\cos t, 0 ) = 0$ there is no tangetial component only normal, thus $a(t) = Na_N = ( -\sin t, -\cos t, 0 )\: and\: \|a(t)\| = 1$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 14, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9076278805732727, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/100753/how-to-show-that-a-measure-is-countably-additive	# How to show that a measure is countably additive. Define $$\alpha(E)=\sup\{\mu(F)-\nu(F)\,\|\,F\subset E~\text{and}~\nu(F)<\infty\}.$$ where $\mu$ and $\nu$ are measures on some $\sigma$-algebra. How do I show that $\alpha$ is countably additive? I've already got one direction. That is $\sum \alpha(E_n)\geq \alpha(\cup E_n).$ I am however struggling with the other direction: showing that $\sum\alpha(E_n)\leq \alpha(\cup E_n)$. thanks. - I guess the sets $E_n$ must be disjoint. Then, for every positive $\varepsilon$, consider $F_n\subset E_n$ such that $\mu(F_n)-\nu(F_n)\geqslant\alpha(E_n)-\varepsilon/2^n$ and their union $F$. – Did Jan 20 '12 at 13:24 2 @Didier: Good start, but of course, it might be that $\nu(F)=\infty$. This objection is not hard to work around, but I thought it worth mentioning. – Harald Hanche-Olsen Jan 20 '12 at 13:38 1 @Harald: Of course, comment $\ne$ answer. – Did Jan 20 '12 at 13:56	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9448731541633606, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/182784-prove-equation-involving-square-roots.html	# Thread: 1. ## Prove an Equation involving Square Roots Proof $<br /> \frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^2-1} <br />$ fankyou. 2. Originally Posted by BabyMilo Proof $<br /> \frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^-1} <br />$ fankyou. I believe that there is a horrible typo in the question: it should be $\frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } =\sqrt{a^2-1}$ I don't know if there is a shorthand, but I would consider writing the LHS as: $\frac{1}{2}\sqrt{\frac{(a+1)(a-1)^2}{a-1}} +\frac{1}{2}\sqrt{\frac{(a-1)(a+1)^2}{a+1}}$ Take it from here. Edit: I see you've corrected the typo in the OP 3. I was trying to correct it, but my internet is so slow. 4. Hello, BabyMilo! Another approach . . . $\text{Prove: }\;\frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } \;=\; \sqrt{a^2-1}$ We have: . $\frac{a-1}{2}\cdot\frac{(a+1)^{\frac{1}{2}}}{(a-1)^{\frac{1}{2}}} + \frac{a+1}{2}\cdot\frac{(a-1)^{\frac{1}{2}}}{(a+1)^{\frac{1}{2}}}$ . . . . . . . $=\;\frac{(a-1)^{\frac{1}{2}}(a+1)^{\frac{1}{2}}}{2} + \frac{(a+1)^{\frac{1}{2}}(a-1)^{\frac{1}{2}}}{2}$ . . . . . . . $=\;\frac{[(a-1)(a+1)]^{\frac{1}{2}}}{2} + \frac{[(a+1)(a-1)]^{\frac{1}{2}}}{2}$ . . . . . . . $=\;\frac{(a^2-1)^{\frac{1}{2}}}{2} + \frac{(a^2-1)^{\frac{1}{2}}}{2}$ . . . . . . . $=\;(a^2-1)^{\frac{1}{2}}$ . . . . . . . $=\;\sqrt{a^2-1}$ 5. ## Equation involving square roots. Originally Posted by BabyMilo Proof $<br /> \frac{a-1}{2 } \sqrt{\frac{a+1}{a-1 } } +\frac{a+1}{2 } \sqrt{\frac{a-1}{a+1 } } = \sqrt{a^2-1} <br />$ fankyou. L.H.S:- $\frac{a-1}{2} \sqrt{\frac{a+1}{a-1}} + \frac{a+1}{2}\sqrt{\frac{a-1}{a+1}}$ $= \sqrt{\frac{(a+1)(a-1)(a-1)}{(a-1)4} } + \sqrt{\frac{(a-1)(a+1)(a+1)}{(a+1)4}}$ $= \sqrt{\frac{a^2-1}{4}} + \sqrt{\frac{a^2-1}{4}} = 2 \times \frac{1}{ 2}\sqrt{a^2-1}$ $= \sqrt{a^2 - 1}$ L.H.S. = R.H.S I hope you have understood. All the best.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 16, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8915137648582458, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/5834?sort=oldest	## Deconvolution of gamma distributions ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If the sum of two independent random variables is gamma distributed does this imply that the individual random variables are also gamma distributed. I suspect that the answer is no, but I do not know how to construct a counter-example. [The convolution of two gamma distributions with the same scale parameter is also a gamma distribution, but I am dealing with the converse.] If this can be answered assuming that the convolution is the exponential distribution f(x) = exp(-x) that would be sufficient for my purposes. Any help with this would be much appreciated. - ## 4 Answers No, take X=0 with probability 1. Y gamma X+Y - Thanks Stella. I should have stated that in the problem I am dealing with the component random variables are non-trivial, well behaved, continuous, positive. Does this make a difference? Or put slightly differently, are there reasonable properties I could specify for the component rvs that would cause the proposition to be true? – Trevor Stewart Nov 17 2009 at 17:11 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Stella gives the example of X = 0 with probability 1 and Y a gamma distribution. I wouldn't count this as a "true" counterexample, because one could think of X = 0 as a Gamma(k,0) random variable. As for a counterexample: the characteristic function of the exponential(1) distribution is 1/(1-it). The characteristic function of Gamma(k, θ) is (1-itθ)-k. So the question is whether you can write 1/(1-it) as the product of two things which are characteristic functions of positive measures (i. e. random variables) other than in the obvious way. - Michael, thanks. This is essentially the way I have been trying to tackle the problem, so far without success. – Trevor Stewart Nov 17 2009 at 17:18 The exponential (with mean 1) can be written as the sum of 2 independent chi-squared 1s. Of course this is just allowing fractional values of the "n" parameter. - Jonathan, Yes this is true and my application contemplates fractional shape parameters. But my question is the converse: given that the sum of the two rvs is gamma are the two components necessarily also gamma. – Trevor Stewart Nov 17 2009 at 20:23 You can decompose the exponential distribution into a sum of two terms, which are not both gamma distributed. Let A,B,ε be independent where A,B are exponentially distributed and ε takes the values 0,1 each with probability 1/2, and set X=A/2, Y=εB. You can calculate the moment generating functions of X and Y, $$E\left[\exp(-\lambda X)\right] = E\left[\exp(-(\lambda/2)A)\right]=1/(1+\lambda/2).$$ $$E\left[\exp(-\lambda Y)\right]=(1/2)E\left[\exp(-\lambda B)\right]+1/2=(2+\lambda)/(2+2\lambda).$$ Then you can check the moment generating function function of X+Y, E[exp(-λ(X+Y)]=E[exp(-λX)]E[exp(-λY)]=1/(1+λ) to see that X+Y has the exponential distribution. Edit: After reading at Michael Lugo's response below, it might be more satisfying to have an answer where neither of X or Y are Gamma distributed. In fact, by iterating my argument above you can get the following example. If A1,A2,... have the exponential distribution and ε1,ε2,... take values 0,1 each with probability 1/2 (and all these rvs are independent), then X=∑n21-nεnAn has the exponential distribution (just check the moment generating function). By splitting this sum up into two smaller sums you can generate a whole load of counterexamples where neither term is gamma distributed. Edit 2: Apologies for keeping coming back to this one, but it seems interesting and my examples above are a special case of the following. For any k>0 and measurable subset A of the interval (0,1], you can define a random variable XA with moment generating function E[exp(-λXA)]=exp(-λk∫Adx/(1+λx)). If you partition (0,1] into two measurable sets A,B and XA,XB are independent, then XA+XB has the Gamma(k) distribution. If A and B are unions of finitely many intervals then the moment generating functions will be kth powers of rational functions of λ and its easy to make sure that XA,XB are not gamma distributed. My first example above is using k=1 and the partition (0,1/2],(1/2,1]. The second one, in the edit, is partitioning (0,1] into the intervals (2-n,21-n]. You can construct XA as follows. Let T1,T2,… be independent with the Exp(k) distribution, and Sn=exp(-T1-…-Tn). The number of Sn in a subset A of (0,1] will be Poisson with parameter ∫Adx/x. If Y1,Y2,… are independent exponentially distributed then XA=∑n1{Sn∈A}SnYn has the correct moment generating function. (I'll leave you work through the details...). Alternatively, the set {(Sn,Yn):n≥1} is a Poisson point process with intensity ke-y dy ds/s. - 1 Some background on where my example came from: I know that the local time at 0 of a Brownian motion B at the first time it hits 1 has the exponential distribution. If you understand these concepts, then it is easy to see that it is the sum of the local time at 0 at which B first hits 1/2 plus the local time at 0 of the BM started at 1/2 when it first hits 1. These are independent, and the second has a prob of 1/2 of being 0, so can't be gamma distributed. I just converted this example into a simple argument using moment generating functions. – George Lowther Nov 17 2009 at 22:59 Actually, A is gamma-distributed here, with k = 1. But B isn't, so this is still a counterexample. – Michael Lugo Nov 17 2009 at 23:01 Thanks, Michael, I fixed my post. I suppose you could further decompose X in a similar way to get a sum of as many independent rvs as you like, and them rearrange them into two terms neither of which are gamma distributed. – George Lowther Nov 17 2009 at 23:05 Further to my comment above, the more general example in my second edit can also be understood in terms of local times of Brownian motion. X_A is the local time at 0 of a Brownian motion B while its maximum process B*(t)=max_{s<=t}B(s) is in A. – George Lowther Nov 18 2009 at 2:08 Thanks, George. I'll work through this. Looks like an ingenious solution. – Trevor Stewart Nov 18 2009 at 12:53	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.900858998298645, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3783970	Physics Forums Page 1 of 2 1 2 > Blog Entries: 1 ## Microeconomics: What is the socially optimum price? It is said in microeconomics that if there is a drought and you're the only one with a few spare water bottles, and you're feeling generous, you shouldn't just hand them out to people on a first-come, first-served basis. That's because you don't know how badly each person needs water, and if you want to maximize social welfare you'll want to give the water to the people who value the water more rather than those that value it less. So the solution to this problem is to sell the water bottles for a price. That way, people will automatically choose to buy the water if the water is worth more to them than the price, and they will choose not to buy the water if the price is worth more to them than the water. So you've classified people based on how much the water is worth to them, and you've given water to those for whom it is worth the most. Microeconomic theory says that you should sell the water for a price even if you're not interested in making money; if you really don't care about the money you should just donate the sales proceeds to charity, but the imposition of a price increases social welfare. My question is, how is the right price determined? I know in the case where you have self-interested buyers and sellers, then the optimum price is the intersection of the supply and demand curves, i.e. the market clearing price. But how would you do it in this case, where the buyers are all self-interested, but the seller doesn't care about himself at all and just wants to help society? The problem is, given a bunch of buyers, each with their own demand curve for water (let's assume for simplicity that they're all linear, but have different slopes), what is the right price to charge them, given a fixed supply of water bottles? Any help would be greatly appreciated. Thank You in Advance. PhysOrg.com social sciences news on PhysOrg.com >> Evolution of lying>> Exhaustive computer research project shows shift in English language>> US white majority to linger if immigration slows Microeconomic theory says that you should sell the water for a price even if you're not interested in making money; if you really don't care about the money you should just donate the sales proceeds to charity, but the imposition of a price increases social welfare. My question is, how is the right price determined? I know in the case where you have self-interested buyers and sellers, then the optimum price is the intersection of the supply and demand curves, i.e. the market clearing price. But how would you do it in this case, where the buyers are all self-interested, but the seller doesn't care about himself at all and just wants to help society? The problem is, given a bunch of buyers, each with their own demand curve for water (let's assume for simplicity that they're all linear, but have different slopes), what is the right price to charge them, given a fixed supply of water bottles Microeconomic theory says that,,,,, Really. I guess someone should tell all the people running the internet and that all the free stuff must now have a charge because social welfare is not being optimized, especially all those free programming people behind Linux and related software. Optimum price for who? the seller or the buyer. The right price. Again that is vague terminology. If someone has no money at all, the "right" price would be \$0.00, if he/she desperately needs the product, in this case water - don't you agree. Someone with loads of money, would not necessarily worry about the price per say, but would nevertheless certainly still feel taken advantage of by price gouging. If you are referring to the market still being in play for someone who wants to do something to benefit society, than he/she would set the price taking into account prevailing market conditions, with referral back to your first paragrah why this is so. At least that is my opinion of the right price. Yes, most of the time, just giving something away or selling at an unreasonable low price does dimminish the value of a product, although with the internet "free" for products such as Facebook, searching, chatting, ... , does not seem to have that affect at all. Just threw that in to explain my first paragraph. Not very vigourous a response, so with luck someone else will climb aboard to answer your question. Blog Entries: 1 Quote by 256bits Really. I guess someone should tell all the people running the internet and that all the free stuff must now have a charge because social welfare is not being optimized, especially all those free programming people behind Linux and related software. Software is different, because one person using Linux does not prevent others from also using it. I was talking about situations where you have a limited supply of some good. Quote by 256bits Optimum price for who? the seller or the buyer. My question was about a seller who is charging a price not to make money, but to optimize social welfare of the people asking for water. Quote by 256bits The right price. Again that is vague terminology. If someone has no money at all, the "right" price would be \$0.00, if he/she desperately needs the product, in this case water - don't you agree. Someone with loads of money, would not necessarily worry about the price per say, but would nevertheless certainly still feel taken advantage of by price gouging. The issue you're talking about is what is known technically as the difference between willingness to pay versus willingness to accept. If a poor man dying of thirst has a water bottle, he may not be willing to sell it for a hundred bucks. Yet if he didn't have it, he wouldn't be willing to buy it for a hundred bucks, because he doesn't have this much money. To avoid this issue, it is assumed that we are dealing with a situation where willingness to pay and willingness to accept are approximately equal. Equivalently, we are assuming that the benefit of getting the water bottle and the cost of not getting it are both relatively small. Quote by 256bits If you are referring to the market still being in play for someone who wants to do something to benefit society, than he/she would set the price taking into account prevailing market conditions, with referral back to your first paragrah why this is so. At least that is my opinion of the right price. No, the market price would be the price that maximizes social welfare of buyers and sellers. I'm talking about an altruistic seller, who is a monopoly, trying to maximize the social welfare of buyers. Quote by 256bits Yes, most of the time, just giving something away or selling at an unreasonable low price does dimminish the value of a product That's a completely different issue. I'm talking about how an unreasonably low price tends to give too much of the limited supply to those who don't really value the product that much. Quote by 256bits although with the internet "free" for products such as Facebook, searching, chatting, ... , does not seem to have that affect at all. Just threw that in to explain my first paragraph. As I said, on the Internet lots of people can use the same product without hindering others, except maybe bandwidth. ## Microeconomics: What is the socially optimum price? Hey Lugita remember me from my questions in the calculus/analysis subforums?? In economics, we do not care about how much a person needs something. If he cannot afford it, his need is meaningless. He only matters if he can afford it at a given price. This is assuming the good has no externalities, which is uncertain in the case of water. If the good has externalities, you must take into account taxes and subsidies, which I won't consider here. To answer your question, even though the seller is not spending the money on himself, as long as at least someone is benefiting from the money, and none of the money is being wasted in the process, (there is zero cost of transporting and distributing the charity money), then it does not matter whether the seller spends the money on himself or some random orphan. In both cases, the producer surplus is still the same. If the money will be used, it does not matter who uses it. Whoever uses it is inevitably a part of "society". If that were not the case, then it is a different story. Otherwise, we will assume that "society" includes the orphans who will receive the charity money. We can treat the problem as an ordinary one assuming that the seller is in fact motivated by self-interest. Your goal is to find the price at which the total surplus is maximized. The total surplus at quantity Q is given by the area between the supply and demand curves, i.e. $\int^{Q}_{0}(D-S) dQ$ It is evident from the graph that this is achieved when supply equals demand, i.e. equilibrium quantity and price. Therefore, the "optimum" price is the one at which supply and demand are at equilibrium. Another way of looking at it is just replace the seller with the charity organization that is receiving the money. That way, you look at "self-interest" as being perfectly present in the problem. Regards, BiP Blog Entries: 1 OK Bipolarity, let me clarify my scenario. The seller is not taking into account the welfare of the orphans, he just wants to maximize the social welfare of those who want his water. And the seller just has a fixed supply he wants to get rid of, not a supply curves. So my question is, given a bunch of buyers, each with their own (known) demand curve, what is the price that optimizes the social welfare of the buyers? Quote by lugita15 OK Bipolarity, let me clarify my scenario. The seller is not taking into account the welfare of the orphans, he just wants to maximize the social welfare of those who want his water. And the seller just has a fixed supply he wants to get rid of, not a supply curves. So my question is, given a bunch of buyers, each with their own (known) demand curve, what is the price that optimizes the social welfare of the buyers? First of all, if the water supply is fixed, just make the supply curve perfectly vertical. It doesn't really change the idea of the problem whether the supply is fixed or not. It just affects the slope of the supply curve. If the seller's money is unimportant, then we can assume that the seller burns all his money. In this case, the producer surplus is always 0, no matter what. The total surplus can be simplified from $\int^{Q}_{0}(D-S) dQ$ to $\int^{Q}_{0} D dQ$ This is maximized when the price is \$0.00 Thus, if the seller's money has no use for society, then the optimum price is \$0.00 Optimally, the water would be "sold" to the people "willing" to pay most for them. In reality the only way to do this is through charging a price. I suppose what you could do is auction the water. The person who is willing to pay most for the water will receive the water, but he will get his money back as well as the water. This way you guarantee the water is actually going to the people who are willing to pay most for it, and yet you aren't charging them anything for it. But this transaction would have to be done out of sight of the other buyers, who may become angry if they find out what is going on. This would be a negative externality, which I won't consider here. You could also give away the water on a first-come first-serve basis. You can think of the early buyers as "willling to pay more" for the water. This argument is not that strong though. But in the end, what matters is that the price must be $0.00. A price of$0.00 guarantees greater consumer surplus than any other price. What may vary is who "buys" the water, which is irrelevant if you are asking only for the optimum price. EDIT: When you talk about supply and demand, you are actually talking about aggregate supply and demand, which is the sum of the supply and demand of each individual seller/buyer respectively. So whether each buyer has his own demand curve makes no difference to the problem. You care only about the aggregate demand at a given price. The individual demands add up to give the aggregate demand. BiP Blog Entries: 1 Let me rephrase my question as follows. Suppose we have N water bottles, and two people who want water, each with their own demand curve for water. Using these individual demand curves, how would you calculate the quantities Q1 and Q2 of water bottles that should be allocated to each person in order to maximize social welfare? And under what conditions does there exist a price P such that (Q1,P) and (Q2,P) are points on the respective demand curves of the two buyers? And finally, how would you generalize this to arbitrary numbers of buyers? Quote by lugita15 Let me rephrase my question as follows. Suppose we have N water bottles, and two people who want water, each with their own demand curve for water. Using these individual demand curves, how would you calculate the quantities Q1 and Q2 of water bottles that should be allocated to each person in order to maximize social welfare? And under what conditions does there exist a price P such that (Q1,P) and (Q2,P) are points on the respective demand curves of the two buyers? And finally, how would you generalize this to arbitrary numbers of buyers? How did you obtain the water bottles - did you pay a price? Blog Entries: 1 Quote by WhoWee How did you obtain the water bottles - did you pay a price? You just happen to have a fixed supply of water bottles, and no ability to get any more. You can assume they fell from the sky or something. Recognitions: Homework Help Science Advisor I feel like this scenario depends a lot on your assumptions. How do you define social welfare? One can cook up all sorts of scenarios where the answer is anything you want. Quote by Physics Monkey I feel like this scenario depends a lot on your assumptions. How do you define social welfare? One can cook up all sorts of scenarios where the answer is anything you want. For the purposes of answering his question, we can assume that social welfare is the same as the total surplus. That's how most microeconomics textbooks define social welfare, assuming no externalities. So it reduces to an optimization problem, whereby our goal is to maximize the total surplus. Terms in economics are generally not defined as vaguely as one might think; most of them are mathematically defined, so we need not argue over definitions. His problem is not really about subjective definitions or anything. It is just complicated mathematically because so many things have to be considered, and first principles come into question. Lugita, I am working on a solution as we speak. I think it may require a bit more mathematics than I had envisioned, so allow me to take my time. What I can assure you is that the price charged to EVERY buyer is definitely \$0.00 no matter the number of bottles or the demand for them. Figuring out the quantities is a little more complicated though, but I'll get back to you on it. BiP I think I have a solution but it assumes that all the demand curves are linear, which is only an approximation of the true general solution. Let's suppose there are N water bottles and M buyers, each with their own individual demand curves. The demand curves for the M buyers can be notated as follows: $$D_{1} = c_{1} - m_{1}P$$ $$D_{2} = c_{2} - m_{2}P$$ $$D_{3} = c_{3} - m_{3}P$$ $$...$$ $$D_{M} = c_{M} - m_{M}P$$ The price charged to all these buyers will be \$0 because any price higher than that results in reduced consumer surplus, as explained in my previous posts. We have to decide on the quantity. Who will receive the N water bottles? Well, when the price is zero, the demand for each of the following buyers can easily be solved by plugging in $P = 0$: Quantity Demanded by each particular buyer when the price is zero: $$D_{1} = c_{1}$$ $$D_{2} = c_{2}$$ $$D_{3} = c_{3}$$ $$...$$ $$D_{M} = c_{M}$$ Thus, the respective quantities that will be sold to each buyer is $(c_{1},c_{2},c_{3}...c_{M})$. This is basically the x-intercept of each buyer's demand curve. This is only possible if $\sum c < N$. If $\sum c > N$, then we do not have enough bottles to feed everyone and so the bottles must be rationed more wisely. I will work on a solution for that case. Blog Entries: 1 Quote by Physics Monkey I feel like this scenario depends a lot on your assumptions. How do you define social welfare? One can cook up all sorts of scenarios where the answer is anything you want. . In economics, welfare is defined as utility measured in dollars (or any currency). If you'd be willing to buy something at $9.99, but willing to sell it at$10.01, then we say that the benefit you get from it is worth \$10 in welfare. Social welfare is defined to be the sum of everyone's individual welfare. Why do economists care about social welfare? Because it is a measure of economic efficiency. A situation is called economically inefficient if someone can be made better off without making anyone worse off, and it's called efficient if all opportunities to mae people better off without hurting anyone have already been taken advantage of. In order to achieve that, we have to maximize social welfare. IF it so happens that If $\sum c > N$, then we must ration the bottles to the buyers willing to pay most for the water (though we will not charge them anything). This we can do through auctioning and looking for the highest bidder. Once we have depleted all our water, we will return the money to them, effectively making the price $0.00 But the actual quantities that each buyer receives can be mathematically derived. For each buyer, we calculate his consumer surplus by integrating the whole area under his demand curve: $$Surplus \ of \ 1st \ buyer = \int^{c_{1}}_{0}D_{1}\ dQ$$ $$Surplus \ of \ 2nd \ buyer = \int^{c_{2}}_{0}D_{2}\ dQ$$ $$Surplus \ of \ 3rd \ buyer = \int^{c_{3}}_{0}D_{3}\ dQ$$ $$...$$ $$Surplus \ of \ last \ buyer = \int^{c_{M}}_{0}D_{M}\ dQ$$ We start by "selling" the bottles for$0.00 to those buyers for whom the surplus is highest, and then continue until we have either run out of bottles, or run out of buyers. In reality, we will not know the demand curves for anybody. We must simply auction them and then refund the money (to prevent confusion, this must be done after all the bottles/buyers have been depleted). I think for now this solution is adequate, but I wonder what we would do for a demand curve that has a more general shape. BiP Blog Entries: 1 Quote by Bipolarity IF it so happens that If $\sum c > N$, then we must ration the bottles to the buyers willing to pay most for the water (though we will not charge them anything). This we can do through auctioning and looking for the highest bidder. Once we have depleted all our water, we will return the money to them, effectively making the price $0.00 But the actual quantities that each buyer receives can be mathematically derived. For each buyer, we calculate his consumer surplus by integrating the whole area under his demand curve: $$Surplus \ of \ 1st \ buyer = \int^{c_{1}}_{0}D_{1}\ dQ$$ $$Surplus \ of \ 2nd \ buyer = \int^{c_{2}}_{0}D_{2}\ dQ$$ $$Surplus \ of \ 3rd \ buyer = \int^{c_{3}}_{0}D_{3}\ dQ$$ $$...$$ $$Surplus \ of \ last \ buyer = \int^{c_{M}}_{0}D_{M}\ dQ$$ We start by "selling" the bottles for$0.00 to those buyers for whom the surplus is highest, and then continue until we have either run out of bottles, or run out of buyers. In reality, we will not know the demand curves for anybody. We must simply auction them and then refund the money (to prevent confusion, this must be done after all the bottles/buyers have been depleted). I think for now this solution is adequate, but I wonder what we would do for a demand curve that has a more general shape. BiP Thanks for all this. I think we can fairly assume that everyone demands all the water bottles at a price of 0 dollars. Once we have integrated the surpluses by integrating each demand curve with respect to quantity, how do we calculate the quantities we should give each person? Is the quantity we should give linearly proportional to the surplus? Or is the relationship more complicated? Recognitions: Homework Help Science Advisor I understand the very narrow definitions in classical microeconomics texts. It just seemed to me that the original question might have been after a more interesting answer. After all, these kinds of scenarios are often used to suggest the serious inadequacies of very simplistic microeconomic models. I made a small mistake regarding the case where you don't have enough bottles to satisfy all the buyers. If you have enough bottles, the problem is very easy. Just calculate the x-intercept for each buyer, and sell each buyer that many bottles. If you don't have enough bottles, which is likely the case in a drought, the problem becomes quite complex. I think you would have to employ computational techniques to solve it. Mere calculus/algebra won't be sufficient, or at least I think a solution would take me (or anyone) a good while to find. You would have to have a program that calculates which buyer is willing to pay the most for the next bottle. You would then sell a single bottle to him. Then the amount he's willing to pay would decrease due to the law of demand. The program would then repeat the process until all the bottles have been sold or until all the buyers have been satisfied. I can't think of any non-computational solution out of the top of my head. BiP Page 1 of 2 1 2 > Thread Tools \| \| \| \| \|--------------------------------------------------------------------------\|-------------------------------------\|---------\| \| Similar Threads for: Microeconomics: What is the socially optimum price? \| \| \| \| Thread \| Forum \| Replies \| \| \| Social Sciences \| 6 \| \| \| Biology, Chemistry & Other Homework \| 0 \| \| \| Relationships \| 69 \| \| \| General Discussion \| 21 \| \| \| General Discussion \| 3 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 20, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9665237069129944, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/19415/yet-another-question-on-the-lindhard-function/24591	# Yet another question on the Lindhard function Here's another question concerning the Lindhard function as used in the physical description of metals. First we define the general Lindhard function in the Random Phase approximation as $\chi(q,\omega)=\frac{\chi_{0}}{1-\frac{4\pi e^2}{q^2}\chi_{0}}$ where $\chi_{0}$ denotes the "bare" Lindhard function" excluding any feedback effects of electrons. A reference is Kittel's book "Quantum Theory of solids" Chapter 6, however he is using the dielectric function instead of the Lindhard function. By writing $\chi_{0}(q,\omega)=\chi_{01}+i\chi_{02}$ as a complex number and using the relation $\frac{1}{z-i\eta} = P(1/z) + i\pi\delta(z)$ in the limit that $\eta$ goes to zero from above and P(..) denotes the principal value it is straight forward to see that the imaginary part of the lindhard function is zero except of the case when we have electron-hole excitations of the form $\hbar\omega=\epsilon_{k+q}-\epsilon_{k}$. Now let's assume we study plasma excitations using the actual Lindhard function (not the bare one). These plasma excitations appear for small q outside of the electron hole continuum. Therefor we have a vanishing imaginary part of $\chi_{0}$ in this case and consequently since also a vanishing part of $\chi$ itself, since its a function of $\chi_{0}$ which is "real" except of the "$\chi_{0}$"-part. However when deriving the plasma excitations using a small q expansion we obtain a none vanishing imaginary part of $\chi$ as given in equation (3.48) of the following document: http://www.itp.phys.ethz.ch/education/lectures_fs10/Solid/Notes.pdf I don't see why this is consistent. I'd be more than happy if you could help me. Thanks in advance. - The relation $z = \epsilon_{k+q} - \epsilon_{k} - \hbar\omega$ does not have to occur between an occupied and an unoccupied state, it can occur between two occupied states (or two unoccupied states). It is the density terms in the numerator that make occupied-occupied or unoccupied-unoccupied contributions zero - not sure if that helps... – Brendan Jan 18 '12 at 23:32 Your definition of the polarizability function was wrong, now it is fixed. – DaniH Apr 29 '12 at 16:36 ## 1 Answer There is no inconsistency. Indeed, when using the full polarizability in the random phase approximation $\chi^{\text{RPA}}(\mathbf{q},\omega)=\dfrac{\chi^{\text{RPA}}_0(\mathbf{q},\omega)}{1-V(q)\chi^{\text{RPA}}_0(\mathbf{q},\omega)}$ you see that the imaginary part of the polarizability has been renormalized when taken into account the [renormalized] electron-electron interactions because $V(q)$ is no longer the bare Coulomb interaction. Note that, from the previous expression, if we set $\text{Im}\,\chi_0(\mathbf{q},\omega)=-i\delta$ $-\text{Im} \,\chi(\mathbf{q},\omega)=\pi \delta(1-V(q)\text{Re}\,\chi_0(\mathbf{q},\omega))$ The fact that the imaginary part of the polarizability function $\text{Im} \,\chi(\mathbf{q},\omega)$ is non-zero is related to the damping of the excitations and it is known as Landau damping. It is possible to understand this linking $\text{Im} \,\chi(\mathbf{q},\omega)$ to the conductivity of the electron gas. Using the definition of the conductivity and the continuity equation we can show that $e^2 \text{Im} \,\chi(\mathbf{q},\omega) = -\dfrac{1}{\omega} \mathbf{q}\cdot[\text{Re}\,\bar\sigma(\mathbf{q},\omega)]\cdot \mathbf{q}$ The real part of the conductivity is related to dissipation in the system [Joule heating] when a current $\mathbf{J}$ is flowing. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9126471281051636, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/133697-integrals-volume.html	# Thread: 1. ## Integrals - Volume Question: The region in the first quadrant bounded by the graphs f(x)=(1/8)(x)^3 and g(x)=2x is rotated about the y-axis. Find the volume of the solid formed. I know the integral basics. I think I have to place the y points and I think I am meant to convert the y= into x= so I have a x equation. This is what I got after integrating etc. ((6144/50)(pi))-((512/12)pi)). This I guess is not right. Thanks for any help. 2. Originally Posted by Awsom Guy Question: The region in the first quadrant bounded by the graphs f(x)=(1/8)(x)^3 and g(x)=2x is rotated about the y-axis. Find the volume of the solid formed. using washers ... $V = \pi \int_0^8 (2\sqrt[3]{y})^2 - \left(\frac{y}{2}\right)^2 \, dy<br />$ using cylindrical shells ... $V = 2\pi \int_0^4 x \left(2x - \frac{x^3}{8}\right) \, dx$ 3. Then I guess I do the normal Solving the function by subbing in the x values. thanks.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8739041686058044, "perplexity_flag": "middle"}
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.em/1317758108	previous :: next ### Elliptic Curves with Surjective Adelic Galois Representations Aaron Greicius Source: Experiment. Math. Volume 19, Issue 4 (2010), 495-507. #### Abstract Let $K$ be a number field. The $\operatorname{Gal}(\bar{K}/K)$-action on the torsion of an elliptic curve $E/K$ gives rise to an adelic representation $ρ_E: \operatorname{Gal}(\bar{K}/K) \to \mathrm{GL}_2(\hat{\mathbb{Z}})$. From an analysis of maximal closed subgroups of $\mathrm{GL}_2(\hat{\mathbb{Z}})$ we derive useful necessary and sufficient conditions for $ρ_E$ to be surjective. Using these conditions, we compute an example of a number field $K$ and an elliptic curve $E/K$ that admits a surjective adelic Galois representation. First Page:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.798326313495636, "perplexity_flag": "head"}
http://mathoverflow.net/questions/15083?sort=oldest	What can be said about the homotopy groups of a CW-complex in terms of its (co)homology? Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) One example is the Hurewicz theorem which tells us that (e.g) a CW-cx with only one 0-cell has a nontrivial fundamental group if H_1 is nontrivial. What other examples are there? (The CW-complexes I have in mind always have exactly one 0-cell, but for the sake of a more wide discussion one could assume this is not generally true. However, I am mostly interested in the higher homotopy groups as I already know everything about the fundamental group.) - The cohomology of $S^n$ doesn't contain much information (nonzero only in degrees 0 and n), but its homotopy groups are really crazy in general, so do you have more specific details about your CW complexes? – Steven Sam Feb 12 2010 at 7:50 2 There's a bunch that can be said. To start getting a feel for your question, look at the Adams spectral sequence and, say, the Kan-Thurston theorem. The first will get your hopes up and the second will (partially) dash them. – Ryan Budney Feb 12 2010 at 7:54 2 Answers Try looking up some references on rational homotopy theory. Rational homotopy theory studies the homotopy groups tensor Q, so basically you kill all torsion information. If we focus only on homotopy groups tensor Q, the question you ask becomes easier. As Steven Sam mentions in the comments, the homotopy groups of spheres are really crazy. But the rational homotopy groups of spheres are quite tractable (in fact completely known, by a theorem of Serre) and can be more or less obtained from cohomology, if I recall correctly. One particularly impressive theorem, of Deligne-Griffiths-Morgan-Sullivan, says that if your space is a compact Kähler manifold (e.g. a smooth complex projective variety), and if you know its rational cohomology ring, then you can compute for instance the ranks of all of its homotopy groups (maybe you need an extra assumption that the space is simply connected or has nilpotent fundamental group). - You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. If you don't want to make any assumptions about $\pi_1$, then I think the question is hard. Maybe Hurewicz is most of what you can say. If assume something like simply connected, you can say a lot more. There are many things you can say rationally, some of which were pointed bout my Kevin, but even integrally or mod p there are a lot of techniques. Can you be more precise about the setup you are interested in? You can always replace a connected CW complex with a homotopy equivalent one that has a single 0-cell, so this is not really an issue. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9471458196640015, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/11820/what-really-cause-light-photons-to-appear-slower-in-media	# What really cause light/photons to appear slower in media? I know that if we solve the maxwell equation, we will end up with the phase velocity of light is related to the permeablity and the permitivity of the material. But this is not what I'm interested in, I want to go deeper than that. We know that the real speed of light is actually not changing, the decrease in speed is just apparent. Material is mostly empty, the light will still travel with c in the spacing. The rare atoms will disturb the light in some way. So I am interested in how the atoms affect the light. Some textbooks that I read explain it in a way kind of like this: In a material the photons is absorbed by atom and then re-emitted a short time later, it then travels at a short distance to the next atom and get absorbed&emitted again and so on. How quickly the atoms in a material can absorb and re-emit the photon and how dense the atoms decides the speed of light in that material. So the light appears slower because it has a smaller “drift speed”. But recently I realize an alternative explanation: Atoms respond to the light by radiating electromagnetic wave. This “new light” interferes with the “old light” in some way that result in delayed light(advanced in phase),this can easily shown by using simple phasor diagram. Consequently effectively the light covers a smaller phase each second, . Which gives the impression of a lower phase velocity . However the group velocity is changing in a complicated way. I think that the first explanation does not explain the change in phase velocity of light. if we consider light travelling into a slab of negative refractive index non-dispersive material, let’s say the light is directed perpendicular to the slab. The phase velocity’s direction will be flipped, but group velocity’s direction in the material will not change. Only the second explanation can explain the flipped phase velocity direction. I guess that the velocity that we get in the first explanation is actually belongs to the group velocity. It makes sense to me that the front most of the photon stream determines the first information that the light delivers. So the question is What really cause the phase velocity of light to be decreased? 1."drift velocity" of photons ( they aren't the same photons, they are re-emitted all the time) 2. phase difference between absorbed and emitted light 3. something else And also, I still don't really understand about the detail of the absorption-emission process for small light's wavelength ( for large lambda compare to the atoms spacing, the photons will be absorbed by the phonons). The dispersion relation that we know is continous and also ome material is non-dispersive, therefore the absorption process must accurs in all frequency for a certain range. So definitely it doesn't involve the atomic transition, otherwise it will be quantized. My guess is that the relevant absorption process gets smooth out by the dipole moment. What makes the spectrum continous? EDIT: link for dispersion relation: http://refractiveindex.info/?group=CRYSTALS&material=Si - I actually read a paper that took this approach at some point. That was decades ago, and I haven't a clue where I found it. – dmckee♦ Jul 2 '11 at 22:34 @Emitabsorb I do not think that the atoms can be absorbing the light because the frequency and phase would change as there is no guarantee that the de-excitation will come with the same frequency ( unless it is a laser setup). IMO it must be higher order QED diagrams playing ball with the electric field of the atoms. Now for x rays, which have a lamda smaller than the crystal/glass spacings, do you have a link that a difference between phase velocity and group velocity has been measured? – anna v Jul 4 '11 at 7:02 – pcr Jul 4 '11 at 16:57 @pcr Inelastic scattering, which is Raman scattering, means change of frequency – anna v Jul 4 '11 at 17:12 @ anna v I don't think it is necessary to be spontaneous emission, if it is spontaneous emission, then the emission will be directed in random direction right. If the emission is random, then the light would be heavily diffused, which is not what usually happen in daily life. In the real life case e.g light emitted by a lamp, the light's intensity can be considered as high. therefore it is still possible that the absorption emission process is primarily governed by stimulated absorption/emission. Is it true? – Emitabsorb Jul 4 '11 at 18:11 show 6 more comments ## 3 Answers Looks like you are already familiar with the classical explanation but are still curious about the quantum version of it. 2.phase difference between absorbed and emitted light Yeah, this is essentially the lowest order contribution to the phase shift in the photon-electron scattering. Here is the sloppy way to visualize it continuously (this is basically the 'classical EM wave scattering' point of view): you can imagine that the "kinetic energy" (-> frequency) of the "photon" increases as it approaches the atom's potential well and then it goes back to its normal frequency upon leaving the atom. This translates to a net increase in the phase ($(n-1)\omega/c$). 1. "drift velocity" of photons ( they aren't the same photons, they are re-emitted all the time) By "drift velocity" do you mean a pinball-like, zigzag motion of the photon? This won't contribute that much because it requires more scattering (basically it is a higher order process). And also, I still don't really understand about the detail of the absorption-emission process. Yes the absorption will still occur in all range of the frequency. The hamiltonian of the atom will be modified by the field (by $- p \cdot E$ where p is the dipole moment of the atom and E is the electric field component of the light). This will give us the required energy level to absorb the photon momentarily, which will be re-emitted again by stimulated+spontaneous emission. edit: clarification, the term 'energy level' is misleading, since the temporarily 'excited' atom is not in an actual energy eigenstate. See the diagram here: http://en.wikipedia.org/wiki/Raman_scattering - ok seems like I don't have any problem with the classical explanation, but I'm still pretty confused about the absorption process. Would u mind to elaborate? where does the phase shift comes from quantum mechanically? and how do u explain for the case of negative index of refraction? Is it due to the same effect but the decrease in phase is larger than the phase that the light gains phase by travelling between atoms. – Emitabsorb Jul 3 '11 at 19:06 In addition to everything said, I'd like to comment on the following: We know that the real speed of light is actually not changing, the decrease in speed is just apparent. Material is mostly empty, the light will still travel with c in the spacing. 1) The speed of light does change. It's the speed of light in vacuum that doesn't. 2) A very short answer to your question: Light is, so to say, ‘larger’ than the inter-atomic space. Therefore I wouldn't speak of it travelling in the spacing. This situation is similar to a human running through bushes as opposed to running in a forest. Because you are larger than the individual branches of a bush, you interact with the bush in a different manner than with trees in the forest. May be an illustration with a subwavelength-diameter optical fibre could help to see the problem from another side. This fibre is thinner than the wavelength of light. For example, half a micron diameter for a 1 um light. As the light propagates through such a fibre (which can be done with basically 100 % transmission), the light field doesn't ‘fit’ into the fibre, and about half of energy propagates actually outside the fibre. However, it is not the case that the outer part of light goes faster than the one inside the fibre. The wave remains single, travelling with the speed of $$v = \frac{c}{n_\text{eff}},$$ where $n_\text{eff}$ is the effective refractive index, which is somewhere between 1 (the apparent refractive index for the outer part of the wave) and $n_\text{glass}$ (for the inner one). So, light is ‘larger’ than the inter-atom distance, and therefore it will continuously ‘see’ the atoms. - I am actually interested in every wavelength, what you are saying is for long light's wavelength or small atoms spacing. In that case, I know that the photons will be scattered by atoms as a group(phonons). But for small enough wavelength, the light will only collide with one atoms at a time. And yet in practice, we didn't see any quantized the efect. I would like to hear your explanation for the case of short wavelength, a quantum mechanics explanation. – Emitabsorb Jul 3 '11 at 19:26 I don't believe it is generally helpful to try and analyze these things in terms of photons, so I'm going to try and point out a few things about the classical picture. The big difficulty from the mathematical perspective is that you're working in a continuous medium where the phase of the wave is changing continuously. It makes the visualisation much easier to start off with if we restrict ourselves to a thin slab, where "thin" means small with respect to the wavelength. We know that there is a dielectric constant which represents the tendency for charges to displace themselves in response to an external field. But how fast to the charges respond? Is it a quasi-static case, where the maximum field strength coincides with the maximum charge displacement? I think we will find that this is the case, for example, when light is travelling through glass. Note that in this case the displacement current is leading the incident field by 90 degrees. This makes sense: as the frequency of light approaches the resonant frequency of the material, the phase lags more and more; when the phase difference goes to zero, you have resonant absorption. (EDIT: To be more clear, I choose to define the phase difference in terms of its far-field relation to the incident field!) In the case of the thin slab, you can see that the transmitted wave is the sum of the incident wave and a wave generated by the displacement current. Because you are absorbing, the phase in the far field must be opposite so that energy is removed from the incident wave. It is instructive to do the energy balance. Let's say the displacement current generates a wave equal to 2% of the incident wave. Then the amplitude of the reflected wave is 2%, and the transmitted wave is 98%. It is easy to calculate (by squaring amplitudes) that almost 4% of the energy is missing. Where does it go? It continuously builds up the amplitude of the displacement current until the resistive losses in the material are equal to the power extracted from the incident wave. Let's now go back to the case of the transparent medium. Take the same value for the displacement current, namely 2%. The reflected wave is the same, but the transmitted wave is different because now you are adding phasors that are at 90 degrees to each other, so the amplitude of the transmitted wave is, to the first order, unchanged. It's the phase that's confusing. Because we are in the quasi-static regime, the phase is leading. In any case it must be leading in comparison to the absorptive case. Don't we want a lagging phase in order to slow down the wave? This is where you have to be very careful. Because we are adding a leading phase, the wave peaks occur sooner than otherwise...in other words, they are close together. This is indeed the condition for a wave to travel slower. It's all very confusing, which is why I took the case of a thin slab so the math would be simpler. Let the incident wave be sin(kx-wt) Then the wave generated by the slab will be 0.02*cos(kx-wt) Note the cosine wave leads the sine wave by 90 degrees. If you draw these two waves on a graph and add them together, you can see that the peaks of the sine wave are pushed slightly to the left. This makes the wave appear slightly delayed. The continous case is harder to do mathematically but you can see that it ought to follow by treating it as a series of slabs. - Actually I'm already using that point of view too, we can understand it more easily with phasor diagram. If the "new wave" is pi/2 degrees lag then if we add the old phasor with the new one, the result is like the old phasor gets rotated a little bit. So how do u explain for the case of negative index of refraction? Is it due to the same effect but the decrease in phase is larger than the phase that the light gains by travelling between atoms? – Emitabsorb Jul 3 '11 at 19:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9417381882667542, "perplexity_flag": "head"}
http://mathoverflow.net/questions/32765/whats-after-natural-transformations/32809	## What’s after natural transformations? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If functors are morphisms between categories, and natural transformations are morphisms between functors, what's a morphism between natural transformations? Is there ever a need for such a notion? - ## 4 Answers (Small) categories form what's called a 2-category, which is a structure that has objects, morphisms (functors), and morphisms between morphisms (natural transformations). There are also n-categories, which have a deeper morphisms structure. A google search will point you to a lot of references about n-categories. But for ordinary categories, the story ends at natural transformations. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The next one is called modification. Modifications between ordinary natural transformations between functors between categories are trivial, but they exist between lax natural transformations between 2-functors between 2-categories. The next one after that is called perturbation : a perturbation goes between modifications between pseudonatural transformations between 3-functors between 3-categories. Beyond that, no established terms exists. Instead one starts numbering things and speaks of transfors. n-Functors are 0-transfors. Transformations are 1-transfors. Modifications are 2-transfors. Perturbations are 3-transfors. And so on. - 9 Urgh. Is transfor an established term? :( – Mariano Suárez-Alvarez Jul 21 2010 at 15:36 4 I prefer to rearrange the terminology, and just have plain functors between n-categories, 2-functors between functors (so, in the n=1 case, 2-functors are the same as natural transformations), 3-functors between 2-functors, and so on. – Scott Morrison♦ Jul 21 2010 at 16:59 7 This sort of reminds me of the naming of higher derivatives of position. We have velocity and acceleration which are common, then after that there are the increasingly obscure 'jerk' for the third derivative, and then 'snap', 'crackle', and 'pop' for the fourth, fifth, and sixth derivatives. – Simon Rose Jul 21 2010 at 17:02 2 Scott, the term n-functor is already widely established to mean a morphism between n-categories. – Urs Schreiber Jul 21 2010 at 19:15 One could also say that an n-transfor is a directed homotopy of order n in the ambient whatever category of whatever categories. Cause in the case where all n-categories here are in fact n-groupoids / infty-groupoids, an n-transfor is nothing but an order n homotopy (under the identification of oo-groupoids with topological spaces). – Urs Schreiber Jul 21 2010 at 19:17 show 3 more comments One advantage of the abstraction of category theory is that one is not constrained to "concrete" objects and morphisms (I mean, made by set with a structure together with functions preserving it), and constructions of new categories from simpler ones are very easily performed. As a result, any further and more general categorical notion can always be read as a particular case of a simpler and more basic one, in a suitable category. Thus in the proper context, a morphism is an object; a natural transformation is a morphism; similarly, a universal arrow is a particular case of an initial object, which of course is a particular case of a universal arrow, and so on. So in a sense, there is no need of the notion of "morphism between natural transformations", just because it is already a particular case of a more basic notion already defined. In practice, several used categories (e.g. algebras; preshaves; chain complexes,...) are themselves categories of functors, where arrows are natural transformations. In this context a morphism between natural transformation naturally arises, even if it will be seen as just an ordinary morphism. - to whom downvoted: it would be more constructive if you exlpain your reasons; this way you'd give everybody the opportunity to learn something. – Pietro Majer Sep 4 2010 at 8:09 This is not really a sophisticated answer as the other ones, but maybe it makes visible why higher structures are needed to get an interesting notion of morphism between natural transformations. Assume $F,G : C \to D$ are functors and $\alpha, \beta$ are natural transformations $F \to G$. What could be a morphism $\alpha \to \beta$? Since $\alpha$ and $\beta$ consist of their components $\alpha(c) : F(c) \to G(c), \beta(c) : F(c) \to G(c)$, the only reasonable way of "connecting" these data in our category $D$ is by means of two morphisms $\gamma(c) : F(c) \to F(c), \delta(c) : G(c) \to G(c)$, so that the resulting diagram becomes commutative. Furthermore, $\gamma$ and $\delta$ should become natural transformations $F \to F, G \to G$. But this comes from a more general concept, namely the arrow category: If $C$ is a category (in the above case, this is a functor category), its arrow category has as objects the morphisms of $C$ and a morphism between two morphisms $x \to y, x' \to y'$ is a pair of morphisms $x \to x', y \to y'$, making the obvious diagram commutative. This may be also described as the functor category $C^{\textbf{2}}$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9309554696083069, "perplexity_flag": "middle"}
http://programmingpraxis.com/2012/05/	# Programming Praxis A collection of etudes, updated weekly, for the education and enjoyment of the savvy programmer ## Streaming Median ### May 29, 2012 The median of a set of numbers is the number in the middle when the items are arranged in sorted order, when the number of items is odd, or the average of the two numbers in the middle, when the number of items is even; for instance, the median of {3 7 4 1 2 6 5} is 4 and the median of {4 2 1 3} is 2.5. The normal algorithm for computing the median considers the entire set of numbers at once; the streaming median algorithm recalculates the median of each successive prefix of the set of numbers, and can be applied to a prefix of an infinite sequence. For instance, the streaming medians of the original sequence of numbers are 3, 5, 4, 3.5, 3, 3.5, and 4. The streaming median is computed using two heaps. All the numbers less than or equal to the current median are in the left heap, which is arranged so that the maximum number is at the root of the heap. All the numbers greater than or equal to the current median are in the right heap, which is arranged so that the minimum number is at the root of the heap. Note that numbers equal to the current median can be in either heap. The count of numbers in the two heaps never differs by more than 1. When the process begins the two heaps are initially empty. The first number in the input sequence is added to one of the heaps, it doesn’t matter which, and returned as the first streaming median. The second number in the input sequence is then added to the other heap, if the root of the right heap is less than the root of the left heap the two heaps are swapped, and the average of the two numbers is returned as the second streaming median. Then the main algorithm begins. Each subsequent number in the input sequence is compared to the current median, and added to the left heap if it is less than the current median or to the right heap if it is greater than the current median; if the input number is equal to the current median, it is added to whichever heap has the smaller count, or to either heap arbitrarily if they have the same count. If that causes the counts of the two heaps to differ by more than 1, the root of the larger heap is removed and inserted in the smaller heap. Then the current median is computed as the root of the larger heap, if they differ in count, or the average of the roots of the two heaps, if they are the same size. Your task is to write a function that computes the streaming medians of a sequence. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Pages: 1 2 Posted by programmingpraxis Filed in Exercises 8 Comments » ## Ackermann’s Function ### May 25, 2012 In the 1920s, Wilhelm Ackermann demonstrated a computable function that was not primitive-recursive, settling an important argument in the run-up to the theory of computation. There are several versions of his function, of which the most common is $A(m,n) = \left\{ \begin{array}{ll} n+1 & \mbox{if $m=0$} \\ A(m-1,1) & \mbox{if $m > 0$ and $n = 0$} \\ A(m-1, A(m,n-1)) & \mbox{if $m > 0$ and $n > 0$} \end{array} \right.$ defined over non-negative integers m and n. The function grows very rapidly, even for very small inputs; as an example, A(4,2) is an integer of about 20,000 digits. Your task is to implement Ackermann’s function. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Pages: 1 2 Posted by programmingpraxis Filed in Exercises 9 Comments » ## Hamming Codes ### May 22, 2012 Hamming codes, which were invented by Richard Hamming in 1950, are a method of transmitting data over a noisy channel that give the recipient the ability to correct simple errors. The sender adds parity bits to the transmission stream so that, when the data bits and parity bits are combined, any single-bit error, in either the data bits or parity bits, can be identified and corrected. The number of parity bits that are required is given by the Hamming rule d + p + 1 ≤ 2p where d is the number of data bits and p is the number of parity bits. The length of the code word c, which combines the data bits and parity bits, is d + p, and a Hamming code is described by (c,d). We will illustrate using a 4-bit data word, which requires 3 parity bits to satisfy the Hamming rule (2 is insufficient because 4+2+1>4, but 3 is sufficient because 4+3+1≤8) and is described by (7,4). A particular instance of a Hamming code uses two matrices, G the generator matrix and H the syndrome matrix. Here are sample generator (left) and syndrome (right) matrices for a (7,4) Hamming code: ```1 0 0 0 1 1 1 1 0 1 1 1 0 0 0 1 0 0 0 1 1 1 1 0 1 0 1 0 0 0 1 0 1 0 1 1 1 1 0 0 0 1 0 0 0 1 1 1 0``` The generator matrix is denoted [ I : A ] and consists of an identity matrix I in the left-most four columns (the number of data bits) and a parity coding matrix A in the right-most three columns (the number of parity bits). There is no formula for the A matrix; it must be constructed so that each data bit is checked by two or more parity bits, in such a way that no combination of parity bits overlaps a data bit. The syndrome matrix is denoted [ AT : I ] and consists of the transpose of the parity coding matrix A in the left-most four columns and the identity matrix in the right-most four columns. A data word is encoded by multiplying it by the generator matrix, with all arithmetic done modulo 2; we gave an algorithm for matrix multiplication in a previous exercise. For instance, the data word [1 0 0 1] is encoded as [1 0 0 1 0 0 1] like this: ``` \| 1 \| \| 0 \| \| 1 0 0 0 1 1 1 \| \| 0 \| \| 1 0 0 1 \| * \| 0 1 0 0 0 1 1 \| = \| 1 \| \| 0 0 1 0 1 0 1 \| \| 0 \| \| 0 0 0 1 1 1 0 \| \| 0 \| \| 1 \|``` Decoding is the inverse operation, with the syndrome matrix multiplied by the encoded data: ``` \| 1 \| \| 0 \| \| 1 0 1 1 1 0 0 \| \| 0 \| \| 0 \| \| 1 1 0 1 0 1 0 \| * \| 1 \| = \| 0 \| \| 1 1 1 0 0 0 1 \| \| 0 \| \| 0 \| \| 0 \| \| 1 \|``` If all of the result bits are zero, then the transmission succeeded without errors, and the input code is the first four bits of the encoding [1 0 0 1]. But if there was a transmission error, the syndrome points to it. For instance, if the recipient received the tranmission as [1 0 1 1 0 0 1], then the syndrome computes as [1 0 1], which matches the third column of the H matrix, indicating that the third bit of the transmission was in error, so instead of [1 0 1 1] the message is [1 0 0 1], which is correct. Your task is to write functions that encode and decode a message using Hamming codes as described above. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Pages: 1 2 Posted by programmingpraxis Filed in Exercises 1 Comment » ## Formatted Numeric Output ### May 18, 2012 It is often necessary for programs to produce numeric output in various formats, and most languages provide libraries for this purpose; for instance, C provides the `printf` function, which includes the `d` and `f` format specifications for decimal numbers (integers) and floating point numbers, respectively. Your task is to write library functions that format integers and floating point numbers; you may follow the formatting conventions of C, or those of some other language, or invent your own. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Pages: 1 2 Posted by programmingpraxis Filed in Exercises 2 Comments » ## Streaming Knapsack ### May 15, 2012 A famous problem of computer science is the knapsack problem, in which you are to find a combination of items from a population that sums to a given target, often with some kind of constraint such as maximizing the value of the items. In today’s problem we want to find the first possible combination of k integers from a stream of positive integers that sum to n. For instance, given the input stream 4, 8, 9, 2, 10, 2, 17, 2, 12, 4, 5, …, we want to find the knapsack containing 4, 2, 10, 2, 2 immediately after reading the third 2, without reading the 12, 4, 5 that follow it. Your task is to write a program that takes parameters k and n and an input stream and returns the first possible knapsack. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Pages: 1 2 Posted by programmingpraxis Filed in Exercises 4 Comments » ## Partitions ### May 11, 2012 The partitions of an integer is the set of all sets of integers that sum to the given integer. For instance, the partitions of 4 is the set of sets ((1 1 1 1) (1 1 2) (1 3) (2 2) (4)). We computed the number of partitions of an integer in a previous exercise. In today’s exercise, we will make a list of the partitions. The process is recursive. There is a single partition of 0, the empty set (). There is a single partition of 1, the set (1). There are two partitions of 2, the sets (1 1) and (2). There are three partitions of 3, the sets (1 1 1), (1 2) and (3). There are five partitions of 4, the sets (1 1 1 1), (1 1 2), (1 3), (2 2), and (4). There are seven partitions of 5, the sets (1 1 1 1 1), (1 1 1 2), (1 2 2), (1 1 3), (1 4), (2 3) and (5). And so on. In each case, the next-larger set of partitions is determined by adding each integer x less than or equal to the desired integer n to all the sets formed by the partition of n − x, eliminating any duplicates. Your task is to write a function that generates the set of all partitions of a given integer. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Pages: 1 2 Posted by programmingpraxis Filed in Exercises 12 Comments » ## Factor Tables ### May 8, 2012 Before the dawn of computers, most computations were done with the aid of tables: logarithm tables, sine tables, and so on. These tables were ubiquitous, indispensable, and riddled with errors. Number theorists who needed to factor numbers used tables of the least prime factor of a number. The oldest such table dates to 1603 (it contained the least prime factor of all numbers to 750), and new tables were being constructed as late as Derrick N. Lehmer’s table of least prime factors to ten million in 1909 (he was the father of Derrick H. Lehmer); Maarten Bullynck gives the history. Here’s a sample page from a large table, showing the least prime factors of all numbers less than a thousand; numbers divisible by 2 and 5 are omitted, and primes are skipped: ``` 0 1 2 3 4 5 6 7 8 9 1 -- 3 7 -- 3 -- -- 3 17 3 -- -- 7 3 13 -- 3 19 11 3 7 -- -- 3 -- 11 3 -- 7 3 -- 9 3 -- 11 3 -- -- 3 -- -- 3 11 -- 3 -- -- 3 7 13 3 -- -- 13 -- -- 3 -- 7 3 -- 23 3 11 17 -- 3 7 -- 3 11 -- 3 19 7 19 -- 7 3 11 -- 3 -- -- 3 -- 21 3 11 13 3 -- -- 3 7 -- 3 23 -- 3 -- 17 3 -- 7 3 -- 13 27 3 -- -- 3 7 17 3 -- -- 3 29 -- 3 -- 7 3 23 17 3 -- -- 31 -- -- 3 -- -- 3 -- 17 3 7 33 3 7 -- 3 -- 13 3 -- 7 3 37 -- -- 3 -- 19 3 7 11 3 -- 39 3 -- -- 3 -- 7 3 -- -- 3 41 -- 3 -- 11 3 -- -- 3 29 -- 43 -- 11 3 7 -- 3 -- -- 3 23 47 -- 3 13 -- 3 -- -- 3 7 -- 49 7 -- 3 -- -- 3 11 7 3 13 51 3 -- -- 3 11 19 3 -- 23 3 53 -- 3 11 -- 3 7 -- 3 -- -- 57 3 -- -- 3 -- -- 3 -- -- 3 59 -- 3 7 -- 3 13 -- 3 -- 7 61 -- 7 3 19 -- 3 -- -- 3 31 63 3 -- -- 3 -- -- 3 7 -- 3 67 -- -- 3 -- -- 3 23 13 3 -- 69 3 13 -- 3 7 -- 3 -- 11 3 71 -- 3 -- 7 3 -- 11 3 13 -- 73 -- -- 3 -- 11 3 -- -- 3 7 77 7 3 -- 13 3 -- -- 3 -- -- 79 -- -- 3 -- -- 3 7 19 3 11 81 3 -- -- 3 13 7 3 11 -- 3 83 -- 3 -- -- 3 11 -- 3 -- -- 87 3 11 7 3 -- -- 3 -- -- 3 89 -- 3 17 -- 3 19 13 3 7 23 91 7 -- 3 17 -- 3 -- 7 3 -- 93 3 -- -- 3 17 -- 3 13 19 3 97 -- -- 3 -- 7 3 17 -- 3 -- 99 3 -- 13 3 -- -- 3 17 29 3``` For example the table shows that the least prime factor of 923, in the column headed 9 and row headed 23, is 13, and 997 is the greatest prime less than a thousand. To factor a number, find its least prime factor in the table, compute the remaining cofactor by division, and repeat until the cofactor is prime. When building the table, the least prime factor is computed by sieving, not by trial division. The setup is the same as the Sieve of Eratosthenes, except that integers are used instead of booleans, and each item in the sieve is initialized to 1. Then each successively-smallest prime p is sieved, but instead of changing `true` to `false`, each 1 in the chain of multiples of p is changed to p (other values are ignored). The same optimizations as the normal sieve — odd numbers only, start at the square of p, and stop when p2 is greater than n — apply here. Your task is to write a function that sieves for least prime factors, and to use that function to write a program that builds factor tables as illustrated above. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Pages: 1 2 Posted by programmingpraxis Filed in Exercises 5 Comments » ## Even-Odd Partition ### May 4, 2012 I’m not sure where this problem comes from; it’s either homework or an interview question. Nonetheless, it is simple and fun: Take an array of integers and partition it so that all the even integers in the array precede all the odd integers in the array. Your solution must take linear time in the size of the array and operate in-place with only a constant amount of extra space. Your task is to write the indicated function. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Pages: 1 2 Posted by programmingpraxis Filed in Exercises 30 Comments » ## Legendre’s Symbol ### May 1, 2012 The Legendre Symbol and its cousin the Jacobi Symbol are used in modular arithmetic to determine if a number a is a quadratic residue to the modulus m. A number a is a quadratic residue if there exists a number x such that x2 ≡ a (mod m) and is defined only when a and m are co-prime. For instance, a2 (mod 7) for a from 0 to 6 is the list 02 (mod 7) = 0, 12 (mod 7) = 1, 22 (mod 7) = 4, 32 (mod 7) = 2, 42 (mod 7) = 2, 52 (mod 7) = 4, and 62 (mod 7) = 1, so the quadratic residues of 7 are 1, 2 and 4 (0 is excluded because it isn’t co-prime to 7). The jacobi symbol considers any odd modulus; the legendre symbol is limited to odd prime moduli. The symbols are usually written in parentheses with a over m, like this: $\left( a \atop m \right)$. Sometimes the symbol is written with a horizontal rule between the a and m, and sometimes it is written on a single line as (a / m). The legendre/jacobi symbol can be calculated according to the following three termination rules: 1. (0 / m) = 0 2. (1 / m) = 1 3. (2 / m) = −1 if m mod 8 ∈ {3, 5} or 1 if m mod 8 ∈ {1, 7} and the following three reduction rules: 4. [reducing factors of 2] (2a / m) = (2 / m) × (a / m) 5. [reducing modulo m] (a / m) = (a (mod m) / m) if a ≥ m or a < 0 6. [reducing odd co-primes a and m] (a /m) = −(m / a) if a ≡ m ≡ 3 (mod 4), or (m / a) otherwise Thus, the legendre/jacobi symbol is 1 if a is a quadratic residue, -1 if a is not a quadratic residue, and 0 if a and m are not co-prime. Our various prime-number programs have used a definition of the legendre/jacobi symbol that doesn’t work; for some values of a and m it returned wrong results, and for other values of a and m it entered an infinite loop. This exercise fixes the problem. Your task is to write a function that calculates the legendre/jacobi symbol using the rules given above. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below. Pages: 1 2 Posted by programmingpraxis Filed in Exercises 13 Comments »	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 2, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9193313717842102, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/123649-factoring-polynomials-help.html	# Thread: 1. ## Factoring Polynomials Help So I have to do a review in my Precalculus textbook from Algebra 2 and their are these 3 questions that I just don't seem to understand. I have the answer in the back of the book but I want to know how to get to the answer. Questions are: 1-8x^2-9x^4 - Answer is -(3x-1)(3x+1)(x^2-1) (x+2)^2 - 5(x+2) - Answer is - (x+2)(x-3) 3(x^2+10x+25) - 4(x+5) - Answer is (x+5)(3x+11) I Just don't undestand how they got to those answers. 2. Originally Posted by Kros So I have to do a review in my Precalculus textbook from Algebra 2 and their are these 3 questions that I just don't seem to understand. I have the answer in the back of the book but I want to know how to get to the answer. Questions are: 1-8x^2-9x^4 - Answer is -(3x-1)(3x+1)(x^2-1) (x+2)^2 - 5(x+2) - Answer is - (x+2)(x-3) 3(x^2+10x+25) - 4(x+5) - Answer is (x+5)(3x+11) I Just don't undestand how they got to those answers. 1) $1 -8x^2 - 9x^4 = -9x^4 - 8x + 1$ We need two numbers that multiply to become $-9$ and add to become $-8$. They are $-9$ and $1$. $-9x^2 - 8x + 1 = -9x^4 - 9x^2 + 1x^2 + 1$ $= -9x^2(x^2 + 1) + 1(x^2 + 1)$ $= (x^2 + 1)(1 - 9x^2)$ $= (x^2 + 1)[1^2 - (3x)^2]$ $= (x^2 + 1)(1 - 3x)(1 + 3x)$. Double check the answer you have been given. 2) $(x + 2)^2 - 5(x + 2)$ Take out a common factor of $x + 2$ $= (x + 2)(x + 2 - 5)$ $= (x + 2)(x - 3)$. 3) $3(x^2+10x+25) - 4(x+5) = 3(x + 5)^2 - 4(x + 5)$ $= (x + 5)[3(x + 5) - 4]$ $= (x + 5)(3x + 15 - 4)$ $= (x + 5)(3x + 11)$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9538902044296265, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=355064	Physics Forums ## Kinetic energy and gravitational potential energy questions 1. An acrobat on skis starts from rest 50.0 m above the ground on a frictionless track and flies off the track at a 45.0 degree angle above the horizontal and at a height of 10.0m. disregard air resistance. what is the skier's speed when leaving the track? 2. K=.5MV^2 and U=Mg*h 3. I don't have a attempt at the solution not because I am trying to get you to do my homework but because I don't understand the question I am having trouble picturing the scene what does it mean by the horizontal and the 45 degree angle. What is needed and what is just fluff... Can someone help? PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Before the skier starts down the hill his potential energy is maximum. The skier converts his potential energy to kinetic as he travels down the incline. So I ask you this: when is the skiers kinetic energy at a maximum and what is the skiers potential the moment he leaves the ramp? It will help you to associate velocity with kinetic energy. When is the velocity the highest, When is it the lowest? also note: potential energy is a function of position, kinetic energy is a function of velocity what I don't get is how do we get the skiers mass with the given info ## Kinetic energy and gravitational potential energy questions When we write our equation we get this: $$.5mv^2=m(9.8)(40)$$ (e.i potential energy is converted to kinetic) If you notice, $$m$$ reduces to 1. That is why we don't need the mass ^^ Thread Tools \| \| \| \| \|----------------------------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Kinetic energy and gravitational potential energy questions \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 5 \| \| \| Introductory Physics Homework \| 9 \| \| \| General Physics \| 4 \| \| \| Introductory Physics Homework \| 1 \| \| \| General Physics \| 1 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9066250920295715, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/31929?sort=oldest	## The consequence of overlap sharing for the length-distribution of rods randomly placed on a line ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Please imagine that one populates a finite line of unit length, or circle with unit length contour (to avoid edge-effects), with $N$ one-dimensional 'rods' such that their LHS-ends, at positions $(p_1, ..., p_k, ..., p_N) \in P$, are placed in accordance with a uniform random distribution over [0, 1]. Here, the rod lengths, $(l_1, ..., l_k, ..., l_N) \in L$, are exponentially distributed according to some rate parameter $\lambda$ - i.e. the random variable $l_k$ has distribution $l_k$ ~ Exp($\lambda$), giving a probability density function for rod length of $\lambda e^{\lambda l}$. One would similarly expect an exponential distribution for the distances between adjacent points in the set $P$. We have the following two rules for handling overlaps between rods: (1) - If the 'contour' of one rod (say, 'Rod A') completely covers another (say, 'Rod B'), i.e. where (Rod A-LHS) < (Rod B-LHS) and (Rod A-RHS) > (Rod B-RHS), we remove 'Rod B' from from the line and no longer consider it. (2) - If there is only a partial overlap in the contours of two rods, 'Rod A' and 'Rod B', the length of this overlap is split evenly and each half is added to the contours of 'Rod A' and 'Rod B', respectively. Starting from our initial exponential distribution of rod lengths, $(l_1, ..., l_k, ..., l_N)$, after this overlap-splitting process what is the new probability distribution for the length of some rod, $l_k$? A few observations: As $\lambda \rightarrow \infty$, the number of rods left on the line (after overlaps are handled) should increase, and the mean rod length should decrease. As $N \rightarrow \infty$, the number of overlap-processed rods left on the line should increase, and the mean rod length should decrease. Intuitively I would expect that the number of rods remaining on the line after overlap processing will increase ever more slowly with $N$ after some threshold/'saturation' value is reached (presumably where the line is completely covered with rods). As $\lambda \rightarrow -\infty$, there should be fewer rods left remaining on the line after overlap processing, and the mean rod length should increase. At some sufficiently large value of $\lambda$, we should be left with only a single rod on the line which has the left-most/smallest LHS-side. If we also have that $N \rightarrow \infty$, the mean length of the rod should approach the unit length of the line. As $N \rightarrow 0$, there should be fewer rods, and an increasing mean rod length. Inspired by Joseph O'Rourke's answer, and some simulation results of mine, if one fixes $\lambda$ and lets $N \rightarrow \infty$, one appears to converge to a rod length distribution centered around a mean value somewhere between $\frac{L}{2}$ and $L$, where $L$ is the original mean length of the rods before overlap processing. However, this distribution appears to be Gaussian, not uniform. Do we actually converge to a Gaussian distribution? How does the distribution and its variance change with increasing $N$? - Am I correct in these two consequences of your rules?: (a) After each step, the rods have disjoint interiors; (b) eventually, the unit interval/circle is entirely covered end-to-end by rods (of various lengths). – Joseph O'Rourke Jul 15 2010 at 1:03 Dear Joseph, Yes about (a), however, (b) is not necessarily true for small 'N' and/or short rod lengths. There can be gaps. – Rob Grey Jul 15 2010 at 1:21 @Rob: I have some trouble understanding the model itself. Assume for instance that some rods A, B and C are such that LHS(A) < LHS(B) < LHS(C) < RHS(A) < RHS(B) < RHS(C). Then what happens? If one applies your rule to A and B first, getting A' and B', and then (if necessary), to B' and C, getting B'' and C', one gets a configuration A', B'' and C' which could be different from the configuration one gets if one applyes the rule to B and C first and then (if necessary) to A and the modified B. And this is only one configuration amongst many where this kind of ambiguity arises. Or am I mistaken? – Didier Piau Dec 13 2010 at 8:42 ## 2 Answers This is not an answer, only a simplification and conjecture concerning that simplification. First, only consider $N$ large enough so that the interval/circle is fully covered (the "eventually" in my comment). Second, rather than your exponential distribution, fix all rods to the same (small) length $L$, perhaps $L < \frac{1}{2}$ suffices. Retain your assumption that the left endpoint of each rod is chosen uniformly in $[0,1]$. Then I conjecture that the limiting distribution is uniform with mean rod length $L/2$. I have only heuristic arguments for this (shorter rods get absorbed by newly added ones, existing longer rods get chopped from the ends). Perhaps you could alter your simulation to this simplified circumstance to see if this is empirically true? If this conjecture holds, then perhaps it holds even for an exponential distribution, with $L$ now the mean length of that distribution. Addendum: I verified this myself, and indeed it seems to hold empirically. Here are results of a simulation adding 10 million rods of length $L=\frac{1}{10}$ to $[0,1]$. Only lengths of rods within $[L,1-L]$ are averaged in the graph (to exclude edge effects). - Dear Joseph, Thanks for your answer! However, the most interesting part of this problem for me (which I accidently stumbled upon while simulating another system) is the dramatic effect this had on smoothing out the exponential distribution of the rods with the right $\lambda$. – Rob Grey Jul 15 2010 at 14:26 Joseph, very cool, thanks for running the simulation! From my own simulations, it appears that if we fix $\lambda$ and let $N \rightarrow \infty$, we converge to a Gaussian-looking distribution centered around a mean of $\frac{L}{2}$, where $L$ is the mean length of the rods before overlap processing. – Rob Grey Jul 16 2010 at 23:19 Actually, to be more accurate, the mean seems to be somewhere between L/2 and L, not strictly at L/2... – Rob Grey Jul 17 2010 at 0:19 @Rob: My simulation converged to $L/2$ after $10^7$ iterations. I used $[0,1]$ but had to remove end-effects. I did not look at the distribution, however, just the mean. Cannot post data now... – Joseph O'Rourke Jul 17 2010 at 0:26 Dear Joseph, It looks like my simulation might converge to L/2 as well... but I'm still looking at higher values of 'N' and wanted to be conservative with my statement. My guess is that you're not going to have a uniform distribution! – Rob Grey Jul 17 2010 at 0:34 show 3 more comments ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I don't have an answer either, except to note that this setup is very similar to the core of my dissertation. This stochastic distribution of random-length intervals models the problem of genomic mapping; if you have blasted many copies of some genetic material into fragments, how do you characterize the fragmentations that can uniquely be reconstructed (up to complete reversal) -- vs. fragmentations that are ambiguous (can be consistently reconstructed in multiple ways)? Then there's the question of, OK, you've sampled fragments of DNA with a distribution that you believe gives you sufficient probability of being unambiguous, so how do you actually figure out how to order the fragments into a map? If you further cut the fragments at all occurrences of a few short DNA sequences (restriction sites for digestion enzymes), overlaping areas will yield common distributions of sub-fragment lengths, and you can begin to see who overlaps who, and start to piece the whole thing together. If you are cool with postscript, you can find two papers at this link (search for Settergren), or I have pdf'd them here and there. Or,you could just download the whole dissertation. - The system I was modeling concerned crosslinks between polyethylene molecules (and a coding mistake leading to this question). However, sonicated/digested/etc. DNA is a really good example of a system where your going to have to deal with an exponential distribution of fragment lengths. This is probably a tangent, but how much of an advantage do you get from a pseudo-deterministic cleavage pattern vs. one that's uniform? People seem to make such a fuss about 'unbiased' cleavage with sonication or endonucleases. Is there an inflection point with longer sequencing read lengths? – Rob Grey Jul 15 2010 at 14:57 I think the answer might be that "longer" is not the right question so much as "more variable". I edited my comment above to put a link to the whole dissertation. My model involved a genome length of L, and clone lengths uniformly sampled from [1,1+delta]. If you look at page 28 of the dissertation, you will see there is something special happening at delta=2 (lengths ~ U(1,3)) Short answer, if you want to avoid ambiguity with constant-length clones, you need n~L(lnL+lnlnL), but for 0<del<1, you can get away with more like n~L/2(lnL). I think; it's been a long time... – RubeRad Jul 15 2010 at 15:25 And I didn't answer your first question; back in the day (mid-90s) when I was in school, that was not a question I ever asked. The model I built upon involved constant- (unit-)length clones, and I analyzed assumptions of either U(1,1+delta), or 1+exp(). I suppose it might be interesting to analyze instead an arbitrary discrete distribution at values 1, 1+d1, 1+d2, ... But probably quite harder; simulation would probably work better for getting answers. You know, now I wish I had added another chapter to my dissertation, doing simulations of my theoretical results, and other distros as well! – RubeRad Jul 15 2010 at 15:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9438819289207458, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/80307/intrinsic-description-of-the-image-of-v-to-v	## Intrinsic description of the image of $V \to V^{*}$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $V$ be a vector space over a field $K$. Call a linear map $F : V^ \to K$ representable if there is some $v \in V$ such that $F(w) = \langle w,v \rangle$ for all $w \in V^$. Here, $\langle w,v \rangle := w(v)$. Remark that this representing vector $v$ is then uniquely determined. Remark the similarity to category theory: A functor $F : C^{\mathrm{op}} \to \mathrm{Set}$ is representable if there is an object $v \in C$ such that $F \cong C(-,v)$. By the Yoneda Lemma, this object is uniquely determined up to canonical isomorphism. Now there are various theorems which say when a certain functor is representable. Often we require that $F$ preserves limits and satisfies some finiteness condition. Now back to linear algebra, is there a criterion when a linear map $V^ \to K$ is representable? In other words, is there an intrinsic description of the image of $V \to V^{*}$? If $V$ is finite-dimensional, then we get representability for free. I'm interested in the general case. I would like to mimic somehow the category theoretic conditions. // Moosbrugger has given below the following nice characterization: Give $K$ the discrete topology, $K^V$ the product topology, and $V^ \subseteq K^V$ the subspace topology. Then a linear map $V^* \to K$ is representable iff it is continuous! However, the proof is rather trivial, except that it uses the result for finite-dimensional vector spaces. So in practise, I doubt that this will be a good criterion. Therefore I would add the following requirement to the criterion: it should be useful in practise (whatever this means) or linear algebraic / geometric: When is a hyperplane in $V^$ cut out by a single vector $v \in V$? - The image is contained, at least, in the annihilator of the annihilator of $V$. – Mariano Suárez-Alvarez Nov 7 2011 at 15:03 3 Continuity is equivalent to representability (in both cases). – Moosbrugger Nov 7 2011 at 15:06 1 @Moosbrugger: Not every continuous functor $C^{op} \to \mathrm{Set}$ is representable. For example, $\lim_i \hom(-,x_i)$ is continuous, but is representable iff $\lim_i x_i \in C$ exists. However, there many categories with this property, called SAFT by Theo Johnson-Freyd (mathoverflow.net/questions/49175/…). When is a linear map $V^ \to K$ called continuous? Do you assume that $K$ is a topological field, or even $\mathbb{R}$ or alike? – Martin Brandenburg Nov 7 2011 at 15:27 @Mariano: The annihilator of $V$ is the zero subspace of $V^$. Perhaps I misunderstand your comment. – Martin Brandenburg Nov 7 2011 at 15:32 1 My answer was meant as a bit of a joke: certainly one requires hypotheses for such representability theorems. The one I know is called "presentable." But anyway, it's a precise statement for vector spaces for any field $K$. You topologize the dual as a subset of $\prod_{v\in{V}}{K}$, where the latter has the usual ("Tychonoff") topology (and the map $V^\to\prod_{v\in{V}}{K}$ is $\lambda\mapsto (\lambda(v))_{v\in{V}}$). – Moosbrugger Nov 7 2011 at 18:33 show 4 more comments ## 1 Answer Moosbrugger has given an excellent answer to the question on which I'll elaborate a bit. A similar result holds in functional analysis: Let $V$ be a Banach space and $V'=hom(V,\mathbb C)$its dual (i.e., the set of continuous linear functionals from $V$ to $\mathbb C$). If you equip $V'$ with the norm topology, then its dual is typically bigger than $V$. However, if you equip it with the weak-star topology, then its dual is exactly $V$. In other words, $V\subset V''$ is the subset of weak-star continuous functionals (and the norm on $V$ is induced by that on $V''$). In that way, you can recover $V$ from $V'$ if you known its weak-star topology. Now, for the sake of completess, let me also say what the weak-star topology on $V'$ is: it's the topology under which which a net $\{\varphi_i\}_{i\in I}$ converges to some element $\varphi$ if and only if $\{\varphi_i(v)\}_{i\in I}$ converges to $\varphi(v)$ for every $v\in V$. - Thanks! So the weak-star topology is just the topology of pointwise-convergence. – Martin Brandenburg Nov 7 2011 at 20:38 @Martin: Correct. – André Henriques Nov 8 2011 at 9:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 53, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9204801321029663, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2011/07/07/identifying-tensor-fields/?like=1&source=post_flair&_wpnonce=78d053c4d8	# The Unapologetic Mathematician ## Identifying Tensor Fields Just as for vector fields, we need a good condition to identify tensor fields in the wild. And the condition we will use is similar: if $T$ is a smooth tensor field of type $(r,s)$, then for any coordinate patch $(U,x)$ in the domain of $T$, we should be able to write out $\displaystyle T\vert_U=\sum\limits_{i_1,\dots,i_r,j_1,\dots,j_s=1}^nT^{i_1\dots i_r}{}_{j_1\dots j_s}\frac{\partial}{\partial x^{i_1}}\otimes\dots\otimes\frac{\partial}{\partial x^{i_r}}\otimes dx^{j_1}\otimes\dots\otimes dx^{j_s}$ for some smooth functions $T^{i_1\dots i_r}{}_{j_1\dots j_s}$ on $U$. Conversely, this formula defines a smooth tensor field on $U$. Indeed, we can find these coefficient functions by evaluation: $\displaystyle T^{i_1\dots i_r}{}_{j_1\dots j_s}=T\left(dx^{i_1},\dots,dx^{i_r},\frac{\partial}{\partial x^{j_1}},\dots,\frac{\partial}{\partial x^{j_s}}\right)$ for using this definition, if we plug these coordinate vector fields and coordinate covector fields into either the left or the right side of the expression above we will get the same answer. Any vector or covector fields on $U$ can be written as a linear combination of these coordinate fields with smooth functions as coefficients, and the multilinear properties of tensors will ensure that both sides get the same value no matter what fields we evaluate them on. Similarly, if $\alpha$ is a differential $k$-form and $(U,x)$ is a coordinate patch within its domain, then we can write $\displaystyle\alpha\vert_U=\sum\limits_{1\leq i_1<\dots<i_k\leq n}\alpha_{i_1\dots i_k}dx^{i_1}\wedge\dots\wedge dx^{i_k}$ for some smooth functions $\alpha_{i_1\dots i_k}$ on $U$. The proof in this case is similar, following from the definition $\displaystyle\alpha_{i_1\dots i_k}=\alpha\left(\frac{\partial}{\partial x^{i_1}},\dots,\frac{\partial}{\partial x^{i_k}}\right)$ In this case we can pick the indices to be strictly increasing because of the antisymmetry of the tensors. ### Like this: Posted by John Armstrong \| Differential Topology, Topology ## 5 Comments » 1. [...] seen that given a local coordinate patch we can decompose tensor fields in terms of the coordinate bases and on and , respectively. But what happens if we want to pass [...] Pingback by \| July 8, 2011 \| Reply 2. [...] for exterior multiplication, we will use the fact that we can write any -form as a linear combination of -fold products of -forms. Thus we only have to check [...] Pingback by \| July 13, 2011 \| Reply 3. [...] show that the result doesn’t actually depend on this choice. Picking a coordinate patch gives us a canonical basis of the space of -forms over , indexed by “multisets” . Any -form over can be written [...] Pingback by \| July 19, 2011 \| Reply 4. [...] check this, we will use our trick: let be a coordinate patch around , giving us the basic coordinate vector fields in the patch. If [...] Pingback by \| September 27, 2011 \| Reply 5. [...] we know that any -form on can be written out as a sum of functions times -fold [...] Pingback by \| October 8, 2011 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 17, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8998980522155762, "perplexity_flag": "head"}
http://mathhelpforum.com/geometry/3452-semi-perimeter-s.html	# Thread: 1. ## The semi-perimeter 's' P is any point inside a triangle ABC. The perimeter of the triangle AB +BC+CA =2s. Prove that s< AP+BP+CP<2s. 2. Originally Posted by malaygoel P is any point inside a triangle ABC. The perimeter of the triangle AB +BC+CA =2s. Prove that s< AP+BP+CP<2s. I did not do this inequality, but since you seem knowledgable this should greatly help you finish it. ------ That point is called the "Fermat Point". Consider triangle ABC and let P be inside this triangle. We want that AP+BP+CP to be minimized. Rotate 60 degrees triangle APB around point B to outside. You now have a new triangle A'P'B Triangle P'BP is equilateral cuz, P'B=PB with <P'BP=60 Thus AP+BP+CP=A'P'+P'P+PC But A'P'+P'P+PC is the path from A' to C It is minimal when A'P'+P'P+PC is a sraight line. Thus, BPC=180-P'PB=180-60=120 Thus, the "Fermat Point" form lines of 120 degrees. 3. I have completed the solution to this problem. As I suspected my first post on this problem was necessary to complete the proof. Thus, to understand my solution you will not to understand my first post denomstrating that the minimized point has the property that, <CPB= <BPA= <CPA= 120 degrees. Below is a picture showing the geometric construction in case you were no able to draw one yourself due to my bad explantion. Attached Thumbnails 4. I will demonstrate the first part of the inequality that, $s<AP+BP+CP$ That is simple to answer. 1)Triangle ABC 3)Point P is inside triangle ABC 2)Draw AP, BP, CP 3)Three triangle, APB, APC, BPC. 4)By the triangular inequality you have, $\left\{ \begin{array}{c}<br /> AP+BP> AB\\<br /> BP+PC> BC\\<br /> AP+PC> AC\\<br />$ Add them, $2AP+2BP+2CP> AB+BC+AC=2s$ Thus, $2(AP+BP+CP)> 2s$ Thus, $AP+BP+CP>s$ And the first part is complete. Note: I made no use of the fact that P was a "Fermat Point". 5. I appologize that I have not been able to find an elegant geometric method to demonstrate the second part of this inequality. You need to be familiar with Lagrange Multipliers. Begin by P being a "Fermat Point", Meaning that the triangles, APB, APC, BPC are have angle P all 120 degress (this is my first post). Take triangle APC and let side $AP=a$ and sides $AP=x,PC=y$ Now, we will find the necessary conditions that $AP+PC=x+y$ is maximized. Since angle P=120 we can use the law of cosines, $a^2=x^2+y^2-2xy\cos 120=x^2+xy+y^2$. In analytic terms we need to, $\mbox{Maximize }x+y$ $\mbox{Subject to }x^2+xy+y^2=a^2, x,y>0$ Thus, maximize $f(x,y)=x+y$ with constraint curve, $g(x,y)=a^2$ where $g(x,y)=x^2+xy+y^2$. By Lagrange's Theorem,if $(x,y)$ is a extrenum point, then $\nabla f(x_0,y_0)=k\nabla g(x_0,y_0)$ But, $\nabla f(x,y)=\frac{\partial (x+y)}{\partial x}\bold{i}+\frac{\partial (x+y)}{\partial y}\bold{j}=\bold{i}+\bold{j}$ Also, $\nabla g(x,y)=\frac{\partial (x^2+xy+y^2)}{\partial x}\bold{i}+\frac{\partial (x^2+xy+y^2)}{\partial y}\bold{j}$= $(2x+y)\bold{i}+(2y+x)\bold{j}$ Thus, $\bold{i}+\bold{j}=k(2x+y)\bold{i}+k(2y+x)\bold{j}$ Thus solve, $\left\{ \begin{array}{c}<br /> 2x+y=1/k\\<br /> 2y+x=1/k\\<br /> x^2+xy+y^2=a^2$ A unique solution exists, $(x,y,k)=\left( \frac{a}{\sqrt{3}},\frac{a}{\sqrt{3}}, a\sqrt{3} \right)$ Thus the extrerum value for $x+y$ is $x=y=\frac{a}{\sqrt{3}}$ (an isoseles triangle with vertex 120). Question is whether this is maximum or minimum. Select any triangle with longest side $a$ and angle $120$ and see whether the sum of its sides exluding $a$ exceeds $x+y=\frac{2a}{\sqrt{3}}$ if it does then we found that the minimized value for $x+y$ otherwise we found the maximized value. For example, select a triangle with sides, $d,2d,a$ Thus, by the law of cosines, $d+2d^2+4d^2=a^2$ thus, $d=\frac{a}{\sqrt{7}}$ and $2d=\frac{2a}{\sqrt{7}}$ in sum you have, $\frac{3a}{\sqrt{7}}<\frac{2a}{\sqrt{3}}$ Thus, $x+y$ is a maximum point! Now, the rest is easy. By this argument we have shown that the maximum value that AP+PC can have is $\frac{2a}{\sqrt{3}}$ Similarly, the max value of the AP+PB and BP+CP is $\frac{2b}{\sqrt{3}}$ and $\frac{2c}{\sqrt{3}}$ where $a,b$ represent the other sides of triangle. Thus, $2(AP+BP+CP)$= $(AP+PB)+(BP+CP)+(AP+PC)=\frac{2a}{\sqrt{3}}+\frac{ 2b}{\sqrt{3}}+\frac{2c}{\sqrt{3}}$= $\frac{2(a+b+c)}{\sqrt{3}}$ Thus,(the largest possible value of) $AP+BP+CP=\frac{a+b+c}{\sqrt{3}}=\frac{2s}{\sqrt{3} }<2s$ Because, $\frac{1}{\sqrt{3}}<1$ Since its max is less than $2s$ then certainly an other lengths (because they are less than max which is less than $2s$ and use transitive property.) 6. ## Try, it is simple Originally Posted by ThePerfectHacker I appologize that I have not been able to find an elegant geometric method to demonstrate the second part of this inequality. You need to be familiar with Lagrange Multipliers. I am not familiar with Langrange Multipliers, nevertheless I have been able able to work out a geometric method to demonstrate the secont part of the inequality, although I don't know whether it is elegant or not. The method makes use of the triangle inequality and the condition stated in the question(could you figure out the condition?). Try to work it out, you will have fun and also it will strenghthen your belief in simplicity as it has done mine. 7. Originally Posted by malaygoel I am not familiar with Langrange Multipliers, nevertheless I have been able able to work out a geometric method to demonstrate the secont part of the inequality, although I don't know whether it is elegant or not. The method makes use of the triangle inequality and the condition stated in the question(could you figure out the condition?). Try to work it out, you will have fun and also it will strenghthen your belief in simplicity as it has done mine. If, $x,y,z$ represents the lengths of AP,BP,CP and $a,b,c$ then, $\left\{ \begin{array}{c}<br /> x^2+xy+y^2=a^2\\<br /> x^2+xz+z^2=b^2\\<br /> y^2+yz+z^2=c^2\\<br />$ But solving these equations is probably messy and the solution is not as simple as through Lagrange Multipliers. 8. Originally Posted by ThePerfectHacker If, $x,y,z$ represents the lengths of AP,BP,CP and $a,b,c$ then, $\left\{ \begin{array}{c}<br /> x^2+xy+y^2=a^2\\<br /> x^2+xz+z^2=b^2\\<br /> y^2+yz+z^2=c^2\\<br />$ But solving these equations is probably messy and the solution is not as simple as through Lagrange Multipliers. I will post the second proof I have worked out, but to do it a diagram a needed.Please post the following diagram and I will follow it with the solution. Draw a triangle ABC Take a point P in the interior of ABC Join AP, BP and CP Extend AP to meet BC at D(this can be done since P is an interior point) Keep Smiling 9. Originally Posted by malaygoel Extend AP to meet BC at D(this can be done since P is an interior point) Two mistakes (since I am a proud formalist ) 1)You need to demonstrate that AP intersects BC. Actually I made a thread about this, but cannot find it 2)You forgot to state the second postulate, how dare you! Attached Thumbnails 10. Originally Posted by ThePerfectHacker Two mistakes (since I am a proud formalist ) 1)You need to demonstrate that AP intersects BC. Actually I made a thread about this, but cannot find it 2)You forgot to state the second postulate, how dare you! Is there any condition in which AP doesn't intersect BC? I didn't forget to state the second postulate, you can see my first post in this thread. Nevertheless, if it not the condition you cannot prove it because it is not valid then. Keep Smiling 11. Originally Posted by malaygoel I will post the second proof I have worked out, but to do it a diagram a needed.Please post the following diagram and I will follow it with the solution. I will use only triangle inequality so I will not state the name of ineequality used in solution. In triangle ABD $AB + BD > AP + PD$ $AC + CD > AP + PD$ $PD + BD> PB$ $PD + CD> PC$ Adding these four equations we get $AB + BC + CA > 2AP + PB + PC$ Let BP intersect AC at E, CP intersect AB at F Similarly we can form three eqution. adding three gives $4(AB + AC + BC) > 4(AP + BP + CP)$ $2s > AP + BP + CP$ Proof complete Do you like it? Keep Smiling Malay	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 59, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9307631850242615, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/34860-algorithm.html	# Thread: 1. ## Algorithm the _ stand for subscript Code: begin Input X_1 , X_2 .....X_n count := - for i:=2 to n do begin for j:=1 to (i-1) do begin if x_i = x_j then begin count := count = 1 end end output count end end i need to see if the algorithm works with the list 2, 3,6,2,6 the theory would be very muc appricated many thanks 2. Originally Posted by Tom101 the _ stand for subscript Code: begin Input X_1 , X_2 .....X_n count := - for i:=2 to n do begin for j:=1 to (i-1) do begin if x_i = x_j then begin count := count = 1 end end output count end end i need to see if the algorithm works with the list 2, 3,6,2,6 the theory would be very muc appricated many thanks It would be a good idea if you told us what you thought this did. As you have written it it should output "false" Ron: 3. Originally Posted by CaptainBlack It would be a good idea if you told us what you thought this did. As you have written it it should output "false" well looking at it you started by inputting a sequecne of numbers set the count to zero the you see if i is equal to 2 - n and then you begin the fsecond loop which check to see if j is equal to 1 all the way to i-1 (which in the first casue is 1) and if that is correct then you add one to the count if x_i it equal to x_j you end but outputting the count that is what i think it does many thanks 4. Originally Posted by Tom101 well looking at it you started by inputting a sequecne of numbers set the count to zero the you see if i is equal to 2 - n and then you begin the fsecond loop which check to see if j is equal to 1 all the way to i-1 (which in the first casue is 1) and if that is correct then you add one to the count if x_i it equal to x_j you end but outputting the count that is what i think it does many thanks You assign count to "count=1" that is count becomed true if "count" is 1 otherwise count is "false" Also you intialised count to "-" RonL 5. sorry that was a typo meant to inilize count to 0 6. Originally Posted by Tom101 sorry that was a typo meant to inilize count to 0 So you mean: Code: begin Input X_1 , X_2 .....X_n count := 0 for i:=2 to n do begin for j:=1 to (i-1) do begin if x_i = x_j then begin count := count + 1 end end output count end end RonL 7. Originally Posted by CaptainBlack So you mean: Code: begin Input X_1 , X_2 .....X_n count := 0 for i:=2 to n do begin for j:=1 to (i-1) do begin if x_i = x_j then begin count := count + 1 end end output count end end RonL Which can be animated so we get a trace: Code: >load "C:\Documents and Settings\Ron\My Documents\temp\test.e" >type test function test () count= 0; x=[2,3,6,2,6]; n=length(x); for i=2 to n for j=1 to (i-1) if x(i) == x(j) count = count + 1; endif [i,j,count] end end return 0 endfunction > > >..output is i, j, count > >test 2 1 0 3 1 0 3 2 0 4 1 1 4 2 1 4 3 1 5 1 1 5 2 1 5 3 2 5 4 2 0 > RonL 8. thanks just wondering, that the 6 is not used, is that becasue it cannot be used inthe sequence... cheers 9. Originally Posted by Tom101 thanks just wondering, that the 6 is not used, is that becasue it cannot be used inthe sequence... cheers The 6 is used it is x(n) (which is x(5) in this case) RonL 10. thanks a lot it has been much help 11. so what you are saying that the algorithm works on the list 2,3,6,2,6. and could you plz tell me what is the time comlexity function for the algorithm by calculating how many times the comarison xi = xj is performed for an input sequence of length n. Originally Posted by CaptainBlack Which can be animated so we get a trace: Code: >load "C:\Documents and Settings\Ron\My Documents\temp\test.e" >type test function test () count= 0; x=[2,3,6,2,6]; n=length(x); for i=2 to n for j=1 to (i-1) if x(i) == x(j) count = count + 1; endif [i,j,count] end end return 0 endfunction > > >..output is i, j, count > >test 2 1 0 3 1 0 3 2 0 4 1 1 4 2 1 4 3 1 5 1 1 5 2 1 5 3 2 5 4 2 0 > RonL 12. Originally Posted by badi6 so what you are saying that the algorithm works on the list 2,3,6,2,6. and could you plz tell me what is the time comlexity function for the algorithm by calculating how many times the comarison xi = xj is performed for an input sequence of length n. For each $i$ the comparison is made $i-1$ times. So the total number of comparisons is: $N(n)=\sum_{i=2}^n (i-1)=\frac{n(n-1)}{2}$ RonL 13. ## Thank you very much! thanks for your quick reply but if one asks you to verify that the algorithm works on the list 2, 3, 6, 2, 6. what would be the best answer then? thanks again 14. Originally Posted by badi6 thanks for your quick reply but if one asks you to verify that the algorithm works on the list 2, 3, 6, 2, 6. what would be the best answer then? thanks again Tracing the execution is one way. RonL 15. Originally Posted by CaptainBlack Which can be animated so we get a trace: Code: >load "C:\Documents and Settings\Ron\My Documents\temp\test.e" >type test function test () count= 0; x=[2,3,6,2,6]; n=length(x); for i=2 to n for j=1 to (i-1) if x(i) == x(j) count = count + 1; endif [i,j,count] end end return 0 endfunction > > >..output is i, j, count > >test 2 1 0 3 1 0 3 2 0 4 1 1 4 2 1 4 3 1 5 1 1 5 2 1 5 3 2 5 4 2 0 > RonL As it says above the output is i, j, count for each trip around the inner loop. The final 0 is the return value of the function to indicate successfull execution, and can be ignored. The only thing that might be considered slightly obscure is the line [i,j,count], which because it does not end in a semicolon echos the vector consisting ot i, j and count to the console. RonL	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9463721513748169, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/152246-solutions-trigonometic-equation.html	# Thread: 1. ## Solutions to Trigonometic Equation Finding all exact solutions to the equation on [0, 2pi) 2(cos^2)x + 3sinx = 3 2. I would suggest rewriting this using the fact that $\cos^2 x + \sin^2 x = 1$. From there you can treat $\sin x$ as a variable and solve the quadratic that results. If you don't understand the second part just post what you have after using the first suggestion.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.944092869758606, "perplexity_flag": "head"}
http://mathoverflow.net/questions/108949/various-definitions-of-recursion-from-ordinal-machines	## Various definitions of recursion from ordinal machines ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Background: I'm trying to get an intuitive understanding of α-recursion and related concepts in higher recursion theory. Once nice book is Peter Hinman's Recursion-Theoretic Hierarchies, available here (open access), and specifically chapter VIII which is intuitively appealing because he presents recursion on ordinals by directly constructing recursive functions rather than going through the constructible hierarchy. However, this also suggests a number of questions which he does not fully answer, and I'm hoping someone can help me get a clearer picture. Let me make the following definitions, which I state informally but should be straightforward to formalize: • A primitive `$(\infty,0)$`-machine is given by a program in a programming language, all of whose variables are ordinals, which can do the following: • Basic operations (assigning variables to other variables or to any natural number, comparing ordinals, computing the successor, predecessor, sum and product, and constructing/deconstructing finite tuples of ordinals as ordinals using the standard bijection between `$\mathit{Ord}^2$` and `$\mathit{Ord}$`; part of this is probably redundant). • Looping over an ordinal `$\alpha$` which has been computed in advance (i.e., in a variable): by this I mean that a subprogram will be iterated for all ordinals `$\beta<\alpha$`; to define the value of a variable at limit times `$\beta$`, we declare it to be the lim sup of variables up to this point. These loops always terminate. (Of course, it doesn't hurt to allow the program to interrupt the loop early.) • Recursive function calls are not allowed (all subroutines must be defined prior to where they are used). Because of this, a primitive `$(\infty,0)$`-machine always terminates, whatever its inputs. Comment: Unless I got my definition horribly wrong, a primitive `$(\infty,0)$`-machine, when given natural numbers as input, can only produce a natural number as output, and the functions `$\omega^r \to \omega$` thus defined are exactly the primitive recursive functions as usually defined. • A general `$(\infty,0)$`-machine is similar to a primitive `$(\infty,0)$`-machine, except that recursive function calls are allowed: unless I am mistaken, this is equivalent to allowing a "universal" machine (i.e., a general `$(\infty,0)$`-machine can compute the value returned by another general `$(\infty,0)$`-machine on any specified input). Programs can now fail to terminate (and the precise (program,input)↦output partial function is defined as the smallest which satisfies the requirements). This definition is supposed to coincide with what Hinman's aforementioned book calls a "`$(\infty,0)$`-partial recursive function" (with or without parameters as are passed to the machine): see p. 377. If it does not coincide, I must have made a mistake. The difference between "general" and "primitive" is supposed to be clause (2) in definition 1.1 on p. 376 of Hinman's book. Comment: Again, when given natural numbers as input, a general `$(\infty,0)$`-machine can only produce natural numbers as output (or fail to terminate), and the partial functions `$\omega^r \to \omega$` thus defined are exactly the usually defined general recursive (partial) functions. • Lastly, if `$\lambda$` is any ordinal (fixed in advance), a primitive `$(\infty,\lambda)$`-machine and a general `$(\infty,\lambda)$`-machine are the same as a primitive `$(\infty,0)$`-machine and a general `$(\infty,0)$`-machine respectively, but with an additional construct: • We can also loop over all ordinals `$\beta<\lambda$`; the difference with the previously defined type of loop (apart from the fact that here `$\lambda$` is fixed in advanced rather than computed from the input) is that if the program does not interrupt the loop early (i.e., issue a `break` at some time `$\beta<\lambda$`), it does not terminate. Again, general `$(\infty,\lambda)$`-machines are supposed to define the same thing as Hinman's "`$(\infty,\lambda)$`-partial recursive function". Note: If `$\kappa$` is recursively regular (=admissible), then the total functions computed by general `$(\infty,\kappa)$`-machines on inputs `$<\kappa$`, and possibly using extra parameters (=constants) `$<\kappa$`, are exactly the `$\kappa$`-recursive total functions `$\kappa^r \to \kappa$` (viz., those which are `$\Delta^1_1$`-definable over `$L_\kappa$`). I believe that primitive `$(\infty,\kappa)$`-machines also compute the same functions in this context (because recursivity can be simulated by a loop over ordinals `$<\kappa$`), but that's not really important. Now basically what I'd like to understand is what can be computed by a general `$(\infty,0)$`-machine (=general 0-machine), or even a primitive `$(\infty,0)$`-machine, if it is allowed a given ordinal `$\alpha$` as input. So here is my MAIN QUESTION: Let `$\alpha$` be an ordinal, and let `$\kappa$` be the smallest recursively regular (=admissible) ordinal `$>\alpha$`. Is it true that any total function `$\kappa^r \to \kappa$` which can be computed using a general `$(\infty,\kappa)$`-machine without extra parameters (=constants) can, in fact, be computed using a general `$(\infty,0)$`-machine from the only parameter `$\alpha$`? This question is motivated by the fact that if given the parameter `$\omega$`, a general `$(\infty,0)$`-machine can clearly compute all hyperarithmetic functions `$\omega^r \to \omega$`, it can compute any ordinal `$<\omega_1^{CK}$`, and it seems as though it can compute the same things as a general `$(\infty,\omega_1^{CK})$`-machine. A positive answer to this question would imply, among other things, that the ordinals stable under general `$(\infty,0)$`-recursive functions are exactly the recursively regular ordinals and limits thereof. BONUS QUESTION: With the same notations, is it true for `$\gamma<\kappa$` that any total function `$\gamma^r \to \gamma$` which can be computed using a general `$(\infty,\kappa)$`-machine without extra parameters (=constants) can, in fact, be computed using a primitive `$(\infty,0)$`-machine from the only parameter `$\alpha$`? More generally, I'm interested in any statements along the same lines that can help clear the relation between these various notions of ordinal machines. - 1 As an aside, if you haven't seen it already, Sacks' book "Higher recursion theory" is also quite good. – Noah S Oct 5 at 23:47 1 I'd hesitate to give a "answer" without seeing the full definitions, but what you've written makes a positive answer extremely plausible. This kind of functional notation for $\alpha$-recursion theory (as in Hinman) has somewhat gone out of fashion (as has $\alpha$-recursion theory itself). But Aczel & Richter's article may be of use here where they discuss the closure of various monotone operators: Inductive definitions and reflecting properties of admissible ordinals, in "Generalised Recursion Theory, Proc. Symp., Oslo", Ed Fenstad & Hinman, 1972, North-Holland, 1974 – Philip Welch Dec 30 at 10:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9002844095230103, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/20837/list	## Return to Question 2 Motivation to question and background Motivation to this question, short version: I feel this is a good new way to look at functions while retaining the time-domain toolset of real analysis. More motivation for this question: I am wondering whether this sort of analysis tool could become a generalization of the time domain. We have a lot of tools for analyzing functions in the frequency domain, and we have a lot of tools in the time domain - however the results they yield are somewhat disjoint. The way this 'spinning graph' effect depends directly on the frequency content of the function at hand. It could make it possible to do time domain based analysis of some form of frequency content, or rather 'slope content': what I am refering to here is the fact that in the time domain it's easier to think in terms of slope rather than the frequency. The two are tied together (a function with a sharp rising edge will probably have a lot of high-frequency content) but not directly (a simple 1 Hz sinusoid with enough amplitude can have higher slope on an interval than a 200 Hz triangle wave). This tool that arises from simply frequency-shifting the function gives a new way to look at time-domain signals, and so I think it could give rise to interesting questions - I am extremely surprised that this is not mentioned anywhere in literature: it feels like a very direct generalization of the time domain display of a signal and can be applied to pretty much every real-valued function out there giving us new information about it. On the one hand it is very useful for periodic functions, on the other hand the way the graph seems to 'coil up' when the frequency shift is increased can make it useful for the analysis of functions which are not periodic. I strongly feel that the analysis of 'where the graph is on that spinning cylinder' can provide new information about functions. Background: I had come across this effect when studying the musical properties of stretched-harmonic waveform synthesis, motivated by the fact that real instruments tend to have slightly inharmonic rather than harmonic frequency content, whereas electronic synthesizers tend to have purely harmonic timbre. This got me asking some questions not really related to music! 1 # What is this effect in Fourier/additive synthesis called? Hi, I have re-synthesized a cyclic function additively, and I added a fixed offset to the frequency of each partial. So if the function was $\sum a_{n} sin(2 \pi x * n)$ and its frequencies were $nf_{B}$ where n is the number of the partial and f_B is the base frequency of the function, then now the frequencies of the partials are $nf_{B}+f_{o}$. The function becomes an almost-periodic function. The waveform seems to be 'rotating' around the time axis. Is this effect described somewhere? If so, where, and what is its name? Here are animations of this happening: rotation.rar, 2MB By the end of the first animation the function is a bit rippled, those ripples are only created by errors in the model. It should in fact look smooth like in the beginning. In the 2nd movie, at the beginning I raise and then lower the offset frequency of the partials (f_o), and then set it to 0. On 0:32 I reset the resynthesis model (this is irrelevant to the effect, but it explains why the waveform suddenly changes). Later I set f_o to +1 Hz and 0 Hz, alternating the setting. In the 3rd and 4th animations I alternate f_o between +1 Hz and 0 Hz. It is as though the function graph is drawn on a 3-dimensional glass cylinder that rotates around the time axis. To avoid confusion, I define that if f_o is positive, then the direction of the rotation is positive. Note that the points of the graph don't seem to be equidistant from the axis of rotation - some seem to be orbiting at smaller and some at bigger radii. This is not really exposed in the movies, but if you start going up very far with f_o, the graph becomes 'twisted', as if the 'cylinder' were becoming twisted around its axis of rotation. Thanks!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9489470720291138, "perplexity_flag": "middle"}
http://mathhelpforum.com/number-theory/139024-trouble-solving-system-linear-congruences.html	# Thread: 1. ## Trouble solving a system of linear congruences. Hi all, I have three related linear congruences, described below, and I'm trying to solve for x and y. 224x + 225y = 226 (mod 228) 3x + 6y = 153 (mod 228) 0x + 226y = 132 (mod 228) The trouble I'm having is that these can have multiple solutions and I don't know of a good way to determine which ones to use without calculating all of them. For example, 226y = 132 (mod 228) has two solutions, y=48 and y=162. The second equation (3x + 6y = 153 mod 228) will have three solutions for x regardless of which of the values for y is chosen (x=31, x=107, x=183). However, the first equation (224x + 225y = 226 mod 228) has no solution for x when y=48 and has 4 solutions when y=162 (x=50, x=107, x=164, x=221). After having calculated all of the combinations, I can tell by looking at it that y=162 (because y=48 doesn't solve the first congruence) and that x=107 (because it is common in the solutions for both the first and second congruence. Is there a formula I can apply to determine which solutions to use for any particular congruence, given the related congruences? Maybe some way to calculate them all simultaneously? While calculating solutions for all of the combinations worked out in this case, I don't think it is practical when the number of unknowns increases. I started with five unknowns across six congruences but used matrix reduction to get it down to this. I'm a computer science major without a very strong number theory background and I'm not even sure what to call the problem I'm trying to solve, which makes it difficult to search for on the internet (though I've tried) or to find what sections of math books might be related (though I've scoured the local book stores). Any help would be appreciated. m 2. The congruence $ax\equiv{b}\ (\text{mod}\ m)$ has a solution if and only if d=gcd(a,m) divides b. The solution is given by $x\equiv{x_0}\ \left(\text{mod}\ \frac{m}{d}\right)$. That much I know. And I see you have put the system into upper triangular form, which is probably a good idea. One idea is to break the modulus down to powers of primes (in your case 228=4319) and solve for each modulus, then use the Chinese Remainder Theorem to get the final solution. I haven't worked out what happens for your system. I would just work backward through the equations (as you did): $226y\equiv{132}\ (\text{mod}\ 228)$ $y\equiv{48}\ (\text{mod}\ 114)$, since gcd(226,228)=2. Or, $y\equiv{48}+114z\ (\text{mod}\ 228)$ where z=0 or 1. Now the next equation up: $3x+6y\equiv{153}\ (\text{mod}\ 228)$ $3x+6(48+114z)\equiv{153}\ (\text{mod}\ 228)$ $3x+60\equiv{153}\ (\text{mod}\ 228)$ (z dropped out of the equation, otherwise we would treat it like a constant) $x\equiv{31}+76w\ (\text{mod}\ 228)$ where w=0,1, or 2. So that gives us six solutions. But there is another equation: $224x+225y\equiv{226}\ (\text{mod}\ 228)$ Since gcd(224,228)=4 and gcd(225,228)=1, substitute for y: $224x+225(48+114z)\equiv{226}\ (\text{mod}\ 228)$ $224x+84+114z\equiv{226}\ (\text{mod}\ 228)$ $224x\equiv{142}-114z\ (\text{mod}\ 228)$ and since gcd(224,228)=4 must divide 142-114z, we must have z=1. $224x\equiv{28}\ (\text{mod}\ 228)$ Now substitute for x: $224(31+76w)\equiv{28}\ (\text{mod}\ 228)$ $104+152w\equiv{28}\ (\text{mod}\ 228)$ and only w=1 works. So only the solution z=1, w=1 is valid, so the solution is: $x\equiv{107}\ (\text{mod}\ 228)$ $y\equiv{162}\ (\text{mod}\ 228)$ - Hollywood 3. It does work to break down the modulus into powers of primes and solve 3 systems of congruences: 224x + 225y = 226 (mod 228) 3x + 6y = 153 (mod 228) 0x + 226y = 132 (mod 228) The three systems are: 0x + 1y = 2 (mod 4) 3x + 2y = 1 (mod 4) 0x + 2y = 0 (mod 4) solution: x=3, y=2 (mod 4) 2x + 0y = 1 (mod 3) 0x + 0y = 0 (mod 3) 0x + 1y = 0 (mod 3) solution: x=2, y=0 (mod 3) 15x + 16y = 17 (mod 19) 3x + 6y = 1 (mod 19) 0x + 17y = 18 (mod 19) solution: x=12, y=10 (mod 19) And applying the CRT gives the solution. - Hollywood 4. ## Thanks Hollywood, I hadn't considered CRT. Thanks Hollywood for spotting that. I was looking into calculating an inverse matrix and using that to coordinate the different congruences, but I was worried because not all matrices have an inverse. I'll try coding your method up and seeing if there are any cases in which it does not work. Thanks again, m	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9330175518989563, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/44464/does-the-algebraic-structure-of-a-lie-group-restrict-the-possible-dimensions-of/44468	# Does the algebraic structure of a Lie group restrict the possible dimensions of other Lie groups isomorphic to it? In a recent question, I initially doubted that $\mathbb{C}^\times\cong S^1$, my intuition being that $\mathbb{C}^\times$ has one more "dimension" than $S^1$ - in rigorous terms, $S^1$ is (or rather, can be given the structure of) a 1-dimensional Lie group, and $\mathbb{C}^\times$ is (can be given the structure of) a 2-dimensional Lie group. My question is rather naive: Given a Lie group $G$ of dimension $n\in\mathbb{N}\cup\{\infty\}$, for which $m\in\mathbb{N}\cup\{\infty\}$ does there exist a Lie group $H$ of dimension $m$ such that $G\cong H$ as groups (forgetting about the manifold structure)? Surely there is at least one $G$ such that the answer isn't all $m$? - I guess your question is motivated by the fact that for all $n,m \in \mathbb{N}$, you have a group ($\mathbb{Q}$-vector space) isomorphism $\mathbb{R}^n \simeq \mathbb{R}^m$, right ? – Joel Cohen Jun 10 '11 at 1:23 I had also thought about that while composing the question, but I was mainly motivated by my incorrect intuition about $\mathbb{C}^\times\cong S^1$. – Zev Chonoles♦ Jun 10 '11 at 2:16 Still informally, re the dimension issue, $C^x$ is homotopic to $S^1$, by retraction. – gary Jun 10 '11 at 16:52 ## 2 Answers I believe all 1-dimensional real lie groups are abelian, so given a non-abelian lie group there exists no lie group of dimension 1 isomorphic to it. - A related question was asked at MathOverflow. The author argues that there is a unique Lie group structure on $SU(2)$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.950722336769104, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Tetrahedron	# Tetrahedron Not to be confused with tetrahedroid. For the academic journal, see Tetrahedron (journal). Regular Tetrahedron Type Platonic solid Elements F = 4, E = 6 V = 4 (χ = 2) Faces by sides 4{3} Schläfli symbol {3,3} and s{2,2} Wythoff symbol 3 \| 2 3 \| 2 2 2 Coxeter diagram Symmetry Td, A3, [3,3], (332) Rotation group T, [3,3]+, (332) References U01, C15, W1 Properties Regular convex deltahedron Dihedral angle 70.528779° = arccos(1/3) 3.3.3 (Vertex figure) Self-dual (dual polyhedron) Net In geometry, a tetrahedron (plural: tetrahedra) is a polyhedron composed of four triangular faces, three of which meet at each vertex. It has six edges and four vertices. The tetrahedron is the only convex polyhedron that has four faces.[1] The tetrahedron is the three-dimensional case of the more general concept of a Euclidean simplex. The tetrahedron is one kind of pyramid, which is a polyhedron with a flat polygon base and triangular faces connecting the base to a common point. In the case of a tetrahedron the base is a triangle (any of the four faces can be considered the base), so a tetrahedron is also known as a "triangular pyramid". Like all convex polyhedra, a tetrahedron can be folded from a single sheet of paper. It has two nets.[1] For any tetrahedron there exists a sphere (the circumsphere) such that the tetrahedron's vertices lie on the sphere's surface. ## Regular tetrahedron Main article: Regular tetrahedron A regular tetrahedron is one in which all four faces are equilateral triangles, and is one of the Platonic solids. ### Formulas for a regular tetrahedron The following Cartesian coordinates define the four vertices of a tetrahedron with edge-length 2, centered at the origin: (±1, 0, -1/√2) (0, ±1, 1/√2) Another set of coordinate are based on an alternated cube. Inverting these coordinates generates another dual tetrahedron, with edge length $\sqrt{2}$; (1,1,1), (1,-1,-1), (-1,1,-1), (-1,-1,1) For a regular tetrahedron of edge length a: Base plane area $A_0={\sqrt{3}\over4}a^2\,$ Surface area[2] $A=4\,A_0={\sqrt{3}}a^2\,$ Height[3] $H={\sqrt{6}\over3}a=\sqrt{2\over3}\,a\,$ Volume[2] $V={1\over3} A_0h ={\sqrt{2}\over12}a^3={a^3\over{6\sqrt{2}}}\,$ Angle between an edge and a face $\arccos\left({1 \over \sqrt{3}}\right) = \arctan(\sqrt{2})\,$ (approx. 54.7356°) Angle between two faces[2] $\arccos\left({1 \over 3}\right) = \arctan(2\sqrt{2})\,$ (approx. 70.5288°) Angle between the segments joining the center and the vertices,[4] also known as the "tetrahedral angle" $\arccos\left({-1\over3}\right ) = 2\arctan(\sqrt{2})\,$ (approx. 109.4712°) Solid angle at a vertex subtended by a face $\arccos\left({23\over27}\right)$ (approx. 0.55129 steradians) Radius of circumsphere[2] $R={\sqrt{6}\over4}a=\sqrt{3\over8}\,a\,$ Radius of insphere that is tangent to faces[2] $r={1\over3}R={a\over\sqrt{24}}\,$ Radius of midsphere that is tangent to edges[2] $r_M=\sqrt{rR}={a\over\sqrt{8}}\,$ Radius of exspheres $r_E={a\over\sqrt{6}}\,$ Distance to exsphere center from a vertex $d_{VE}={\sqrt{6}\over2}a=\sqrt{3\over2}\,a\,$ Note that with respect to the base plane the slope of a face ($\scriptstyle 2 \sqrt{2}$) is twice that of an edge ($\scriptstyle \sqrt{2}$), corresponding to the fact that the horizontal distance covered from the base to the apex along an edge is twice that along the median of a face. In other words, if C is the centroid of the base, the distance from C to a vertex of the base is twice that from C to the midpoint of an edge of the base. This follows from the fact that the medians of a triangle intersect at its centroid, and this point divides each of them in two segments, one of which is twice as long as the other (see proof). ### Orthogonal projections The regular tetrahedron has two special orthogonal projections, one centered on a vertex, or equivalently on a face, and one centered on an edge. The first corresponds to the A2 Coxeter plane. Orthogonal projection Centered by Edge Face/vertex Image Projective symmetry [4] [3] ## Other special cases An isosceles tetrahedron, also called a disphenoid, is a tetrahedron where all four faces are congruent isosceles triangles. A space-filling tetrahedron packs with congruent copies of itself to tile space, like the disphenoid tetrahedral honeycomb. Name Coxeter Diagram Faces Symmetry Schönflies Coxeter Orbifold Order Digonal disphenoid Two types of isosceles triangles D1h [2] (22) 4 Rhombic disphenoid (scalene tetrahedron) Identical scalene triangles D2 [2,2]+ (222) 4 Tetragonal disphenoid (isosceles tetrahedron) Identical isosceles triangles D2d [2+,4] (22) 8 Regular tetrahedron Equilateral triangles Td [3,3] (332) 24 In a trirectangular tetrahedron the three face angles at one vertex are right angles. If all three pairs of opposite edges of a tetrahedron are perpendicular, then it is called an orthocentric tetrahedron. When only one pair of opposite edges are perpendicular, it is called a semi-orthocentric tetrahedron. An isodynamic tetrahedron is one in which the cevians that join the vertices to the incenters of the opposite faces are concurrent, and an isogonic tetrahedron has concurrent cevians that join the vertices to the points of contact of the opposite faces with the inscribed sphere of the tetrahedron. ## Volume The volume of a tetrahedron is given by the pyramid volume formula: $V = \frac{1}{3} A_0\,h \,$ where A0 is the area of the base and h the height from the base to the apex. This applies for each of the four choices of the base, so the distances from the apexes to the opposite faces are inversely proportional to the areas of these faces. For a tetrahedron with vertices a = (a1, a2, a3), b = (b1, b2, b3), c = (c1, c2, c3), and d = (d1, d2, d3), the volume is (1/6)·\|det(a − d, b − d, c − d)\|, or any other combination of pairs of vertices that form a simply connected graph. This can be rewritten using a dot product and a cross product, yielding $V = \frac { \|(\mathbf{a}-\mathbf{d}) \cdot ((\mathbf{b}-\mathbf{d}) \times (\mathbf{c}-\mathbf{d}))\| } {6}.$ If the origin of the coordinate system is chosen to coincide with vertex d, then d = 0, so $V = \frac { \|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\| } {6},$ where a, b, and c represent three edges that meet at one vertex, and a · (b × c) is a scalar triple product. Comparing this formula with that used to compute the volume of a parallelepiped, we conclude that the volume of a tetrahedron is equal to 1/6 of the volume of any parallelepiped that shares three converging edges with it. The triple scalar can be represented by the following determinants: $6 \cdot V =\begin{vmatrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \end{vmatrix}$ or $6 \cdot V =\begin{vmatrix} \mathbf{a} \\ \mathbf{b} \\ \mathbf{c} \end{vmatrix}$ where $\mathbf{a} = (a_1,a_2,a_3) \,$ is expressed as a row or column vector etc. Hence $36 \cdot V^2 =\begin{vmatrix} \mathbf{a^2} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{a} \cdot \mathbf{b} & \mathbf{b^2} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{a} \cdot \mathbf{c} & \mathbf{b} \cdot \mathbf{c} & \mathbf{c^2} \end{vmatrix}$ where $\mathbf{a} \cdot \mathbf{b} = ab\cos{\gamma}$ etc. which gives $V = \frac {abc} {6} \sqrt{1 + 2\cos{\alpha}\cos{\beta}\cos{\gamma}-\cos^2{\alpha}-\cos^2{\beta}-\cos^2{\gamma}}, \,$ where α, β, γ are the plane angles occurring in vertex d. The angle α, is the angle between the two edges connecting the vertex d to the vertices b and c. The angle β, does so for the vertices a and c, while γ, is defined by the position of the vertices a and b. Given the distances between the vertices of a tetrahedron the volume can be computed using the Cayley–Menger determinant: $288 \cdot V^2 = \begin{vmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & d_{12}^2 & d_{13}^2 & d_{14}^2 \\ 1 & d_{12}^2 & 0 & d_{23}^2 & d_{24}^2 \\ 1 & d_{13}^2 & d_{23}^2 & 0 & d_{34}^2 \\ 1 & d_{14}^2 & d_{24}^2 & d_{34}^2 & 0 \end{vmatrix}$ where the subscripts $i,\,j\in\{1,\,2,\,3,\,4\}$ represent the vertices {a, b, c, d} and $\scriptstyle d_{ij}$ is the pairwise distance between them – i.e., the length of the edge connecting the two vertices. A negative value of the determinant means that a tetrahedron cannot be constructed with the given distances. This formula, sometimes called Tartaglia's formula, is essentially due to the painter Piero della Francesca in the 15th century, as a three dimensional analogue of the 1st century Heron's formula for the area of a triangle.[5] ### Heron-type formula for the volume of a tetrahedron If U, V, W, u, v, w are lengths of edges of the tetrahedron (first three form a triangle; u opposite to U and so on), then[6] $\text{volume} = \frac{\sqrt {\,( - a + b + c + d)\,(a - b + c + d)\,(a + b - c + d)\,(a + b + c - d)}}{192\,u\,v\,w}$ where $\begin{align} a & = \sqrt {xYZ} \\ b & = \sqrt {yZX} \\ c & = \sqrt {zXY} \\ d & = \sqrt {xyz} \\ X & = (w - U + v)\,(U + v + w) \\ x & = (U - v + w)\,(v - w + U) \\ Y & = (u - V + w)\,(V + w + u) \\ y & = (V - w + u)\,(w - u + V) \\ Z & = (v - W + u)\,(W + u + v) \\ z & = (W - u + v)\,(u - v + W). \end{align}$ ## Distance between the edges Any two opposite edges of a tetrahedron lie on two skew lines, and the distance between the edges is defined as the distance between the two skew lines. Let d be the distance between the skew lines formed by opposite edges a and b − c as calculated here. Then another volume formula is given by $V = \frac {d \|(\mathbf{a} \times \mathbf{(b-c)})\| } {6}.$ ## Properties of a general tetrahedron The tetrahedron has many properties analogous to those of a triangle, including an insphere, circumsphere, medial tetrahedron, and exspheres. It has respective centers such as incenter, circumcenter, excenters, Spieker center and points such as a centroid. However, there is generally no orthocenter in the sense of intersecting altitudes. The circumsphere of the medial tetrahedron is analogous to the triangle's nine-point circle, but does not generally pass through the base points of the altitudes of the reference tetrahedron.[7] Gaspard Monge found a center that exists in every tetrahedron, now known as the Monge point: the point where the six midplanes of a tetrahedron intersect. A midplane is defined as a plane that is orthogonal to an edge joining any two vertices that also contains the centroid of an opposite edge formed by joining the other two vertices. If the tetrahedron's altitudes do intersect, then the Monge point and the orthocenter coincide to give the class of orthocentric tetrahedron. An orthogonal line dropped from the Monge point to any face meets that face at the midpoint of the line segment between that face's orthocenter and the foot of the altitude dropped from the opposite vertex. A line segment joining a vertex of a tetrahedron with the centroid of the opposite face is called a median and a line segment joining the midpoints of two opposite edges is called a bimedian of the tetrahedron. Hence there are four medians and three bimedians in a tetrahedron. These seven line segments are all concurrent at a point called the centroid of the tetrahedron.[8] The centroid of a tetrahedron is the midpoint between its Monge point and circumcenter. These points define the Euler line of the tetrahedron that is analogous to the Euler line of a triangle. The nine-point circle of the general triangle has an analogue in the circumsphere of a tetrahedron's medial tetrahedron. It is the twelve-point sphere and besides the centroids of the four faces of the reference tetrahedron, it passes through four substitute Euler points, 1/3 of the way from the Monge point toward each of the four vertices. Finally it passes through the four base points of orthogonal lines dropped from each Euler point to the face not containing the vertex that generated the Euler point.[9] The center T of the twelve-point sphere also lies on the Euler line. Unlike its triangular counterpart, this center lies 1/3 of the way from the Monge point M towards the circumcenter. Also, an orthogonal line through T to a chosen face is coplanar with two other orthogonal lines to the same face. The first is an orthogonal line passing through the corresponding Euler point to the chosen face. The second is an orthogonal line passing through the centroid of the chosen face. This orthogonal line through the twelve-point center lies midway between the Euler point orthogonal line and the centroidal orthogonal line. Furthermore, for any face, the twelve-point center lies at the midpoint of the corresponding Euler point and the orthocenter for that face. The radius of the twelve-point sphere is 1/3 of the circumradius of the reference tetrahedron. There is a relation among the angles made by the faces of a general tetrahedron given by [10] $\begin{vmatrix} -1 & \cos{(\alpha_{12})} & \cos{(\alpha_{13})} & \cos{(\alpha_{14})}\\ \cos{(\alpha_{12})} & -1 & \cos{(\alpha_{23})} & \cos{(\alpha_{24})} \\ \cos{(\alpha_{13})} & \cos{(\alpha_{23})} & -1 & \cos{(\alpha_{34})} \\ \cos{(\alpha_{14})} & \cos{(\alpha_{24})} & \cos{(\alpha_{34})} & -1 \\ \end{vmatrix} = 0\,$ where $\alpha_{ij}$ is the angle between the faces i and j. ## More vector formulas in a general tetrahedron This section does not cite any references or sources. Please help improve this section by adding citations to reliable sources. Unsourced material may be challenged and removed. (May 2013) If OABC forms a general tetrahedron with a vertex O as the origin and vectors a, b and c represent the positions of the vertices A, B, and C with respect to O, then the radius of the insphere is given by[citation needed]: $r= \frac {6V} {\|\mathbf{b} \times \mathbf{c}\| + \|\mathbf{c} \times \mathbf{a}\| + \|\mathbf{a} \times \mathbf{b}\| + \|(\mathbf{b} \times \mathbf{c}) + (\mathbf{c} \times \mathbf{a}) + (\mathbf{a} \times \mathbf{b})\|} \,$ and the radius of the circumsphere is given by: $R= \frac {\|\mathbf{a^2}(\mathbf{b} \times \mathbf{c}) + \mathbf{b^2}(\mathbf{c} \times \mathbf{a}) + \mathbf{c^2}(\mathbf{a} \times \mathbf{b})\|} {12V} \,$ which gives the radius of the twelve-point sphere: $r_T= \frac {\|\mathbf{a^2}(\mathbf{b} \times \mathbf{c}) + \mathbf{b^2}(\mathbf{c} \times \mathbf{a}) + \mathbf{c^2}(\mathbf{a} \times \mathbf{b})\|} {36V} \,$ where: $6V= \|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})\|. \,$ In the formulas throughout this section, the scalar a2 represents the inner vector product a·a; similarly b2 and c2. The vector positions of various centers are as follows: The centroid $\mathbf{G} = \frac{\mathbf{a} + \mathbf{b} + \mathbf{c}}{4}. \,$ The incenter $\mathbf{I}= \frac{ \|\mathbf{b}\times \mathbf{c}\| \, \mathbf{a} + \|\mathbf{c}\times \mathbf{a}\| \, \mathbf{b} + \|\mathbf{a}\times \mathbf{b}\| \, \mathbf{c} }{ \|\mathbf{b}\times \mathbf{c}\| + \|\mathbf{c}\times \mathbf{a}\| + \|\mathbf{a}\times \mathbf{b}\| + \|\mathbf{b}\times \mathbf{c} + \mathbf{c}\times \mathbf{a} + \mathbf{a}\times \mathbf{b}\| }. \,$ The circumcenter $\mathbf{O}= \frac {\mathbf{a^2}(\mathbf{b} \times \mathbf{c}) + \mathbf{b^2}(\mathbf{c} \times \mathbf{a}) + \mathbf{c^2}(\mathbf{a} \times \mathbf{b})} {2\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}. \,$ The Monge point $\mathbf{M} = \frac {\mathbf{a} \cdot (\mathbf{b} + \mathbf{c})(\mathbf{b} \times \mathbf{c}) + \mathbf{b}\cdot (\mathbf{c} + \mathbf{a})(\mathbf{c} \times \mathbf{a}) + \mathbf{c} \cdot (\mathbf{a} + \mathbf{b})(\mathbf{a} \times \mathbf{b})} {2\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}. \,$ The Euler line relationships are: $\mathbf{G} = \mathbf{M} + \frac{1}{2} (\mathbf{O}-\mathbf{M})\,$ $\mathbf{T} = \mathbf{M} + \frac{1}{3} (\mathbf{O}-\mathbf{M})\,$ where T is twelve-point center. Also: $\mathbf{a} \cdot \mathbf{O} = \frac {\mathbf{a^2}}{2} \quad\quad \mathbf{b} \cdot \mathbf{O} = \frac {\mathbf{b^2}}{2} \quad\quad \mathbf{c} \cdot \mathbf{O} = \frac {\mathbf{c^2}}{2}\,$ and: $\mathbf{a} \cdot \mathbf{M} = \frac {\mathbf{a} \cdot (\mathbf{b} + \mathbf{c})}{2} \quad\quad \mathbf{b} \cdot \mathbf{M} = \frac {\mathbf{b} \cdot (\mathbf{c} + \mathbf{a})}{2} \quad\quad \mathbf{c} \cdot \mathbf{M} = \frac {\mathbf{c} \cdot (\mathbf{a} + \mathbf{b})}{2}.\,$ ## Geometric relations A tetrahedron is a 3-simplex. Unlike the case of the other Platonic solids, all the vertices of a regular tetrahedron are equidistant from each other (they are the only possible arrangement of four equidistant points in 3-dimensional space). A tetrahedron is a triangular pyramid, and the regular tetrahedron is self-dual. A regular tetrahedron can be embedded inside a cube in two ways such that each vertex is a vertex of the cube, and each edge is a diagonal of one of the cube's faces. For one such embedding, the Cartesian coordinates of the vertices are (+1, +1, +1); (−1, −1, +1); (−1, +1, −1); (+1, −1, −1). This yields a tetrahedron with edge-length $\scriptstyle 2 \sqrt{2}$, centered at the origin. For the other tetrahedron (which is dual to the first), reverse all the signs. These two tetrahedra's vertices combined are the vertices of a cube, demonstrating that the regular tetrahedron is the 3-demicube. The stella octangula. The volume of this tetrahedron is 1/3 the volume of the cube. Combining both tetrahedra gives a regular polyhedral compound called the compound of two tetrahedra or stella octangula. The interior of the stella octangula is an octahedron, and correspondingly, a regular octahedron is the result of cutting off, from a regular tetrahedron, four regular tetrahedra of half the linear size (i.e. rectifying the tetrahedron). The above embedding divides the cube into five tetrahedra, one of which is regular. In fact, 5 is the minimum number of tetrahedra required to compose a cube. Inscribing tetrahedra inside the regular compound of five cubes gives two more regular compounds, containing five and ten tetrahedra. Regular tetrahedra cannot tessellate space by themselves, although this result seems likely enough that Aristotle claimed it was possible. However, two regular tetrahedra can be combined with an octahedron, giving a rhombohedron that can tile space. However, several irregular tetrahedra are known, of which copies can tile space, for instance the disphenoid tetrahedral honeycomb. The complete list remains an open problem.[11] If one relaxes the requirement that the tetrahedra be all the same shape, one can tile space using only tetrahedra in many different ways. For example, one can divide an octahedron into four identical tetrahedra and combine them again with two regular ones. (As a side-note: these two kinds of tetrahedron have the same volume.) The tetrahedron is unique among the uniform polyhedra in possessing no parallel faces. ### Related polyhedra A truncation process applied to the tetrahedron produces a series of uniform polyhedra. Truncating edges down to points produces the octahedron as a rectified tetrahedron. The process completes as a birectification, reducing the original faces down to points, and producing the self-dual tetrahedron once again. Family of uniform tetrahedral polyhedra Symmetry: [3,3], (332) [3,3]+, (332) {3,3} t0,1{3,3} t1{3,3} t1,2{3,3} t2{3,3} t0,2{3,3} t0,1,2{3,3} s{3,3} Duals to uniform polyhedra V3.3.3 V3.6.6 V3.3.3.3 V3.6.6 V3.3.3 V3.4.3.4 V4.6.6 V3.3.3.3.3 This polyhedron is topologically related as a part of sequence of regular polyhedra with Schläfli symbols {3,n}, continuing into the hyperbolic plane. {3,3} {3,4} {3,5} {3,6} {3,7} {3,8} {3,9} ... (3,∞} The tetrahedron is topologically related to a series of regular polyhedra and tilings with order-3 vertex figures. Polyhedra Euclidean Hyperbolic tilings {2,3} {3,3} {4,3} {5,3} {6,3} {7,3} {8,3} ... (∞,3} Compounds: ### Intersecting tetrahedra An interesting polyhedron can be constructed from five intersecting tetrahedra. This compound of five tetrahedra has been known for hundreds of years. It comes up regularly in the world of origami. Joining the twenty vertices would form a regular dodecahedron. There are both left-handed and right-handed forms, which are mirror images of each other. ## Isometries ### Isometries of regular tetrahedra The proper rotations, (order-3 rotation on a vertex and face, and order-2 on 2 edges) and reflection plane (through 2 faces and one edge) in the symmetry group of the regular tetrahedron Tetrahedral symmetry subgroup relations The vertices of a cube can be grouped into two groups of four, each forming a regular tetrahedron (see above, and also animation, showing one of the two tetrahedra in the cube). The symmetries of a regular tetrahedron correspond to half of those of a cube: those that map the tetrahedra to themselves, and not to each other. The tetrahedron is the only Platonic solid that is not mapped to itself by point inversion. The regular tetrahedron has 24 isometries, forming the symmetry group Td, [3,3], (332), isomorphic to the symmetric group, S4. They can be categorized as follows: • T, [3,3]+, (332) is isomorphic to alternating group, A4 (the identity and 11 proper rotations) with the following conjugacy classes (in parentheses are given the permutations of the vertices, or correspondingly, the faces, and the unit quaternion representation): • identity (identity; 1) • rotation about an axis through a vertex, perpendicular to the opposite plane, by an angle of ±120°: 4 axes, 2 per axis, together 8 ((1 2 3), etc.; (1 ± i ± j ± k) / 2) • rotation by an angle of 180° such that an edge maps to the opposite edge: 3 ((1 2)(3 4), etc.; i, j, k) • reflections in a plane perpendicular to an edge: 6 • reflections in a plane combined with 90° rotation about an axis perpendicular to the plane: 3 axes, 2 per axis, together 6; equivalently, they are 90° rotations combined with inversion (x is mapped to −x): the rotations correspond to those of the cube about face-to-face axes ### Isometries of irregular tetrahedra The isometries of an irregular (unmarked) tetrahedron depend on the geometry of the tetrahedron, with 7 cases possible. In each case a 3-dimensional point group is formed. Two other isometries (C3, [3]+), and (S4, [2+,4+]) can exist if the face or edge marking are included. Tetrahedral diagrams are included for each type below, with edges colored by isometric equivalence, and are gray colored for unique edges. Tetrahedron name Edge Equivalence diagram Description Symmetry Schön. Cox. Orb. Ord. Regular Tetrahedron Four equilateral triangles, forming the symmetry group Td, isomorphic to the symmetric group, S4. Td T [3,3] [3,3]+ 332 332 24 12 Trigonal pyramid An equilateral triangle base and three isosceles triangle sides gives 6 isometries, corresponding to the 6 isometries of the base. As permutations of the vertices, these 6 isometries are the identity 1, (123), (132), (12), (13) and (23), forming the symmetry group C3v, isomorphic to the symmetric group, S3. C3v C3 [3] [3]+ 33 33 6 3 Tetragonal disphenoid Isosceles tetrahedron Four congruent isosceles triangles gives 8 isometries. If edges (1,2) and (3,4) are of different length to the other 4 then the 8 isometries are the identity 1, reflections (12) and (34), and 180° rotations (12)(34), (13)(24), (14)(23) and improper 90° rotations (1234) and (1432) forming the symmetry group D2d. D2d S4 [2+,4] [2+,4+] 22 2× 8 4 Rhombic disphenoid Scalene tetrahedron Four congruent scalene triangles gives 4 isometries. The isometries are 1 and the 180° rotations (12)(34), (13)(24), (14)(23). This is the Klein four-group V4 or Z22, present as the point group D2. D2 [2,2]+ 222 4 Digonal disphenoid Two pairs of congruent isosceles triangles. This gives two opposite edges (1,2) and (3,4) that are perpendicular but different lengths, and then the 4 isometries are 1, reflections (12) and (34) and the 180° rotation (12)(34). The symmetry group is C2v, isomorphic to the Klein four-group V4. C2v =D1h [2] 22 4 Sphenoid Two unequal isosceles triangles with a common base edge. This has two pairs of equal edges (1,3), (1,4) and (2,3), (2,4) and otherwise no edges equal. The only two isometries are 1 and the reflection (34), giving the group Cs, also isomorphic to the cyclic group, Z2. Cs =C1h =C1v [ ] * 2 Half turn tetrahedron Two pairs of congruent scalene triangles. This has two pairs of equal edges (1,3), (2,4) and (1,4), (2,3) but otherwise no edges equal. The only two isometries are 1 and the rotation (12)(34), giving the group C2 isomorphic to the cyclic group, Z2. C2 =D1 [2]+ 22 2 Irregular tetrahedron No edges equal, four unequal scalene triangles, so that the only isometry is the identity, and the symmetry group is the trivial group. C1 [ ]+ 1 1 ## A law of sines for tetrahedra and the space of all shapes of tetrahedra A corollary of the usual law of sines is that in a tetrahedron with vertices O, A, B, C, we have $\sin\angle OAB\cdot\sin\angle OBC\cdot\sin\angle OCA = \sin\angle OAC\cdot\sin\angle OCB\cdot\sin\angle OBA.\,$ One may view the two sides of this identity as corresponding to clockwise and counterclockwise orientations of the surface. Putting any of the four vertices in the role of O yields four such identities, but in a sense at most three of them are independent: If the "clockwise" sides of three of them are multiplied and the product is inferred to be equal to the product of the "counterclockwise" sides of the same three identities, and then common factors are cancelled from both sides, the result is the fourth identity. One reason to be interested in this "independence" relation is this: It is widely known that three angles are the angles of some triangle if and only if their sum is 180° (π radians). What condition on 12 angles is necessary and sufficient for them to be the 12 angles of some tetrahedron? Clearly the sum of the angles of any side of the tetrahedron must be 180°. Since there are four such triangles, there are four such constraints on sums of angles, and the number of degrees of freedom is thereby reduced from 12 to 8. The four relations given by this sine law further reduce the number of degrees of freedom, not from 8 down to 4, but only from 8 down to 5, since the fourth constraint is not independent of the first three. Thus the space of all shapes of tetrahedra is 5-dimensional.[12] ## Applications The ammonium+ ion is tetrahedral ### Numerical analysis In numerical analysis, complicated three-dimensional shapes are commonly broken down into, or approximated by, a polygonal mesh of irregular tetrahedra in the process of setting up the equations for finite element analysis especially in the numerical solution of partial differential equations. These methods have wide applications in practical applications in computational fluid dynamics, aerodynamics, electromagnetic fields, civil engineering, chemical engineering, naval architecture and engineering, and related fields. ### Chemistry Main article: Tetrahedral molecular geometry The tetrahedron shape is seen in nature in covalent bonds of molecules. All sp3-hybridized atoms are surrounded by atoms lying in each corner of a tetrahedron. For instance in a methane molecule (CH4) or an ammonium ion (NH4+), four hydrogen atoms surround a central carbon or nitrogen atom with tetrahedral symmetry. For this reason, one of the leading journals in organic chemistry is called Tetrahedron. See also tetrahedral molecular geometry. The central angle between any two vertices of a perfect tetrahedron is $\arccos{\left(-\tfrac{1}{3}\right)}$, or approximately 109.47°. Water, H2O, also has a tetrahedral structure, with two hydrogen atoms and two lone pairs of electrons around the central oxygen atoms. Its tetrahedral symmetry is not perfect, however, because the lone pairs repel more than the single O-H bonds. Quaternary phase diagrams in chemistry are represented graphically as tetrahedra. However, quaternary phase diagrams in communication engineering are represented graphically on a two-dimensional plane. ### Electricity and electronics Main articles: Electricity and Electronics If six equal resistors are soldered together to form a tetrahedron, then the resistance measured between any two vertices is half that of one resistor.[13][14] Since silicon is the most common semiconductor used in solid-state electronics, and silicon has a valence of four, the tetrahedral shape of the four chemical bonds in silicon is a strong influence on how crystals of silicon form and what shapes they assume. ### Games Main article: Game The Royal Game of Ur, dating from 2600 BC, was played with a set of tetrahedral dice. Especially in roleplaying, this solid is known as a 4-sided dice, one of the more common polyhedral dice, with the number rolled appearing around the bottom or on the top vertex. Some Rubik's Cube-like puzzles are tetrahedral, such as the Pyraminx and Pyramorphix. The net of a tetrahedron also makes the famous Triforce from Nintendo's The Legend of Zelda franchise. ### Color space Main article: Color space Tetrahedra are used in color space conversion algorithms specifically for cases in which the luminance axis diagonally segments the color space (e.g. RGB, CMY).[15] ### Contemporary art Main article: Contemporary art The Austrian artist Martina Schettina created a tetrahedron using fluorescent lamps. It was shown at the light art biennale Austria 2010.[16] It is used as album artwork, surrounded by black flames on The End of All Things to Come by Mudvayne. ### Popular culture Stanley Kubrick originally intended the monolith in 2001: A Space Odyssey to be a tetrahedron, according to Marvin Minsky, a cognitive scientist and expert on artificial intelligence who advised Kubrick on the Hal 9000 computer and other aspects of the movie. Kubrick scrapped the idea of using the tetrahedron as a visitor who saw footage of it did not recognize what it was and he did not want anything in the movie regular people did not understand.[17] In Season 6, Episode 15 of Futurama, aptly named Möbius Dick, the Planet Express crew pass through an area in space known as the Bermuda Tetrahedron. Where many other ships passing through the area have mysteriously disappeared, including that of the first Planet Express crew. In the 2013 film Oblivion the large structure in orbit above the Earth is of a tetrahedron design and referred to as the Tet. ### Geology Main article: Geology The tetrahedral hypothesis, originally published by William Lowthian Green to explain the formation of the Earth,[18] was popular through the early 20th century.[19][20] ### Structural engineering A tetrahedron having stiff edges is inherently rigid. For this reason it is often used to stiffen frame structures such as spaceframes. ### Aviation At some airfields, a large frame in the shape of a tetrahedron with two sides covered with a thin material. It is mounted on a rotating pivot and always points into the wind. It's construction to be big enough to be seen from the air and is sometimes illuminated. Its purpose is to serve as a reference to pilots indicating wind direction.[21][22] ## See also • Boerdijk–Coxeter helix • Caltrop • Demihypercube and simplex - n-dimensional analogues • Hill tetrahedron • Schläfli orthoscheme • Tetra Pak • Tetrahedral kite • Tetrahedral number • Tetrahedron packing • Triangular dipyramid – constructed by joining two tetrahedra along one face • Trirectangular tetrahedron ## References 1. ^ a b 2. Coxeter, Harold Scott MacDonald; , Methuen and Co., 1948, Table I(i) 3. Havlicek, Hans; Weiß, Gunter (2003). "Altitudes of a tetrahedron and traceless quadratic forms". 110 (8): 679–693. doi:10.2307/3647851. JSTOR 3647851. 4. Leung, Kam-tim; and Suen, Suk-nam; "Vectors, matrices and geometry", Hong Kong University Press, 1994, pp. 53–54 5. 6. 7. Senechal, Marjorie (1981). "Which tetrahedra fill space?". (Mathematical Association of America) 54 (5): 227–243. doi:10.2307/2689983. JSTOR 2689983 8. Rassat, André; Fowler, Patrick W. (2004). "Is There a "Most Chiral Tetrahedron"?". Chemistry: A European Journal 10 (24): 6575–6580. doi:10.1002/chem.200400869 9. Klein, Douglas J. (2002). "Resistance-Distance Sum Rules" (PDF). Croatica Chemica Acta 75 (2): 633–649. Retrieved 2006-09-15. 10. Vondran, Gary L. (April 1998). "Radial and Pruned Tetrahedral Interpolation Techniques" (PDF). HP Technical Report. HPL-98-95: 1–32. 11. "Marvin Minsky: Stanley Kubrick Scraps the Tetrahedron". Web of Stories. Retrieved 20 February 2012. 12. Green, William Lowthian (1875). Vestiges of the Molten Globe, as exhibited in the figure of the earth, volcanic action and physiography. Part I. London: E. Stanford. OCLC 3571917. 13. Holmes, Arthur (1965). Principles of physical geology. Nelson. p. 32. 14. Hitchcock, Charles Henry (January 1900). "William Lowthian Green and his Theory of the Evolution of the Earth's Features". In Winchell, Newton Horace. The American Geologist XXV (Geological Publishing Company). pp. 1–10. 15. "Tetrahedron (wind Direction Indicator)". datwiki.net. Retrieved 2013-03-19. 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http://crypto.stackexchange.com/questions/tagged/discrete-logarithm?sort=faq&pagesize=30	# Tagged Questions In cryptography, a discrete logarithm is the number of times a generator of a group must be multiplied by itself to produce a known number. By choosing certain groups, the task of finding a discrete logarithm can be made intractable. learn more… \| top users \| synonyms 3answers 429 views ### Would the ability to efficiently find Discrete Logs have any impact on the security of RSA? This answer makes the claim that the Discrete Log problem and RSA are independent from a security perspective. RSA labs makes a similar statement: The discrete logarithm problem bears the same ... 2answers 267 views ### Are safe primes $p=2^k \pm s$ with $s$ small less recommandable than others as a discrete log modulus? I take the definition of safe prime as: a prime $p$ is safe when $(p-1)/2$ is prime. Safe primes of appropriate size are the standard choice for the modulus of cryptosystems related to the discrete ... 1answer 69 views ### iterated discrete log problem Consider the following problem: given $g_1 \ldots g_i,h_1 \ldots h_i \in G$, $\forall i$ find $x_i$ such that $g_i^{x_i}=h_i$ For $i=1$ this is the discrete log problem and is assumed to to have ... 3answers 398 views ### What is the relation between Discrete Log, Computational Diffie-Hellman and Decisional Diffie-Hellman? How are the three problems Discrete Logarithm, Computational Diffie-Hellman and Decisional Diffie-Hellman related? From my understanding, since the Discrete Log (DL) Problem is considered hard, then ... 0answers 78 views ### RSA security assumptions - does breaking the DLP also break RSA? [duplicate] Possible Duplicate: Would the ability to efficiently find Discrete Logs have any impact on the security of RSA? I'm wondering if breaking the DLP, that is the basis for ElGamal and DSA, ... 1answer 69 views ### How hard are discrete logarithms problems in $\mathbb Z^{}_{n}$ and $\mathbb Z^{}_{n^2}$, where $n$ is the RSA $n=pq$ Use the notations form the Wikipedia article Paillier Cryptosystem , assume that the chipertext $c$ and $c^{\lambda} \mod n^2$ are both given, is it possible to compute $\lambda$ easily? 1answer 178 views ### A discrete-log-like problem, with matrices: given $A^k x$, find $k$ Let $p$ be a large prime; we will work in $GF(p)$. Let $A$ be a $n\times n$ matrix. Also, let $x$ be a $n$-vector and $k$ a positive integer. Suppose we are given $p$, $A$, $x$, and $y$. The goal ... 0answers 124 views ### Finding where I am in a linear recurrence relation Suppose I have a linear recurrence relation $$a(n) = c_1 a(n-1) + \dots + c_k a(n-k) + d,$$ where the constants $c_1,\dots,c_k,d$ are given and the initial values $a(0),\dots,a(k-1)$ are given as ... 0answers 163 views ### Is there a practical zero-knowledge proof for this special discrete log equation? We have a multiplicative cyclic group $G$ with generators $g$ and $h$, as in El Gamal. Assume $G$ is a subgroup of $(\mathbb{Z}/n\mathbb{Z})^*$. There are two parties, Alice and Bob: Alice knows: ... 1answer 214 views ### Why are elliptic curve variants of RSA “chiefly of academic interest”? Yesterday I was thinking about elliptic curve variants of popular protocols/algorithms (ECDH, ECES[1], etc) and the thought occured that I had never seen an elliptic curve variant of RSA. My ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9148797988891602, "perplexity_flag": "middle"}
http://climateaudit.org/2008/10/21/resolving-the-santer-problem/	by Steve McIntyre Resolving the Santer Problem In today’s post, I think that I’ve developed an interesting approach to the Santer problem, which represents a substantial improvement to the analyses of either the Santer or Douglas posses. I think that the approach proposed here is virtually identical to Jaynes’ approach to analyzing the difference between two means, as set out in the article recommended by beaker. As it happens, I’d done all of the calculations shown today prior to reading this article. While my own calculations were motivated primarily by trying to make sense of the data rather than anything philosophical, academics like to pigeonhole approaches and, to that extent, the approach shown below would perhaps qualify as a “bayesian” approach to the Santer problem, as opposed to the “frequentist” approach used both by Santer and Douglass. I had the post below pretty much in hand, when I was teasing beaker about being a closet frequentist. The first problem discussed in Jaynes 1976 was the following question about the difference of two means (and keep in mind that the issue in Santer v Douglass is the difference between two trends): The form of result arrived at by Jaynes was the following – that there was a 92% probability that B’s components have a greater mean life. The form of conclusion that I’m going to arrive at today is going to be identical – ironically even the number is identical: there is a 92% probability that a model trend will exceed the true observed trend. Now as a caveat, my own terminology here may be a little homemade and/or idiosyncratic; I place more weight on the structure of the calculations which are objective. I think that the calculations below are identical to the following equation in Jaynes 1976 (though I don’t swear to this and, as noted before, I’d done the calculations below before I became aware of Jaynes 1976). Let me start with the following IMO instructive diagram, which represents my effort to calculate a probability distribution of the slope of the trend, given the observations to 1999 (Santer), 2004 (Douglass) and up to 2008 (most of which were available to Santer et al 2008). It looks like this corresponds to Jaynes $P_n(a)$, i.e. what bayesians call a “posterior distribution”, but I’m just learning the lingo. The dotted vertical lines represent 95% confidence intervals from the profile likelihood calculations shown in a prior post, color coded by endpoint. The colored dots represent 95% confidence intervals calculated using the Neff rule of thumb for AR1 autocorrelation in Santer. The black triangle at the bottom shows ensemble mean trend (0.214 deg C/decade). Figure 1. See explanation in text immediately preceding figure. A couple of obvious comments. The 95% confidence intervals derived from profile likelihoods are pretty much identical to the 95% confidence intervals derived from the AR1 rule of thumb – that’s reassuring in a couple of ways. It reassured me at least that my experimental calculations had not gone totally off the rails. Second, the illustration of probability distributions shown here is vastly more informative than anything in either Santer or Douglass both for the individual years and showing the impact of over 100 more months of data in reducing the CI of the trend due to AR1 autocorrelation. (A note here: AR1 autocorrelation wears off fairly quickly – if LTP is really in play, then the story would be different.) Thirdly and this is perhaps a little unexpected, inclusion of data from 1999 to 2008 has negligible impact on the maximum likelihood trend of the MSU tropical data; the most distinct impact is on the confidence intervals, which narrow a great deal. The CIs inclusive of 2008 data are about 50% narrower than the ones using data only up to 1999. Whether this is apples-and-apples or the extent to which this is apples and apples, I’ll defer for now. Fourth and this may prove important, although the ensemble mean triangle relative to 95% CI intervals is unaffected by the inclusion of data up to 2004, it is not unaffected by the inclusion of data up to 2008. With 2008 data in, the CI is narrowed such that ensemble mean is outside the 95% CI interval of the observations – something that seems intuitively relevant to an analyst. I feel somewhat emboldened by Jaynes 1976′s strong advocacy of pointing out things that are plain to analysts, if not frequentists. My procedure for calculating the distribution was, I think, interesting, even if a little homemade. The likelihood diagrams that I derived a couple of days ago had a function that yielded confidence intervals (this is what yielded the 95% CI line segment.) I repeated the calculation for confidence levels ranging from 0 to 99%, which with a little manipulation gave a cumulative distribution function at irregular intervals. I fitted a spline function to these irregular intervals and obtained values at regular intervals and from these obtained the probability density functions shown here. The idea behind the likelihood diagrams was “profile likelihood” along the lines of Brown and Sundberg – assuming a slope and then calculating the likelihood of the best AR1 fit to the residuals. I don’t know how this fits into Bayesian protocols, but it definitely yielded results, that seem to be far more “interesting” than the arid table of Santer et al. My next diagram is a simple histogram of LT2T trends derived from data in Douglass et al 2007 Table 1, which, for the moment, I’m assuming is an accurate collation of Santer’s 20CEN results. In Douglass’ table, he averaged results within a given model; I’ll revisit the analysis shown here if and when I get digital versions of the 49 runs from Santer (Santer et al did not archive digital information for their 49 runs; I’ve requested the digital data from Santer – see comments below). [Note - beaker in a comment below states that this comparison is pointless. I remain baffled by his arguments as to why this sort of comparison is pointless and have requested a reference illuminating the statistical philosophy of taking a GCM ensemble and will undertake to revisit the matter if and when such a reference arrives.] I derived LT2T trends by applying weights for each level (sent to me by John Christy) to the trends at each level in the Douglass table. In the diagram, I’ve carried forward the 95% CI intervals from the observations for location and scale comparison, as well as the black triangle for the ensemble mean. The \$64 dollar question is then the one that Jaynes asks. [Note - I think that this needs to be re-worked a bit more as a "posterior" distribution to make it more like Figure 1; I'm doing that this afternoon - Oct 21 - and will report on this.] Figure 2. Histogram of LT trends from Douglass et al 2007 data. My next diagram is the proportion of models with LT2T trends that exceed a given x-value. I think that this corresponds to Jaynes $P_m(b)$. The dotted lines are as before; the solid color-coded vertical lines are the maximum likelihood trend estimates for the respective periods. Again for an analyst, the models seem to the the right of these estimates. Figure 3. Proportion of Model Trends (per Douglass Table 1 information) exceeding a given value. I then did a form of integration which I think is along the lines of the Jaynes recipe. I made a data frame with x-values in 0.01 increments (Data[[i]]\$a, with i subscripting the three periods 1999, 2004 and 2008. For each value of x, I made a column of values representing the results of Figure 3: for(i in 1:3) Data[[i]]\$fail=sapply(Data[[i]]\$a, function(x) sum(douglass\$trend_LT>=x )/22) Having already calculated the distribution $P_n(a)$ in column Data[[i]]\$d, I calculated the product as follows: for(i in 1:3) Data[[i]]\$density=Data[[i]]\$d*Data[[i]]\$fail Then simply add up the columns to get the integral (which should be an answer to Jaynes question): sapply(Data,function(A) sum(A\$density)) # 1999 2004 2008 #0.8638952 0.8530540 0.9164450 On this basis, there is a 91.6% probability that the model trend will exceed the observed trend. I leave it to climate philosophers to decide whether this is “significant” or not. Going back to Jaynes 1976 – I think that this calculation turns out to be an uncannily exact implementation of how Jaynes approached the problem. In that respect, it was very timely for me to read this reference while the calculation was fresh in my mind. It was also reassuring that, merely by implementing practical analysis methods, I’d in effect got to the same answer as Jaynes had long ago. Like this: This entry was written by , posted on Oct 21, 2008 at 9:37 AM, filed under Santer and tagged CI, Confidence interval, Cumulative distribution function, Probability density function, Probability distribution, Rule of thumb, Statistics, United States. Bookmark the permalink. Follow any comments here with the RSS feed for this post. Post a comment or leave a trackback: Trackback URL. 59 Comments 1. Steve McIntyre I asked one of Santer’s coauthors for a digital version of the 49 runs, but he said that he didn’t have the data, that I would have to write a letter to Santer. I wrote a letter to Santer yesterday saying that I’d been a good boy and could I please have the data. I received an automated email saying that Santer was occupied at a workshop. Of course. It’s only a couple of months till Christmas. Yeah, yeah, I know it’s lame, but it seemed funny at the time. (And yes, the obvious joke about a collective name for Santer’s coauthors crossed my mind.) 2. beaker I think that the approach proposed here is virtually identical to Jaynes’ approach to analyzing the difference between two means, as set out in the article recommended by beaker. Steve, while I would recommend the tools of Bayesian probability that Jaynes recommends, a test of a difference between the mean of the ensemble and the mean of the observations is not appropriate. This is because the ensemble mean is an estimate of the average tropical trend for a frequentist style hypothetical population of Earths, not the trend on a specific member of that population (i.e. the Earth we actually occupy). When you understand the parallel Earth thought experiment, you will realise why this is the case. There is no point in talking about statistics until you appreciate what the modellers claim the models are actually capable of (i.e. the significance of the “forced” in “forced climate change”). Sadly, I doubt that Jaynes would support your test as it there is a disconnect with the question. P.S. I suspect Santer may just have heard that one before ;o) Steve: beaker, I would very much like to “appreciate” what the modelers claim that the models are capable of. On previous occasions, I’ve asked you to provide a citation to a relevant authority and you’ve failed to do so and I wasn’t able to see a connection between your own opus (beaker has cordially identified himself to me offline) and use of ensembles in models. If I am having trouble grasping this point and there is no usable reference material, then you can hardly blame me. The fault lies with the climate community for not providing reference material on an important point. I know that you’ve tried hard to educate me on this, but honestly I don’t get the point. In any event, it’s not your responsibility. But again, if you can think of a statistical reference that deals appropriately with the concept of a model ensemble, I’m all ears. • RomanM Re: beaker (#2), This is because the ensemble mean is an estimate of the average tropical trend for a frequentist style hypothetical population of Earths, not the trend on a specific member of that population (i.e. the Earth we actually occupy). Beaker, give it up! Your straw man of a “hypothetical population of Earths” is just hogwash. Hey, the past really happened. There IS an actual set of “trends” to the temperature that can be defined in a unique fashion. One set. A single set of values , one for each pressure. The next group studying what happened will be considering that same set of values. The only uncertainty in the measured trends comes from the measuring and the processing of the data. The real question is how well can the models guess those values. Nobody is asking them to match every wiggle. All we want in this case is that the models exhibit a certain general type of behaviour as measured by the trend values. Calculate a summary of 9 or 10 trend values for each model. Get a handle on how variable those values from each model are when random components are included in the model. Now compare (using the appropriate statistical methodology) to see how close they came to the real world values with a view to whether they might be able to do this consistently (no, not your specious definition) in the future. It’s plain ordinary common sense! Your personal version of what constitutes “testing consistency” is pretty much a nothing exercise and can provide no insight to the situation. • DeWitt Payne Re: RomanM (#7), I think you have it exactly backwards. The variability within and between models is only important if it tells us something about the variability of the real climate, the weather noise as opposed to measurement error. Beaker’s parallel universes represent the ideal case of an ensemble of perfect physical models. Our planet, however, is only one realization. Even if you could measure past trends on this planet with zero error and plug the numbers into a perfect physical model, the future trend would still be uncertain. To represent this uncertainty you would still need to put non-zero confidence (or whatever you want to call them) limits on the past trend to see if any model or collection of models fit the data. • Alan Wilkinson Posted Oct 22, 2008 at 1:45 PM \| Permalink Re: DeWitt Payne (#43), If you can’t predict the future with any model because of chaotic uncertainty, what on earth (or off it) is the point of the IPCC projections? And why should anyone believe that an ensemble of arbitrarily chosen runs is any more powerful than the best single model? • Kenneth Fritsch Posted Oct 22, 2008 at 1:46 PM \| Permalink Re: DeWitt Payne (#43), Beaker’s parallel universes represent the ideal case of an ensemble of perfect physical models. Our planet, however, is only one realization. Even if you could measure past trends on this planet with zero error and plug the numbers into a perfect physical model, the future trend would still be uncertain. You have summarized the ongoing debate that I have had with Beaker. If a single realization of the earth’s climate has this uncertain value attached and if that uncertain value varies with individual realizations and further if the models cannot capture that uncertain value and its variation, then how could a model or ensemble of models make a prediction about future earth climates. If one could estimate that uncertain value then one could compare a climate model (or ensemble of models average) result to that of the observed earth (our experienced single rendition that is) plus/minus that uncertain value and proceed with that result in doing the statistical testing. If one cannot estimate that uncertain value of the single rendition then climate models results would become prone to subjective interpretation as they would say, our perfect models say this will be the future trend due to AGW, or what ever, but unfortunately our single earth rendition will have this plus/minus uncertain value that is, well uncertain and evidentally not capable of being estimated. • RomanM Posted Oct 22, 2008 at 2:22 PM \| Permalink Re: DeWitt Payne (#43), We are interested in examining what is happening on this earth (there is only one). The “perfectly measured” temperature record is what we are interested in. The talk of parallel universes is an unnecessary construct which offers nothing tangible to the situation. I would also be interested in knowing on what scientific basis you would model the distribution of the characteristics of these worlds – some sort of invented Bayesianism? Surely, you must also realize that a set of nonrandomly selected models each consisting of a single value description cannot be scientifically analyzed in a meaningful way without an extraordinary amount of assumed baggage. The only reason this has been done by these authors is that the IPCC chose to do this in their reports. The only common sense way to assess the models is individually. Models have many different aspects to them – the trends are just a single part. If a model is deterministic, its ability to “model” the trend does not lend itself to statistical analysis. You need to look at it in terms of evaluating how different input conditions affect the output. If there are random components then you also need many runs to evaluate how the model responds to the random factors. Look at how variable the predicted trends are and where they are centered. These runs are all under the same (known) conditions that existed on this earth. In the case that these results are not centered near the measured trends, then there is a bias in the models. That’s not good. If the variability is too high, then they have no practical value. Both of these last two features are where statistical techniques are useful. And, yes, regardless of how the models perform, there is no guarantee that they will do that in the future. However, they must first be able to hindcast in a sufficiently accurate fashion at the same time demonstrating they are reacting in an appropriate fashion to changes in the various input factors for us to have any faith in their future ability. • DeWitt Payne Posted Oct 22, 2008 at 10:31 PM \| Permalink Re: RomanM (#46), However, they must first be able to hindcast in a sufficiently accurate fashion at the same time demonstrating they are reacting in an appropriate fashion to changes in the various input factors for us to have any faith in their future ability. I agree completely. I don’t think we’re all that far apart on the general principles. I found Beaker’s parallel Earths a useful concept for gaining a better understanding of the problem. YMMV. • Kenneth Fritsch Posted Oct 23, 2008 at 10:21 AM \| Permalink Re: RomanM (#46), The only common sense way to assess the models is individually. I agree, Roman, and that is why I was hoping someone would apply Ross McKitrick’s suggested method to the data. His approach as I understand it would take into account the individual model results and the individual observation results and the results compared over the entire spectrum of temperature trends over the altitude range. I can, on the other hand, see Douglass et al. attempting to comply with their impression of the prevailing view on the handling of the ensemble of model results. One must also be careful to note that the claim of Santer et al. is that in bringing the new adjusted observed results into the calculations that the model and observed results become more compatible (or some such spin for future IPCC reviews). To me this is revealing in two ways: One, the authors recognize a difference in the model and observed results. Two, that the newer (and adjusted) observed results that show on some parts of the altitude range even higher tropospheric temperature trends than the models are what the authors use in their claim, but at the same time not discussing the fit of the model results and newer and adjusted observed results over the entire altitude range. I think a McKitrick-like analysis across the altitude spectrum would address this cherry picking tendency. The discussion here at CA, in my view, has skipped over what cancels when a difference or ratio for the tropical tropospheric to surface temperature trends is used. Would that include ENSO effects and/or the chaotic content of one of beakers parallel earths? Those effects, if not canceling, would be relevant to a statistical comparison whether one used averages of model ensembles or individual model results. Further we are losing the critical issue of the differences between the adjusted observed results both for MSU and radio sondes. Steve M uses the UAH MSU adjustments and perhaps with some good a priori reasons, but if one has no good preconceived basis these differences have to be considered in the analysis and discussions. If one looks hard at the all the observed and model results (from the Santer and Douglass papers) and over the entire altitude range, I think one has to conclude that overall the picture is a mess and that, further, this area of climate science has a lot work to do. • Lance Re: beaker (#2), …a test of a difference between the mean of the ensemble and the mean of the observations is not appropriate. I agree except that this is the metric by which various defenders of the IPCC’s climate models claim that these models have predictive skill. 3. tolkein There’s a couple of questions. (1) What do the modellers claim the models are actually capable of? (2) How does global climate actual compare with the models forecasts in the various IPCC ? We’ve had 17 years since 1989 and a few since 2004 and a comparison with how good the models were in forecasting would be very good, and I don’t mean confidence intervals, I mean visual look and see. 4. Jeff Alberts I received an automated email saying that Santer was occupied at a workshop. Of course. It’s only a couple of months till Christmas. Yeah, yeah, I know it’s lame, but it seemed funny at the time. (And yes, the obvious joke about a collective name for Santer’s coauthors crossed my mind.) Sorry, but ROTFL! Steve: I thought that the workshop part made the joke rise slightly above the usual situation, notwithstanding beaker’s turning up his nose. 5. Steve McIntyre #2. beaker, I didn’t use the ensemble mean at any point in the above calculations other than as a location point on the graph. Had Santer provided individual runs in a usable format, I would have used those. In the absence of Santer providing such information, I used the available information from Douglass Table 1. I’ll revisit the calculation if and when I get the info that IMO Santer should already have archived. (Yeah, yeah, the information is doubtless somewhere at PCMDI, but the exercise of replicating his collation is not germane to the present tests.) 6. Lune This is because the ensemble mean is an estimate of the average tropical trend for a frequentist style hypothetical population of Earths, not the trend on a specific member of that population (i.e. the Earth we actually occupy). What’s the difference? The hypothetical population of Earths is of value only if it can provide us with meaningful information about the Earth we actually occupy. Presumably one must fit some kind of distribution curve to the hypothetical Earths in order for them to be predictive with respect to our actual Earth. (If they were generated randomly, that wouldn’t be very interesting.) Steve’s analysis says, if nothing else, that from the standpoint of the methodology that was used to generate all the hypothetical Earths, our actual observed Earth is ‘low’ probability. Certainly that tells us we need to better understand the assumptions underlying that methodology. • srp Re: Lune (#6), As far as the “philosophical” debate about the appropriate comparison statistics goes, I think this is the key point. Beaker’s parallel Earths thought experiment only works if we know the degree of variation (caused by senstitive dependence on initial conditions) of the various Earths with respect to the property we are trying to predict/explain. If the distribution of the property over the parallel Earths is very tight, then it is reasonable to treat the realized data on our earth as the benchmark to which any model or ensemble mean must converge. If not, then a looser standard should be applied. But simple hand-waving about variability due to dependence on initial conditions is not adequate for a quantitative prescription of test statistics. 7. Craig Loehle To my mind there are two ways of looking at the model results. In the first, we want to see if the population of models looks like the population of data we have. This leads to various analyses, of which Steve’s in this thread is very nice. The second says the ensemble mean is the best estimate of model wisdom, so we use that as a SINGLE POINT ESTIMATE of what the models say about reality and compare that to the data (and it is outside the ci of the data by any possible test). • MC Re: Craig Loehle (#8), I haven’t read Santer in full yet but will do. I agree though with what you say in that the reference data (RSS, UAH) should be combined as a reference with errors and the model data combined as a reference with errors and the two compared. Then each model should be compared with the RSS/UAH etc data to see what model assumptions are agreeing with measurements. Obviously if both data sets have large errors (50% plus) then I would be less confident in overlap and would state the null hypothesis and ask for better resolved data. I normally would accept data with 10% error or less for a reasonable estimate between the two. As for in general, Steve this is a good post but I have to disagree with yourself and Jaynes for that matter. You both appear to have made the Central Limit Theorem assumption in that you regard a measurement x ± y as an expected value ± error. In a pure scientific sense (or more exactly in an empirical measurement sense) the convention of x ± y is used with the understanding that the ‘real’ or expected value has equal probability of lying within x -y to x +y. Some may disagree but this is making an assumption. Now how the errors y are decided can be a bit finger in the air but as long as assumptions and rationale are stated and the intention is to be as truthful and conservative as possible then the errors are accepted. There can be devil in some details admittedly but in general good scientific method should be clear enough. So for Jaynes example, I would have to say there is no difference because a) the uncertainty is large and b) there are only two data points. Get me more data with less uncertainty and then I’ll be able to resolve the issue better. Otherwise it is just a ‘gut feeling’ guess and no more different than cherry picking proxies for that sports implement people like to talk about. 8. Jeff Alberts We’ve had 17 years since 1989 I’m no math whiz or anything, but I’m pretty sure it’s been 19 years since 1989 • Posted Oct 21, 2008 at 11:18 AM \| Permalink \| Reply Re: Jeff Alberts (#9), I’m no math whiz or anything, but I’m pretty sure it’s been 19 years since 1989 What’s that in model years? (It appears the answer may be that we’ve had 10 years since 1989!) • Jeff Alberts Re: lucia (#10), Well, as stated it said “we’ve had 17 years since 1989″, not 17 model runs. • Dave Dardinger Re: Jeff Alberts (#9), Jeff & others, We now have the handy “reply and paste link” below the displayed comment number. I’d urge everyone to use it regularly as it’s a savings in comment construction time and of real value to the reader as well. In this case it saves having to look at several messages to see who used “17″ In other cases it will take you to the proper prior message even if a message has been deleted in between. If used regularly it allows a person to easily go back through a thread of responses at the click of a button. • Jeff Alberts Re: Dave Dardinger (#11), Point taken. I’m used to Blockquote… 9. Jedwards Beaker, I think I see what you are getting at. Steve’s calculation of the probability distribution Pn(A) for the MSU Tropical Trends appears to be a correct application of Jaynes96, but the comparison should be against the probability distributions Pn(B) of each model run individually in order to give a true apples to apples comparison. In other words, the methodology that Steve has illustrated here may be a reasonable qualitative test for the ability of individual models to properly hindcast the “real Earth”. Craig’s point in #8 appears to get to the crux of the problem in that the ensemble mean is used “politically” by the IPCC as a single point estimator without regard (apparently) to the CI of the individual model runs which have their own distributions. So what we really need are the distributions of those individual runs, and we could then determine whether there are any of those models which are better/worse at hindcasting, which would then give us “confidence” in that particular models ability to forecast. Steve: I think that the “posterior” for the MSU trends is probably OK, but I’m going to re-work the “posterior” for the model trends. • Craig Loehle Re: Jedwards (#12), I agree that the multiple runs of a single model make a proper ensemble. It is not clear, as others have stated, that it makes any sense to combine the output of a bunch of models and call this an “ensemble” but that is what IPCC does. 10. Posted Oct 21, 2008 at 11:58 AM \| Permalink \| Reply It’s a great post. I love to read the most. The graphical represent is not bad at all. It’s simply the great. Thanks for sharing this in your post with us. 11. SteveSadlov Probably the first time someone has applied these several-decades-old, completely main stream quality methods to the models. The modelers are unlikely to have encountered their practical use previously, since most of them never worked in manufacturing. • Jedwards Re: SteveSadlov (#18), I’m actually in QA, and this method makes more sense than anything else I’ve seen (maybe because we actually use it,…a lot). 12. tolkein re 9. Whoops! Slipped when typing, honest. 19 years since 1989. But the point is, how did the models do compared to actual? A visual would do best. • Jeff Alberts Re: tolkein (#20), Hehe, no harm no foul! 13. Henry Steve You are going to have to be very careful with language if you delve into Bayesian statistics: you lose phrases like “confidence interval” and “statistical significance”; instead you get things like “credible interval” and “this decision is better than that”. And while Jaynes wrote in an attractive style, he was often pressing an unconventional line (even among Bayesians), for example with his determination to use “maximum entropy”. You will even have to consider again your conclusion “there is a 92% probability that a model trend will exceed the true observed trend”. Since the actual trend has turned out lower than the model, there is little surprise there, but what you need to deal with is how big the difference probably is (if you were absolutely certain that the actual trend was exactly 0.0001° less than the model then the probability would be 100%, but the difference would not be seen as substantial enough to care about, since any loss associated with the error would be minimal); Bayesian methods provide a way into this, but the price to pay is that you are no longer talking the same language as those you are auditing. 14. Steve McIntyre #22. This sort of enterprise is more prone to problems than working with the proxies. The stalemate between Santer and Douglass does seem frustrating and sometimes it’s a little fun experimenting. 15. sky SUMMUS SUMMARUM: We are interested in the collective skill of models in estimating the actual observations. While the prospect of some model (or individual run) showing some skill cannot be statistically foreclosed, their pitiful lack of skill in the mean is unmistakable, no matter what statistical formalism is applied. 16. Nick Moon Usually, I reckon I can follow the postings on CA. And usually I can just about keep up with the maths. But this posting makes me realise I know far too little statistics. So on behalf of for those of us with only average sized brains, can I ask a hypothetical question? Supposing a bunch of skeptics got themselves a few super-computers and came up with a few models for predicting global temperatures. And, further suppose, they ran these models over and over again to produce an ensemble of ensembles of different model runs. And now lets suppose a really big suppose. Lets suppose that the spread of the individual runs in this skeptics ensemble had pretty much the same variance/SD/error as the Santer ensemble. And say, in some eerily spooky way that when Steve McI plotted a triangular point to represent the mean of the skeptics ensemble, it came out at around -0.1C. i.e. The same distance from the mean of the 2008 plot but on the opposite side. Now is there any mathematical technique which would allow one to say that this ensemble mean is more or less consistent with reality than the Santer one? 17. Roger Pielke. Jr. You’ll find some interesting parallels to this discussion here: http://rankexploits.com/musings/2008/ipcc-projections-do-falsify-or-are-swedes-tall/ and here: http://sciencepolicy.colorado.edu/prometheus/archives/prediction_and_forecasting/001428comparing_distrubuti.html 18. Christopher Steve, NOAA has an model ensemble walkthru with sources. Might be fruitful? http://www.hpc.ncep.noaa.gov/ensembletraining/ 19. Alan Wilkinson I am still stuck on my original issue with beaker’s parallel earths. Although there may be an infinite number of them, the only ones which can conceivably model and predict our own earth’s future climate are ones that are compatible(?) with the observed data on this earth. The others (including non-complying runs) must all be excluded from IPCC prediction methods. Otherwise, I am entirely in agreement with the strategy of focussing on the real observed data and measuring the models’ ability to match it rather than the absurdity of using model scatter as an input to judging compatibility. 20. Dishman It seems to me that at least some of the model runs within this ensemble have been falsified. I’m curious whether IPCC or anyone else puts any effort into investigating why. Are we running open-loop here, or is feedback from reality taken into consideration? • Pat Keating Re: Dishman (#31), or is feedback from reality taken into consideration? No, if the reality doesn’t agree with model results, reality must be “adjusted”. The feedback is in the other direction. 21. Christopher And this looks promising. Sorry for the linkfarm-esque posts but I’ve been looking at this for another reason and simply wanted to share. 22. James Smyth I received an automated email saying that Santer was occupied at a workshop. Of course. It’s only a couple of months till Christmas. Yeah, yeah, I know it’s lame, but it seemed funny at the time. (And yes, the obvious joke about a collective name for Santer’s coauthors crossed my mind.) I don’t get it. Are you saying they are short, have point ears and do all the work; or that they have four legs and antlers and pull him around? 23. Paul29 Has anyone treated the standard deviation of the model runs as an estimate of the standard error of the average model (since each model run provides an estimate of the average). To get an estimate of the population’s standard deviation, multiply this standard error by the square root of “n”. Could this standard deviation and the ensemble average be used to define the population distribution that can be compared to actual data’s distribution? 24. Steve McIntyre I re-did this calculation assuming a normal posterior (?) distribution for the model trends centered on the sample mean of my estimate of the mean of the model trends ( 0.208) and sd of the model trends (0.091). This yielded the following probabilities that the model trend exceeded the observed trend. I think that the posterior distribution may be a t-distribution of some sort, but I’d be surprised if the results varied a whole lot of the assumption made here. # 1999 2004 2008 #0.858 0.840 0.904 • eilert Re: Steve McIntyre (#35), Thanks Steve. I like your approach of analysing the data far better, since is designed to highlight the differences and not hidden behind some porbabilty values, which are more designed to obfuscate a problem. From the Fig.2 we can clearly see that there is a systematic shift in the two data sets. If I were a scientist who is modelling such data I would be very much more interested in what these shifts represents to actually improve my modelling. 25. eilert Your green (2008) curve in Fig. 1 shows a closer distribution (the base is narrower). Which would mean that the curve fitted to the data has less outliers. Do you perhapse have a plot of that curve? It schould be interesting to see how the trends have evolved with time. • Alan Wilkinson Re: eilert (#36), No, it means the line is longer thereby giving a more certain definition to the trend (slope) even if the scatter around the trend line remains similar. 26. Clark I still don’t see how one can use an “ensemble” of models, and their mean or variance for anything. These measurements are completely dependent on the choices one makes in what models to include, and how many runs of each. Variance too small to include current observations? Simply include a few more runs of the high- and low-end models. Presto – larger uncertainty ranges. Mean not trending high-enough? Add in a few models with strong positive feedback. Oh look, the mean is now higher. The ensemble values say more about the ensembler than the predictive power of any individual model. • JamesG Re: Clark (#40), Douglass and Gavin Schmidt clashed about the reasoning behind the use of an ensemble on Matt Briggs’s blog and Gavin’s response was simply an Italian phrase meaning “it just works”. Hence I think that’s admitting that there isn’t any actual theory behind it. It was just a discovery that they couldn’t match any individual model to the surface temperatures but they could get a match using an ensemble. Which is really quite cheeky of them. Since you clearly don’t get a match at the troposphere then I guess it actually “just doesn’t work” so the practice should be abandoned henceforth. But nobody in the IPCC worries about that too much I’m sure. Plenty of time for more data adjustments before the next report. • Pat Frank Re: JamesG (#41), Reading that debate, I got the feeling that Gavin used the Italian phrase to analogize himself to Galileo, who allegedly said, concerning Earth, “Eppur si muove (Yet it moves)” as he was threatened by the Inquisition. Gavin apparently likens himself to a martyr for science. What an irony. Some time ago, I watched the Iq^2 debate on global warming, here, featuring Gavin on one side and Richard Lindzen on the other. Of the six principals in that debate, only Gavin accused his opponents of lying. Some martyr. 27. tolkein Am I understanding this correctly? The IPCC central tendency is not just the mean of lots of runs of the same model but averages of runs of lots of different models, each with different assumptions. So how do we (relatively simply) flex the model forecasts so as to see what happens if assumptions change? I can’t see the financial regulators going a bundle on this if I was trying this approach to justify my capital requirements under Basel 2, so how can this approach be valid for spending \$trillions on AGW mitigation? How does this make any sense? 28. Dave Andrews There’s an interesting discussion of climate models, involving Gavin and others, over a period of months at:- http://www.thebulletin.org/web-edition/roundtables/the-uncertainty-climate-modeling# • Willis Eschenbach Re: Dave Andrews (#49), thanks for the links to the discussion. On the page you cited, Gavin Schmidt says: So how should one interpret future projections from climate models? I suggest a more heuristic approach. If models agree that something (global warming and subtropical drying for instance) is relatively robust, then it is a reasonable working hypothesis that this is a true consequence of our current understanding. If the models fail to agree (as they do in projecting the frequency of El Niño) then little confidence can be placed in their projections. Additionally, if there is good theoretical and observational backup for the robust projections, then I think it is worth acting under the assumption that they are likely to occur. This is statistical handwaving of the highest order. How can models agree that something is “relatively robust”? To start with, what on earth does “relatively robust” mean? How would we measure it? Is Santer showing a “relatively robust” result? Do the Santer models “agree”? How would we know if they do or don’t? He says that if the results are “relatively robust” (whatever that means), that a reasonable hypothesis is that this is a “true consequence of our current understanding.” As opposed to a “false consequence”? … I think that he is trying to say that if the models agree, the programmers of the models likely agree … but so what? Why should we pay a lot of attention to which climate principles computer programmers might happen to agree on? The fact that two models agree, or two hundred models agree, does not make a result “robust” in any sense of the word. It only reflects the knowledge, ideas, biases and errors of the computer programmers who wrote the models. It has no more weight that having two hundred fundamentalist Christians tell me that the Earth is only 6,700 years old. Yes, the result with the Christians is “relatively robust” … but that does not mean that it is right. Let me restate the problem: we have no general theory of climate. We don’t know what all the feedbacks, forcings, and inter-relationships are. We don’t know if the earth has an “equilibrium” temperature, and if so, what mechanisms might maintain it. We don’t know many of the feedbacks’ size, and in some cases we don’t even know their sign. We are trying to model the most complex system we have ever attempted to model, and we have not been at it for long. All of the models have hideous errors which are covered up by parameterization. Given all of that, even “relatively robust” doesn’t mean anything except that the models agree. But do they agree because they are correct, or because they are all built on a common base of incorrect assumptions? As a man who has made models of varying complexity for a variety of systems including the earth’s climate, I strongly support choice B, that they are built on common, incorrect assumptions. For example, most models see increasing clouds as warming the earth. To me, that’s nuts, that’s an incorrect assumption shared by most models. It doesn’t even accord with our everyday experience. We all know that a cloud passing overhead on a hot day cools us more than the same cloud passing overhead warms us at night. In addition, a cloud near the horizon can block the sun entirely, but it has almost no effect on the IR radiation. In short, clouds cool the earth more than they warm the earth. However, most of the models agree on this positive cloud feedback … does that make it a “robust result”, or merely a “robust error”? For me, the fact that the most recent, up-to-date you-beaut models give results that are no better than the models of 20 years ago means that the modelers are on a fundamentally wrong path … w. PS – he says that if “future projections from climate models” have good “observational backup”, then we should do something or other … but me, I got lost at the part where he claims to have “observational backup” for future projections … me, I find it a bit difficult to make observations in the future, but then I don’t work for NASA, maybe it is rocket surgery after all … • henry Re: Willis Eschenbach (#56), I think the “observational backup” he’s referring to is the list of observations that all show the AGW “result”: the melting glaciers, the shrinking sea ice, the change in growing seasons – you know, “the list”. They seem to pull out “the list” every time one of the observations don’t show what they expected. • Alan Wilkinson Posted Oct 26, 2008 at 2:36 AM \| Permalink Re: henry (#58), Except that “the list” may suggest GW but cannot distinguish, measure or imply AGW. 29. Martin Lewitt > On this basis, there is a 91.6% probability that the model trend will exceed the observed trend. > I leave it to climate philosophers to decide whether this is “significant” or not. The IPCC has spoken on this issue in the AR4. For most of their conclusions they used 90% as “very likely”. 30. ED Attached is a link to an article about a substantial attempt to eliminate an area of uncertainty in the Climate models. “Scientists Go Cloud-hopping In The Pacific To Improve Climate Predictions” linked at http://www.sciencedaily.com/releases/2008/10/081021190646.htm What type of date will the they be able to collect to determine how a cloud forms and how reflective it is? 31. Craig Loehle Teacher: What are all these answers to my questions? And none of them are right! Student to teacher: I applied different computational techniques and got a range of answers. The variance in the answers is such that you can not preclude an overlap in their distributions. Therefore I pass the exam. 32. Bob North Craig — I think you just summarized the viewpoint of many of the modellers perfectly. 33. BarryW Four modelers were playing golf. On a par 3 hole the first hit his ball even with the hole but ten feet to the right. The second hit it even with the hole but ten feet to the left. The third hit his ten feet past the hole but dead in line, while the last hit his ten feet short of the hole but also in line. They all started cheering and each marked his score card with a hole in one, since the average of the ball’s locations was in the hole. 34. Jeff Norman Steve, Excellent post! As an engineer I understand this practical QA approach better than the more hypothetical discussions. Regarding your Santer workshop comment. This would go a long way in explaining why the “climate community” appears to ignore what’s happening in the Antarctic/South Pole. Thanks yet again. 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Richardson on PAGES2K Online “Journal Club” • TerryMN on PAGES2K Online “Journal Club” • Duster on PAGES2K Online “Journal Club” %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 3, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9404163956642151, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/5225-proof-question-print.html	# Proof Question Printable View • August 29th 2006, 09:18 AM nortonKM Proof Question Hi, I have been trying to solve this problem but I cannot figure out a way to solve it. Show that: For every n in Z+, n can be written as the power of 2 and an odd integer. Thanks • August 29th 2006, 09:56 AM Glaysher Suppose $n$ is even Then $n = 2^0 + (n - 1)$ where $n - 1$ is odd Suppose $n$ is odd Suppose positive integer $x \ne 0$ is such that $2^x < n$ Then $2^x$ is even and you need to add an odd number to get equal to $n$ All times are GMT -8. The time now is 11:56 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9562470316886902, "perplexity_flag": "middle"}
http://mathdl.maa.org/mathDL/23/?pa=content&sa=viewDocument&nodeId=3321&pf=1	# Thinking Outside the Box -- or Maybe Just About the Box by David Meel (Bowling Green State Univ.) and Thomas Hern (Bowling Green State Univ.) ### Abstract We present a fresh look at an age-old calculus problem, the box problem. The box problem is traditionally used as one of the typical exemplars for the study of optimization. However, this paper illustrates that another box problem, tied much closer to reality, can be used to investigate optimization. In particular, we present an improved box problem and its real-world context along with activities that blend hands-on and applet-based investigations with a strong dose of analysis that culminates in other possible extensions and investigations. ### Table of Contents This article uses Java Sketchpad to embed Geometer's Sketchpad figures. In order to view Java Sketchpad content, you will need to install the Java plug-in for your browser. Readers interested in Java Sketchpad and its capabilities may wish to read our Developers Area article, "Quick Interactive Web Pages with Java Sketchpad," by Michael E. Mays. #### Appendix: Student Module (Condensed from the article for student and classroom exploration) ### Introduction Perhaps one of the deepest entrenched calculus problem is the canonical box problem that students encounter when discussing applied extrema problems. In fact, Friedlander and Wilker (1980) commented, ''This question must be answered nearly a million times a year by calculus students from every corner of the globe'' (p. 282). Without a doubt, almost every recently published calculus text contains a problem similar to the following: A sheet of cardboard is rectangular, 14 inches long and 8.5 inches wide. Congruent squares are to be cut from its four corners. The resulting piece of cardboard is to be folded and its edges taped to form an open-topped box (see figure below). How should this be done to get a box of largest possible volume? Figure 1: Canonical Box Problem In particular, this problem (or some derivative of it) occurs in a classic text such as Thomas (1953) as well as modern texts such as Stewart (2008), Larson, Hostetler and Edwards (2007) and Smith and Minton (2007). But, what about this problem makes it so common and prevalent to the calculus experience? This paper takes a closer look at this common box problem and asks some difficult questions: • Does this box problem really reflect reality? • Is there a better box problem more consistent with modern box building techniques? And • Can a better box problem be explored by Calculus 1 students in a single variable course? If you enjoy this canonical box problem, we developed a Java Sketchpad applet, called OpenBox, for this problem. For this applet, the yellow points are moveable and control the size of the sheet of cardboard and the position of the cut. A dynamical graphic contained in a grey box shows the graph of the Volume to cut length (x) function. In addition, a dynamical picture of the open box is provided to illustrate how the changes in cut length impacts the configuration of the open box. (Warning: The OpenBox page is best viewed in 1024 x 768 resolution or greater and may take up to a minute to load.) ### Background #### Historical References The Box problem is so timeless that it even goes back almost a hundred years to the classic Granville and Smith (1911) text. Except for the fact that it is made out of tin, doesn't this look familiar? Figure 2. The box problem on page 115 in Granville and Smith (1911) However, this instance was not the first appearance of the box problem in either an academic publication or in popular culture. A derivation of the box problem first appeared in popular culture in the 1903 Henry Dudeny's puzzle column in the Weekly Dispatch and again in 1908, it was rephrased for his puzzle column in Cassell's Magazine. Below is the graphic and phrasing of the problem contained in Cassell's Magazine: Figure 3. Illustration from Cassell's Magazine. ''No. 525 -- HOW TO MAKE CISTERNS. Here is a little puzzle that will elucidate a point of considerable importance to cistern makers, ironmongers, plumbers, cardboard-box makers, and the public generally. Our friend the cistern-maker has an interesting task before him. He has a large sheet of zinc, measuring eight feet by three feet, and he proposes to cut out square pieces from the four corners (all, of course, of the same size), then fold up the sides, join them with solder, and make a cistern. So far, the work appears to be pretty obvious and easy. But the point that puzzles him is this: What is the exact size for the square pieces that he must cut out if the cistern is to contain the greatest possible quantity of water? Call the feet inches, and take a piece of cardboard or paper eight inches long and three inches wide. By experimenting with this you will soon see that a great deal depends on the size of those squares. To get the greatest contents you have to avoid cutting those squares too small on the one hand and too large on the other. How are you going to get at the right dimensions? I shall award our weekly half-guinea prize for a correct answer. State the dimensions of the squares and try to find a rule that the intelligent working man may understand.'' (Dudeney, 1908, p. 430) But just in case one might think that the problem is only a century old, a Treatise on Differential Calculus by Todhun ter (1855) contains the following related problem: Figure 4. The box problem on page 213 of Todhunter (1855). Consequently, this problem has appeared in calculus texts for at least the last one and a half centuries. It has become so entrenched in the calculus curriculum and articles are still being written about it. For instance, an article by Miller and Shaw (2007) breathed new life into this problem for it served as the backdrop to explore a conventional problem in unconventional ways or an article by Fredrickson (2003) reconceptualizing the constraints on the ''cutouts'' while striving for a container of maximum volume. But what is it about this problem that makes it so universal and mathematics teachers keep coming back to it? Does it really motivate student learning of essential calculus concepts? We concur with Underwood Dudley (2002) when he pointed out in his talk on Calculus Books, "when have you encountered a cardboard box constructed in such a way?" and the problem is ''quite silly'' according to Friedlander and Wilker (1980), Dundas (1984) and Pirich (1996) since the corners are wasted. As previously identified, this particular calculus problem predates modern use of cardboard packaging. But, most would agree that constructing a cardboard box in this manner is quite unusual. So, why do calculus teachers keep including it as part of their repertoire of examples other than the fact that it is straightforward and easily comprehensible? Doesn't the inclusion make students begin to wonder about the ability of calculus to model reality? In fact, if one attempts to rip apart the classic open box used for the typical shirt box that one receives at Christmas, it is not constructed in this manner. It is interesting to note that both the Granville and Smith text and Dudeney's puzzle used tin, a little more practical material than cardboard for such a construction. A box of cardboard might use the squares as tabs to connect the sides together and increase stability and strength. In fact, we have often asked students to construct boxes from index cards we gave them and they tend to do use these tabs naturally to make the connections with tape, glue or a staple. The students can typically answer the mathematical questions of the corner-removed box problem but at the same time they recognize the lack of realism when the corners are removed since boxes just are not constructed that way. #### Introducing the Regular Slotted Container It is our contention that there is a better box problem to be examined - a box problem that students will see as realistic as well as to be able to explore with tools from single-variable calculus. This adventure began a little over 25 years ago when one of the authors was curious about how to model realistic shipping boxes and tore one apart to find out. His first exploration was with a box known in the shipping industry as a Regular Slotted Container (RSC). The RSC is perhaps one of the most commonly utilized class of boxes in the shipping industry because it is one of the most economical boxes to manufacture and adapt for the shipment of most commodities. At first exploration, the construction of a RSC appears to require two variables but a simple but practical constraint makes it a problem appropriate for single variable calculus. Let's first explain what exactly a RSC is. As shown below in figures 5 and 6, a Regular Slotted Container is constructed from a large rectangular sheet of corrugated cardboard. Depending on the construction details, it may be manufactured with a tab used to either glue or stitch the joint (the inclusion of the tab results in waste cardboard during production) or may rely on a taped joint making it a minimal shipping container (i.e. a box where there is ''no'' scrap corrugated cardboard generated in the manufacturing process). In either case, the lengthwise (normally outer) flaps meet at the center of the box allowing it to be affixed by tape or staples. Figure 5: Construction schematics for the RSC (Safeway Packaging, n.d.) The box is closed Top flaps opened All flaps opened Completely disassembled Figure 6. Various stages of a RSC being disassembled #### Movie of the construction of the RSC It is difficult to conceive of how a blank sheet of cardboard with cuts at appropriate points and to particular lengths is transformed into a box that is commonly used across the shipping industry. In order to get a clear sense of this, we provide a link to a movie that shows the SWF Companies (n.d.) CE-451 manufacturing a Regular Slotted Container. Click here to see a movie of the CE-451 in action Warning: The CE-451 movie is 12 MB and may take more than a minute to load. What one sees in this movie are the blanks being loaded into a hopper and the robotic elements of the CE-451 taking those blanks and converting them into RSCs. What it does not show is how the cardboard was sliced, the positioning of those slices and how the slice lengths help maximize the volume capacity of the RSC. ### Discussing Dundas's (1984) analysis In a paper by Dundas (1984), he looked at a variety of box problems and one of them was configured like the RSC. The tactic that Dundas (1984) took was to look at the problem from the perspective of a piece of cardboard of fixed area rather than a fixed dimensions. We find this choice mathematically sound but problematic to the exploration of the real-world situation since cardboard is not typically sold by area but rather constrained by manufacturable lengths and widths. Consequently, when Dundas (1984) explored the function, $V(l) = T\left( {{1 \over 2} - T} \right)\left( {{A \over l} - T} \right)$ there was no way to convey to the reader information about the length and width of the cardboard (since the cardboard area was being restricted to A square inches). In addition, a critical piece of the problem is what the lengths of the cuts should be, since that transforms the problem from a multiple-variable problem to a single-variable problem. Unfortunately for a student reading the paper, it was never motivated or explained. It is these particular elements that we wish to revisit and illustrate using text and a set of applet-based activities to lead students to similar conclusions without making the activity all about pushing algebra around on a page. ### The First Box Problem #### Introduction Being able to step back and consider the problem without many constraints allows one to see the interdependence of the cut length, positioning of cuts, and the volume of the box. However, will students recognize that if they encounter the following box problem? A sheet of cardboard is rectangular, 14 inches long and 8.5 inches wide. If six congruent cuts (denoted in black) are to be made into the cardboard and five folds (denoted by dotted lines) made to adjoin the cuts so that the resulting piece of cardboard is to be folded to form a closeable rectangular box (see figures below). How should this be done to get a box of largest possible volume? Figure 7: The box problem graphics Before interacting with the Box Problem applet, students should be given the opportunity to physically build their own boxes out of paper or cardboard using scissors and tape or dissect a provided RSC. In doing so, students' physical constructions or destructions will help them recognize the particular physical constraints and how elements of the RSC come together prior to exploring the virtual world presented in the applets. Questions that can be asked of the students during this initial phase of exploration, if they don't naturally generate them on their own, are: 1. What does a ''closeable rectangular box'' infer about the characteristics of the box? 2. What impact does "closeable" have with respect to the length of at least a pair of the flaps? 3. Should the flaps overlap? and 4. If they do overlap, is that the best use of the cardboard? After students have constructed boxes in this manner then we suggest that students begin to interact with a set of Java applets designed to help students interact with the problem in ways that would not be possible when building boxes out of cardboard. But where does one go from there? Is moving expressly toward the algebraic solution the only path? It is our contention that considerably more learning can occur if this path is at least delayed a little while longer so students can be asked to explore the problem with greater depth. The pertinent question here is how does one encourage students to search for deeper insights that can be drawn from such a seemingly simple problem. We suggest the use of a pair of applets to help students investigate the problem in increasingly more complex ways and open their eyes to conditions and constraints that were not obvious through their initial encounters. #### The first Box Problem applet One element in the applets that we have designed are the multiple representations and alternative ways of conveying information. For instance, the image shown below is a screen shot from the ClosedBox applet, which allows students to explore various scenarios. The student can manipulate the lengths A and B by just pulling on the yellow points A' and B'. The yellow points P and Q also move. In doing so, the other components of the applet change in accordance with these manipulations. The power here is that students can interact with a wide variety of examples and see if the conjectures they make hold up to empirical investigation. In addition, the student can move the corners of the box, in the lower right-hand corner of the applet, and see if the box will actually close or not, an important aspect if you want the box to hold something. The last elements in this applet are the two different graphical indicators of maximal volume. The one graph shows the volume with respect to P or Q while the other is held constant and the bar graph next to it displays the percentage of maximal volume obtained by the current configuration. If the volume is too large, one can resize the vertical unit, a yellow point denoted by U, to get the graphs comfortably into the grey viewing window. As students play with these to attempt to improve their maximal value score, they will be led to ask a variety of questions, such as: • Is there a relationship between the cut length and the position of the cut that maximizes the volume? • Under what conditions does the volume drop to zero? • Is there a best way to orient P and Q so that the maximum is achieved? • How do the two graphics on the bottom left-hand side interact with each other? Figure 8: The first Box Problem applet Click Here to open the first Box Problem applet Warning: The first Box Problem applet page, entitled ClosedBox, is best viewed in 1024 x 768 resolution or greater and may take up to a minute to load. The goal of this exploration is to help students recognize that even though this problem may at first appear to be a problem involving two variables, the cut length and the position of the cut, focusing on maximizing the volume allows one to turn the problem into a problem of one variable. By exploring the applet, students should find that if the cut length is held constant and the position of the cut is allowed to vary, then the volume of the corresponding box is described by a constrained quadratic equation (expressed by the blue graph), and if the position of the cut is held constant while the length of the cut is allowed to vary, the volume of the box is described by a constrained linear equation (expressed by the red graph). The constraints result from the physical constraints on the box as well as whether, when the box is constructed, it will hold its contents, i.e. the flaps meet or overlap so there are no holes in the box. All it takes is some imagination after working through multiple examples and noticing the relationship between these two variables that maximizes the volume for any particular condition. From our perspective, being able to explore, quantify, and utilize is an important aspect of learning mathematics. #### The first Box Problem activity This applet was designed to help students come to a realization that for any positioning of the fold, the maximum volume can be best achieved when the cut length is set to one-half the smallest width of the box. Not only can students disassemble the box by pulling on the corner points, the applet helps students investigate how constraints affect the box problem. For instance, if the cut length exceeds the minimum of the length and width of the box, then the box will not close because the flaps created by the cuts will interfere with each other. Alternately, if the cut length is less than half of the minimum of the length and width of the box, then the box will not close because the flaps created will leave a hole in the box, thereby allowing the contents to fall out. So, how does one accomplish this? In the applet, the points denoted with a yellow dot are those that are moveable. Perhaps, the first question that students should be asked concerns the midpoint fold. In particular, why does one of the folds need to occur at the midpoint? Another question should focus on the symmetry. In fact, asking ''what happens if the two folds aren't symmetric about the midpoint fold?'' can yield interesting conversations amongst the students. Getting back to the movable points, a student can set a position for where the cut is to occur and then vary the length of the cut. We suggest students carefully record the data from their various trials where the cut position, $$m( \overline{AP})$$, is set and the cut length, $$m( \overline{BQ})$$, is changed to maximize the volume for that particular cut position. The last row of each Trial is for the student to search for the maximal cut length and the corresponding cut position. The final trial is for an open exploration of a piece of cardboard of the student's choosing. \| $$m( \overline {BQ})$$ \| $$m( \overline {AP})$$ \| $$m( \overline {PM})$$ \| $$m( \overline {QQ'})$$ \| $$m( \overline {BB'})$$ \| $$m( \overline {AA'})$$ \| % of max volume \| \|----------------------------\|----------------------------\|----------------------------\|-----------------------------\|-----------------------------\|-----------------------------\|-------------------\| \| \| \| \| \| \| \| \| \| 1.5 \| 3.0 \| 4.0 \| 5.5 \| 8.5 \| 14.0 \| 97.75% \| \| \| 3.5 \| 3.5 \| \| 8.5 \| 14.0 \| \| \| \| 4.5 \| 2.5 \| \| 8.5 \| 14.0 \| \| \| \| \| \| \| 8.5 \| 14.0 \| \| \| \| 2.5 \| 2.5 \| \| 10.0 \| 10.0 \| \| \| \| 1.5 \| 3.5 \| \| 10.0 \| 10.0 \| \| \| \| 4.25 \| 0.75 \| \| 10.0 \| 10.0 \| \| \| \| \| \| \| 10.0 \| 10.0 \| \| \| \| 3.0 \| 2.5 \| \| 8.5 \| 11.0 \| \| \| \| 2.75 \| 2.75 \| \| 8.5 \| 11.0 \| \| \| \| 4.0 \| 1.5 \| \| 8.5 \| 11.0 \| \| \| \| \| \| \| 8.5 \| 11.0 \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| \| Exploring various positions of a cut, should lead the students' to a conjecture that for any position of a cut, the maximum volume of the corresponding box is obtained when the cut length is half the length of the shorter length between $$m( \overline {AP})$$ and $$m( \overline {PM})$$. There are so many different questions that can be asked when students are interacting with the first Box Problem applet. In addition to those already mentioned, one can ask: • Under what conditions is the blue graph connected above the x-axis? • Under those conditions and for a particular length and width, what are the dimensions of the box with maximum volume? • Why does the red graph always appear as a slanted portion? What mathematical meaning does this red graph have? Does this slanted portion ever change slope? If so, why and if not, why not? It should be mentioned that we have found that if students have experienced the open box problem, there is a tendency of students to either focus on the length of the cut or the position of the cut but not typically both. Some students want to make the controlling variable the position of the cut and the resulting calculus computations are more difficult but not impossible. However, if we were to focus on the computations and not a deep exploration of the problem, students would miss a perfect opportunity to look beyond the numbers and see relationships that are intrinsically interwoven in this problem. ### The Second Box Problem #### Introduction After exploring the first Box Problem applet, some essential elements of moving a seemingly two-variable problem into a single variable problem have come to light. In particular, students are now prepared to explore the single-variable box problem. Specifically, the introduction provided by the first box problem applet has provided a means of reducing the complexity of the problem. Instead of having two variables with which to contend, we have found that without loss of generality, we can set the position of the cut to be twice the length of the cut. What exactly does this provide? We can now express the volume of the box with respect to a single variable thereby opening this problem to the techniques from Calculus 1. In particular, we can turn our attention to maximizing the volume of a box based upon the cut length. Now, the second Box Problem applet, will focus its attention on looking at $V(l) = \left( {B - 2l} \right)\left( {{1 \over 2}A - 2l} \right)\left( {2l} \right)$ where A and B are the correspondent length and width of the rectangular piece of cardboard and l corresponds to the length of the cut. We should note that this description is equivalent to the formula presented in Dundas (1984) of $V(l) = T\left( {{1 \over 2} - T} \right)\left( {{A \over l} - T} \right)$ as long as the following adjustments are made: • $$T = 2l$$ • $$l_{Dundas} = A$$ • $$B = w = {{A_{Dundas} } \over {l_{Dundas} }}$$ In addition, the l of Dundas (1984) needs to correspond to the position of the cut instead of the length of the cut. #### The second Box Problem applet At first glance, this applet, ClosedBox2, contains many of the same components as the first Box Problem applet; however, the cut length determines the positioning of the cut so that in each case the box volume is relatively maximized. Figure 9: The second Box Problem applet Click Here to open the second Box Problem applet Warning: The second Box Problem applet page, entitled ClosedBox2, is best viewed in 1024 x 768 resolution or greater and may take up to a minute to load. In this applet there are a variety of elements that can be seen. First, the point P is no longer adjustable but is rather determined by the length of the cut defined by the segment BQ. In addition, the grey box in the lower left-hand of the applet contains a dynamic graphical depiction of the functional relationship between cut length and volume. That is, it contains a graphical depiction of $V(l) = \left( {B - 2l} \right)\left( {{1 \over 2}A - 2l} \right)\left( {2l} \right)$ where l corresponds to $$m( \overline {BQ} )$$ and currently B = 8.5 and A = 14.0. One element that students need to grapple with when working with this particular applet is the graphical depiction of the function. In particular, graphing $$V(l)$$ using a graphing calculator or computer algebra system yields a figure similar to the following: Figure 10: Graph of the Box Problem Function The graphical depiction of the function presented in the applet is a truncated version of the one in figure 10. Students will need to come to grips with the fact that the applet is only concerned with the volume of boxes that are physically constructible whereas the graph of the function shown in figure 10, does not necessarily concern itself with the constructability of the box. Instead, it provides a graph of the functional relationship between an independent variable l and a dependent variable V. In essence, this graph in figure 10 is less concerned with cut length and volume and more concerned with expressing the relationship for all possible values of l, irrespective if these values are possible cut lengths or if those cut lengths yield appropriate volumes. Too often, we see students focused on the algebraic elements of a problem without considering the physical (or mathematical) constraints on that problem. We seek in this applet to guide students to grapple with the interplay of these two seemingly disparate forces. In turn, we hope to lead them to reconcile for themselves how functional relationships that model real-world phenomena require a careful examination of the domain for which that relationship actually does model the phenomena. For instance, students might at first think that l's only restriction is that it must be less than $${1 \over 2}m( {\overline {BB'} } )$$ since a cut cannot exceed half the width of the cardboard and maintain its connectedness. However, there is another constraint: the cut length, l, cannot exceed $${1 \over 4}m( {\overline {AA'} } )$$ . This ''hidden'' constraint comes directly from the relationship of cut length and position of the cut and depends on the relationship between length and width of the rectangular piece of cardboard. #### The second Box Problem activity This applet was designed to help investigate the interrelationship between cut length and volume. In particular, the previous interactions with the first Box Problem applet allowed the student to recognize the relationship between cut length and position of the cut. Using this, the student can build a functional relationship, based on a single variable, relating cut length and volume. This applet is designed to explore that functional relationship and help students come to a realization that in modeling real-world phenomena, a careful examination of the domain of a the functional relationship is necessary. How can one ask questions to help students interact with the second Box Problem applet in meaningful ways? From our experience, the first question that probably should be asked involves the relationship between the first Box Problem applet and this second Box Problem applet. In particular, we can see in this new applet that the point P varies as the point Q moves, what information drawn from the first Box Problem applet allows us to control the point P? Another question should focus on constructing the volume of the box based upon the length of cut. Essentially, a series of questions to help lead students to developing a functional relationship between volume and cut length, such as: • How does one determine the volume of a rectangular box? • What would be a description for the length of the box? • What would be a description for the width of the box? • What would be a description for the height of the box? • How can you use these descriptions to build a functional relationship between volume and cut length? Inherent to these questions are questions about what are the constants and what are the variables? Such a question is important when students are faced with multiple letter-based descriptions such as those expressed in the formula for $$V(l)$$ : $V(l) = \left( {B - 2l} \right)\left( {{1 \over 2}A - 2l} \right)\left( {2l} \right)$ Once students have developed appropriate functional relationships between cut length and volume, a variety of questions relating to that function can be asked. In general, we typically ask students to investigate the graph of the function on a hand-held graphing calculator. And then, we start asking various questions to help the students compare and contrast the graphical representation produced by the graphing calculator and that of the applet. For instance, we might ask: • Why does the graphing calculator seem to show you more of the graph than the applet does? • Why does the applet truncate the graph? • What conditions should be on the domain of the function and how do they relate to physically constructing a box? So far, these questions have focused primarily on a static rectangular sheet of cardboard. That is, we have not asked questions that forced students to think about varying the size of the cardboard. • For different lengths or widths, the applet's graph seems to change shape near the right-hand terminus, what mathematical reason can you provide for this change or provide an argument that it does, in fact, not change? • Are there two (or more) non-isomorphic sheets of cardboard, so the maximal volume is the same? If so, identify them and if not, explain why not. • Are there two (or more) non-isomorphic sheets of cardboard, so the placement of the maximal cut is the same? If so, identify them and if not, explain why not. Here, the questions are designed to force students to think beyond the original question and attempt to distill commonality and explore generalizations. Stretching students to think beyond and to ask ''what if'' questions is an important aspect of interacting with this applet. We sought to design the applet to support both the investigation of a specific scenario as well as a broad range of extended questions related to the original problem. ### Reflections and Extensions This paper has attempted to illustrate how applets can broaden the investigation of mathematical relationships and reach beyond mere algebraic manipulations. The ability to investigate, conjecture, test, and pose new questions in an electronic environment is important to the development of understanding. Beyond this, such investigations are enjoyable for students. For instance, students mentioned the box problem in their weekly Calculus 1 journals and two examples are provided below: ''This last week we learned of Indeterminate terms and L' Hopital's rule. Did quite a few examples on how to perform the rule, contionued [sic] this for four days, then I think we made boxes. I believe that I learned how to do L' Hop's rule fairly well, hopefully I could do well on a test, and the box making was fun and surprised me in that you fit it into the lesson. I enjoyed the interactivity.'' - Charles ''This week we learned about maximum volume of boxes and solved a pretty cool problem involving James Bond and Bombs. I love these sorts of problem-solving and I hope that we do more of it in the future.'' - Dan Our goal has always been to get students to think that problem-solving is fun and the interaction between building physical boxes and the computer applets helps students draw essential mathematical connections. The lessons that can be drawn from these activities bolster students' perceptions that mathematics is eminently useful and applicable to real-world phenomena much more than the typical box problem shown in many of the calculus texts available on the market. But is that as far as one can go into the world of boxes? We should also mention that this problem of optimal box building could be further extended by including the tab to affix the sides of the box to each other without the use of a taped joint. At one level, our discussion has focused its attention on construction of a minimal shipping container of the Regular Slotted Container type; however, if we were to introduce the requirement of a tab it does impact the problem. It should be noted that there are general governmental guidelines for the width of such a tab but 1 1/4 - 1 1/2 inches is typical. For specific guidelines about tab regulations, one should check the Fiber Box Handbook (Fiber Box Association, 2005), the National Motor Freight Classification (NMFC) Item 222 (American Trucking Association, 1970) for common carrier motor freight shipment, or the Uniform Freight Classification (UFC) Rule 41 (Dolan, 1991) for rail shipment. In any case, we think that after exploring the box problem through these applets, students might be better prepared to think how to construct a RSC containing tabs (as shown in figure 5) with maximum volume? Beyond even looking at a tabbed RSC, the world of boxes is much more encompassing. In this paper, we have focused our attention on the most popular box, a regular slotted container, but there are a host of other types of boxes fabricated from a single piece of corrugated cardboard that students can investigate, including: • A Full Overlap Container (FOL), see figure 10a, that is resistant to rough handling since all flaps are of the same length (the width of the box) and when closed, the outer flaps come within one inch of complete overlap, • A Five Panel Folder (FPF), see figure 10b, that features a fifth panel used as a closing flap completely covering a side panel and includes end pieces composed of multiple layers that provide stacking strength and protection of long articles with small diameter, or • A Once Piece Folder (OPF), see figure 10c, typically used books and printed materials since it has a flat bottom with flaps forming the sides and ends along with extended flaps that meet toform the top (GoPackaging.com, 2006). \| \| \| \| \|-----\|-----\|-----\| \| \| \| \| \| (a) \| (b) \| (c) \| Figure 11: Other types of boxes manufactured from a single piece of corrugated cardboard Each of these boxes would provide an opportunity to extend investigation and gain a better sense of the issues behind box construction. We certainly hope that this paper hasn't boxed you in but rather opened your eyes to the ways boxes can be used to help students see the ability of calculus to play a meaningful role in examining real-world problems. ### References American Trucking Association. (1970). National motor freight classification Washington, DC: National Motor Freight Traffic Association, Inc. Dolan, J.J. (1991). Uniform Freight Classification Rule 41. Chicago, IL: National Railroad Freight Committee. Dudeney, H.E. (1908, September). The puzzle realm. Cassell's Magazine. p. 430. Dudley, U. (2002, October). ''Calculus Books''. Paper presented at the meeting of the MAA Ohio Section with OhioMATYC at Kent State University Trumbull Campus, Warren, OH. (many of the same remarks were published in: Dudley, U. (1988). Calculus with Analytic Geometry by George F. Simmons. The American Mathematical Monthly, 95(9), 888-892. Available at: http://www.jstor.org/stable/2322923). Dundas, K. (1984). To build a better box. College Mathematics Journal, 15(4), 30-36. Available at: http://www.jstor.org/stable/3027427 Fredrickson, G.N. (2003). A New Wrinkle on an Old Folding Problem. The College Mathematics Journal, 34(4), 258-263. Friedlander, J.B. & Wilker, J.B. (1980). A budget of boxes. Mathematics Magazine, 53(5), 282-286. Fiber Box Association. (2005). Fiber Box Handbook. Rolling Meadows, IL: Fiber Box Association. GoPackaging.com (2006). Custom Corrugated Shipping Boxes. Retrieved October 23, 2008, from http://www.gopackaging.com/custom-boxes.aspx. Granville, W.A. & Smith, P.F. (1911). Elements of the Differential and Integral Calculus. Boston, MA: Ginn & Co. Larson, R., Hostetler, B. & Edwards, B. (2007) Calculus: Early Transcendental Functions, 4th Edition. Belmont, CA: Houghton Mifflin. Miller, C.M. & Shaw, D. (2007). What else can you do with an open box? Mathematics Teacher, 100(7), 470-474. Pirich, D.M. (1996). A new look at the classic box problem. PRIMUS, 6(1), 35-48. Safeway Packaging. (n.d.). RSC.bmp. Retrieved October 23, 2008, from http://www.safewaypkg.com/images/Design_Catalog/RSC.bmp. Smith, R.T. & Minton, R.B. (2007). Calculus: Early Transcendentals, 3rd Edition. New York, NY: McGraw Hill. Stewart, J. (2008). Calculus, 6th Edition. Belmont, CA: Brooks/Cole Publishing Company. SWF Companies. (n.d.). CE 451 Case Erector. Retrieved October 23, 2008, from http://www.swfcompanies.com/CE451.htm. Thomas, G.B. (1953). Calculus and Analytic Geometry, 2nd Edition. Reading, MA: Addison Wesley. Todhunter, I. (1855) A treatise on the differential calculus, 2nd Edition, Revised. London, UK: McMillan & Co.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 18, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9322487115859985, "perplexity_flag": "middle"}
http://cs.stackexchange.com/questions/8952/is-there-a-sometimes-efficient-algorithm-to-solve-sat	# Is there a sometimes-efficient algorithm to solve #SAT? Let $B$ be a boolean formula consisting of the usual AND, OR, and NOT operators and some variables. I would like to count the number of satisfying assignments for $B$. That is, I want to find the number of different assignments of truth values to the variables of $B$ for which $B$ assumes a true value. For example, the formula $a\lor b$ has three satisfying assignments; $(a\lor b)\land(c\lor\lnot b)$ has four. This is the #SAT problem. Obviously an efficient solution to this problem would imply an efficient solution to SAT, which is unlikely, and in fact this problem is #P-complete, and so may well be strictly harder than SAT. So I am not expecting a guaranteed-efficient solution. But it is well-known that there are relatively few really difficult instances of SAT itself. (See for example Cheeseman 1991, "Where the really hard problems are".) Ordinary pruned search, although exponential in the worst case, can solve many instances efficiently; resolution methods, although exponential in the worst case, are even more efficient in practice. My question is: Are any algorithms known which can quickly count the number of satisfying assignments of a typical boolean formula, even if such algorithms require exponential time in the general instance? Is there anything noticeably better than enumerating every possible assignment? - I tried to add a tag for #p-completeness, but the Stack Exchange software doesn't like the # sign. – Mark Dominus Jan 16 at 0:24 I'd be careful with claiming that "there are relatively few really difficult instances of SAT itself". I believe the paper you link actually talks about random $k$-SAT. Furthermore, the phase transition phenomenon only applies to the random instances. There are many really tough handcrafted, industrial etc. instances of SAT. – Juho Jan 16 at 0:32 Thanks. Do you think this tends to make my question less clear? Do you understand what I am asking for? – Mark Dominus Jan 16 at 0:35 It is clear to me. It's only important to remember what instances exhibit the phase transition :) – Juho Jan 16 at 0:42 ## 1 Answer ### Counting in the general case The problem you are interested in is known as #SAT, or model counting. In a sense, it is the classical #P-complete problem. Model counting is hard, even for $2$-SAT! Not surprisingly, the exact methods can only handle instances with around hundreds of variables. Approximate methods exist too, and they might be able to handle instances with around 1000 variables. Exact counting methods are often based on DPLL-style exhaustive search or some sort of knowledge compilation. The approximate methods are usually categorized as methods that give fast estimates without any guarantees and methods that provide lower or upper bounds with a correctness guarantee. There are also other methods that might not fit the categories, such as discovering backdoors, or methods that insist on certain structural properties to hold on the formulas (or their constraint graph). There are practical implementations out there. Some exact model counters are CDP, Relsat, Cachet, sharpSAT, and c2d. The sort of main techniques used by the exact solvers are partial counts, component analysis (of the underying constraint graph), formula and component caching, and smart reasoning at each node. Another method based on knowledge compilation converts the input CNF formula into another logical form. From this form, the model count can be deduced easily (polynomial time in the size of the newly produced formula). For example, one might convert the formula to a binary decision diagram (BDD). One could then traverse the BDD from the "1" leaf back to the root. Or for another example, the c2d employs a compiler that turns CNF formulas into deterministic decomposable negation normal form (d-DNNF). If your instances get larger or you don't care about being exact, approximate methods exist too. With approximate methods, we care about and consider the quality of the estimate and the correctness confidence associated with the estimate reported by our algorithm. One approach by Wei and Selman [2] uses MCMC sampling to compute an approximation of the true model count for the input formula. The method is based on the fact that if one can sample (near-)uniformly from the set of solution of a formula $\phi$, then one can compute a good estimate of the number of solutions of $\phi$. Gogate and Dechter [3] use a model counting technique known as SampleMinisat. It's based on sampling from the backtrack-free search space of a boolean formula. The technique builds on the idea of importance re-sampling, using DPLL-based SAT solvers to construct the backtrack-free search space. This might be done either completely or up to an approximation. Sampling for estimates with guarantees is also possible. Building on [2], Gomes et al. [4] showed that using sampling with a modified randomized strategy, one can get provable lower bounds on the total model count with high probabilistic correctness guarantees. There is also work that builds on belief propagation (BP). See Kroc et al. [5] and the BPCount they introduce. In the same paper, the authors give a second method called MiniCount, for providing upper bounds on the model count. There's also a statistical framework which allows one to compute upper bounds under certain statistical assumptions. ### Algorithms for #2-SAT and #3-SAT If you restrict your attention to #2-SAT or #3-SAT, there are algorithms that run in $O(1.3247^n)$ and $O(1.6894^n)$ for these problems respectively [1]. There are slight improvements for these algorithms. For example, Kutzkov [6] improved upon the upper bound of [1] for #3-SAT with an algorithm running in time $O(1.6423^n)$. As is in the nature of the problem, if you want to solve instances in practice, a lot depends on the size and structure of your instances. The more you know, the more capable you are in choosing the right method. [1] Vilhelm Dahllöf, Peter Jonsson, and Magnus Wahlström. Counting Satisfying Assignments in 2-SAT and 3-SAT. In Proceedings of the 8th Annual International Computing and Combinatorics Conference (COCOON-2002), 535-543, 2002. [2] W. Wei, and B. Selman. A New Approach to Model Counting. In Proceedings of SAT05: 8th International Conference on Theory and Applications of Satisfiability Testing, volume 3569 of Lecture Notes in Computer Science, 324-339, 2005. [3] R. Gogate, and R. Dechter. Approximate Counting by Sampling the Backtrack-free Search Space. In Proceedings of AAAI-07: 22nd National Conference on Artificial Intelligence, 198–203, Vancouver, 2007. [4] C. P. Gomes, J. Hoffmann, A. Sabharwal, and B. Selman. From Sampling to Model Counting. In Proceedings of IJCAI-07: 20th International Joint Conference on Artificial Intelligence, 2293–2299, 2007. [5] L. Kroc, A. Sabharwal, and B. Selman. Leveraging Belief Propagation, Backtrack Search, and Statistics for Model Counting. In CPAIOR-08: 5th International Conference on Integration of AI and OR Techniques in Constraint Programming, volume 5015 of Lecture Notes in Computer Science, 127–141, 2008. [6] K. Kutzkov. New upper bound for the #3-SAT problem. Information Processing Letters 105(1), 1-5, 2007. - +1 Nice answer. – Kaveh Jan 21 at 9:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9093813300132751, "perplexity_flag": "middle"}
http://www.reference.com/browse/Logarithm+function	Definitions Nearby Words # Logarithm [law-guh-rith-uhm, -rith-, log-uh-] /ˈlɔgəˌrɪðəm, -ˌrɪθ-, ˈlɒgə-/ In mathematics, the logarithm of a number to a given base is the power or exponent to which the base must be raised in order to produce the number. For example, the logarithm of 1000 to the base 10 is 3, because 10 raised to the power of 3 is 1000; the base 2 logarithm of 32 is 5 because 2 to the power 5 is 32. The logarithm of x to the base b is written logb(x) or, if the base is implicit, as log(x). So, for a number x, a base b and an exponent y, $text\left\{ if \right\}b^y = x,text\left\{ then \right\}y = log_b \left(x\right),.$ An important feature of logarithms is that they reduce multiplication to addition, by the formula: $log \left(x times y\right) = log x + log y ,.$ That is, the logarithm of the product of two numbers is the sum of the logarithms of those numbers. The use of logarithms to facilitate complex calculations was a significant motivation in their original development. ## Properties of the logarithm When x and b are restricted to positive real numbers, logb(x) is a unique real number. The magnitude of the base b must be neither 0 nor 1; the base used is typically 10, e, or 2. Logarithms are defined for real numbers and for complex numbers. The major property of logarithms is that they map multiplication to addition. This ability stems from the following identity: $b^x times b^y = b^\left\{x+y\right\} ,$ which by taking logarithms becomes $log_b left\left(b^x times b^y right\right) = log_b left\left(b^\left\{x+y\right\} right\right)$ $= x + y = log_b left\left(b^x right\right) + log_b left\left(b^y right\right).$ For example, $4=2^2 , Rightarrow , log_2\left(4\right)=2 , ,$ $8=2^3 , Rightarrow , log_2\left(8\right)=3 , ,$ $log_2\left(32\right) = log_2\left(4 times 8\right) = log_2\left(4\right) + log_2\left(8\right) = 2 + 3 = 5 , .$ A related property is reduction of exponentiation to multiplication. Using the identity: $c = b^\left\{log_b \left(c \right)\right\} ,$ it follows that c to the power p (exponentiation) is: $c^p = left\left(b^\left\{log_b \left(c \right)\right\}right\right)^p = b^\left\{p log_b \left(c \right)\right\} ,$ or, taking logarithms: $log_b left\left(c^p right\right) = p log_b \left(c \right) .$ In words, to raise a number to a power p, find the logarithm of the number and multiply it by p. The exponentiated value is then the inverse logarithm of this product; that is, number to power = bproduct. For example, $log_2\left(64\right) = log_2\left(4^3\right) = 3log_2\left(4\right) = 6 , .$ Besides reducing multiplication operations to addition, and exponentiation to multiplication, logarithms reduce division to subtraction, and roots to division. For example, $log_2\left(16\right) = log_2 left \left(frac\left\{64\right\}\left\{4\right\} right \right) = log_2\left(64\right) - log_2\left(4\right) = 6 - 2 = 4 , ,$ $log_2\left(sqrt\left[3\right]4\right) = frac \left\{1\right\}\left\{3\right\} log_2 \left(4\right) = frac \left\{2\right\}\left\{3\right\} , .$ Logarithms make lengthy numerical operations easier to perform. The whole process is made easy by using tables of logarithms, or a slide rule, antiquated now that calculators are available. Although the above practical advantages are not important for numerical work today, they are used in graphical analysis (see Bode plot). ## The logarithm as a function Though logarithms have been traditionally thought of as arithmetic sequences of numbers corresponding to geometric sequences of other (positive real) numbers, as in the 1797 Britannica definition, they are also the result of applying an analytic function. The function can therefore be meaningfully extended to complex numbers. The function logb(x) depends on both b and x, but the term logarithm function (or logarithmic function) in standard usage refers to a function of the form logb(x) in which the b is fixed and so the only argument is x. Thus there is one logarithm function for each value of the base b (which must be positive and must differ from 1). Viewed in this way, the base-b logarithm function is the inverse function of the exponential function bx. The word "logarithm" is often used to refer to a logarithm function itself as well as to particular values of this function. ## Logarithm of a negative or complex number There is no real valued logarithm for negative or complex numbers. The logarithm function can be extended to the complex logarithm. which does apply to these cases. The value is not unique though, since e2πi = e0 = 1 then both 2πi and 0 are equally valid logarithms to base e of 1. When z is a complex number, say z = x + i y, the logarithm of z is found by putting z in polar form that is, z = reiθ = r cos(θ) + i r sin(θ), where r = \|z\| = sqrt(x2 + y2) and θ = arg(z) is any angle such that and . Arg is a multi-valued function. If the base of the logarithm is chosen as e , that is, using loge (denoted by ln and called the natural logarithm), the complex logarithm is: $ln\left(z\right) = ln\left(r\right) + i left \left(theta + 2 pi k right \right) .$ The principal value is defined by setting k = 0. The principal value has imaginary part in the range (-π,π] and equals the natural logarithm for the positive reals. The principal value of the logarithm of a negative number is: $ln\left(-r\right) = ln\left(r\right) + i pi .$ For a base b other than e the complex logarithm Logb(z) can be defined as ln(z)/ln(b), the principal value of which is given by the principal values of ln(z) and ln(b). Note that ln(zp) is not in general the same as p ln(z), see failure of power and logarithm identities. ## Group theory From the pure mathematical perspective, the identity $log\left(cd\right) = log\left(c\right) + log\left(d\right) ,$ is fundamental in two senses. First, the remaining three arithmetic properties can be derived from it. Furthermore, it expresses an isomorphism between the multiplicative group of the positive real numbers and the additive group of all the reals. Logarithmic functions are the only continuous isomorphisms from the multiplicative group of positive real numbers to the additive group of real numbers. ## Bases The most widely used bases for logarithms are 10, the mathematical constant ≈ 2.71828... and 2. When "log" is written without a base (b missing from logb), the intent can usually be determined from context: • natural logarithm (log, ln, log, or Ln) in mathematical analysis, statistics, economics and some engineering fields. The reasons to consider e the natural base for logarithms, though perhaps not obvious, are numerous and compelling. • common logarithm (log10 or simply log; sometimes lg) in various engineering fields, especially for power levels and power ratios, such as acoustical sound pressure, and in logarithm tables to be used to simplify hand calculations • binary logarithm (log2; sometimes lg, lb, or ld), in computer science and information theory • indefinite logarithm (Log or [log ] or simply log) when the base is irrelevant, e.g. in complexity theory when describing the asymptotic behavior of algorithms in big O notation. To avoid confusion, it is best to specify the base if there is any chance of misinterpretation. ### Other notations The notation "ln(x)" invariably means loge(x), i.e., the natural logarithm of x, but the implied base for "log(x)" varies by discipline: • Mathematicians understand "log(x)" to mean loge(x). Calculus textbooks will occasionally write "lg(x)" to represent "log10(x)". • Many engineers, biologists, astronomers, and some others write only "ln(x)" or "loge(x)" when they mean the natural logarithm of x, and take "log(x)" to mean log10(x) or, in computer science, log2(x). • On most calculators, the LOG button is log10(x) and LN is loge(x). • In most commonly used computer programming languages, including C, C++, Java, Fortran, Ruby, and BASIC, the "log" function returns the natural logarithm. The base-10 function, if it is available, is generally "log10." • Some people use Log(x) (capital L) to mean log10(x), and use log(x) with a lowercase l to mean loge(x). • The notation Log(x) is also used by mathematicians to denote the principal branch of the (natural) logarithm function. • In some European countries, a frequently used notation is blog(x) instead of logb(x). This chaos, historically, originates from the fact that the natural logarithm has nice mathematical properties (such as its derivative being 1/x, and having a simple definition), while the base 10 logarithms, or decimal logarithms, were more convenient for speeding calculations (back when they were used for that purpose). Thus natural logarithms were only extensively used in fields like calculus while decimal logarithms were widely used elsewhere. As recently as 1984, Paul Halmos in his "automathography" I Want to Be a Mathematician heaped contempt on what he considered the childish "ln" notation, which he said no mathematician had ever used. The notation was in fact invented in 1893 by Irving Stringham, professor of mathematics at Berkeley. In computer science, the base 2 logarithm is sometimes written as lg(x), as suggested by Edward Reingold and popularized by Donald Knuth. However, lg(x) is also sometimes used for the common log, and lb(x) for the binary log. In Russian literature, the notation lg(x) is also generally used for the base 10 logarithm. In German, lg(x) also denotes the base 10 logarithm, while sometimes ld(x) or lb(x) is used for the base 2 logarithm. The clear advice of the United States Department of Commerce National Institute of Standards and Technology is to follow the ISO standard Mathematical signs and symbols for use in physical sciences and technology, ISO 31-11:1992, which suggests these notations: • The notation "ln(x)" means loge(x); • The notation "lg(x)" means log10(x); • The notation "lb(x)" means log2(x). As the difference between logarithms to different bases is one of scale, it is possible to consider all logarithm functions to be the same, merely giving the answer in different units, such as dB, neper, bits, decades, etc.; see the section Science and engineering below. Logarithms to a base less than 1 have a negative scale, or a flip about the x axis, relative to logarithms of base greater than 1. ### Change of base While there are several useful identities, the most important for calculator use lets one find logarithms with bases other than those built into the calculator (usually loge and log10). To find a logarithm with base b, using any other base k: $log_b\left(x\right) = frac\left\{log_k\left(x\right)\right\}\left\{log_k\left(b\right)\right\}.$ Moreover, this result implies that all logarithm functions (whatever the base) are similar to each other. So to calculate the log with base 2 of the number 16 with a calculator: $log_2\left(16\right) = frac\left\{log\left(16\right)\right\}\left\{log\left(2\right)\right\}.$ ## Uses of logarithms Logarithms are useful in solving equations in which exponents are unknown. They have simple derivatives, so they are often used in the solution of integrals. The logarithm is one of three closely related functions. In the equation bn = x, b can be determined with radicals, n with logarithms, and x with exponentials. See logarithmic identities for several rules governing the logarithm functions. ### Science Various quantities in science are expressed as logarithms of other quantities; see logarithmic scale for an explanation and a more complete list. • In chemistry, the negative of the base-10 logarithm of the activity of hydronium ions (H3O+, the form H+ takes in water) is the measure known as pH. The activity of hydronium ions in neutral water is 10−7 mol/L at 25 °C, hence a pH of 7. • The bel (symbol B) is a unit of measure which is the base-10 logarithm of ratios, such as power levels and voltage levels. It is mostly used in telecommunication, electronics, and acoustics. The Bel is named after telecommunications pioneer Alexander Graham Bell. The (dB), equal to 0.1 bel, is more commonly used. The is a similar unit which uses the natural logarithm of a ratio. • The Richter scale measures earthquake intensity on a base-10 logarithmic scale. • In spectrometry and optics, the absorbance unit used to measure optical density is equivalent to −1 B. • In astronomy, the apparent magnitude measures the brightness of stars logarithmically, since the eye also responds logarithmically to brightness. • In psychophysics, the Weber–Fechner law proposes a logarithmic relationship between stimulus and sensation. • In computer science, logarithms often appear in bounds for computational complexity. For example, to sort N items using comparison can require time proportional to the product N × log N. Similarly, base-2 logarithms are used to express the amount of storage space or memory required for a binary representation of a number—with k bits (each a 0 or a 1) one can represent 2k distinct values, so any natural number N can be represented in no more than (log2 N) + 1 bits. • Similarly, in information theory logarithms are used as a measure of quantity of information. If a message recipient may expect any one of N possible messages with equal likelihood, then the amount of information conveyed by any one such message is quantified as log2 N bits. • In geometry the logarithm is used to form the metric for the half-plane model of hyperbolic geometry. • Many types of engineering and scientific data are typically graphed on log-log or semilog axes, in order to most clearly show the form of the data. • In inferential statistics, the logarithm of the data in a dataset can be used for parametric statistical testing if the original data does not meet the assumption of normality. • Musical intervals are measured logarithmically as semitones. The interval between two notes in semitones is the base-21/12 logarithm of the frequency ratio (or equivalently, 12 times the base-2 logarithm). Fractional semitones are used for non-equal temperaments. Especially to measure deviations from the equal tempered scale, intervals are also expressed in cents (hundredths of an equally-tempered semitone). The interval between two notes in cents is the base-21/1200 logarithm of the frequency ratio (or 1200 times the base-2 logarithm). In MIDI, notes are numbered on the semitone scale (logarithmic absolute nominal pitch with middle C at 60). For microtuning to other tuning systems, a logarithmic scale is defined filling in the ranges between the semitones of the equal tempered scale in a compatible way. This scale corresponds to the note numbers for whole semitones. (see microtuning in MIDI). ### Exponential functions One way of defining the exponential function ex, also written as exp(x), is as the inverse of the natural logarithm. It is positive for every real argument x. The operation of "raising b to a power p" for positive arguments b and all real exponents p is defined by $b^p = left\left(e^\left\{ln b\right\} right\right) ^p = e^\left\{p ln b \right\}.,$ The antilogarithm function is another name for the inverse of the logarithmic function. It is written antilogb(n) and means the same as bn. ### Easier computations Logarithms can be used to replace difficult operations on numbers by easier operations on their logs (in any base), as the following table summarizes. In the table, upper-case variables represent logs of corresponding lower-case variables: Operation with numbers Operation with exponents Logarithmic identity $!, c d$ $!, C + D$ $!, log\left(c d\right) = log\left(c\right) + log\left(d\right)$ $!, c / d$ $!, C - D$ $!, log\left(c / d\right) = log\left(c\right) - log\left(d\right)$ $!, c ^ d$ $!, Cd$ $!, log\left(c ^ d\right) = d log\left(c\right)$ $!, sqrt\left[d\right]\left\{c\right\}$ $!, C / d$ $!, log\left(sqrt\left[d\right]\left\{c\right\}\right) = frac\left\{log\left(c\right)\right\}\left\{d\right\}$ These arithmetic properties of logarithms make such calculations much faster. The use of logarithms was an essential skill until electronic computers and calculators became available. Indeed the discovery of logarithms, just before Newton's era, had an impact in the scientific world that can be compared with that of the advent of computers in the 20th century because it made feasible many calculations that had previously been too laborious. As an example, to approximate the product of two numbers one can look up their logarithms in a table, add them, and, using the table again, proceed from that sum to its antilogarithm, which is the desired product. The precision of the approximation can be increased by interpolating between table entries. For manual calculations that demand any appreciable precision, this process, requiring three lookups and a sum, is much faster than performing the multiplication. To achieve seven decimal places of accuracy requires a table that fills a single large volume; a table for nine-decimal accuracy occupies a few shelves. Similarly, to approximate a power cd one can look up log c in the table, look up the log of that, and add to it the log of d; roots can be approximated in much the same way. One key application of these techniques was celestial navigation. Once the invention of the chronometer made possible the accurate measurement of longitude at sea, mariners had everything necessary to reduce their navigational computations to mere additions. A five-digit table of logarithms and a table of the logarithms of trigonometric functions sufficed for most purposes, and those tables could fit in a small book. Another critical application with even broader impact was the slide rule, an essential calculating tool for engineers. Many of the powerful capabilities of the slide rule derive from a clever but simple design that relies on the arithmetic properties of logarithms. The slide rule allows computation much faster still than the techniques based on tables, but provides much less precision, although slide rule operations can be chained to calculate answers to any arbitrary precision. ## Related operations ### Cologarithms The cologarithm of a number is the logarithm of the reciprocal of the number, meaning cologb(x) = logb(1/x) = −logb(x). ### Antilogarithms The antilogarithm is the logarithmic inverse of the logarithm, meaning that the antilogb(logb(x)) = x. Thus, setting by = x implies that logb(x) = y. By taking the antilogb of both sides, antilogb(logb(x)) = antilogby, thus x = antilogby. Therefore, by = antilogby. ## Calculus The natural logarithm of a positive number x can be defined as $ln \left(x\right) equiv int_\left\{1\right\}^\left\{x\right\} frac\left\{dt\right\}\left\{t\right\}.$ The derivative of the natural logarithm function is $frac\left\{d\right\}\left\{dx\right\} ln\left(x\right) = frac\left\{1\right\}\left\{x\right\}.$ By applying the change-of-base rule, the derivative for other bases is $frac\left\{d\right\}\left\{dx\right\} log_b\left(x\right) = frac\left\{d\right\}\left\{dx\right\} frac \left\{ln\left(x\right)\right\}\left\{ln\left(b\right)\right\} = frac\left\{1\right\}\left\{x ln\left(b\right)\right\} = frac\left\{log_b\left(e\right)\right\}\left\{x\right\}.$ The antiderivative of the natural logarithm ln(x) is $int ln\left(x\right) ,dx = x ln\left(x\right) - x + C,$ and so the antiderivative of the logarithm for other bases is $int log_b\left(x\right) ,dx = x log_b\left(x\right) - frac\left\{x\right\}\left\{ln\left(b\right)\right\} + C = x log_b left\left(frac\left\{x\right\}\left\{e\right\}right\right) + C.$ See also: Table of limits, list of integrals of logarithmic functions. ## Series for calculating the natural logarithm There are several series for calculating natural logarithms. The simplest, though inefficient, is: $ln \left(z\right) = sum_\left\{n=1\right\}^infty -frac\left\{\left(1-z\right)^n\right\}\left\{n\right\}$ when $\|1-z\|<1 !.$ To derive this series, start with ($\|x\|<1 !.$) $frac\left\{1\right\}\left\{1-x\right\} = 1 + x + x^2 + x^3 + cdots.$ Integrate both sides to obtain $-ln\left(1-x\right) = x + frac\left\{x^2\right\}\left\{2\right\} + frac\left\{x^3\right\}\left\{3\right\} + cdots$ $ln\left(1-x\right) = -x - frac\left\{x^2\right\}\left\{2\right\} - frac\left\{x^3\right\}\left\{3\right\} - frac\left\{x^4\right\}\left\{4\right\} - cdots.$ Letting $z = 1-x !$ and thus $x = \left(1-z\right) !$, we get $ln z = -\left(1-z\right) - frac\left\{\left(1-z\right)^2\right\}\left\{2\right\} - frac\left\{\left(1-z\right)^3\right\}\left\{3\right\} - frac\left\{\left(1-z\right)^4\right\}\left\{4\right\} + cdots$ A more efficient series is $ln \left(z\right) = 2 sum_\left\{n=0\right\}^infty frac\left\{1\right\}\left\{2n+1\right\} \left\{left \left(frac\left\{z-1\right\}\left\{z+1\right\} right \right) \right\}^\left\{2n+1\right\}$ for z with positive real part. To derive this series, we begin by substituting −x for x and get $ln\left(1+x\right) = x - frac\left\{x^2\right\}\left\{2\right\} + frac\left\{x^3\right\}\left\{3\right\} - frac\left\{x^4\right\}\left\{4\right\} + cdots.$ Subtracting, we get $ln frac\left\{1+x\right\}\left\{1-x\right\} = ln\left(1+x\right) - ln\left(1-x\right) = 2x + 2frac\left\{x^3\right\}\left\{3\right\} + 2frac\left\{x^5\right\}\left\{5\right\} + cdots.$ Letting $z = frac\left\{1+x\right\}\left\{1-x\right\} !$ and thus $x = frac\left\{z-1\right\}\left\{z+1\right\} !$, we get $ln z = 2 left \left(frac\left\{z-1\right\}\left\{z+1\right\} + frac\left\{1\right\}\left\{3\right\}\left\{left\left(frac\left\{z-1\right\}\left\{z+1\right\}right\right)\right\}^3 + frac\left\{1\right\}\left\{5\right\}\left\{left\left(frac\left\{z-1\right\}\left\{z+1\right\}right\right)\right\}^5 + cdots right \right).$ For example, applying this series to $z = frac\left\{11\right\}\left\{9\right\},$ we get $frac\left\{z-1\right\}\left\{z+1\right\} = frac\left\{frac\left\{11\right\}\left\{9\right\} - 1\right\}\left\{frac\left\{11\right\}\left\{9\right\} + 1\right\} = frac\left\{1\right\}\left\{10\right\},$ and thus $ln \left(1.2222222dots\right) = frac\left\{2\right\}\left\{10\right\} left \left(1 + frac\left\{1\right\}\left\{3cdot 100\right\} + frac\left\{1\right\}\left\{5 cdot 10000\right\} +$ frac{1}{7 cdot 1000000} + cdots right ) $= 0.2 cdot \left(1.0000000dots + 0.0033333dots + 0.0000200dots + 0.0000001dots + cdots\right)$ $= 0.2 cdot 1.0033535dots = 0.2006707dots$ where we factored 1/10 out of the sum in the first line. For any other base b, we use $log_b \left(x\right) = frac\left\{ln \left(x\right)\right\}\left\{ln \left(b\right)\right\}.$ ## Computers Most computer languages use log(x) for the natural logarithm, while the common log is typically denoted log10(x). The argument and return values are typically a floating point (or double precision) data type. As the argument is floating point, it can be useful to consider the following: A floating point value x is represented by a mantissa m and exponent n to form $x = m2^n.,$ Therefore $ln\left(x\right) = ln\left(m\right) + nln\left(2\right).,$ Thus, instead of computing $ln\left(x\right)$ we compute $ln\left(m\right)$ for some m such that 1 ≤ m < 2. Having m in this range means that the value $u = frac\left\{m - 1\right\}\left\{m+1\right\}$ is always in the range $0 le u < frac13$. Some machines use the mantissa in the range $0.5 le m < 1$ and in that case the value for u will be in the range $-frac13 < u le 0$ In either case, the series is even easier to compute. To compute a base 2 logarithm on a number between 1 and 2 in an alternate way, square it repeatedly. Every time it goes over 2, divide it by 2 and write a "1" bit, else just write a "0" bit. This is because squaring doubles the logarithm of a number. The integer part of the logarithm to base 2 of an unsigned integer is given by the position of the left-most bit, and can be computed in O(n) steps using the following algorithm: int log2(int x){ ` int r = 0;` ` while((x >> r) != 0){` ` r++;` } ` return r-1; // returns -1 for x==0, floor(log2(x)) otherwise` } However, it can also be computed in O(log n) steps by trying to shift by powers of 2 and checking that the result stays nonzero: for example, first >>16, then >>8, ... (Each step reveals one bit of the result) ## Generalizations The ordinary logarithm of positive reals generalizes to negative and complex arguments, though it is a multivalued function that needs a branch cut terminating at the branch point at 0 to make an ordinary function or principal branch. The logarithm (to base e) of a complex number z is the complex number ln(\|z\|) + i arg(z), where \|z\| is the modulus of z, arg(z) is the argument, and i is the imaginary unit; see complex logarithm for details. The discrete logarithm is a related notion in the theory of finite groups. It involves solving the equation bn = x, where b and x are elements of the group, and n is an integer specifying a power in the group operation. For some finite groups, it is believed that the discrete logarithm is very hard to calculate, whereas discrete exponentials are quite easy. This asymmetry has applications in public key cryptography. The logarithm of a matrix is the inverse of the matrix exponential. It is possible to take the logarithm of a quaternions and octonions. A double logarithm, $ln\left(ln\left(x\right)\right)$, is the inverse function of the double exponential function. A or is the inverse function of tetration. The super-logarithm of x grows even more slowly than the double logarithm for large x. For each positive b not equal to 1, the function logb (x) is an isomorphism from the group of positive real numbers under multiplication to the group of (all) real numbers under addition. They are the only such isomorphisms that are continuous. The logarithm function can be extended to a Haar measure in the topological group of positive real numbers under multiplication. ## History The method of logarithms was first publicly propounded in 1614, in a book entitled Mirifici Logarithmorum Canonis Descriptio, by John Napier, Baron of Merchiston, in Scotland. (Joost Bürgi independently discovered logarithms; however, he did not publish his discovery until four years after Napier.) Early resistance to the use of logarithms was muted by Kepler's enthusiastic support and his publication of a clear and impeccable explanation of how they worked. Their use contributed to the advance of science, and especially of astronomy, by making some difficult calculations possible. Prior to the advent of calculators and computers, they were used constantly in surveying, navigation, and other branches of practical mathematics. It supplanted the more involved method of prosthaphaeresis, which relied on trigonometric identities as a quick method of computing products. Besides the utility of the logarithm concept in computation, the natural logarithm presented a solution to the problem of quadrature of a hyperbolic sector at the hand of Gregoire de Saint-Vincent in 1647. At first, Napier called logarithms "artificial numbers" and antilogarithms "natural numbers". Later, Napier formed the word logarithm to mean a number that indicates a ratio: λόγος () meaning proportion, and ἀριθμός (arithmos) meaning number. Napier chose that because the difference of two logarithms determines the ratio of the numbers they represent, so that an arithmetic series of logarithms corresponds to a geometric series of numbers. The term antilogarithm was introduced in the late 17th century and, while never used extensively in mathematics, persisted in collections of tables until they fell into disuse. Napier did not use a base as we now understand it, but his logarithms were, up to a scaling factor, effectively to base 1/e. For interpolation purposes and ease of calculation, it is useful to make the ratio r in the geometric series close to 1. Napier chose r = 1 - 10−7 = 0.999999 (Bürgi chose r = 1 + 10−4 = 1.0001). Napier's original logarithms did not have log 1 = 0 but rather log 107 = 0. Thus if N is a number and L is its logarithm as calculated by Napier, N = 107(1 − 10−7)L. Since (1 − 10−7)107 is approximately 1/e, this makes L/107 approximately equal to log1/e N/107. ## Tables of logarithms Prior to the advent of computers and calculators, using logarithms meant using tables of logarithms, which had to be created manually. Base-10 logarithms are useful in computations when electronic means are not available. See common logarithm for details, including the use of characteristics and mantissas of common (i.e., base-10) logarithms. In 1617, Henry Briggs published the first installment of his own table of common logarithms, containing the logarithms of all integers below 1000 to eight decimal places. This he followed, in 1624, by his Arithmetica Logarithmica, containing the logarithms of all integers from 1 to 20,000 and from 90,000 to 100,000 to fourteen places of decimals, together with a learned introduction, in which the theory and use of logarithms are fully developed. The interval from 20,000 to 90,000 was filled up by Adriaan Vlacq, a Dutch mathematician; but in his table, which appeared in 1628, the logarithms were given to only ten places of decimals. Vlacq's table was later found to contain 603 errors, but "this cannot be regarded as a great number, when it is considered that the table was the result of an original calculation, and that more than 2,100,000 printed figures are liable to error. An edition of Vlacq's work, containing many corrections, was issued at Leipzig in 1794 under the title Thesaurus Logarithmorum Completus by Jurij Vega. François Callet's seven-place table (Paris, 1795), instead of stopping at 100,000, gave the eight-place logarithms of the numbers between 100,000 and 108,000, in order to diminish the errors of interpolation, which were greatest in the early part of the table; and this addition was generally included in seven-place tables. The only important published extension of Vlacq's table was made by Mr. Sang in 1871, whose table contained the seven-place logarithms of all numbers below 200,000. Briggs and Vlacq also published original tables of the logarithms of the trigonometric functions. Besides the tables mentioned above, a great collection, called Tables du Cadastre, was constructed under the direction of Gaspard de Prony, by an original computation, under the auspices of the French republican government of the 1700s. This work, which contained the logarithms of all numbers up to 100,000 to nineteen places, and of the numbers between 100,000 and 200,000 to twenty-four places, exists only in manuscript, "in seventeen enormous folios," at the Observatory of Paris. It was begun in 1792; and "the whole of the calculations, which to secure greater accuracy were performed in duplicate, and the two manuscripts subsequently collated with care, were completed in the short space of two years." Cubic interpolation could be used to find the logarithm of any number to a similar accuracy.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 67, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9210798144340515, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/7812-integrals-exam.html	# Thread: 1. ## Integrals for exam Hey I have an exam in 2 days and I really need some help gettin these questions under control! integral (x^2e^-x)dx I am not quite sure where to start, I was thinkin that I need to use the substitution where u=x^2 but then I am just confused after that! integral (2x+4)/(x^3-2x^2)dx For this one I factorised and bottom line so I had integral (2x+4)/(x^2)(x-2) dx Then I sed that (2x+4)/(x^2)(x-2)= A/x+b/x+c/x-2 Then I didnt really know wat to do after that! 3. The functions S and T are defined by setting S(x):= (integral x^2 to 0) sqrt(1+t^2)dt I tried doin this one but when I substituted for 0 later on I got that it was undefined... If someone could show the workings or at least the start of these problems I would be much appreciative... then I would be able to use these as examples for the other practice exams! 2. Originally Posted by taryn Hey I have an exam in 2 days and I really need some help gettin these questions under control! integral (x^2e^-x)dx I am not quite sure where to start, I was thinkin that I need to use the substitution where u=x^2 but then I am just confused after that! Integration by parts. $\int x^2 e^{-x} dx$ Let, $u=x^2$ and $v'=e^{-x}$ Thus, $u'=2x$ and $v=-e^{-x}$ Thus, $uv-\int u'vdx$ Thus, (note the signs change ) $-x^2e^{-x}+2\int xe^{-x} dx$ Do the same integration by parts on this integral. $u=x$ and $v'=e^{-x}$ Thus, $u'=1$ and $v=-e^{-x}$ Thus, (note the parantheses ) $-x^2e^{-x}+2\left( -xe^{-x}+\int e^{-x} dx\right)$ Thus, $-x^2e^{-x}-2xe^{-x}-2e^{-x}+C$ Thus, $-e^{-x}(x^2+2x+2)+C$ 3. Originally Posted by ThePerfectHacker Integration by parts. $\int x^2 e^{-x} dx$ Let, $u=x^2$ and $v'=e^{-x}$ Thus, $u'=2x$ and $v=-e^{-x}$ Thus, $uv-\int u'vdx$ Thus, (note the signs change ) $-x^2e^{-x}+2\int xe^{-x} dx$ [/tex] This is the only bit I am a little unsure of, Is there a reason y u can just take the uv part out of the integral? $uv-\int u'vdx$ Thanks for ur help! 4. Originally Posted by taryn This is the only bit I am a little unsure of, Is there a reason y u can just take the uv part out of the integral? $uv-\int u'vdx$ Thanks for ur help! I do not understand. If you are asking whether you can do this, $\int x^2 e^x dx=x^2\int e^x dx$ No because $x^2$ is not a constant function throughout the interval of integration. 5. Originally Posted by ThePerfectHacker Integration by parts. $\int x^2 e^{-x} dx$ Let, $u=x^2$ and $v'=e^{-x}$ Thus, $u'=2x$ and $v=-e^{-x}$ Thus, $uv-\int u'vdx$ Thus, (note the signs change ) $-x^2e^{-x}+2\int xe^{-x} dx$ Do the same integration by parts on this integral. $u=x$ and $v'=e^{-x}$ Thus, $u'=1$ and $v=-e^{-x}$ Thus, (note the parantheses ) $-x^2e^{-x}+2\left( -xe^{-x}+\int e^{-x} dx\right)$ Thus, $-x^2e^{-x}-2xe^{-x}-2e^{-x}+C$ Thus, $-e^{-x}(x^2+2x+2)+C$ Originally Posted by ThePerfectHacker I do not understand. If you are asking whether you can do this, $\int x^2 e^x dx=x^2\int e^x dx$ No because $x^2$ is not a constant function throughout the interval of integration. No I know thats not possible... I just mean how can u just take the uv out of the integral! Y are u able to do that! Obviously I know that x^2 is not a constant function throughout the interval... so does that mean u are sayin that uv is so thats y u can take it out? Is there a web site or somethin that I can read to help m e understand wat u have done? 6. Originally Posted by taryn No I know thats not possible... I just mean how can u just take the uv out of the integral! Y are u able to do that! Obviously I know that x^2 is not a constant function throughout the interval... so does that mean u are sayin that uv is so thats y u can take it out? Is there a web site or somethin that I can read to help m e understand wat u have done? Wikipedia explains it here. But be warned it sometimes like to overcomplicate things (I was told I do that too). Basically it is a special rule. You can take out that factor but then you need to change the integral. So you are not really taking the function out of the integral you are used making the integral simpler to work with.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 41, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9392473697662354, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/157633/determine-the-galois-group-of-the-splitting-field-of-x3-1x2-5-over-mat	# Determine the Galois group of the splitting field of $(x^3-1)(x^2-5)$ over $\mathbb{Q}$ Determine the Galois group of the splitting field of $(x^3-1)(x^2-5)$ over $\mathbb{Q}$ I've been struggling with some of these Galois group questions. - 1 Welcome to math.stackexchange! What have you tried? – talmid Jun 13 '12 at 5:16 ## 2 Answers Let $f=(x^3-1)(x^2-5)\in\mathbb{Q}[x]$. Note that $$\begin{align}f&=(x-1)(x^2+x+1)(x^2-5)\\ &=(x-1)(x-\zeta_3)(x-\zeta_3^2)(x-\sqrt{5})(x+\sqrt{5})\end{align}$$ where $\zeta_3=\frac{-1+\sqrt{3}}{2}$ is a primitive cube root of unity. Thus, the splitting field of $f$ over $\mathbb{Q}$ is $$K=\mathbb{Q}(\zeta_3,\sqrt{5})=\{a+b\zeta_3+c\sqrt{5}+d\zeta_3\sqrt{5}\mid a,b,c,d\in\mathbb{Q}\}.$$ Suppose that $\phi:K\to K$ is an automorphism of $K$ fixing $\mathbb{Q}$. Then $$\begin{align}\phi(a+b\zeta_3+c\sqrt{5}+d\zeta_3\sqrt{5})&=\phi(a)+\phi(b)\phi(\zeta_3)+\phi(c)\phi(\sqrt{5})+\phi(d)\phi(\zeta_3)\phi(\sqrt{5}) \\ &=a+b\phi(\zeta_3)+c\phi(\sqrt{5})+d\phi(\zeta_3)\phi(\sqrt{5})\end{align}$$ so that $\phi$ is determined entirely by the values of $\phi(\zeta_3)$ and $\phi(\sqrt{5})$; that is, if $\psi$ is an automorphism of $K$ such that $\psi(\zeta_3)=\phi(\zeta_3)$ and $\psi(\sqrt{5})=\phi(\sqrt{5})$, then in fact $\psi=\phi$. Because $\zeta_3^2+\zeta_3+1=0$ and $(\sqrt{5})^2-5=0$, we must have that $$0=\phi(0)=\phi(\zeta_3^2+\zeta_3+1)=\phi(\zeta_3)^2+\phi(\zeta_3)+1$$ and $$0=\phi(0)=\phi((\sqrt{5})^2-5)=\phi(\sqrt{5})^2-5$$ so that $\phi(\zeta_3)$ must be one of the two roots of $x^2+x+1$, namely $\zeta_3$ and $\zeta_3^2$, and $\phi(\sqrt{5})$ must be one of the two roots of $x^2-5$, namely $\sqrt{5}$ and $-\sqrt{5}$. Thus, there are at most four elements of $\operatorname{Gal}(K/\mathbb{Q})$ (there are two choices for where $\zeta_3$ goes, and two choices for where $\sqrt{5}$ goes): $$\begin{align} \operatorname{id}_K(\zeta_3)&=\zeta_3 & \operatorname{id}_K(\sqrt{5})&=\sqrt{5}\\ \phi(\zeta_3) &= \zeta_3^2 & \phi(\sqrt{5})&=\sqrt{5}\\ \rho(\zeta_3) &= \zeta_3 & \rho(\sqrt{5})&=-\sqrt{5}\\ \sigma(\zeta_3) &= \zeta_3^2 & \sigma(\sqrt{5})&=-\sqrt{5}\\ \end{align}$$ Check that all four do in fact define valid automorphisms, and see what the group structure on $\operatorname{Gal}(K/\mathbb{Q})$ is by composing them and seeing what comes up. For example, we have that $\phi\circ \rho=\sigma$, because $$(\phi\circ\rho)(\zeta_3)=\phi(\rho(\zeta_3))=\phi(\zeta_3)=\zeta_3^2=\sigma(\zeta_3)$$ $$(\phi\circ\rho)(\sqrt{5})=\phi(\rho(\sqrt{5}))=\phi(-\sqrt{5})=-\phi(\sqrt{5})=-\sqrt{5}=\sigma(\sqrt{5})$$ Are there any elements of order 4? - The splitting field of $(x^3-1)(x^2-5) = (x-1)(x^2+x+1)(x^2-5)$ is the compositum of the splitting fields of $x^2+x+1$ and of $x^2-5$; it is not hard to check that it is of degree $4$ over $\mathbb{Q}$. Hence, the Galois group must be either cyclic of order $4$, or isomorphic to the Klein $4$-group. Now, which is it? By the Fundamental Theorem of Galois Theory, the subgroups of the Galois group correspond to intermediate fields of the extension. We can distinguish the cyclic group of order $4$ from the Klein $4$-group in many ways, but one of them is that the former has a unique subgroup of index $2$, whereas the latter has three distinct subgroups of index $2$. A subgroup of index $2$ corresponds to an intermediate extension of degree $2$. So... does the splitting field contain a unique quadratic extension of $\mathbb{Q}$? Or does it contain more than one? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 48, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9433709383010864, "perplexity_flag": "head"}
http://terrytao.wordpress.com/tag/stationary-process/	What’s new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao # Tag Archive You are currently browsing the tag archive for the ‘stationary process’ tag. ## Rohlin’s problem on strongly mixing systems 9 March, 2011 in expository, math.DS, math.PR, question \| Tags: ergodic theory, stationary process, strong mixing \| by Terence Tao \| 7 comments Let ${G = (G,+)}$ be an abelian countable discrete group. A measure-preserving ${G}$-system ${X = (X, {\mathcal X}, \mu, (T_g)_{g \in G})}$ (or ${G}$-system for short) is a probability space ${(X, {\mathcal X}, \mu)}$, equipped with a measure-preserving action ${T_g: X \rightarrow X}$ of the group ${G}$, thus $\displaystyle \mu( T_g(E) ) = \mu(E)$ for all ${E \in {\mathcal X}}$ and ${g \in G}$, and $\displaystyle T_g T_h = T_{g+h}$ for all ${g, h \in G}$, with ${T_0}$ equal to the identity map. Classically, ergodic theory has focused on the cyclic case ${G={\bf Z}}$ (in which the ${T_g}$ are iterates of a single map ${T = T_1}$, with elements of ${G}$ being interpreted as a time parameter), but one can certainly consider actions of other groups ${G}$ also (including continuous or non-abelian groups). A ${G}$-system is said to be strongly ${2}$-mixing, or strongly mixing for short, if one has $\displaystyle \lim_{g \rightarrow \infty} \mu( A \cap T_g B ) = \mu(A) \mu(B)$ for all ${A, B \in {\mathcal X}}$, where the convergence is with respect to the one-point compactification of ${G}$ (thus, for every ${\epsilon > 0}$, there exists a compact (hence finite) subset ${K}$ of ${G}$ such that ${\|\mu(A \cap T_g B) - \mu(A)\mu(B)\| \leq \epsilon}$ for all ${g \not \in K}$). Similarly, we say that a ${G}$-system is strongly ${3}$-mixing if one has $\displaystyle \lim_{g,h,h-g \rightarrow \infty} \mu( A \cap T_g B \cap T_h C ) = \mu(A) \mu(B) \mu(C)$ for all ${A,B,C \in {\mathcal X}}$, thus for every ${\epsilon > 0}$, there exists a finite subset ${K}$ of ${G}$ such that $\displaystyle \|\mu( A \cap T_g B \cap T_h C ) - \mu(A) \mu(B) \mu(C)\| \leq \epsilon$ whenever ${g, h, h-g}$ all lie outside ${K}$. It is obvious that a strongly ${3}$-mixing system is necessarily strong ${2}$-mixing. In the case of ${{\bf Z}}$-systems, it has been an open problem for some time, due to Rohlin, whether the converse is true: Problem 1 (Rohlin’s problem) Is every strongly mixing ${{\bf Z}}$-system necessarily strongly ${3}$-mixing? This is a surprisingly difficult problem. In the positive direction, a routine application of the Cauchy-Schwarz inequality (via van der Corput’s inequality) shows that every strongly mixing system is weakly ${3}$-mixing, which roughly speaking means that ${\mu(A \cap T_g B \cap T_h C)}$ converges to ${\mu(A) \mu(B) \mu(C)}$ for most ${g, h \in {\bf Z}}$. Indeed, every weakly mixing system is in fact weakly mixing of all orders; see for instance this blog post of Carlos Matheus, or these lecture notes of myself. So the problem is to exclude the possibility of correlation between ${A}$, ${T_g B}$, and ${T_h C}$ for a small but non-trivial number of pairs ${(g,h)}$. It is also known that the answer to Rohlin’s problem is affirmative for rank one transformations (a result of Kalikow) and for shifts with purely singular continuous spectrum (a result of Host; note that strongly mixing systems cannot have any non-trivial point spectrum). Indeed, any counterexample to the problem, if it exists, is likely to be highly pathological. In the other direction, Rohlin’s problem is known to have a negative answer for ${{\bf Z}^2}$-systems, by a well-known counterexample of Ledrappier which can be described as follows. One can view a ${{\bf Z}^2}$-system as being essentially equivalent to a stationary process ${(x_{n,m})_{(n,m) \in {\bf Z}^2}}$ of random variables ${x_{n,m}}$ in some range space ${\Omega}$ indexed by ${{\bf Z}^2}$, with ${X}$ being ${\Omega^{{\bf Z}^2}}$ with the obvious shift map $\displaystyle T_{(g,h)} (x_{n,m})_{(n,m) \in {\bf Z}^2} := (x_{n-g,m-h})_{(n,m) \in {\bf Z}^2}.$ In Ledrappier’s example, the ${x_{n,m}}$ take values in the finite field ${{\bf F}_2}$ of two elements, and are selected at uniformly random subject to the “Pascal’s triangle” linear constraints $\displaystyle x_{n,m} = x_{n-1,m} + x_{n,m-1}.$ A routine application of the Kolmogorov extension theorem allows one to build such a process. The point is that due to the properties of Pascal’s triangle modulo ${2}$ (known as Sierpinski’s triangle), one has $\displaystyle x_{n,m} = x_{n-2^k,m} + x_{n,m-2^k}$ for all powers of two ${2^k}$. This is enough to destroy strong ${3}$-mixing, because it shows a strong correlation between ${x}$, ${T_{(2^k,0)} x}$, and ${T_{(0,2^k)} x}$ for arbitrarily large ${k}$ and randomly chosen ${x \in X}$. On the other hand, one can still show that ${x}$ and ${T_g x}$ are asymptotically uncorrelated for large ${g}$, giving strong ${2}$-mixing. Unfortunately, there are significant obstructions to converting Ledrappier’s example from a ${{\bf Z}^2}$-system to a ${{\bf Z}}$-system, as pointed out by de la Rue. In this post, I would like to record a “finite field” variant of Ledrappier’s construction, in which ${{\bf Z}^2}$ is replaced by the function field ring ${{\bf F}_3[t]}$, which is a “dyadic” (or more precisely, “triadic”) model for the integers (cf. this earlier blog post of mine). In other words: Theorem 2 There exists a ${{\bf F}_3[t]}$-system that is strongly ${2}$-mixing but not strongly ${3}$-mixing. The idea is much the same as that of Ledrappier; one builds a stationary ${{\bf F}_3[t]}$-process ${(x_n)_{n \in {\bf F}_3[t]}}$ in which ${x_n \in {\bf F}_3}$ are chosen uniformly at random subject to the constraints $\displaystyle x_n + x_{n + t^k} + x_{n + 2t^k} = 0 \ \ \ \ \ (1)$ for all ${n \in {\bf F}_3[t]}$ and all ${k \geq 0}$. Again, this system is manifestly not strongly ${3}$-mixing, but can be shown to be strongly ${2}$-mixing; I give details below the fold. As I discussed in this previous post, in many cases the dyadic model serves as a good guide for the non-dyadic model. However, in this case there is a curious rigidity phenomenon that seems to prevent Ledrappier-type examples from being transferable to the one-dimensional non-dyadic setting; once one restores the Archimedean nature of the underlying group, the constraints (1) not only reinforce each other strongly, but also force so much linearity on the system that one loses the strong mixing property. Read the rest of this entry » ### Recent Comments Sandeep Murthy on An elementary non-commutative… Luqing Ye on 245A, Notes 2: The Lebesgue… Frank on Soft analysis, hard analysis,… andrescaicedo on Soft analysis, hard analysis,… Richard Palais on Pythagoras’ theorem The Coffee Stains in… on Does one have to be a genius t… Benoît Régent-Kloeck… on (Ingrid Daubechies) Planning f… Luqing Ye on 245B, Notes 7: Well-ordered se… Luqing Ye on 245B, Notes 7: Well-ordered se… Arjun Jain on 245B, Notes 7: Well-ordered se… %anchor_text% on Books Luqing Ye on 245B, Notes 7: Well-ordered se… Arjun Jain on 245B, Notes 7: Well-ordered se… Luqing Ye on 245A, Notes 2: The Lebesgue…	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 91, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9250329732894897, "perplexity_flag": "head"}
http://www.bci2000.org/wiki/index.php/Contributions:HilbertFilter	# Contributions:HilbertFilter From BCI2000 Wiki ## Synopsis This filter computes the envelope or the phase of a signal using Hilbert transform. The discrete input signal x(n) is first transformed to its analytic representation (i.e., analytic signal), which is composed of real and imaginary parts. $x_{a}(n) = x(n) + j \operatorname{H}(x(n))$ The real part is the same input signal, and the imaginary part is the Hilbert transform of the input signal. The Hilbert transform is implemented as the convolution of the input signal with the filter h(n). $h(n) = \begin{cases} \ \ {2 \over \pi n}, & \mbox{for } n \mbox{ odd}\\ \ \ 0, & \mbox{for } n \mbox{ even}\\ \end{cases}$ To get an ideal Hilbert transform, n must be infinitely long $(-\infty < n < \infty)$. However, for real time implementations, h(n) must be truncated and delayed to guarantee a causal filter. Thus, the FIR filter is defined as $h(n) = \begin{cases} \ \ {2 \over \pi (n-\delta)}, & \mbox{for } n \mbox{ odd}\\ \ \ 0, & \mbox{for } n \mbox{ even},\\ \end{cases}$ for $0\leq n \leq N-1$, where N is an odd number representing the length of the filter in samples, and the resulting Hilbert transform is delayed by a number of samples $\delta = (N-1)/2$. The real part of the analytic signal must also be delayed by the same amount in order to estimate the envelope and phase of the input signal. ## Location http://www.bci2000.org/svn/trunk/src/contrib/SignalProcessing/HilbertSignalProcessing/HilbertFilter.cpp ## Versioning ### Authors Cristhian Potes, Jeremy Hill ## Parameters ### OutputSignal This parameter determines which quantity is output from the filter. It may be one of: 0 - Copy input signal no processing, 1 - Magnitude Hilbert envelope amplitude, 2 - Phase Hilbert phase, 3 - Real part original input signal, but with a delay to match its timing to the imaginary part. 4 - Imaginary part original signal filtered with an FIR-Hilbert transformer. ### Delay As for most BCI2000 parameters expressing time, this parameter should be expressed either as a number of SampleBlocks (a bare number without unit) or as a physical length of time with a unit appended (e.g. "0.1s"). The length of the Hilbert filter itself is $N = 2 \delta + 1$ once the delay has been converted to a number of samples $\delta$. Conversely the delay is $\delta = ( N - 1 ) / 2$, where N is the (odd) number of samples in the filter. Empirically, we have found that N ≥ 201 samples (i.e. $\delta$ ≥ 100 samples) is advisable for a reasonable-quality approximation to the infinite Hilbert transform. A warning is issued if the delay is shorter than this. None. ## See also User Reference:Filters, Contributions:SignalProcessing, Contributions:HilbertSignalProcessing	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 10, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8443549275398254, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/7369-intersecting-intervals.html	# Thread: 1. ## intersecting intervals Question attatched Attached Thumbnails 2. None of the intervals can be unbounded, right or left rays. Otherwise there is a counter-example. Name the interval with the least left end point $I_1$. There are at most six intervals that may share a point with it. So we have at least 43 which have no point common with $I_1$. Of those, name the interval with the least left end point $I_2$. Again are at most six intervals that may share a point with $I_2$, leaving 36 that have no point either of the two named. We continue getting $I_3$ from 28 and so fourth until we get $I_8$ from 8. Now you have 8 intervals that are pairwise disjoint.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.92674320936203, "perplexity_flag": "head"}
http://mathoverflow.net/questions/54074/how-many-consecutive-composite-integers-follow-k1-closed	## How many consecutive composite integers follow k!+1? [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I originally thought this question was too basic for MathOverflow, so I tried Math.StackExchange, but no one there seemed to have a solution. Here goes. How many consecutive composite integers follow `k!+1`? I wrote a program in Mathematica to compute explicit answers for the first 300, but there doesn't seem to be much of a pattern. The results of that are here. This is a problem in Underwood Dudley's Elementary Number Theory (section 23.2 part (b)), but for the LIFE of me, I can't figure it out! My initial thought was this: Let `m` be the smallest prime such that `k<m`. Then any number of the form `k!+i`, where `1<i<m` will clearly be composite. Hence, there are at least `m-2` composite numbers which follow `k!+1`. However, past that it seems hard to say. For example, `11!+13` factorizes as `199*200587`, which seems to be unpredictable behavior. I'm leaning toward thinking this was a misprint in the book and it's a much harder problem than something that should be in Elementary Number Theory. Perhaps I'm overlooking a simple, elegant solution. The other thing is, if there were a closed-form solution, then we would have an easy formula for calculating arbitrarily large primes. Take the largest known prime `p` and suppose we determine that `q` composite numbers follow `p!+1`. Then `p!+2+q` is prime. - 6 I suspect that all Dudley expects by way of an answer is something like, "at least $k-1$". – Gerry Myerson Feb 2 2011 at 8:46 That's what I'm leaning toward. It seems pretty dang hard to predict anything past `m-2`, where `m` is as I defined it (the least prime greater than or equal to `k`). – Willy Feb 2 2011 at 9:08 3 Note that this question has been cross-posted at math.SE (and I have answered it there, essentially elaborating on Gerry's comment). Please note that undergraduate homework questions are off-topic for this site, c.f. the FAQ. – Pete L. Clark Feb 2 2011 at 10:02 Noted. My mistake. – Willy Feb 2 2011 at 10:14 Link to math.se question: math.stackexchange.com/questions/20001/… – François G. Dorais♦ Feb 2 2011 at 15:06 show 1 more comment ## 1 Answer This seems like an interesting problem. Prime gap problems are notoriously difficult (Compare with Cramer's conjecture) and I do not expect this to be any easier. I will therefore not give a definitive answer, but rather try to give some heuristics. The Cramer model (see for example the paper of Granville "Harald Cramer and the distribution of prime numbers") where we assume that the probability that $n$ is prime is about $1/\log n$ gives that the probability that $k!+j$ should be prime should be about $1/\log(k!)$ which by Stirling's formula is about $1/k \log k$. This would give an average prime gap of about $k\log k$. One might reason a little bit further. Of course when we consider $k!+j$ for $2 \leq j \leq k$ we have forced these number to be composite. In general the number $k!+j$ will be composite if some prime factor of $j$ is less than $k$. We may then ask what is the probability that the number $j$ has all prime factors greater than $k$? If $k < j < k^2$ this is true if and only if $j$ is prime which has probability about $1/\log j$ which if $k < j < k (\log k)^N$ is about $1/\log k$. One might then expect the probability that the number $k^2+j$ to be prime should be the product of these probabilities or $1/(k(\log k)^2)$. However then we have not recognized the fact that we have forced the number $k!+j$ to have no prime factor less than $k$. This means that we should look at the conditional probability that an integer $n$ is prime given that it has no prime factors less than $k$. A simple sieving argument shows the number is odd should give us $1/2$ of the numbers, forcing that the number is not divisable by 3 should give us about $2/3$ of the numbers, and so forth. In general we should have about $$\prod_{p \le k} \left(1-\frac 1 p \right) \sim \frac{e^{-\gamma}}{\log k}$$ (By Merten's formula) numbers remaining after the sieving process. This is also suggested by the use of Dirichlet's theorem. This should give us the probability $e^{\gamma}/k \log k$ that the number $k!+j$ is prime, which is up to a constant the same as suggested by the Cramer model. In both cases the average prime gap can be expected to be of the order $k \log k$, albeit with different constants. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9629501700401306, "perplexity_flag": "head"}
http://mathhelpforum.com/math-topics/37283-sinusoidal-help.html	# Thread: 1. ## Sinusoidal Help? Determine the values of a and d in the sinusoidal function y=asin(bx+c) +d if the maximum value is 12 and the minimum value is 4. Find the value of b in the sinusoidal function if the period is 5. That is what I'm being asked. And yet again the crappy online course hasn't even mentioned it. 2. Originally Posted by MathIsRelativelyUseless Determine the values of a and d in the sinusoidal function y=asin(bx+c) +d if the maximum value is 12 and the minimum value is 4. Find the value of b in the sinusoidal function if the period is 5. That is what I'm being asked. And yet again the crappy online course hasn't even mentioned it. First part: Ok, for this you need to know that $\sin{(bx+c)}$ ranges from -1 to 1. You are given that the minimum value of y is 4 and the maximum value of y is 12. Since the sine function is the only value of the function that is going to be changing, this is a relatively easy problem to solve. When the sine is at its least, the function will be at its least. So when the sine is -1, the function is 4. Similarly, when the sine is at its greatest, the function will be at its greatest. So when the sine is 1, the function is 12. Hence we have: $4 = -a + d$ $12 = a + d$ Adding these two yields $16 = 2d$ from which we determine $d = 8, a = 4$. 3. ah ok and what about finding b? 4. Originally Posted by MathIsRelativelyUseless ah ok and what about finding b? To find b, note that the period of the sin function (without being modified by b) is $2\pi$. The period is therefore going to be equal to $\frac{2\pi}{b}$. So you just need to solve the equation $\frac{2\pi}{b} = 5$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9066334962844849, "perplexity_flag": "head"}
http://alanrendall.wordpress.com/2008/12/14/shock-waves-part-2/	# Hydrobates A mathematician thinks aloud ## Shock waves, part 2 In a previous post I mentioned the recent work of Demetrios Christodoulou on shock waves. On 08.12 I heard a talk by Christodoulou on this subject and since then he has explained some of the most important points of this work to me in more detail. Here I will present a little of what I learned. These results concern solutions of the relativistic Euler equations in Minkowski space. According to Christodoulou analogous results could be obtained in the non-relativistic case but no details of this have been published. The initial data is given on a hyperplane and is assumed to coincide outside a compact set with a constant state where the fluid is at rest with constant density. On the compact set the data is close to the constant state in a suitable sense. The object of study is the maximal smooth solution evolving from the given data and its future boundary. This boundary will be non-empty exactly in the case when a shock is formed. Necessary and sufficient conditions are given for there to be a shock. Most of the results concern a fluid which is isentropic and irrotational. The two conditions are intimately connected and cannot be assumed independently of each other. These results do have consequences for the general case since a sufficiently large region exists where the extra conditions are satisfied. Here I will concentrate on the isentropic and irrotational case. A central point is that while the solution becomes singular at the future boundary it is actually smooth up to and including the boundary with respect to a non-standard differential structure. Key computations are done in coordinates which define this different kind of smoothness. One of these coordinates is constant on sound cones of the solution being considered. The condions for shock formation depend very much on the sign of the quantity $\rho f''(\rho)+2f'(\rho)$ at the constant state, where $p=f(\rho)$ is the equation of state of the fluid. This is the same as the sign of the quantity $H$ introduced in the book. If this quantity actually vanishes on the constant state then there are no shocks. For an irrotational and isotropic flow the evolution equations of the fluid can be written as a quasilinear wave equation for a scalar function $\phi$. In the case that $H$ is identically zero for a given equation of state this coincides with a geometric equation which is obtained as follows. Suppose that a timelike hypersurface in five-dimensional Minkowski space can be expressed as the graph of a function $\phi$ on ${\bf R}^4$. Suppose in other words that the hypersurface is of the form $(x_0,x_1,x_2,x_3,\phi(x_0,x_1,x_2,x_3))$. Then the condition that this hypersurface has vanishing mean curvature is equivalent to the equation of motion for this particular type of fluid. The fluid is related to the Chaplygin gas which has been studied in cosmology in recent years. It has equation of state $p=-A\rho^{-1}$ for a constant $A$ and satisfies $H=0$. This type of fluid originally came up in aerodynamics in the early years of the twentieth century. Sergei Chaplygin, after whom this fluid is named, seems to have been quite a prominent figure since the town he grew up in is now named after him, as is a crater on the moon. For fluids under normal physical conditions $\rho f''(\rho)+2f'(\rho)>0$ but, as pointed out in the book, there are physical situations where the opposite sign occurs. ### Like this: This entry was posted on December 14, 2008 at 9:28 am and is filed under hydrodynamics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site. ### 2 Responses to “Shock waves, part 2” 1. Uwe Brauer Says: January 13, 2009 at 6:36 pm \| Reply Hello, right now I see the document with some formula does not parse in yellow! Besides this, it looks extremely interesting what you are talking about. “A central point is that while the solution becomes singular at the future boundary it is actually smooth up to and including the boundary with respect to a non-standard differential structure.” could you outline what this non-standard differential structure is supposed to be? thanks Uwe Brauer 2. hydrobates Says: January 13, 2009 at 8:33 pm \| Reply I do not know what is the problem with parsing. When I view the post it looks OK. As far as the non-standard differential structure is concerned perhaps the following is helpful. One idea is to follow a family of sound cones starting on the initial surface and watch their density in space. This density blows up at the shock. Now this sounds strange since it seems to depend so much on which family you started with on the initial surface. If I understood Christodoulou correctly the answer is: it does not matter – whichever choice you make the density of hypersurfaces blows up at the same place. He seemed to find this very remarkable. If you tried to define a coordinate which is constant on these sound cones then it would become singular when the shock is reached. The variables defining the fluid also become singular. The statement is now that if you change to a suitable new coordinate system where one of the coordinates is constant on these sound cones then something amazing happens. Of course the family of sound cones remains smooth in the new coordinate system, it was built that way, but the fluid variables are also smooth when expressed in these new coordinates. The statement about the differential structure is just a way of saying that in a coordinate-free way. %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 12, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9543819427490234, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/23186/maximum-range-of-a-projectile-launched-from-an-elevation	# Maximum range of a projectile (launched from an elevation) If a projectile is launched at a speed $u$ from a height $H$ above the horizontal axis, and air resistance is ignored, the maximum range of the projectile is $R_{max}=\frac ug\sqrt{u^2+2gH}$, where $g$ is the acceleration due to gravity. The angle of projection to achieve $R_{max}$ is $\theta = \arctan \left(\frac u{\sqrt{u^2+2gH}} \right)$. Can someone help me derive $R_{max}$ as given above? I have tried substituting $y=0$ and $x=R$ into the trajectory equation $$y=H+x \tan\theta -x^2\frac g{2u^2}(1+\tan^2\theta),$$ then differentiating with respect to $\theta$ so that we can let $\frac {dR}{d\theta}=0$ (so that $R=R_{max}$), but this would eliminate the $H$, so it won't lead to the expression for $R_{max}$ that I want to derive. - – Qmechanic♦ Apr 2 '12 at 20:55 Differentiating will not eliminate H. You need the derivative of x with respect to $\theta$. I can read off from what you have that $H$ is divided by $\tan\theta$ when you solve for $x$. (I didn't check everything else, though.) – Mark Eichenlaub Apr 2 '12 at 21:23 H is a constant, so it gets eliminated, no? – Ryan Apr 2 '12 at 21:51 No. Take $\frac{d}{dx} 5x$. 5 is a constant but does not get eliminated. – Mark Eichenlaub Apr 3 '12 at 0:00 @Mark, perhaps we're talking at cross-purposes. Please refer to leongz's answer below to see why our H gets eliminated. – Ryan Apr 3 '12 at 0:36 ## 1 Answer As you described, we substitute $y=0$ and $x=R$ into the trajectory equation: $$0=H+R\tan{\theta}-R^2\frac{g}{2u^2}\sec^2\theta.\tag{1}$$ Then, differentiating with respect to $\theta$ and setting $\frac{dR}{d\theta}=0$: $$0=R_{max}\sec^2\theta-R_{max}^2\frac{g}{2u^2}2\sec^2\theta\tan\theta,$$ which simplifies to $$R_{max}=\frac{u^2}{g}\cot\theta.\tag{2}$$ Solving $(1)$ and $(2)$ will yield the desired expressions for $\theta$ and $R_{max}$. - Beautiful! THANK YOU! ps. I've cleaned up this page to make it more general and useful. – Ryan Apr 2 '12 at 23:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 27, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.938731849193573, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/4683/continuous-and-bounded-variation-does-not-imply-absolutely-continuous?answertab=oldest	# Continuous and bounded variation does not imply absolutely continuous I know that a continuous function which is a BV may not be absolutely continuous. Is there an example of such a function? I was looking for a BV whose derivative is not Lebesgue integrable but I couldn't find one. - My apologies; I only saw Byron's when I posted that comment. – Akhil Mathew Sep 15 '10 at 2:56 ## 3 Answers The Devil's staircase function does the trick. Its derivative is almost surely zero with respect to Lebesgue measure, so the function is not absolutely continuous. See http://mathworld.wolfram.com/DevilsStaircase.html - 2 – Pete L. Clark Sep 14 '10 at 23:00 Byron already answered your main question, but your last sentence is another matter. You want a BV function whose derivative is not integrable, but such things don't exist. In particular, if $f$ is monotone on $[a,b]$, then $f'$ exists a.e., is Lebesgue integrable, and $\int_a^b f' \leq f(b)-f(a)$. Thus half of the fundamental theorem of calculus holds, so to speak. General BV functions are differences of monotone functions, so their derivatives are also Lebesgue integrable. - Thanks for pointing out my error. – jennifer Sep 14 '10 at 22:14 f(x)=[x] is Of bounded variation on [0,1] but noncontinuous and not abs. cont. - 2 The question asks for a continuous function. – rschwieb Dec 7 '12 at 17:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9339377880096436, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/104283-distance-point-line.html	# Thread: 1. ## Distance from a point to a line So, the point given is A = ( 4, 2 ) and the line given is p: (x,y) = (1,2) + s[-1 3] Okay.. so I re-write the line in form ax + by = c which would be.. y - 2 = (-1/3)(x-1) y = (-1/3)x + (7/3) (1/3)x - y = (7/3) so a = (1/3) b = (-1) so a normal to the line p would be n = [1/3 -1] I find a point on line p, I pick Q = (1,2) I find a new vector by subtracting Q from A.. so (4,2) - (1,2) would be the new vector r. now for me to find the distance of the point, I project n (r).. so I take a unit vector of n and take the dot product of it with r and to find the scalar. then I take the scalar and multiple it by n again. then i find the distance of the new vector and that is the distance of the point? Does that make sense, or am I completely wrong? 2. The distance from the point $(p,q)$ to the line $Ax+By+C=0$ is given as $\frac{{\left\| {Ap + Bq + C} \right\|}}{{\sqrt {A^2 + B^2 } }}$ Your line is standard form is $3x+y-5=0$ 3. okay, i see now i messed up converting from slope to standard form.. i understand the formula but not so much how it works.. my question is, how do you get to the formula? I thought it involved the dot product of a unit vector which is a normal to the line given and the vector from the point given to any point on the line then once you have the scalar, you multiple it by the unit vector of the normal and find the length of that vector.. 4. Originally Posted by zodiacbrave okay, i see now i messed up converting from slope to standard form.. i understand the formula but not so much how it works.. my question is, how do you get to the formula? I thought it involved the dot product of a unit vector which is a normal to the line given and the vector from the point given to any point on the line then once you have the scalar, you multiple it by the unit vector of the normal and find the length of that vector.. Here is a general solution.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.945731520652771, "perplexity_flag": "head"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Abelian_integral	# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Abelian integral In mathematics, an abelian integral in Riemann surface theory is a function related to the indefinite integral of a differential of the first kind. Suppose given a Riemann surface S and on it a differential 1-form ω that is everywhere on S holomorphic, and fixing a point P on S from which to integrate. We can regard $\int_P^Q \omega$ as a multi-valued function f(Q), or (better) an honest function of the chosen path C drawn on S from P to Q. Since S will in general be multiply-connected, one should specify C, but the value will in fact only depend on the homology class, of C modulo cycles on S. In the case of S a compact Riemann surface of genus 1, i.e. an elliptic curve, such functions are the elliptic integrals. Logically speaking, therefore, an abelian integral should be a function such as f. Such functions were first introduced to study hyperelliptic integrals, i.e. for the case where S is a hyperelliptic curve. This is a natural step in the theory of integration to the case of integrals involving algebraic functions √A, where A is a polynomial of degree > 4. The first major insights of the theory were given by Niels Abel; it was later formulated in terms of the Jacobian variety J(S). Choice of P gives rise to a standard holomorphic mapping S → J(S) of complex manifolds. It has the defining property that the holomorphic 1-forms on J(S), of which there are g independent ones if g is the genus of S, pull back to a basis for the differentials of the first kind on S. Last updated: 10-15-2005 16:23:59 03-10-2013 05:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8577905297279358, "perplexity_flag": "middle"}
http://mathhelpforum.com/math-topics/105763-speed-fan.html	# Thread: 1. ## Speed of a fan A large mine ventilation fan is rotating at a speed of 120 RPM when the electricity is turned off. The fan then decelerates due to friction at a rate of -0.05ω rad/s2, where ω is the angular velocity in rad/s. How long does it take for the fan speed to come down to N RPM? N[rpm] = 6; Could someone please show me how to solve this? 2. Originally Posted by m_i_k_o A large mine ventilation fan is rotating at a speed of 120 RPM when the electricity is turned off. The fan then decelerates due to friction at a rate of -0.05ω rad/s2, where ω is the angular velocity in rad/s. How long does it take for the fan speed to come down to N RPM? N[rpm] = 6; Could someone please show me how to solve this? first, you need to convert $\omega_o = 120$ RPM and $\omega_f = 6$ RPM to rad/sec $\alpha = \frac{d\omega}{dt} = -0.05\omega$ $\frac{d \omega}{\omega} = -0.05 \, dt$ $\ln{\omega} = -0.05t + C$ $\omega = e^{-0.05t + C} = e^C \cdot e^{-0.05t} = Ae^{-0.05t}$ sub in the initial condition ... $\omega_o = Ae^{-0.05 \cdot (0)} = A \cdot 1$ ... $A = \omega_o$ $\omega = \omega_o e^{-0.05t}$ sub in $\omega_o$ and $\omega_f$ in rad/sec and solve for t	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9026421308517456, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/13871/list	## Return to Question 2 Added circular permutations I'm working on the interaction between equivalence relations on permutations and pattern avoidance. I've only considered Knuth equivalence and cyclic shifts until now and I'm looking for other equivalence relations to test some conjectures on. So my question is: What interesting equivalence relations on permutations are there? ## Background/Motivation By a permutation of $n$ I mean a bijection $\lbrace 1,2,\dots,n\rbrace \to \lbrace 1,2,\dots,n \rbrace$. The set of all permutations of $n$ will be denoted $S_n$. I'll use the one-line notation for permutations, e.g., $132$ means the permutation $1\to1$, $2\to3$, $3\to2$. A pattern will also be a permutation, but we are interested in how patterns occur in permutations. E.g., the pattern $132$ occurs in the permutation $215314$ as the letters $2$, $5$, $3$, because these have the same relative order as the pattern. Note that the letters do not have to be adjacent in the permutation. When a pattern does not occur in a permutation we say that the permutation avoids that pattern. Now, fix an equivalence relation on $S_n$. For any pattern $p$ we define two subsets of $S_n$: $X_n(p) = \lbrace \sigma \in S_n \phantom{i}\|\phantom{i} \sigma \text{ and every equivalent permutation avoids } p \rbrace$ $Y_n(p) = \lbrace \sigma \in S_n \phantom{i}\|\phantom{i} \sigma \text{ avoids } p \text{ and every equivalent pattern}\rbrace$ What I am mostly interested in is how these two sets are related. Example: Assume our equivalence relation is cyclic shifts, meaning that two permutations (or patterns) $p$, $q$ are equivalent if we can write $p = r_1 * r_2$ and $q = r_2 * r_1$ where $$ means concatenation. E.g., $1234$ is equivalent to $3412$. Here the two sets $X_n(p)$ and $Y_n(p)$ are equal for any $p$. (This is also true if we generalize our patterns to bivincular patterns, whose definition I'll omit here) [Note considering permutations up to cyclic shifts is equivalent to looking at circular permutations.] Example: If our equivalence relation is Knuth equivalence, meaning that two permutations (or patterns) are equivalent if they have the same insertion tableaux under the RSK-correspondence, then the two sets are not always the same. They are the same for any pattern of length $3$, but $X_4(1324) \neq Y_4(1324)$. When the two sets are equal then counting $Y_n(p)$ can sometimes be done more easily by looking at $X_n(p)$. When the relation is Knuth equivalence one can use the hook-length formula for instance. So I would very happy if someone could tell me about other interesting equivalence relations on permutations so I could see how $X_n(p)$ and $Y_n(p)$ are related in other cases. 1 # Equivalence relations on permutations and pattern avoidance I'm working on the interaction between equivalence relations on permutations and pattern avoidance. I've only considered Knuth equivalence and cyclic shifts until now and I'm looking for other equivalence relations to test some conjectures on. So my question is: What interesting equivalence relations on permutations are there? ## Background/Motivation By a permutation of $n$ I mean a bijection $\lbrace 1,2,\dots,n\rbrace \to \lbrace 1,2,\dots,n \rbrace$. The set of all permutations of $n$ will be denoted $S_n$. I'll use the one-line notation for permutations, e.g., $132$ means the permutation $1\to1$, $2\to3$, $3\to2$. A pattern will also be a permutation, but we are interested in how patterns occur in permutations. E.g., the pattern $132$ occurs in the permutation $215314$ as the letters $2$, $5$, $3$, because these have the same relative order as the pattern. Note that the letters do not have to be adjacent in the permutation. When a pattern does not occur in a permutation we say that the permutation avoids that pattern. Now, fix an equivalence relation on $S_n$. For any pattern $p$ we define two subsets of $S_n$: $X_n(p) = \lbrace \sigma \in S_n \phantom{i}\|\phantom{i} \sigma \text{ and every equivalent permutation avoids } p \rbrace$ $Y_n(p) = \lbrace \sigma \in S_n \phantom{i}\|\phantom{i} \sigma \text{ avoids } p \text{ and every equivalent pattern}\rbrace$ What I am mostly interested in is how these two sets are related. Example: Assume our equivalence relation is cyclic shifts, meaning that two permutations (or patterns) $p$, $q$ are equivalent if we can write $p = r_1 r_2$ and $q = r_2 * r_1$ where $*$ means concatenation. E.g., $1234$ is equivalent to $3412$. Here the two sets $X_n(p)$ and $Y_n(p)$ are equal for any $p$. (This is also true if we generalize our patterns to bivincular patterns, whose definition I'll omit here) Example: If our equivalence relation is Knuth equivalence, meaning that two permutations (or patterns) are equivalent if they have the same insertion tableaux under the RSK-correspondence, then the two sets are not always the same. They are the same for any pattern of length $3$, but $X_4(1324) \neq Y_4(1324)$. When the two sets are equal then counting $Y_n(p)$ can sometimes be done more easily by looking at $X_n(p)$. When the relation is Knuth equivalence one can use the hook-length formula for instance. So I would very happy if someone could tell me about other interesting equivalence relations on permutations so I could see how $X_n(p)$ and $Y_n(p)$ are related in other cases.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 68, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9261963963508606, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=381864	Physics Forums ## Estimating Pion mass using the uncertainty principle 1. The problem is based uponYukawa's original prediction for pion mass-energy. suppose the force between nucleons is due to the emission of a particle mass m from one nucleon and the absorption by another. given the range of the nuclear force is R=(1.4)$$\times$$10$$^{-15}$$ use $$\Delta$$E$$\Delta$$t ~ h bar to make an order of magnitude estimate of the mass energy in MeV 2 x=position p=momentum E=energy m=pion rest mass h bar = reduced planck constant t=time c=speed of light $$\Delta$$=uncertainty in _ the equation required by the question is $$\Delta$$E$$\Delta$$t ~ h bar i also attempted using E$$^{2}$$=p$$^{2}$$c$$^{2}$$ +m$$^{2}$$c$$^{4}$$ and the momentum form of the uncertainty principle $$\Delta$$p$$\Delta$$x ~ h I was pretty stumped because i couldn't fathom what the delta t represented, and how to use the required equation with just one variable. i used dimensional analysis as i knew it would have to be made up of constants h and c and the given value R (taken to be delta x) and got E~$$\frac{hc}{R}$$ But this does not take into account any prefactors and doesn't use the required equation. i also tried subbing the momentum form of the uncertainty principle into thye relativistic energy equation, but to no avail as the Delta t is still there This is one of the first questions from my introductory quantum mech. course, so i'm not particularly clue up on the physical principle, can someone please point me in the right direction? Help! PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Mentor Think of the speed of light as a conversion factor between length and time. You're given a length scale R; it'll correspond to a time scale, which gives you an estimate for $\Delta t$. The uncertainty principle $\Delta E\Delta t \approx \hbar$ says that over short time scales, the energy of a system doesn't have a definite value but has a spread of values, so as long as a process, like exchanging a meson, occurs over a short enough period of time, the energy required for the process, like the energy of the meson, is available. Tags particle theory, pion, quantum, rest mass, uncertainty Thread Tools \| \| \| \| \|---------------------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Estimating Pion mass using the uncertainty principle \| \| \| \| Thread \| Forum \| Replies \| \| \| Advanced Physics Homework \| 3 \| \| \| Quantum Physics \| 5 \| \| \| Introductory Physics Homework \| 1 \| \| \| General Astronomy \| 2 \| \| \| Quantum Physics \| 8 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 15, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8990646600723267, "perplexity_flag": "middle"}
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http://unapologetic.wordpress.com/2009/02/10/transformations-with-all-eigenvalues-distinct/?like=1&source=post_flair&_wpnonce=8cb43e0c56	# The Unapologetic Mathematician ## Transformations with All Eigenvalues Distinct When will a linear transformation $T$ have a diagonal matrix? We know that this happens exactly when we can find a basis $\{e_i\}_{i=1}^d$ of our vector space $V$ so that each basis vector is an eigenvector $T(e_i)=\lambda_ie_i$. But when does this happen? The full answer can be complicated, but one sufficient condition we can describe right away. First, remember that we can always calculate the characteristic polynomial of $T$. This will be a polynomial of degree $d$ — the dimension of $V$. Further, we know that a field element $\lambda$ is an eigenvalue of $T$ if and only if $\lambda$ is a root of the characteristic polynomial. Since the degree of the polynomial is $d$, we can expect to find no more than $d$ distinct roots of the polynomial — no more than $d$ distinct eigenvalues of $T$. I want to consider what happens in this generic case. Now we have $d$ field elements $\lambda_i$, and for each one we can pick a vector $e_i$ so that $T(e_i)=\lambda_ie_i$. I say that they form a basis of $V$. If we can show that they are linearly independent, then they span some subspace of $V$. But since there are $d$ distinct vectors here, they span a subspace of dimension $d$, which must be all of $V$. And we know, by an earlier lemma that a collection of eigenvectors corresponding to distinct eigenvalues must be linearly independent! Thus, the $e_i$ form a basis for $V$ consisting of eigenvectors of $T$. With respect to this basis, the matrix of $T$ is diagonal. This is good, but notice that there are still plenty of things that can go wrong. It’s entirely possible that two (or more!) of the eigenvalues are not distinct. Worse, we could be working over a field that isn’t algebraically closed, so there may not be $d$ roots at all, even counting duplicates. But still, in the generic case we’ve got a diagonal matrix with respect to a well-chosen basis. ### Like this: Posted by John Armstrong \| Algebra, Linear Algebra ## 2 Comments » 1. [...] space of finite dimension . We take its characteristic polynomial to find the eigenvalues. If all of its roots are distinct (and there are of them, as there must be if we’re working over an algebraically closed [...] Pingback by \| February 11, 2009 \| Reply 2. [...] we’ve established that distinct eigenvalues allow us to diagonalize a matrix, but repeated eigenvalues cause us problems. We need to generalize [...] Pingback by \| February 16, 2009 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 28, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9152717590332031, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/30654/calculating-probability-of-project-tasks-and-most-likely-ending-date	# Calculating probability of project tasks, and most likely ending date? Say I have a project (you know the typical one you can represent using a Gantt Chart ) If I where to assign a particula probability to each of the tasks in the Gantt... say for example: • Task1 (2 days, i am 75% sure of that) • Task2 (4 days, i am 50% sure of that) for it to start, Task 1 most be finished • Task3 (8 days, i am 25% sure of that) for it to start, Task 2 most be finished How can I calculate how likely it is that I will finish this 14 day project? (In case you are wondering why I am asking this, I am a software developer, and I am constantly creating project plans and I would like to learn more on this subject so that I can project estimation better) - 1 One conventional approach to this problem, which has been around for at least a quarter century (it was in Primavera in the late '80s), is to specify Beta distributions for task durations in the natural but somewhat indirect way of giving an "expected" (modal) duration, a minimum, a maximum, and (perhaps) additional information. (Without additional information, many people assume triangular distributions.) You should be able to read about this in the project management literature, such as routledge.com/books/details/9781420083170. – whuber♦ Jun 18 '12 at 21:15 ## 2 Answers The joint probability of two events is just their product: $P(A \& B) = P(A) \cdot P(B)$, assuming the events are independent. While the probabilities of each of your events aren't independent, the conditional probabilities P(A) and P(B\|A) presumably are, so you can multiply them together. In this case, you need to complete all three tasks on time, so the probability of success is $$P(\textrm{Success}) = P(\textrm{Task1}) \cdot P(\textrm{Task2} \| \textrm{Task1}) \cdot P(\textrm{Task3} \| \textrm{Task1 & Task2}).$$ You've already provided these numbers, so just plug them in and you get $0.75 \cdot 0.5 \cdot 0.25=0.09375$, or roughly 9 percent. Alternately, we can instead enumerate the ways in which you might fail.The probability of failing is $1-P(success)$, since you can only either succeed at a task or fail it. For the first task, $$P(\textrm{Task1}=fail)= 1 - P(\textrm{Task1=success})=0.25$$ In order to fail Task2, you first have to succeed at Task1. There's a 75% chance you've succeeded at Task1 and thus get to attempt Task2, coupled with a 50% chance you've failed Task2, which gives you $$P(\textrm{Task2}=fail) = P(\textrm{Task1}=success) \cdot P(\textrm{Task2}=failure) = 0.75 \cdot 0.5=0.375$$ To compute the probability of failing Task3, you must succeed at Task1 and Task2, then fail Task3: the probability of therefore is $0.75 \cdot 0.5 \cdot 0.75 = 0.28125$. Since failing any one task derails the entire project, we add the probability of failures together, and get $P(failure) = 0.25 + 0.375 + 0.28125 = 0.90625$. The probability of success is $1-P(failure)$, which is 0.09375, matching the earlier answer. I should also note that, in the general case, you haven't quite provided enough information. I've assumed that you can't finish a task early. Furthermore, given your constraints, there's absolutely no "slack" in the schedule: failing one task fails the project. If neither of these are true, then you need more information to solve this problem. Specifically, you need to specify a distribution of times (e.g., 25% chance it takes 1 day, 50% chance it takes 2, 25% it takes 3). - We both get the same answer but as I said we do need to assume that if p is the probability that task i takes x days then 1-p is the probability that it will take longer than x days. – Michael Chernick Jun 18 '12 at 19:25 You can't compute the probability because you haven't been specific enough. You say task 1 will be finished in 2 days with probability 0.75 but do not specify the distribution of the number of days required with the remaining probaility 0.25 when it can't be done in two days. The same is true for the way you specify how long tasks 2 and 3 will take. Based on your assumptions the project will be done in exactly 14 days if task 1 takes 2 days, task 2 takes 4 days after that and task 3 takes 8 days after that. If the time for each task is independent of the time it takes to do the others then this probability will be the product namely 0.75x0.5x0.25=3/32=0.09375. Now this will be the answer but only if otherwise the tasks will take longer which may be what you intended to say but did not specify. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9392474889755249, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/5291/randomized-algorithms-and-the-one-time-pad/5292	# Randomized algorithms and the one time pad The way I understand it, an algorithm is said to be randomized if it uses randomness as part of its logic (quoting Wikipedia). Now, in the case of encryption algorithms, I assume this means that for the same input, different outputs (i.e. ciphertexts) will be produced. The question is, what does "input" mean in this context? In particular, does it include only the plain text to be encrypted, or does it also include the key? To give a concrete example, is the one time pad a randomized algorithm? If it is so, then "input" must mean plain text only. To make matters more confusing (to me at least), again quoting Wikipedia: "To be semantically secure, that is, to hide even partial information about the plaintext, an encryption algorithm must be probabilistic.", where the last word is linked to the "Randomized Algorithm" page I link to in the first paragraph. So, given that the one time pad is semantically secure, is must be a randomized algorithm... right? To add the cherry at the top of the confusion cake, there are still algorithms like, for instance, AES in CBC mode with a random IV, which will produce different outputs, when run multiple time with the same plain text, and the same key. Of course, this means that when you consider the input to one such algorithm as being the plain text only, you get the same result: different ciphertexts for the same plain text. So it would seem that the only "coherent" (lacking a better word) definition would require that input == plain text. Am I surmising correctly? - ## 4 Answers You indeed need to define what you understand by an encryption algorithm. The common way of doing this is to consider and encryption scheme as a tuple of three primitives: a key generation algorithm (upon a security level, output a randomly chosen pair of keys), an encryption algorithm to be instantiated with the generated key (that maps an input plaintext to an output ciphertext) and a corresponding decryption algorithm to be instantiated with the corresponding key. To be semantically secure for an encryption scheme means that given an a priori information about a plaintext and the corresponding ciphertext under the encryption algorithm, no information from the plaintext can be derived that could not have been derived from the a priori information alone. (This has to be understood complexity theoretic-wise.) Now the one-time pad encryption as you describe it does not fit the above scheme as it does not have this set of three primitives: the key (one-time pad) is to be changed for each plaintext. One could however construct a close encryption scheme this way (I assume fixed length plaintexts for simplicity, but it's not a limitation): one can see the key generation algorithm as a big table of pads indexed to somehow correspond to each one of the possible plaintexts. The encryption algorithm then lookup this table and xor it with the plaintext to produce the ciphertext. This scheme is semantically secure. Here's is the bottom-line: an asymmetric encryption scheme has to be probabilistic since otherwise, given the encryption key (which is not the decryption key) it would be possible to distinguish, say, the encryption of the "hi bob!" message from that of the "hi alice!" message. But a symmetric encryption algorithm does not have to be probabilistic for the scheme to be semantically secure (the encryption scheme defined above is). However, if the security goal is to achieve the same for pairs of ciphertexts instead of single ciphertexts, then scheme defined above to fit the framework would not be sufficient as it would allow an adversary to distinguish between the pairs of ciphertexts corresponding to the same messages from the pairs of ciphertexts corresponding to distinct messages. So, yes, a probabilistic encryption algorithm inputs a plaintext and outputs a ciphertext that will vary according to the random coins generated internally. - 1 What you described is not a one-time pad. It's stream encryption using the key generation algorithm as the stream generator, and uses XOR to combine the key material with the plaintext. – John Deters Nov 8 '12 at 21:21 @John Deters: You're correct. This is why I stressed that one-time pad does not fit the framework. I'll change the wording accordingly. Thanks. – bob Nov 8 '12 at 21:28 You would not index by the set of plaintexts (the receiver wouldn't know which one to use), but by some external message index. – Paŭlo Ebermann♦ Nov 8 '12 at 22:45 @PaŭloEbermann: Indeed, thank you. – bob Nov 8 '12 at 23:03 @bob, thank you for your answer, it has been insightful. However, one doubt remains: what's the problem of "the key (one-time pad) is to be changed for each plaintext"? Isn't this the way one time pads are supposed to work? I think you I are referring to the fact that "my" one time pad does not include (explicitly) a key generation algorithm. Say I added one; borrowing from another answer, the key generation is done throwing dice. Would it then be semantically secure? – wmnorth Nov 9 '12 at 8:28 show 1 more comment No, the encryption algorithm used in one-time pad encryption does not need to be either probabilistic or random. The most common historical OTP algorithm is a simple substitution cypher. In digital encryption, the OTP algorithm most commonly used is XOR. Neither algorithm introduces any uncertainty. It's the key material that must be random. And how that key material is generated is critical. It must be derived from a cryptographically secure source of randomness. Acceptable sources of randomness are surprisingly difficult to come by. Computers, despite their reputations, are designed to be precisely accurate and repeatable. The most secure random numbers come from hardware using stochastic processes such as the timing of radioactive decay, thermal component noise, and other such sources. Most commercial computers don't contain a hardware random number generator, however, so they collect different bits of things that are considered hard to predict, and combine them together into a large collection of hard-to-predict bits. This process is often called "gathering entropy". To stretch a small number of hard-to-predict bits into a usefully large number of random bits, pseudo-random number generation algorithms are sometimes used, which accept the small number of bits as a seed and produce a larger number of bits of output. Pseudo-random number generators are often built from proven secure cryptographic routines such as AES or SHA-2. And if you're using AES to stretch a small number of bits into a long string of bits, you are essentially encrypting with AES, starting with a random number as the key. That's why a "one time pad" using a computer-generated key stream is rarely as effective as an actual one-time pad. The good news is that it's still as effective as AES, which is considered strong. The bad news is that because people don't understand randomness very well, they think that any series of values they themselves can't predict will serve as an adequate key for a one-time pad. To help understand why one-time pads work the way they do, I recommend studying real world attacks. One of the best documented attacks is the Venona project, recently declassified by the NSA, where they deciphered the one-time pads used by Soviet spies. The reason they were able to decipher them is that Soviet agents actually reused the keys, turning them into two-time pads. It is generally accepted that because generating the key material was tedious, time consuming, and expensive, and distributing it securely was extremely risky and difficult, that they economized by reusing the keys. That led to the break in the code, which in turn identified such notorious spies as Julius and Ethel Rosenberg and David Greenglass, and provided absolute proof of their guilt in delivering the secrets of the atomic bomb to the Soviet Union. That cryptanalysis of course revealed the keys of the one-time pads, and those were also studied. It was determined from the distribution of the letters used that a typist simply banged back and forth on a keyboard, from one side to the other, to generate the keys. It's hard to imagine a more tedious job. - Others did already answer about the terms of a randomized (encryption) algorithm, but one thing is missing: The one-time pad is not a semantically secure encryption algorithm. Semantic security in essence means that you can encrypt multiple messages with the same key, and the attacker has no way of finding out which message was sent, even in the case of chosen-plaintext attacks. While the one-time pad provides perfect secrecy when used correctly, it is broken whenever you use a key for more than one message (i.e. the "two-time pad"). - In cryptography, I have typically seen "randomized algorithm" used for a randomized search algorithm, that is employing randomness to organize a search for a solution (or an optimal solution) to some problem. It opposes to "deterministic (search) algorithm". For example, an algorithm that attempts to factor $N$ by picking a random number $R$ about the same size as $N$ and computing $\gcd(N,R)$ until that is neither $1$ nor $N$, is a randomized algorithm; but trial division by primes up to $\sqrt N$ is a deterministic algorithm. I would personally not use just "randomized algorithm" for anything that use randomness essentially to generate unpredictable data; like generating a pad to be used with the One-Time-Pad method, generating a prime for RSA, or concatenating random data and plaintext before encryption (one of few techniques to transform a deterministic encryption scheme into a semantically secure encryption scheme). Bob's first comment pointed that the right term for that technique is "probabilistic encryption", as coined by Goldwasser and Micali. I could have used "randomized encryption scheme" for that, and would understand "randomized scheme", or "randomized encryption algorithm" (because encryption dissipates the aforementioned difficult goal randomized search default assumption). In the context of an encryption scheme, input includes the plaintext and key, and only that, unless otherwise stated. If OTP encryption and OTP decryption are seen as algorithms with the random single-use pad as the key input, then neither is a randomized algorithm. Update: as explained in Bob's other answer, the OTP does not fit the definition of a semantically secure cipher, even though, assuming safe generation and transmission of the pad, it is semantically secure. Caveat: I'm not a native English speaker, nor in academia. - 1 The standard naming, coined by Goldwasser and Micali, is probabilistic encryption. – bob Nov 8 '12 at 7:34 @bob probabilistic encryption refers to an algorithm receiving the same PT and key, and output varying output, right? – wmnorth Nov 8 '12 at 16:07 @fgrieu: I fail to see what you mean by "difficult goal". A randomized algorithm can be used for any kind of goal. Although it is true that they are often the only known algorithms to efficiently solve some difficult problems, it is also not known that there cannot always be a deterministic algorithm solving the same problem at least as efficiently. Also, in your example, the randomized version for factorization you give is less efficient than the deterministic one. I guess this comes from the typical representation of an attacker as a probabilistic polynomial time algorithm. – bob Nov 8 '12 at 21:12 @bob: removed "difficult goal", thanks – fgrieu Nov 8 '12 at 22:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9375731945037842, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=579711	Physics Forums ## Speed of sound in a vacuum Does anyone have an explanation for why kinetic energy carried by electromagnetic radiation travells at the speed of light? My understanding of the speed of sound is that the denser the medium, the faster the wave velocity. Since this is just kinetic energy propagating through a medium, how is it different when applied to electromagnetic radiation? If it is no different, does that mean that a vacuum has infinite density? PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus you seem to have confusion between sound waves and light waves. when you say electromagnetic radiation, you are talking about light wave. light can travel in vacuum. sound cannot travel in vacuum, it needs medium to propagate. True, sound waves (kinetic energy dispersing through a medium) requires something to travel through. So how can photons transfer kinetic energy (and at such velocity) without a medium. Correct me if I am wrong (very possible) but photons carry kinetic energy as shown by the photo electric effect and the fact that they have mass when not in a rest state. Also, if a laser is directerd onto a suspended metal surface the surface will rotate? Recognitions: Gold Member ## Speed of sound in a vacuum Quote by David Burke True, sound waves (kinetic energy dispersing through a medium) requires something to travel through. So how can photons transfer kinetic energy (and at such velocity) without a medium. Correct me if I am wrong (very possible) but photons carry kinetic energy as shown by the photo electric effect and the fact that they have mass when not in a rest state. Also, if a laser is directerd onto a suspended metal surface the surface will rotate? That's just one of the wild things about light, it requires no medium to travel through. Photons do not have mass, but they do have an energy and momentum (they simply are not given by $KE = {{1}\over{2}}mv^2$ and $p = mv$). Blog Entries: 1 You might have been confused. Sound propagate in vacuum but light can(300,000m/s) Quote by David Burke Also, if a laser is directerd onto a suspended metal surface the surface will rotate? Yes, it will. There's something called radiation pressure. This pressure of light will be doubled for all angles of incidence if the light is both absorbed and reflected. However, I think the formula of solar radiation pressure on that page is wrong.It should be F= -p(c+1)Ar where p is the force per unit area, c is the coefficient of reflectivity Thread Tools \| \| \| \| \|-------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Speed of sound in a vacuum \| \| \| \| Thread \| Forum \| Replies \| \| \| Introductory Physics Homework \| 2 \| \| \| Introductory Physics Homework \| 2 \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 5 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9133171439170837, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/2431/can-a-proof-be-constructed-to-show-there-is-no-distinguisher?answertab=votes	# Can a proof be constructed to show there is no distinguisher? Let's assume a simple algorithm like the Skein hash function. Is it possible, given the algorithm, to construct a proof that it does not have a particular distinguisher, something like: $P(xyz)$ is the probability that $xyz$ is truly random over some alphabet, Given $\vert y \vert = l$, for some fixed length l, $z = f(x)$ (i.e., $z$ is dependent on $x$). Not in general, of course, but for a particular such distinguisher. - Welcome to crypto.se; your Q has been moved here on account of being more on topic here than on SO - do feel free to register an account to pick up your rep and more importantly responses here :) – Antony Vennard Apr 23 '12 at 9:12 Whats the policy on here on helping people answer what are pretty homework questions/ facilitating cheating? – imichaelmiers Apr 24 '12 at 1:59 3 1 - This isn't a homework question. I asked because I was wondering if such proofs are constructed for hash functions. 2 - If it was a homework, the homework would have been turned in long ago - the question was first asked over a year ago! – Vanwaril Apr 24 '12 at 2:13 – CodesInChaos Apr 25 '12 at 12:18 The hash function Skein is not exactly what I would call "simple" from the point of view of understanding the way it maps its inputs to outputs. Otherwise, one would "simply" find pre-images and collisions for it! – bob Oct 29 '12 at 12:46 ## 1 Answer There are a number of distinguishers that it it would be easy to prove are not present in a hash function. For example, I can easily prove that Skein does not have the distinguisher "the 2nd bit in the output is equal to the first bit of the output with probability 1". The proof would be a simple example of a message whose digest does not have this property (which should be fairly easy to find). For more interesting distinguishers, the problem becomes much harder (on a side note, is the distinguisher really interesting if I can prove it doesn't hold for a particular hash function?). My gut feeling is that yes, one could come up with an "interesting" distinguisher and then show that it doesn't hold for the hash function, but I can not offer more than a gut feeling. - Yeah, I was wondering if there were 'real' arguments for 'real' hashing functions that showed one did not contain a distinguisher that another is susceptible to. – Vanwaril Apr 25 '12 at 1:16	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.954214870929718, "perplexity_flag": "middle"}
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Thanks in advance. $$x = \cos^2 t, \quad y = 1 - \sin t, \quad 0 \leq t \leq 2\pi.$$ 1answer 47 views ### Find a function f(x) such that the parametric curve could be obtained by flipping the graph Find a function f(x) such that the parametric curve could be obtained by flipping the graph of f across the line with slope 1 that goes through the origin. parametric curve with coordinates (t^14, ... 3answers 109 views ### How do we prove that two parametric equations are drawing the same thing? For example, if I have $$\begin {align} x(t) &= r\sin t\cos t\\ y(t) &= r\sin^2 t\\ \end {align}$$ and \begin {align} x(t) &= \frac r 2 \cos t\\ y(t) &= \frac r 2 (\sin t + 1) ... 1answer 115 views ### Arc Length Of Parametric Curve I attached the problem as a file: Where did the trig functions go? I sifted through the different trig identities and formulas, but couldn't find anything that I could use. 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Are the graphs the same? Are the orientations the same? Are the curves smooth? Explain." (a) \$x=t;\quad ... 1answer 62 views ### Restriction Of Parametric Functions Domain The problem I am working on is, "Sketch the curve represented by the parametric equations (indicate the orientation of the curve), and write the corresponding rectangular equation by eliminating the ... 2answers 31 views ### Need help with this parametric equation $s(t)=(\frac{2}{t^2+1},\frac{2t}{t^2+1})$ I need to calculate a line integral along this path. But I have trouble understanding what it is. I did some googling and it looks that it is a parabola, but ... 2answers 37 views ### How do I find the 2 slopes at which this parametric function crosses itself? I have a parametric function. If you graph it, you'll find that it looks like a figure 8. x(t) = 2sin(2t) y(t) = 8sin(t) How do I find the slopes of the function ... 1answer 251 views ### Parametric equation of a cone I usually use the following parametric equation to find the surface area of a regular cone $z=\sqrt{x^2+y^2}$: $$x=r\cos\theta$$ $$y=r\sin\theta$$ $$z=r$$ And make $0\leq r \leq 2\pi$, \$0 \leq \theta ... 0answers 63 views ### complex circular motion Assume the planet orbits its star in a circular orbit of 100 units with period 360 days. The moon in turn orbits its planet at a radius of 10 units and period 28 days. Finally the moon rotating, we ... 1answer 69 views ### Can (x(t), y(t)) generate a surface? If so, can the surface be continuous? Intuitively, the parametric equation $z = (x(t), y(t))$ seems to only be able to generate one-dimensional objects, i.e. curves. However... Let $x(t)$ be "the odd-indexed digits of the real number ... 2answers 86 views ### Parameter values that make function values side lengths of a triangle I have been trying to solve the following problem for more than a week without any success. Given the function: $$f(x)=\frac{x^2+mx+4}{x^2+x+4}$$ Find all possible values of the parameter $m$ such ... 1answer 55 views ### How can I re-write an equation (or system of equations) in parametric form? For the equation $y = 3x$ I need to re-write $x$ and $y$ in terms of a variable $t$. How can I find the value of each variable in terms of $t$? 1answer 73 views ### How to find a parametric equation? I want to find an equation for a race track, so I could get the position of a point with respect to time. Let's say I have this track and here are a few points on it: Could it be possible to model ... 1answer 56 views ### parametrizing quarter of a circle I am given the circle whose equation is: $(x-\frac{1}{2})^{2}+(y+\frac{1}{2})^{2}=\frac{1}{2}$. So, the coordinates of the origin of the circle are: $(\frac{1}{2},-\frac{1}{2})$ and the radius of the ... 1answer 141 views ### converting a parametric R5 vector into a Cartesian form How do you solve a problem like this. I'm completely stumped. it seems like there should be an easy solution but I'm obviously over looking it. any help would be greatly appreciated. 2answers 78 views ### Constant velocity of a sine function I am defining the location of an object based on the sine function. The position of the object at s seconds along the x-axis is defined as x=s and its position along the y-axis is defined as y=sin(x). ... 1answer 221 views ### Find the area bounded by the parametric curve… Find the area bounded by the parametric curve $x = \cos(t)$, $y = e^t, 0 < t < \pi/2$, and the lines $y = 1$ and $x = 0$. I do not even know where to start with this problem. I know that I need ... 3answers 1k views ### How to find a parametric equation for the tangent line to the curve of intersection of the cylinders? How can i find a parametric equation for the tangent line to the curve of intersection of the cylinders $x^2 + y^2 = 4$ and and $x^2 + z^2 = 1$ at the point $P_0(1,\sqrt{3}, 0)$? 1answer 78 views ### How do we find the length of the line (parametric curve)? A curve in the $xy$-plane is given parametrically by $$x(t) = e^{2t}, \quad y(t) = e^{2t} \sin(2t), \quad t \in [0, \pi/2].$$ What is the length of this curve? Ok, actually I know what to do, ... 1answer 61 views ### Converting parametric equations in a numerical equation Is it possible convert this parametric equations in a numerical equation? $$\begin{cases} \displaystyle x(t)=tv_0\cos(\theta)\\ \displaystyle y(t)=tv_0\sin(\theta)-\frac{1}{2}gt^2+h \end{cases}$$ ... 0answers 41 views ### Maximum value for parameter I am facing the following problem: A number of a adults, b children older than 12, and c children younger than twelve attend an event. The sum of all people a+b+c=100. The prices are \\$6 per adult, ... 1answer 783 views ### Equation for making a circle in 3D space I have a 3D space with axis $(x, y ,z)$ and I can make a circle in the $xy$-plane. To make a circle in the xy-plane I currently use spherical coordinates $(r, \theta, \phi)$ where $r = 1$, \$\theta = ... 1answer 207 views ### Parameterize a straight line using polar coordinates… without angle. I had to parameterize a straight line with starting point in $A=(-3,7)\\$ and endpoint in $B=(4,1)$. My idea was to use the equation for the line that goes through two points. That is: \frac { ... 1answer 73 views ### Parametric Equation of a Particle Movement inside a Vortex in a Rectangular Box I am trying to simulate the movement of a particle in a vortex in a rectangular box, I am currently using an ellipse but that causes the particle to collide with the walls more that I want. The ... 2answers 101 views ### Finding Tangent line from Parametric I need to find an equation of the tangent line to the curve $x=5+t^2-t$, $y=t^2+5$ at the point $(5,6)$. Setting $x=5$ and $y = 6$ and solving for $t$ gives me $t=0,1,-1$. I know I have to do ... 1answer 376 views ### Parametric Equation Let $P_1$ be the plane through the origin containing the vectors $[1,2,-1]$ and $[0,1,1]$. Let $P_2$ be the plane through the point $(1,1,1)$ parallel to the vectors $[-1,2,2]$ and $[3,4,-2]$ I know ... 1answer 31 views ### Represent sorting position by a parametric form Given a set of random integers {0,5,100,65,...,0,1,2}, is there a mathematical method existing to construct a parametric form $f$ (the number of parameters $<<$ the number of integers) so that ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 77, "mathjax_display_tex": 17, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8932328224182129, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/70750-3d-distance-problem.html	# Thread: 1. ## 3d distance problem The question is what is the equation that includes the points T so that the distance from T to O = (4,0,1) is twice the distance from T to C = (1, -2, 4)? I set the equation up like this $sqrt((x-4)^2 + (y-0)^2 + (z-4)^2) = 2sqrt((x-1)^2 + (y+2)^2 + (z-4)^2)$ I get it down to this: $x^2 + (y + 16/6)^2 + (z - 5)^2 = -137/9$ The problem is the equation doesn't work because the second side is negative 2. Originally Posted by collegestudent123 The question is what is the equation that includes the points T so that the distance from T to O = (4,0,1) is twice the distance from T to C = (1, -2, 4)? I set the equation up like this $sqrt((x-4)^2 + (y-0)^2 + (z-{\color{red}4})^2) = 2sqrt((x-1)^2 + (y+2)^2 + (z-4)^2)$ I get it down to this: $x^2 + (y + 16/6)^2 + (z - 5)^2 = -137/9$ The problem is the equation doesn't work because the second side is negative See your error in red above. 3. Sorry the 4 was actually a typo, in the work I did it was a 1 I tried it again and still come up with that answer Edit: I made a algebra error, I figured it out	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9773637056350708, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Coefficient_of_determination	# Coefficient of determination Not to be confused with Coefficient of variation. Ordinary least squares regression of Okun's law. Since the regression line does not miss any of the points by very much, the R2 of the regression is relatively high. Comparison of the Theil–Sen estimator (black) and simple linear regression (blue) for a set of points with outliers. Because of the many outliers, neither of the regression lines fits the data well, as measured by the fact that neither gives a very high R2. In statistics, the coefficient of determination, denoted R2 and pronounced R squared, is a statistic used in the context of statistical models whose main purpose is either the prediction of future outcomes or the testing of hypotheses, on the basis of other related information. It provides a measure of how well observed outcomes are replicated by the model, as the proportion of total variation of outcomes explained by the model.[1] There are several different definitions of R2 which are only sometimes equivalent. One class of such cases includes that of linear regression. In this case, if an intercept is included then R2 is simply the square of the sample correlation coefficient between the outcomes and their predicted values; in the case of simple linear regression, it is the squared correlation between the outcomes and the values of the single regressor being used for prediction. If an intercept is included and the number of explanators is more than one, R2 is also referred to as the coefficient of multiple correlation. In such cases, the coefficient of determination ranges from 0 to 1. Important cases where the computational definition of R2 can yield negative values, depending on the definition used, arise where the predictions which are being compared to the corresponding outcomes have not been derived from a model-fitting procedure using those data, and where linear regression is conducted without including an intercept. Additionally, negative values of R2 may occur when fitting non-linear functions to data.[2] In cases where negative values arise, the mean of the data provides a better fit to the outcomes than do the fitted function values, according to this particular criterion. ## Definitions $R^2 = 1 - \frac{\color{blue}{SS_\text{err}}}{\color{red}{SS_\text{tot}}}$ The better the linear regression (on the right) fits the data in comparison to the simple average (on the left graph), the closer the value of $R^2$ is to one. The areas of the blue squares represent the squared residuals with respect to the linear regression. The areas of the red squares represent the squared residuals with respect to the average value. A data set has values yi, each of which has an associated modelled value fi (also sometimes referred to as ŷi). Here, the values yi are called the observed values and the modelled values fi are sometimes called the predicted values. In the below $\bar{y}$ is the mean of the observed data: $\bar{y}=\frac{1}{n}\sum_{i=1}^n y_i$ where n is the number of observations. The "variability" of the data set is measured through different sums of squares: $SS_\text{tot}=\sum_i (y_i-\bar{y})^2,$ the total sum of squares (proportional to the sample variance); $SS_\text{reg}=\sum_i (f_i -\bar{y})^2,$ the regression sum of squares, also called the explained sum of squares. $SS_\text{err}=\sum_i (y_i - f_i)^2\,$, the sum of squares of residuals, also called the residual sum of squares. The notations $SS_{R}$ and $SS_{E}$ should be avoided, since in some texts their meaning is reversed to Residual sum of squares and Explained sum of squares, respectively. The most general definition of the coefficient of determination is $R^2 \equiv 1 - {SS_{\rm err}\over SS_{\rm tot}}.\,$ ### Relation to unexplained variance In a general form, R2 can be seen to be related to the unexplained variance, since the second term compares the unexplained variance (variance of the model's errors) with the total variance (of the data). See fraction of variance unexplained. ### As explained variance In some cases the total sum of squares equals the sum of the two other sums of squares defined above, $SS_{\rm err}+SS_{\rm reg}=SS_{\rm tot}. \,$ See partitioning in the general OLS model for a derivation of this result for one case where the relation holds. When this relation does hold, the above definition of R2 is equivalent to $R^{2} = {SS_{\rm reg} \over SS_{\rm tot} } = {SS_{\rm reg}/n \over SS_{\rm tot}/n }.$ In this form R2 is expressed as the ratio of the explained variance (variance of the model's predictions, which is SSreg / n) to the total variance (sample variance of the dependent variable, which is SStot / n). This partition of the sum of squares holds for instance when the model values ƒi have been obtained by linear regression. A milder sufficient condition reads as follows: The model has the form $f_i=\alpha+\beta q_i \,$ where the qi are arbitrary values that may or may not depend on i or on other free parameters (the common choice qi = xi is just one special case), and the coefficients α and β are obtained by minimizing the residual sum of squares. This set of conditions is an important one and it has a number of implications for the properties of the fitted residuals and the modelled values. In particular, under these conditions: $\bar{f}=\bar{y}.\,$ ### As squared correlation coefficient Similarly, in linear least squares regression with an estimated intercept term, R2 equals the square of the Pearson correlation coefficient between the observed and modeled (predicted) data values of the dependent variable. Under more general modeling conditions, where the predicted values might be generated from a model different than linear least squares regression, an R2 value can be calculated as the square of the correlation coefficient between the original and modeled data values. In this case, the value is not directly a measure of how good the modeled values are, but rather a measure of how good a predictor might be constructed from the modeled values (by creating a revised predictor of the form α + βƒi). According to Everitt (2002, p. 78), this usage is specifically the definition of the term "coefficient of determination": the square of the correlation between two (general) variables. ## Interpretation R2 is a statistic that will give some information about the goodness of fit of a model. In regression, the R2 coefficient of determination is a statistical measure of how well the regression line approximates the real data points. An R2 of 1 indicates that the regression line perfectly fits the data. Values of R2 outside the range 0 to 1 can occur where it is used to measure the agreement between observed and modeled values and where the "modeled" values are not obtained by linear regression and depending on which formulation of R2 is used. If the first formula above is used, values can never be greater than one. If the second expression is used, there are no constraints on the values obtainable. In many (but not all) instances where R2 is used, the predictors are calculated by ordinary least-squares regression: that is, by minimizing SSerr. In this case R2 increases as we increase the number of variables in the model (R2 is monotone increasing with the number of variables included, i.e. it will never decrease). This illustrates a drawback to one possible use of R2, where one might keep adding variables (Kitchen sink regression) to increase the R2 value. For example, if one is trying to predict the sales of a model of car from the car's gas mileage, price, and engine power, one can include such irrelevant factors as the first letter of the model's name or the height of the lead engineer designing the car because the R2 will never decrease as variables are added and will probably experience an increase due to chance alone. This leads to the alternative approach of looking at the adjusted R2. The explanation of this statistic is almost the same as R2 but it penalizes the statistic as extra variables are included in the model. For cases other than fitting by ordinary least squares, the R2 statistic can be calculated as above and may still be a useful measure. If fitting is by weighted least squares or generalized least squares, alternative versions of R2 can be calculated appropriate to those statistical frameworks, while the "raw" R2 may still be useful if it is more easily interpreted. Values for R2 can be calculated for any type of predictive model, which need not have a statistical basis. ### In a linear model Consider a linear model of the form ${Y_i = \beta_0 + \sum_{j=1}^p {\beta_j X_{i,j}} + \varepsilon_i},$ where, for the ith case, ${Y_i}$ is the response variable, $X_{i,1},\dots,X_{i,p}$ are p regressors, and $\varepsilon_i$ is a mean zero error term. The quantities $\beta_0,\dots,\beta_p$ are unknown coefficients, whose values are estimated by least squares. The coefficient of determination R2 is a measure of the global fit of the model. Specifically, R2 is an element of [0, 1] and represents the proportion of variability in Yi that may be attributed to some linear combination of the regressors (explanatory variables) in X. R2 is often interpreted as the proportion of response variation "explained" by the regressors in the model. Thus, R2 = 1 indicates that the fitted model explains all variability in $y$, while R2 = 0 indicates no 'linear' relationship (for straight line regression, this means that the straight line model is a constant line (slope=0, intercept=$\bar{y}$) between the response variable and regressors). An interior value such as R2 = 0.7 may be interpreted as follows: "Seventy percent of the variation in the response variable can be explained by the explanatory variables. The remaining thirty percent can be attributed to unknown, lurking variables or inherent variability." A caution that applies to R2, as to other statistical descriptions of correlation and association is that "correlation does not imply causation." In other words, while correlations may provide valuable clues regarding causal relationships among variables, a high correlation between two variables does not represent adequate evidence that changing one variable has resulted, or may result, from changes of other variables. In case of a single regressor, fitted by least squares, R2 is the square of the Pearson product-moment correlation coefficient relating the regressor and the response variable. More generally, R2 is the square of the correlation between the constructed predictor and the response variable. With more than one regressor, the R2 can be referred to as the coefficient of multiple determination. ### Inflation of R2 In least squares regression, R2 is weakly increasing with increases in the number of regressors in the model. Because increases in the number of regressors increase the value of R2, R2 alone cannot be used as a meaningful comparison of models with very different numbers of independent variables. For a meaningful comparison between two models, an F-test can be performed on the residual sum of squares, similar to the F-tests in Granger causality, though this is not always appropriate. As a reminder of this, some authors denote R2 by Rp2, where p is the number of columns in X (the number of explanators including the constant). To demonstrate this property, first recall that the objective of least squares regression is: $\min_b SS_\text{err}(b) \Rightarrow \min_b \sum_i (y_i - X_ib)^2\,$ The optimal value of the objective is weakly smaller as additional columns of $X$ are added, by the fact that less constrained minimization leads to an optimal cost which is weakly smaller than more constrained minimization does. Given the previous conclusion and noting that $SS_{tot}$ depends only on y, the non-decreasing property of R2 follows directly from the definition above. The intuitive reason that using an additional explanatory variable cannot lower the R2 is this: Minimizing $SS_\text{err}$ is equivalent to maximizing R2. When the extra variable is included, the data always have the option of giving it an estimated coefficient of zero, leaving the predicted values and the R2 unchanged. The only way that the optimization problem will give a non-zero coefficient is if doing so improves the R2. ### Notes on interpreting R2 R² does not indicate whether: • the independent variables are a cause of the changes in the dependent variable; • omitted-variable bias exists; • the correct regression was used; • the most appropriate set of independent variables has been chosen; • there is collinearity present in the data on the explanatory variables; • the model might be improved by using transformed versions of the existing set of independent variables. ## Adjusted R2 The use of an adjusted R2 (often written as $\bar R^2$ and pronounced "R bar squared") is an attempt to take account of the phenomenon of the R2 automatically and spuriously increasing when extra explanatory variables are added to the model. It is a modification due to Theil[3] of R2 that adjusts for the number of explanatory terms in a model relative to the number of data points. The adjusted R2 can be negative, and its value will always be less than or equal to that of R2. Unlike R2, the adjusted R2 increases when a new explanator is included only if the new explanator improves the R2 more than would be expected in the absence of any explanatory value being added by the new explanator. If a set of explanatory variables with a predetermined hierarchy of importance are introduced into a regression one at a time, with the adjusted R2 computed each time, the level at which adjusted R2 reaches a maximum, and decreases afterward, would be the regression with the ideal combination of having the best fit without excess/unnecessary terms. The adjusted R2 is defined as $\bar R^2 = {1-(1-R^{2}){n-1 \over n-p-1}} = {R^{2}-(1-R^{2}){p \over n-p-1}}$ where p is the total number of regressors in the linear model (not counting the constant term), and n is the sample size. Adjusted R2 can also be written as $\bar R^2 = {1-{SS_\text{err} \over SS_\text{tot}}}{df_t \over df_e}$ where dft is the degrees of freedom n– 1 of the estimate of the population variance of the dependent variable, and dfe is the degrees of freedom n – p – 1 of the estimate of the underlying population error variance. The principle behind the adjusted R2 statistic can be seen by rewriting the ordinary R2 as ${R^{2} = {1-{\textit{VAR}_\text{err} \over \textit{VAR}_\text{tot}}}}$ where ${\textit{VAR}_\text{err} = SS_\text{err}/n}$ and ${\textit{VAR}_\text{tot} = SS_\text{tot}/n}$ are the sample variances of the estimated residuals and the dependent variable respectively, which can be seen as biased estimates of the population variances of the errors and of the dependent variable. These estimates are replaced by statistically unbiased versions: ${\textit{VAR}_\text{err} = SS_\text{err}/(n-p-1)}$ and ${\textit{VAR}_\text{tot} = SS_\text{tot}/(n-1)}$. Adjusted R2 does not have the same interpretation as R2—while R2 is a measure of fit, adjusted R2 is instead a comparative measure of suitability of alternative nested sets of explanators.[citation needed] As such, care must be taken in interpreting and reporting this statistic. Adjusted R2 is particularly useful in the feature selection stage of model building. ## Generalized R2 Nagelkerke (1991) generalizes the definition of the coefficient of determination: 1. A generalized coefficient of determination should be consistent with the classical coefficient of determination when both can be computed; 2. Its value should also be maximised by the maximum likelihood estimation of a model; 3. It should be, at least asymptotically, independent of the sample size; 4. Its interpretation should be the proportion of the variation explained by the model; 5. It should be between 0 and 1, with 0 denoting that model does not explain any variation and 1 denoting that it perfectly explains the observed variation; 6. It should not have any unit. The generalized R² has all of these properties. $R^{2} = 1 - \left({ L(0) \over L(\hat{\theta})}\right)^{2/n}$ where L(0) is the likelihood of the model with only the intercept, ${L(\hat{\theta})}$ is the likelihood of the estimated model (i.e., the model with a given set of parameter estimates) and n is the sample size. However, in the case of a logistic model, where $L(\hat{\theta})$ cannot be greater than 1, R² is between 0 and $R^2_\max = 1- (L(0))^{2/n}$: thus, it is possible to define a scaled R² as R²/R²max.[4] ## Notes 1. Steel, R.G.D, and Torrie, J. H., Principles and Procedures of Statistics with Special Reference to the Biological Sciences., McGraw Hill, 1960, pp. 187, 287.) 2. Colin Cameron, A.; Windmeijer, Frank A.G.; Gramajo, H; Cane, DE; Khosla, C (1997). "An R-squared measure of goodness of fit for some common nonlinear regression models". Journal of Econometrics 77 (2): 1790–2. doi:10.1016/S0304-4076(96)01818-0. PMID 11230695. 3. Theil, Henri (1961). Economic Forecasts and Policy. Holland, Amsterdam: North. [] 4. Nagelkerke, N. J. D. (1991). "A Note on a General Definition of the Coefficient of Determination". Biometrika 78 (3): 691–2. doi:10.1093/biomet/78.3.691. JSTOR 2337038. ## References • Draper, N.R. and Smith, H. (1998). Applied Regression Analysis. Wiley-Interscience. ISBN 0-471-17082-8 • Everitt, B.S. (2002). Cambridge Dictionary of Statistics (2nd Edition). CUP. ISBN 0-521-81099-X • Nagelkerke, Nico J.D. (1992) Maximum Likelihood Estimation of Functional Relationships, Pays-Bas, Lecture Notes in Statistics, Volume 69, 110p ISBN 0-387-97721-X. • Glantz, S.A. and Slinker, B.K., (1990). Primer of Applied Regression and Analysis of Variance. McGraw-Hill. ISBN 0-07-023407-8.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 37, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9056942462921143, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/129168-comparison-test.html	# Thread: 1. ## Comparison Test $\int_1^\infty \frac{1}{\sqrt{\theta+1}}d\theta$ My professor tells us to create an intuition and then create a proof. What I have so far is $Let f(\theta) = \frac{1}{\sqrt{\theta+1}}d\theta$ $When \theta \rightarrow \infty; f(\theta)$ acts like $\frac{1}{\sqrt{\theta}}$ So I guess that $f(\theta)$ diverges because p = $\frac{1}{2}$ Now I need an easy, smaller function that I can prove that if that smaller function diverges, then $f(\theta)$ diverges, am missing something here? 2. Originally Posted by Latszer $\int_1^\infty \frac{1}{\sqrt{\theta+1}}d\theta$ My professor tells us to create an intuition and then create a proof. What I have so far is $Let f(\theta) = \frac{1}{\sqrt{\theta+1}}d\theta$ $When \theta \rightarrow \infty; f(\theta)$ acts like $\frac{1}{\sqrt{\theta}}$ So I guess that $f(\theta)$ diverges because p = $\frac{1}{2}$ Now I need an easy, smaller function that I can prove that if that smaller function diverges, then $f(\theta)$ diverges, am missing something here? $\int \dfrac{dx}{\sqrt{x+x}}= \dfrac{1}{\sqrt{2}}\int \dfrac{dx}{\sqrt{x}} < \int \dfrac{dx}{\sqrt{x+1}}$ I think now its correct 3. $\frac{1}{x}$ is not less than $\frac{1}{x+1}$ as x approaches infinity, so while your logic makes sense it technically does not "prove" the integral. 4. Is it a sufficient proof if I just sub the integral like.... $\int_1^\infty\frac{d\theta}{\sqrt{\theta+1}} = \lim_{c\to\infty}\int_1^c \frac{d\theta}{\sqrt{\theta+1}}$ let $u=\theta + 1$ ; $du = d\theta$ when $\theta = c \rightarrow u = c + 1$ ; when $\theta = 1 \rightarrow u = 2$ Let $d = c+1$ When $c \rightarrow\infty$ , $d \rightarrow\infty$ $\lim_{d\to\infty}\int_2^d\frac{du}{\sqrt{u}}$ = $\lim_{d\to\infty}\int_1^\infty\frac{du}{\sqrt{u}}d u$ - $\int_1^2\frac{du}{\sqrt{u}}du$ And then since $\lim_{d\to\infty}\int_1^d\frac{du}{\sqrt{u}}$ diverges, then $\int_1^\infty\frac{d\theta}{\sqrt{\theta+1}}$ diverges. Did I do anything "illegal"? Is that a sufficient proof? -Tyler 5. Originally Posted by Latszer Is it a sufficient proof if I just sub the integral like.... $\int_1^\infty\frac{d\theta}{\sqrt{\theta+1}} = \lim_{c\to\infty}\int_1^c \frac{d\theta}{\sqrt{\theta+1}}$ let $u=\theta + 1$ ; $du = d\theta$ when $\theta = c \rightarrow u = c + 1$ ; when $\theta = 1 \rightarrow u = 2$ Let $d = c+1$ When $c \rightarrow\infty$ , $d \rightarrow\infty$ $\lim_{d\to\infty}\int_2^d\frac{du}{\sqrt{u}}$ = $\lim_{d\to\infty}\int_1^\infty\frac{du}{\sqrt{u}}d u$ - $\int_1^2\frac{du}{\sqrt{u}}du$ And then since $\lim_{d\to\infty}\int_1^d\frac{du}{\sqrt{u}}$ diverges, then $\int_1^\infty\frac{d\theta}{\sqrt{\theta+1}}$ diverges. Did I do anything "illegal"? Is that a sufficient proof? -Tyler Why not just realise that $\frac{1}{\sqrt{\theta + 1}} = (\theta + 1)^{-\frac{1}{2}}$? So $\int{\frac{1}{\sqrt{\theta + 1}}\,d\theta} = \int{(\theta + 1)^{-\frac{1}{2}}\,d\theta}$ $= 2(\theta + 1)^{\frac{1}{2}} + C$ $= 2\sqrt{\theta + 1} + C$. So $\int_1^{\infty}{\frac{1}{\sqrt{\theta + 1}}\,d\theta} = \lim_{\epsilon \to \infty}\left[2\sqrt{\theta + 1}\right]_1^{\epsilon}$ $= \lim_{\epsilon \to \infty}2\sqrt{\epsilon + 1} - 2\sqrt{1 + 1}$ This is clearly divergent, as the square root function is always increasing.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 49, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.836306095123291, "perplexity_flag": "middle"}
http://gilkalai.wordpress.com/2008/12/07/test-your-intuition-1/?like=1&source=post_flair&_wpnonce=8ac41143de	Gil Kalai’s blog ## Test Your Intuition (1) Posted on December 7, 2008 by Question: Suppose that we sequentially place $n$ balls into $n$ boxes by putting each ball into a randomly chosen box. It is well known that when we are done, the fullest box has with high probability $(1+o(1))\ln n/\ln \ln n$ balls in it. Suppose instead that for each ball we choose two boxes at random and place the ball into the one which is less full at the time of placement. What will be the maximum occupancy? Test your intuition before reading the rest of the entry. Answer: A beautiful theorem of Yossi Azar, Andrei Broder, Anna Karlin, and Eli Upfal asserts that with high probability, the fullest box contains only $\ln \ln n/\ln 2+O(1)$ balls—exponentially less than before. (The descripion follows Xueliang Li’s mathscinet review.) And here is a link to the paper. Here is a related post “Balls and Bins on Graphs” on Windows on Theory. ### Like this: This entry was posted in Computer Science and Optimization, Probability, Test your intuition and tagged Test. Bookmark the permalink. ### 11 Responses to Test Your Intuition (1) 1. Gil says: Actually, I am not aware of a simple explanation (even a heuristic one) why we go down to log log n. I will be happy to hear such an explanation. Also, this theorem is related to many developments in theoretical CS and probabilistic combinatorics and remarks on these are most welcomed. 2. kunal says: There is in fact more to this! Suppose now you throw m balls in to n bins, where m is much larger than n. For one choice, the imbalance is easily seen to grow as \tilde{O}(\sqrt{m/n}). For the two choice process, the imbalance is in fact independent of n. It stays at O(log log n) w.h.p. See this paper: Petra Berenbrink, Artur Czumaj, Angelika Steger, and Berthold Vöcking. Balanced allocations: The heavily loaded case. In Proceedings of the 32th ACM Symposium on Theory of Computing (STOC’00), pages 745-754, 2000. 3. Gil Kalai says: Dear Kunal, indeed this is part of a large story both in TCS and probabilistic combinatorics. I was motivated by a lecture of Michael Krivelevich on “Achlioptas processes”. In these processes you build a random graph by choosing at each round one out of r random edges in order to postpone or embrace a certain graph proprty. Another connection that comes to mind (which perhaps fits a series of posts under the name: “difficult proofs for easy theorems”) is to find a probabilistic proof that if m is not too large compare to n, it is possible to put n balls into m boxes so that no box contains more than one ball. A simple union bound works for something like $m \ge n^2$ and using Lovasz local lemma you can get it down to $m \ge 6n$ or so. I do not know what is the world record. Anyway, what I am most curious about is a simple conceptual explanation in a few words for the loglogn in the theorem of Azar, Broder, Karlin, and Upfal. 4. Omer says: Ori pointed your question out and we came up with some explanation. Suppose you place the ball in the less populated set. At time t you have $\sim t$ occupied boxes, so the rate of introducing doubly occupied boxes is $(t/n)^2$. This shows that at time \$t\$ you expect $\approx t^3/n^2$ boxes with two balls. The first appears at time $\approx n^{2/3}$ Repeating the argument, the rate at which boxes with three balls are created is $(t^3/n^3)^2$, so the number of such boxes is $\approx t^t/n^6$. In general, the first box with $k$ balls appears at time $n\cdot n^{-1/(2^k-1)}$. If $k=\log\log n/\log 2$ then this is of order $n$, giving the claimed asymptotics. More formally, if $X_i(t)$ is the number of boxes with at least k balls at time t then in expectation, $X_i' = (X_{i-1}/n)^2$. If there are $M$ choices at each step the same scheme gives $\log\log n/\log M$. 5. Gil Kalai says: Many thanks, Omer! 6. arnab says: Thanks for this post! So, as suggested by the informal explanation given above, is it really true that taking M = log n gives a constant number of balls in the fullest bin with high probability? 7. Ori says: aranb: taking anything asymptotically less then $\sqrt{n}$ will give you only 1 ball in the fullest bin with high probability, even if you have not choice (bin selected randomly). In the case you have a choice between 2 bins, this goes up to $n^{1/3}$. 8. Ori says: Gil, I thought you might be interested in the following paper (http://arxiv.org/abs/0901.4056), by Noga Alon, Eyal Lubetzky and yours truly, in which we consider the above problem under memory constraint. • Gil Kalai says: Many thanks, Ori 9. Pingback: Test Your Intuition (4) « Combinatorics and more 10. Pingback: Answer To Test Your Intuition (4) « Combinatorics and more • ### Blogroll %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 22, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9083889722824097, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/46658/localized-ideal-of-a-variety-that-is-smooth-at-a-point	# localized ideal of a variety that is smooth at a point Let $X \subset \mathbb{A}^n$ be an affine variety that is smooth in $a$ of dimension n-k and I its ideal in $K[x_1,...,x_n]$. Is it true that the localized ideal $I_{m_a} \subset \mathcal{O}_{X,a}$ is generated by k elements? - ## 1 Answer Yes, smooth at a point implies it is a local complete intersection at that point, i.e. the defining ideal is generated by the minimum number of elements. See theorem 8.17 on page 178 of Hartshorne. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9477487206459045, "perplexity_flag": "head"}
http://mathoverflow.net/questions/42617/functions-whose-gradient-descent-paths-are-geodesics	## Functions whose gradient-descent paths are geodesics ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $f(x,y)$ define a surface $S$ in $\mathbb{R}^3$ with a unique local minimum at $b \in S$. Suppose gradient descent from any start point $a \in S$ follows a geodesic on $S$ from $a$ to $b$. (Q1.) What is the class of functions/surfaces whose gradient-descent paths are geodesics? Certainly if $S$ is a surface of revolution about a $z$-vertical line through $b$, its "meridians" are geodesics, and these would be the paths followed by gradient descent down to $b$. So the class of surfaces includes surfaces of revolution. But surely it is wider than that? (Q2.) One could ask the same question about paths followed by Newton's method, which in general are different from gradient-descent paths, as indicated in this Wikipedia image: Gradient descent: green. Newton's method: red. (Q3.) These questions make sense in arbitrary dimensions, although my primary interest is for surfaces in $\mathbb{R}^3$. Any ideas on how to formulate my question as constraints on $f(\;)$, or pointers to relevant literature, would be appreciated. Thanks! - 5 Let me add a fourth side to the question. (Q4.) Let $h:{\mathbb R}\rightarrow{\mathbb R}$ be smooth and increasing. Let $f$ be a function as in (Q1). Is $h\circ f$ such a function too ? – Denis Serre Oct 18 2010 at 13:55 @Denis: I see your motivation. Excellent question! – Joseph O'Rourke Oct 18 2010 at 14:21 ## 2 Answers For (Q1). The tangent space of $S$ is generated by the gradient flow vector field $v = (\|\nabla f\|^2, \nabla f)$ and the tangents to the level sets $w= (0, \nabla^\perp f)$. The geodesic constraint can be imposed as the condition "no sideways acceleration", which means that $[(\nabla f \cdot \nabla )v] \cdot w = 0$. This implies that $\nabla^2_{ij} f \nabla^if \nabla^{(\perp)j}f = 0$. In other words, the eigendirections of the Hessian of $f$ must be $\nabla f$ and its orthogonal, or that $\nabla f$ is parallel to $\nabla \|\nabla f\|^2$. So this means that $f$ and $\|\nabla f\|^2$ share the same level sets. (This same characterization is valid for any dimension; so also answers (Q3). ) In particular, this answers Denis Serre's (Q4) in the positive. - 1 @Willie: That is a beautiful characterization, that $f$ and the square gradient share the same level sets! I do not yet see what this implies in terms of the global geometric shape of $f$, but it certainly is a succinct encapsulation. Thanks! – Joseph O'Rourke Oct 18 2010 at 18:23 2 Two functions share the same level sets iff the values of each of them only depend on the values of the other. So Willie's condition writes as an eikonal equation $\|\nabla f(x)\|^2=h(f(x)),$ and I guess that solutions are all of the form $f(x)=g(\mathbb{dist}(x,C)),$ at least locally ($g$ and $h$ being related by $1/h= (g^{-1})^')$. Here $\mathbb{dist}(x,C)$ is the point-set Euclidean distance from $C$. – Pietro Majer Oct 18 2010 at 19:18 @Pietro: And $C$ is ... ? – Joseph O'Rourke Oct 18 2010 at 19:21 1 (I guess) $C$ could be any subset, even not smooth, for the distance function from $C$ is 1-Lipschitz, thus differentiable a.e., and $\|\nabla d_C\|=1$, so in any case one gets a solution. A smooth, convex $C$ should give solutions defined everywhere. Note that $C$ a point gives the surfaces of revolutions you mentioned in the questions. – Pietro Majer Oct 18 2010 at 19:31 (sorry, I changed notation: $d_C(x)=\mathbb{dist}(x,C)$). Another way to characterize the functions $f$ should be, that sublevel sets {f<c} are uniform neighborhoods of $C$ (or of any sublevel set {f<b}, with b<c). – Pietro Majer Oct 18 2010 at 19:46 show 3 more comments ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Here is a function $f(x,y)$ which is 0 inside the square $C=[\pm1,\pm1]$, and outside that square has value equal to the Euclidean distance $d( p, C )$ from $p=(x,y)$ to the boundary of $C$. [I am trying to follow Pietro's suggestion, as far as I understand it.] It is not a surface of revolution (but it is centrally symmetric). Are its gradient descent paths geodesics? I think so... Left above: $f(x,y)$. Right above: Level sets of $f$. Below: $\nabla f$. And here (below) is a closeup of the function defined using squared distance $[d( p, C )]^2$, as per Will's suggestion: - Hmmm, like Jeopardy, :) your answer is in the form of a question... I think they just might be, but now in this case, there is no point that is a unique minimum. Everything within the square $C$ is the minimum. So what is your gradient descending to? The closest point in $C$ to your starting point? So you've got 4 voronoi cells defined by (0.5,0) (0,0.5) (-0.5,0) and (0,-0.5) for E/N/W/S directions, which delineate the four regions closest to the 4 edges of $C$,except for any point within $C$ in which case its gradient descent is the point itself. – sleepless in beantown Oct 19 2010 at 0:53 @sleepless: Yes, no unique local min, but still one can define descent paths for all points exterior to the central square. Your description is accurate, including answer$=$question! – Joseph O'Rourke Oct 19 2010 at 1:02 Or, you've got 8 regions, with the 4 quadrants translated (+/-0.5, +/-0.5) in the xy plane with gradient descents for any point in those planes mapping to the corners of the square $C$, e.g. (x≥0.5,y≥0.5)→(+0.5,+0.5), and the other three quadrants where you swap in less-than signs for -0.5 values all having gradient descents as lines going to the corners of the square $C$; and the regions for x > 0.5 and $-0.5 \le y \le 0.5$ having a gradient descent to $(y,0.5)$, and three other similar regions mapping to the edges of the region $C$ – sleepless in beantown Oct 19 2010 at 1:02 So the answer in this case, is a definite YES, the gradient descents for this figure for any point starting outside of $C$ are definitely geodesics. I'd call this figure a circle fractured into quadrants intercalated by strips of the plane defined by a square at the center of the fractured circle quadrants. (instead of drawing the contours of the distance, draw the gradient of the function and you'll see it clearly.) – sleepless in beantown Oct 19 2010 at 1:09 By the way, this function certainly is rotationally symmetric about the $z$-axis with 4-fold symmetry. – sleepless in beantown Oct 19 2010 at 1:24 show 4 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 67, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9212725758552551, "perplexity_flag": "middle"}
http://nrich.maths.org/7302	### Chocolate There are three tables in a room with blocks of chocolate on each. Where would be the best place for each child in the class to sit if they came in one at a time? ### Plants Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this? ### Gran, How Old Are You? When Charlie asked his grandmother how old she is, he didn't get a straightforward reply! Can you work out how old she is? # Count the Digits ##### Stage: 1 and 2 Challenge Level: We can do all sorts of things with numbers - add, subtract, multiply, divide ... Most of us start with counting when we are very little. We usually count things, objects, people etc. In this activity we are going to count the number of digits that are the same. Rule 1 - The starting number has to have just three different digits chosen from $1, 2, 3, 4$. Rule 2 - The starting number must have four digits - so thousands, hundreds, tens and ones. For example, we could choose $2124$ or $1124$. So when we've got our starting number we'll do some counting. Here is a worked example. Starting Number We then count in order the number of $1$s, then the number of $2$s, then $3$s and lastly $4$s, and write it down as shown here. So the first count gave one $1$, one $3$ and two $4$s. You may see that this has continued so the third line shows that the line above had three $1$s, one $2$, one $3$ and one $4$. The fourth line counts the line above giving four $1$s, one $2$, two $3$s and one $4$. And so it goes on until ... until when? Your challenge is to start with other four digit numbers which satisfy the two rules and work on it in the way I did. Tell us what you notice. What happens if you have five digits in the starting number? The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9587201476097107, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Schr%C3%B6dinger's_cat	# Schrödinger's cat Schrödinger's cat: a cat, a flask of poison, and a radioactive source are placed in a sealed box. If an internal monitor detects radioactivity (i.e. a single atom decaying), the flask is shattered, releasing the poison that kills the cat. The Copenhagen interpretation of quantum mechanics implies that after a while, the cat is simultaneously alive and dead. Yet, when one looks in the box, one sees the cat either alive or dead, not both alive and dead. This poses the question of when exactly quantum superposition ends and reality collapses into one possibility or the other. Quantum mechanics Introduction Glossary · History Background Fundamental concepts Experiments Formulations Equations Interpretations Advanced topics Scientists Schrödinger's cat is a thought experiment, sometimes described as a paradox, devised by Austrian physicist Erwin Schrödinger in 1935. It illustrates what he saw as the problem of the Copenhagen interpretation of quantum mechanics applied to everyday objects, resulting in a contradiction with common sense. The scenario presents a cat that may be both alive and dead, depending on an earlier random event. Although the original "experiment" was imaginary, similar principles have been researched and used in practical applications. The thought experiment is also often featured in theoretical discussions of the interpretations of quantum mechanics. In the course of developing this experiment, Schrödinger coined the term Verschränkung (entanglement). ## Origin and motivation Schrödinger intended his thought experiment as a discussion of the EPR article—named after its authors Einstein, Podolsky, and Rosen—in 1935.[1] The EPR article highlighted the strange nature of quantum entanglement, which is a characteristic of a quantum state that is a combination of the states of two systems (for example, two subatomic particles), that once interacted but were then separated and are not each in a definite state. The Copenhagen interpretation implies that the state of the two systems collapses into a definite state when one of the systems is measured. Schrödinger and Einstein exchanged letters about Einstein's EPR article, in the course of which Einstein pointed out that the state of an unstable keg of gunpowder will, after a while, contain a superposition of both exploded and unexploded states. To further illustrate, Schrödinger describes how one could, in principle, transpose the superposition of an atom to large-scale systems. He proposed a scenario with a cat in a sealed box, wherein the cat's life or death depended on the state of a subatomic particle. According to Schrödinger, the Copenhagen interpretation implies that the cat remains both alive and dead (to the universe outside the box) until the box is opened. Schrödinger did not wish to promote the idea of dead-and-alive cats as a serious possibility; quite the reverse, the paradox is a classic reductio ad absurdum.[2] The thought experiment illustrates quantum mechanics and the mathematics necessary to describe quantum states. Intended as a critique of just the Copenhagen interpretation (the prevailing orthodoxy in 1935), the Schrödinger cat thought experiment remains a typical touchstone for limited interpretations of quantum mechanics. Physicists often use the way each interpretation deals with Schrödinger's cat as a way of illustrating and comparing the particular features, strengths, and weaknesses of each interpretation. ## The thought experiment Schrödinger wrote:[3][2] One can even set up quite ridiculous cases. A cat is penned up in a steel chamber, along with the following device (which must be secured against direct interference by the cat): in a Geiger counter, there is a tiny bit of radioactive substance, so small that perhaps in the course of the hour, one of the atoms decays, but also, with equal probability, perhaps none; if it happens, the counter tube discharges, and through a relay releases a hammer that shatters a small flask of hydrocyanic acid. If one has left this entire system to itself for an hour, one would say that the cat still lives if meanwhile no atom has decayed. The psi-function of the entire system would express this by having in it the living and dead cat mixed or smeared out in equal parts. It is typical of these cases that an indeterminacy originally restricted to the atomic domain becomes transformed into macroscopic indeterminacy, which can then be resolved by direct observation. That prevents us from so naively accepting as valid a "blurred model" for representing reality. In itself, it would not embody anything unclear or contradictory. There is a difference between a shaky or out-of-focus photograph and a snapshot of clouds and fog banks. —Erwin Schrödinger, Die gegenwärtige Situation in der Quantenmechanik (The present situation in quantum mechanics), Naturwissenschaften (translated by John D. Trimmer in Proceedings of the American Philosophical Society) Schrödinger's famous thought experiment poses the question, when does a quantum system stop existing as a superposition of states and become one or the other? (More technically, when does the actual quantum state stop being a linear combination of states, each of which resembles different classical states, and instead begins to have a unique classical description?) If the cat survives, it remembers only being alive. But explanations of the EPR experiments that are consistent with standard microscopic quantum mechanics require that macroscopic objects, such as cats and notebooks, do not always have unique classical descriptions. The thought experiment illustrates this apparent paradox. Our intuition says that no observer can be in a mixture of states—yet the cat, it seems from the thought experiment, can be such a mixture. Is the cat required to be an observer, or does its existence in a single well-defined classical state require another external observer? Each alternative seemed absurd to Albert Einstein, who was impressed by the ability of the thought experiment to highlight these issues. In a letter to Schrödinger dated 1950, he wrote: You are the only contemporary physicist, besides Laue, who sees that one cannot get around the assumption of reality, if only one is honest. Most of them simply do not see what sort of risky game they are playing with reality—reality as something independent of what is experimentally established. Their interpretation is, however, refuted most elegantly by your system of radioactive atom + amplifier + charge of gunpowder + cat in a box, in which the psi-function of the system contains both the cat alive and blown to bits. Nobody really doubts that the presence or absence of the cat is something independent of the act of observation.[4] Note that the charge of gunpowder is not mentioned in Schrödinger's setup, which uses a Geiger counter as an amplifier and hydrocyanic poison instead of gunpowder. The gunpowder had been mentioned in Einstein's original suggestion to Schrödinger 15 years before, and apparently Einstein had carried it forward to the present discussion. ## Interpretations of the experiment Since Schrödinger's time, other interpretations of quantum mechanics have been proposed that give different answers to the questions posed by Schrödinger's cat of how long superpositions last and when (or whether) they collapse. ### Copenhagen interpretation Main article: Copenhagen interpretation The most commonly held interpretation of quantum mechanics is the Copenhagen interpretation.[5] In the Copenhagen interpretation, a system stops being a superposition of states and becomes either one or the other when an observation takes place. This experiment makes apparent the fact that the nature of measurement, or observation, is not well-defined in this interpretation. The experiment can be interpreted to mean that while the box is closed, the system simultaneously exists in a superposition of the states "decayed nucleus/dead cat" and "undecayed nucleus/living cat," and that only when the box is opened and an observation performed does the wave function collapse into one of the two states. However, one of the main scientists associated with the Copenhagen interpretation, Niels Bohr, never had in mind the observer-induced collapse of the wave function, so that Schrödinger's cat did not pose any riddle to him. The cat would be either dead or alive long before the box is opened by a conscious observer.[6] Analysis of an actual experiment found that measurement alone (for example by a Geiger counter) is sufficient to collapse a quantum wave function before there is any conscious observation of the measurement.[7] The view that the "observation" is taken when a particle from the nucleus hits the detector can be developed into objective collapse theories. The thought experiment requires an "unconscious observation" by the detector in order for magnification to occur. In contrast, the many worlds approach denies that collapse ever occurs. ### Many-worlds interpretation and consistent histories The quantum-mechanical "Schrödinger's cat" paradox according to the many-worlds interpretation. In this interpretation, every event is a branch point. The cat is both alive and dead—regardless of whether the box is opened—but the "alive" and "dead" cats are in different branches of the universe that are equally real but cannot interact with each other. Main article: Many-worlds interpretation In 1957, Hugh Everett formulated the many-worlds interpretation of quantum mechanics, which does not single out observation as a special process. In the many-worlds interpretation, both alive and dead states of the cat persist after the box is opened, but are decoherent from each other. In other words, when the box is opened, the observer and the possibly-dead cat split into an observer looking at a box with a dead cat, and an observer looking at a box with a live cat. But since the dead and alive states are decoherent, there is no effective communication or interaction between them. When opening the box, the observer becomes entangled with the cat, so "observer states" corresponding to the cat's being alive and dead are formed; each observer state is entangled or linked with the cat so that the "observation of the cat's state" and the "cat's state" correspond with each other. Quantum decoherence ensures that the different outcomes have no interaction with each other. The same mechanism of quantum decoherence is also important for the interpretation in terms of consistent histories. Only the "dead cat" or "alive cat" can be a part of a consistent history in this interpretation. Roger Penrose criticises this: "I wish to make it clear that, as it stands, this is far from a resolution of the cat paradox. For there is nothing in the formalism of quantum mechanics that demands that a state of consciousness cannot involve the simultaneous perception of a live and a dead cat",[8] however, the mainstream view (without necessarily endorsing many-worlds) is that decoherence is the mechanism that forbids such simultaneous perception.[9][10] A variant of the Schrödinger's cat experiment, known as the quantum suicide machine, has been proposed by cosmologist Max Tegmark. It examines the Schrödinger's cat experiment from the point of view of the cat, and argues that by using this approach, one may be able to distinguish between the Copenhagen interpretation and many-worlds. ### Ensemble interpretation The ensemble interpretation states that superpositions are nothing but subensembles of a larger statistical ensemble. The state vector would not apply to individual cat experiments, but only to the statistics of many similarly prepared cat experiments. Proponents of this interpretation state that this makes the Schrödinger's cat paradox a trivial non-issue. This interpretation serves to discard the idea that a single physical system in quantum mechanics has a mathematical description that corresponds to it in any way. ### Relational interpretation The relational interpretation makes no fundamental distinction between the human experimenter, the cat, or the apparatus, or between animate and inanimate systems; all are quantum systems governed by the same rules of wavefunction evolution, and all may be considered "observers." But the relational interpretation allows that different observers can give different accounts of the same series of events, depending on the information they have about the system.[11] The cat can be considered an observer of the apparatus; meanwhile, the experimenter can be considered another observer of the system in the box (the cat plus the apparatus). Before the box is opened, the cat, by nature of it being alive or dead, has information about the state of the apparatus (the atom has either decayed or not decayed); but the experimenter does not have information about the state of the box contents. In this way, the two observers simultaneously have different accounts of the situation: To the cat, the wavefunction of the apparatus has appeared to "collapse"; to the experimenter, the contents of the box appear to be in superposition. Not until the box is opened, and both observers have the same information about what happened, do both system states appear to "collapse" into the same definite result, a cat that is either alive or dead. ### Objective collapse theories According to objective collapse theories, superpositions are destroyed spontaneously (irrespective of external observation) when some objective physical threshold (of time, mass, temperature, irreversibility, etc.) is reached. Thus, the cat would be expected to have settled into a definite state long before the box is opened. This could loosely be phrased as "the cat observes itself," or "the environment observes the cat." Objective collapse theories require a modification of standard quantum mechanics to allow superpositions to be destroyed by the process of time evolution. ## Applications and tests The experiment as described is a purely theoretical one, and the machine proposed is not known to have been constructed. However, successful experiments involving similar principles, e.g. superpositions of relatively large (by the standards of quantum physics) objects have been performed.[12] These experiments do not show that a cat-sized object can be superposed, but the known upper limit on "cat states" has been pushed upwards by them. In many cases the state is short-lived, even when cooled to near absolute zero. • A "cat state" has been achieved with photons.[13] • A beryllium ion has been trapped in a superposed state.[14] • An experiment involving a superconducting quantum interference device ("SQUID") has been linked to theme of the thought experiment: "The superposition state does not correspond to a billion electrons flowing one way and a billion others flowing the other way. Superconducting electrons move en masse. All the superconducting electrons in the SQUID flow both ways around the loop at once when they are in the Schrödinger's cat state."[15] • A piezoelectric "tuning fork" has been constructed, which can be placed into a superposition of vibrating and non vibrating states. The resonator comprises about 10 trillion atoms.[16] • An experiment involving a flu virus has been proposed.[17] In quantum computing the phrase "cat state" often refers to the special entanglement of qubits wherein the qubits are in an equal superposition of all being 0 and all being 1; e.g., $\| \psi \rangle = \frac{1}{\sqrt{2}} \bigg( \| 00...0 \rangle + \|11...1 \rangle \bigg).$ ## Extensions Wigner's friend is a variant on the experiment with two external observers: the first opens and inspects the box and then communicates his observations to a second observer. The issue here is, does the wave function "collapse" when the first observer opens the box, or only when the second observer is informed of the first observer's observations? In another extension, prominent physicists have gone so far as to suggest that astronomers observing dark energy in the universe in 1998 may have "reduced its life expectancy" through a pseudo-Schrödinger's cat scenario, although this is a controversial viewpoint.[18][19] ## In popular culture Main article: Schrödinger's cat in popular culture ## References 1. ^ a b 2. Hermann Wimmel (1992). Quantum physics & observed reality: a critical interpretation of quantum mechanics. World Scientific. p. 2. ISBN 978-981-02-1010-6. Retrieved 9 May 2011. 3. Faye, J (2008-01-24). "Copenhagen Interpretation of Quantum Mechanics". . The Metaphysics Research Lab Center for the Study of Language and Information, Stanford University. Retrieved 2010-09-19. 4. Carpenter RHS, Anderson AJ (2006). "The death of Schroedinger's cat and of consciousness-based wave-function collapse". 31 (1): 45–52. Archived from the original on 2006-11-30. Retrieved 2010-09-10. 5. Penrose, R. The Road to Reality, p 807. 6. Wojciech H. Zurek, Decoherence, einselection, and the quantum origins of the classical, Reviews of Modern Physics 2003, 75, 715 or [1] 7. Wojciech H. Zurek, "Decoherence and the transition from quantum to classical", Physics Today, 44, pp 36–44 (1991) 8. Rovelli, Carlo (1996). "Relational Quantum Mechanics". International Journal of Theoretical Physics 35 (8): 1637–1678. arXiv:quant-ph/9609002. Bibcode:1996IJTP...35.1637R. doi:10.1007/BF02302261. 9. Chown, Marcus (2007-11-22). "Has observing the universe hastened its end?". . Retrieved 2007-11-25. 10. Krauss, Lawrence M.; James Dent (April 30, 2008). "Late Time Behavior of False Vacuum Decay: Possible Implications for Cosmology and Metastable Inflating States". Phys. Rev. Lett. (US: APS) 100 (17). arXiv:0711.1821. Bibcode:2008PhRvL.100q1301K. doi:10.1103/PhysRevLett.100.171301.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9114349484443665, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/54669/null-geodesic-given-metric	# Null geodesic given metric I (desperately) need help with the following: What is the null geodesic for the space time $$ds^2=-x^2 dt^2 +dx^2?$$ I don't know how to transform a metric into a geodesic...! There is no need to start from the Lagrangian. I know that $$0=g_{ij}V^iV^j$$ where $V^i={dx^i\over d\lambda}$ where $\lambda$ is some parameter. But I don't know what that parameter is nor how to find the geodesic. - A geodesic is just a special type of curve, and the parameter is, well, just a parameter parametrizing that curve, and so doesn't matter ultimately. Also, geodesics are only unique if you specify more about them, like a point on the curve and a direction on that curve. Make sure to understand what these objects are before diving into symbol manipulation. – Chris White Feb 22 at 2:43 ## 2 Answers As you mentioned a null geodesic implies: $$g_{\mu \nu} U^{\mu} U^{\nu}=\left (\frac{ds}{d \lambda} \right )^2=0$$ where $\lambda$ is some affine parameter. If you take $\lambda=t$, then this implies: $$-x^2+\left ( \frac{dx}{dt} \right )^2=0$$ So $~dx/dt=\pm x$ is a null geodesic. (Take a second time derivative to get the actual geodesic.) - Thanks, elfmotat. What do you mean by the "actual" geodesic? – hetherson Feb 22 at 0:06 I mean the second-order differential equation, which is called the "geodesic equation." – elfmotat Feb 22 at 0:18 The easiest way to solve this is to pretend that it IS a Lagrangian: $$I = \frac{1}{2}\int d\lambda \left(-x^{2}{\dot t}^{2} + {\dot x}^{2}\right)$$ Where both $t$ and $x$ are functions of $\lambda$, and ${\dot x} \equiv \frac{dx}{d\lambda}$.. Take the variation of the action, find the minimum, and then set your constants so that your geodesic is null. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8836607336997986, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/247959/shortest-distance-between-clothoid-spline-and-point	# Shortest distance between clothoid spline and point Is there any way to analytically decide the shortest distance between a spline of clothoids and a point? Both lies in XY-plane. The clothoid spline has G2 continuity. The result should be used in geometric optimzation, so squared distance (xp - xc)2 + (yp - yc)2 may be used. If there is no analytic solution, can anyone give a good suggestion to an iterative solution. Definition of clothoid and splines: Clothoid is also called Cornu spiral or Euler spiral. http://mathworld.wolfram.com/CornuSpiral.html A spline is a piecewise-defined function: http://en.wikipedia.org/wiki/Spline_(mathematics) - 1 As a general rule, if a Google search for a term doesn't yield a Wikipedia or MathWorld or PlanetMath article as one of the first hits, and instead your own question appears as one of the first hits, it's a good idea to include a definition of that term in the question. – joriki Nov 30 '12 at 10:50 1 Even if the definition is googleable, it is a good idea to make it easy for people to answer the question by providing the definition. – Phira Nov 30 '12 at 10:56 Two links added for more information on Clothoid and Splines. – Laban Källgren Nov 30 '12 at 11:49 In many practical applications, you only use rather short or rather straight parts of a clothoid, such that any local distance minimum would be a global minimum as well. Is this the case for the individual pieces of your spline as well? If so, you could compute the distance with respect to every piece using some iterative solution, like bisection in the simplest of cases. Otherwise you'd probably need to find the correct “layer” first. – MvG Nov 30 '12 at 16:41 This is true for my problem: we use rather short and straight parts of the clothoid. The problem domain is road and rail geometry; a road/rail is built of straight lines, arcs and "non-aggressive" clothoids. We need this to be solved as a part of an extensive optimization algorithm, hence the method need to be as fast as possible. Analytic solution is preferrable if such exists? Bisection seems to be slow, but may be used to generate a starting guess for a faster algorithm. – Laban Källgren Nov 30 '12 at 17:41 ## 1 Answer There is no analytical solution, since that would include solving integral equations. There are however very fast numerical solutions. Those involve minizing distance function $d(t)=\left(\vec{r}(t)-\vec{p}\right)^2$ where $\vec{p}$ is the point you want to calculate distance to and $t$ parametrizes your clothoid which is given by $\vec{r}(t)$. You have mentioned that your clothoids are "non-agressive" which I presume means that tangent from the start till the end of the clothoid does not rotate by a large angle. That will make $d(t)$ well behaved function of the parameter along the clothoid. What you can do then is to subdivide your spline in sections where tangent at the beginning and the end make up angle not larger than 60 degrees. You are guaranteed that there will be no more than one extremum of $d(t)$ on that section. You can then check if $d'(t_0)d'(t_1) < 0$, where $t_0$ and $t_1$ are parameters of endpoints of your section. If this condition is satisfied you can apply for example Dekker's method to find a zero of $d'(t)=2\left(\vec{r}(t)-\vec{p}\right)\cdot\vec{r}'(t)$ (you can compute $\vec{r}'(t)$ analytically). It will converge in few iterations and you can then check whether you have minimum or maximum there by checking in final iteration whether $\left(d'(a_{k})-d'(b_{k})\right)/(a_k-b_k)>0$ (see Wikipedia article for the notation). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9261741042137146, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/21274/how-to-solve-schrodinger-equation-tunnelling/21296	# How to solve Schrodinger Equation - Tunnelling I have to solve analitically the Schrodinger equation in one-dimension with a barrier of potential (tunnel effect): $$ih \frac{d}{dt} U(x,t) = \left[ \left(-h^2 \frac{d^2}{dx^2} \right) + q V(x) \right] U(x,t)$$ where: $i$ is the imaginary unit, ($d/dt$) is the time derivative, $h$ is the Plank constant, ($d^2/dx^2$) is the second derivative in space, $V(x)$ is an external potential function of $x$, $U(x,t)$ is the wave function of time and place. The barrier of potential is: $$V(x) = \begin{cases} 0, & \mbox{if } -d<x<-L \\ V_0, & \mbox{if } -L<x<L \\ 0, & \mbox{if } L<x<d \end{cases}$$ with $d=10L$ and $V_0>0$; Also the boundary condition are: $U(-d,t)=U(d,t)=0$; and the initial condition is $$U(x,t_{0})=\frac{1}{\sqrt{Dx}} exp \left(i P_0 \frac{x}{h} \right)$$ if $-d<x<-d+Dx$ and $U(x,t_{0})=0$ if $-d+Dx<x<d$; where $Dx<<L$ and $P_0$ is the quantum moment at t0. Also I know that at time $t_0$, the Fourier Transform of $U(x,t_{0})$ is a sinc centered in $P_0$, the aspectated value of position is $-d+Dx/2$ and the aspectated value of velocity is $P_0/m$, where $m$ is the mass of the particle. Then I have to compare analytical results with results from FINITE ELEMENTS and FINITE DIFFERENCE method. I hope that someone can help me to solve this problem. - 3 Welcome luca82 to physics.SE! At the moment there is not really a question in your text. Maybe you could elaborate a bit what you already tried and where you got stuck. Nobody here will solve homework/assignment problems but many people are more than willing to help. – Alexander Feb 21 '12 at 20:35 4 This appears to be copied largely from your assignment, which brings us to some questions for you. What have you done? What do you understand here, and what do you not understand? Where exactly are you stuck? In particular, I see at least three ways of attempting to solve the problem listed---surely you have been able to make some progress with some of them. – dmckee♦ Feb 21 '12 at 21:13 try \hbar: $\hbar$ – Emilio Pisanty Apr 22 '12 at 8:04 You should be careful: the solutions will be very different depending on whether $P_0^2$ is greater than $qV_0$ or not. – Emilio Pisanty Apr 22 '12 at 8:08 ## 1 Answer Without complete solution, just a road map. In principle, there is a standard way how this kind of problems is solved. First solve stationary problem: $$\left[ \left(-h^2 \frac{d^2}{dx^2} \right) + q V(x) \right] U_i(x) = E_i U(x).$$ As long as the potential is symmetric, I would recommend use this and search for even and odd solutions. For $E<V_0$ $$U_{i+}(x) = \begin{cases} A\cos{k(\|x\|-x_0)}, & \mbox{if } L<\|x\|<d \\ B\mathrm{ch}{\varkappa x}, & \mbox{if } -L<x<L \\ \end{cases},$$ $$U_{i-}(x) = \begin{cases} A\sin{k(\|x\|-x_0)}, & \mbox{if } L<\|x\|<d \\ B\mathrm{sh}{\varkappa x}, & \mbox{if } -L<x<L \\ \end{cases}$$ with $k=\sqrt{\frac{2mE}{\hbar^2}}$, $\varkappa = \sqrt{\frac{2m(V_0-E)}{\hbar^2}}$ For energies above $V_0$ replace hyperbolic functions with $\cos{k_2x}$, $\sin{k_2x}$ with $k=\sqrt{\frac{2m(E-V_0)}{\hbar^2}}$. If you substitute these functions into equation, you'll get equation for energy. Note that as long as this form of functions automatically satisfies equation in the regions of constant potential, you should just take care of boundary conditions. Solve this equation and get a spectrum $\left\{ E_{i\pm} \right\}_{i=1}^{\infty}$ and wavefunctions $U_i(x)$. Obviously, these solutions have very nice feature when you consider non-stationary problem: $$U_i(x,t)=e^{-i\frac{\hbar}{E_i} t}U_i(x),$$ which means that if you decompose your initial condition into these solutions (which you may do because eigenfunctions of a Hermitian operator is a complete set) $$U(x,t_0) = \sum_1^{\infty} C_i U_i(x)$$ then time evolution of wavefunction is given by $$U(x,t) = \sum_1^{\infty} C_i U_i(x) e^{-i\frac{\hbar}{E_i} t}$$ - If one needs an analytic solution to the problem, won't the lack of elemental expressions for the energies (which will be roots of a transcendental equation, I should think) severely harm this approach? Without any hope of evaluating the sum, this is arguably a semi-numerical, spectral method to solve the initial-value problem – Emilio Pisanty Apr 22 '12 at 8:10 @episanty No, it will not. There is an enormous difference between solving stranscendent equation and finding solutions of differential equation numerically. The first is called analytic, the second is not. Anyway, there is no better solution. – Misha Apr 22 '12 at 18:40 Yes, it's clear that there isn't a better solution, and that the finite-element and finite-difference methods are on a different scale altogether. My point was that I'm reluctant to call this a fully analytic solution since any attempt at visualising it (and comparing with any other method) will involve numerical work (on this side). – Emilio Pisanty Apr 22 '12 at 22:19 Sure, one can write down an explicit (but not elemental!) formula for the solution in a way no existence theorem can give. However, I see strong parallels between this solution and numerical spectral methods - this one simply has a particularly well chosen spectral basis set, and propagating any basis function is equivalent (at least in any reasonable numerical sense) to solving the transcendental equation. – Emilio Pisanty Apr 22 '12 at 22:19 1 @episanty, there is some misconception in your point of view. You could bring the same arguments if the solution was logriphm or bessel functions or any other non-trivial function: you should had use a computer to get the numbers in that case too. Just call the solution of that transcendent equation a new function, say, $Q_n(e)$ and it will not be much differnent from Bessel or Hankel functions which are also hard to find in calculator. – Misha Apr 23 '12 at 7:17 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9329975247383118, "perplexity_flag": "head"}
http://cstheory.stackexchange.com/questions/2484/are-oracles-associative/2494	# Are Oracles Associative? This question may have an obvious answer ... but here's the question anyway. Intuitively, it is the following plausible statement - "a machine with a subroutine A which in turn has a subroutine B is the same as a machine with a subroutine A which has access to subroutine B". To define this problem formally, I'll use some unconventional notation. When I say $A^B$, I am giving $A$ an oracle for a $B-Complete$ problem. e.g $NP^{NP}=NP^{SAT}=\Sigma_2$. With this "new" notation, it's possible to define $A^{B^C}$, and so on. My question is, is • is this a valid way of thinking about oracles? • is $(A^B)^C = A^{(B^C)}$ For example, $(NP^{NP})^{NP} = \Sigma_2^{NP} = NP^{\Sigma_2}=NP^{({NP}^{NP})}$ I can't think of any obvious counterexamples to this rule. Anyone? - – Sadeq Dousti Oct 26 '10 at 17:30 1 this is a slightly more general question, but Sadeq's question is quite relevant, and especially the comments about the ill-formed-ness of A^B^C if A^B is not a computational model – Suresh Venkat♦ Oct 26 '10 at 17:43 no, but that was the exact thing I was hitting my head on the wall last night about which motivated this question! – gabgoh Oct 26 '10 at 17:49 – Kaveh♦ Oct 27 '10 at 1:41 ## 2 Answers As Venkat told in comments, it seems hard to understand your definition for oracle which aren't defined as some machine characterisation. Let $A^{(B^C)}$ would be the set of TM in $A$ with an oracle which is a machine in ($B$ with an oracle in a machine in $C$). It is obvious that a machine in $A$ can call $C$: it just calls the machine in $B$ and asks it to carry the message directly to $C$. I guess ${(A^B)}^C$ would be a machine in $A$ that can call an oracle in $C$ or an oracle which is (a machine in $B$ that can call a machine in $C$) so it's exactly the same definition. Finally, you may want $A^{B,C}$, which is certainly different from the other two (just take $B=C=NP$ and $A=P$, then $A^{B,C}=NP\cup coNP$ whereas $A^{(B^C)}=\Sigma_2^P\cup\Pi_2^p$. - 3 – Tsuyoshi Ito Oct 26 '10 at 23:00 @Tsuyoshi, thank you for the remark, I don't know why I thought this. In fact if $NP\not=coNP$ it is clear that $P^{NP}$. Let $A$ and $B$ be NPcomplte and coNPcomplete problems, then the problem which take input $(x,y)$ and answer true if $x\in A$ and $y\in B$ is in $P^{NP}$ but not in $NP\cup coNP$. – Arthur MILCHIOR Oct 28 '10 at 15:26 I would write the following as a comment, but it was too long to fit in. Let's first describe the meaning of “algorithms in class $\bf C$ with an oracle for a language A.” (The need for this was pointed out by Tsuyoshi Ito). We will use the same convention used by Ladner and Lynch. The convention is well-described by Bennett & Gill: $\bf{LOGSPACE}^A$ can be defined in various ways, depending on how the query tape is handled. We follow the conventions of Ladner and Lynch [LL]: The query tape is not charged against the space bound, but to keep it from being used as a work tape, the query tape is one-way and write-only, and is erased automatically following each query. (Simon [Si] treats the query tape as one of the work tapes, a two-way read/write tape that is charged against the space bound. The Ladner-Lynch definition is less restrictive and perhaps more natural, since for a random oracle $A \in \bf{LOGSPACE}^A$ holds with probability 1 for [LL] but not for [Si]) [LL] R. E. LADNER AND N. A. LYNCH, Relativization of questions about log space computability, Math. Systems Theory, 10 (1976), pp. 19-32. [Si] J. SIMON, On Some Central Problems in Computational Complexity, Tech. Rep TR 75-224, Dept. of Computer Science, Cornell University, Ithaca, NY, 1975. The standard definition of complexity classes of oracle machines is as follows: Let B and C be complexity classes. Then, $X = B^C$ is a legitimate complexity class, defined as $X = \bigcup_{L\in C}{B^L}$. Here, $B^L$ represents the complexity class of decision problems solvable by an algorithm in class B with an oracle for a language L. Since X is a legitimate complexity class, for any complexity class A, we can speak of complexity classes $A^X = A^{(B^C)}$ and $X^A = (B^C)^A$. • $A^X$ refers to the complexity class of decision problems solvable by an algorithm in class A with an oracle for any language $L' \in X = \bigcup_{L\in C}{B^L}$. In other words, $A ^ X = \bigcup\limits_{L'\in \{\bigcup\limits_{L\in C}{B^L}\}}{A^{L'}}$. • $X^A$ refers to the complexity class of decision problems solvable by an algorithm in class $X = \bigcup_{L\in C}{B^L}$ with an oracle for any language $L' \in A$. In other words, $X ^ A = \bigcup_{L' \in A} {{X^{L'}}} = \bigcup_{L' \in A} {{{\left( {\bigcup_{L \in C} {{B^L}} } \right)}^{L'}}}$. Claim: $(B^{L_1})^{L'} \cup (B^{L_2})^{L'} = (B^{L'})^{L_1 \cup L_2}$. `Side Note: Since it's 3:00 AM now, I'm too sleepy to check the validity of the above claim! I think it's valid & elementary to prove, yet it's nice to see the actual proof.` Therefore, one can write $X ^ A = \bigcup_{L' \in A} {{{\left( {\bigcup_{L \in C} {{B^L}} } \right)}^{L'}}} = \bigcup_{L \in C, L' \in A} {{{\left( B^{L'} \right)}^L}}$. ### Example Let $\bf X = \bf{P^{NP}}$. We know that $\bf{coNP} \subseteq \bf X$. Giving both sides oracle access to $\bf {NP}$, one gets: $\bf{coNP^{NP}} \subseteq \bf{X^{NP}} = (\bf{P^{NP}})^{\bf{NP}}$. ## Epilogue A fruitful discussion with Tsuyoshi Ito revealed (for me) that it is not easy to doubly relativize a complexity class. In fact, even defining one seems to be problematic. I should definitely study more to see if any satisfactory definition is ever given. Meanwhile, I appreciate any comment which can be used to solve this problem. - 3 – Tsuyoshi Ito Oct 27 '10 at 0:11 @Tsuyoshi: I edited the answer to represent which definition I'm using. Does it remove the ill-formed-ness? – Sadeq Dousti Oct 27 '10 at 5:24 No. The added part only defines what LOGSPACE^A means, not what B^A means for something like B=NP^NP. – Tsuyoshi Ito Oct 27 '10 at 5:31 @Tsuyoshi: Thanks. I just added an example to clarify what I mean. I want a definition such that if $A \subseteq X$, then $A^C \subseteq X^C$ (A very natural requirement). Can you help me figure out how this must be defined when X is an oracle class, like NP^NP? – Sadeq Dousti Oct 27 '10 at 5:37 3 – Tsuyoshi Ito Oct 27 '10 at 5:57 show 1 more comment	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 57, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9404066801071167, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/5180/messages-of-different-lengths-and-one-time-computationally-secret/5229	# Messages of different lengths and one-time computationally-secret I'm preparing myself to exam, but I have a lot of troubles with rigorous proofs. Let $\Pi=(Gen,Enc,Dec)$ be an efficient secret-key encryption scheme that is not fixed-length. That is, for any $n$ and any $k \leftarrow Gen(1^n)$ the encryption algorithm $Enc_k(\cdot)$ can encrypt arbitrary length messages $m \in \lbrace 0, 1 \rbrace^$. Prove that $\Pi$ cannot satisfy definition of being one-time computationally-secret when the adversary $\mathcal{A}$ in $PrivK^{eav}_{\mathcal{A}}$ may output messages $m_0$ and $m_1$, that are NOT of the same length. I just can (I think so) start this proof - if $Enc_k(\cdot)$ is some PPT algorithm, then there exists a polynomial $p(x) \in \mathbb{Z}[x]$ such that $\forall n,k \leftarrow Gen(1^n), m \in \lbrace 0, 1 \rbrace^: \|\|Enc_k(m)\|\|<p(\|\|k\|\|+\|\|m\|\|)$. Unfortunately I have no idea how to do the rest of the proof - by contradiction, or not? If you could help me with the rigorous proof, I'd be really grateful for your time. P.S. Reminder (one-time computationally-secret). An (efficient secret-key) encryption scheme $(Gen,Enc,Dec)$ is one-time computationally-secret if for any PPT adversary $\mathcal{A}$ it holds that $Pr[PrivK^{eav}_{\mathcal{A}}(n)=1]-\frac{1}{2}$ is negligible function, where $PrivK^{eav}_{\mathcal{A}}(n)$ denotes the output of the following experiment: (a) The adversary $\mathcal{A}$ on input $1^n$ outputs a pair of messages $m_0,m_1$. (b) Let $k \leftarrow Gen(1^n)$ and let $b \in \lbrace 0,1 \rbrace$ be chosen uniformly at random. Then a ciphertext $c \leftarrow Enc_k(m_b)$ is computed and given to $\mathcal{A}$. (c) $\mathcal{A}$ on input $c$ outputs a bit $b'$. (d) The output of the experiment is $1$ if $b'=b$ and $0$ otherwise. - Hint: The adversary needs to construct two messages such that their ciphertexts are of different length. Use the efficiency condition to show that this is always possible. – Paŭlo Ebermann♦ Oct 28 '12 at 20:11 If this is not enough, have a look at our recent questions Why is a non fixed-length encryption scheme worse than a fixed-length one? and How to construct a variable length IND-CPA cipher from a fixed length one? (including the discussion in the comments). – Paŭlo Ebermann♦ Oct 28 '12 at 20:24 What is efficiency condition? Could you tell me a little bit more how should the formal proof be looking like? I think I know what is the idea, but I have big problem with writing it down rigorously. Thank you. – BiggBen1989 Oct 30 '12 at 0:01 Look up the meaning of the word "efficient" in your definition. (It means something like "a short input will only take a short time", and as such it can also only have a short output. But look it up in your literature.) – Paŭlo Ebermann♦ Nov 1 '12 at 13:07 ## 1 Answer This is already explained well, at a conceptual level, at a different question: Why is a non fixed-length encryption scheme worse than a fixed-length one? To turn it into a formal proof, write down an adversary $A$ that breaks the security of $\Pi$. You may want to first work out the attack at a conceptual level to make sure you are clear on how to attack $\Pi$; then review the definition of what it means for an adversary to break $\Pi$; then turn your conceptual attack into a fully specified algorithm $A$, and check that it does indeed break $\Pi$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 36, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9139277338981628, "perplexity_flag": "head"}
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Because by the time gravity has got from the dimension in ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9306890964508057, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/125517-radius-convergence-power-series.html	# Thread: 1. ## Radius of Convergence of Power Series Hello, I need a little bit help with the following power series. The exercise is to calculate the radius of convergence. $\sum\limits_{n=1}^{\infty} (\mathrm{ln}(n))^{n} x^{n}$ I know the criterion of Euler and the stronger criterion of Hadamard. Euler: $r= \frac{\mid a_{n} \mid}{ \mid a_{n+1} \mid }$ Hadamard: $r= \frac{1}{\mathrm{limsup} ~ \sqrt[n]{\mid a_{n} \mid} }$ and of course $a_{n}$ is $a_{n} = (\mathrm{ln(n)})^{n}$ My problem now, is that I don't know how to calculate this expressions with this logarithm Thanks for help 2. Originally Posted by Besserwisser Hello, I need a little bit help with the following power series. The exercise is to calculate the radius of convergence. $\sum\limits_{n=1}^{\infty} (\mathrm{ln})^{n} x^{n}$ I know the criterion of Euler and the stronger criterion of Hadamard. Euler: $r= \frac{\mid a_{n} \mid}{ \mid a_{n+1} \mid }$ Hadamard: $r= \frac{1}{\mathrm{limsup} ~ \sqrt[n]{\mid a_{n} \mid} }$ and of course $a_{n}$ is $a_{n} = \mathrm{ln}^{n}$ My problem now, is that I don't know how to calculate this expressions with this logarithm Thanks for help Is it $\sum_{n = 1}^{\infty}(\ln{x})^nx^n$? 3. Originally Posted by Prove It Is it $\sum_{n = 1}^{\infty}(\ln{x})^nx^n$? I made typing errors. Sorry. The Power Seris is $\sum_{n = 1}^{\infty}(\mathrm{ln}(n))^{n} x^{n}$ and $a_{n} = (\mathrm{ln}(n))^{n}$ 4. What is the neccesary condition for the convergence of an 'infinite sum'?... Kind regards $\chi$ $\sigma$ 5. Originally Posted by Prove It Is it $\sum_{n = 1}^{\infty}(\ln{x})^nx^n$? But that would not be a power series. 6. Originally Posted by Besserwisser Hello, I need a little bit help with the following power series. The exercise is to calculate the radius of convergence. $\sum\limits_{n=1}^{\infty} (\mathrm{ln}(n))^{n} x^{n}$ I know the criterion of Euler and the stronger criterion of Hadamard. Euler: $r= \frac{\mid a_{n} \mid}{ \mid a_{n+1} \mid }$ Hadamard: $r= \frac{1}{\mathrm{limsup} ~ \sqrt[n]{\mid a_{n} \mid} }$ and of course $a_{n}$ is $a_{n} = (\mathrm{ln(n)})^{n}$ My problem now, is that I don't know how to calculate this expressions with this logarithm Thanks for help For it to converge you clearly need $\ln^n(n)\cdot x^n=\left(x\ln(n)\right)^n\to0$. For this to happen you need only prove that $x\ln(n)$ goes to zero for some $x$..but...there is a problem with that.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 25, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9018070697784424, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/81890?sort=oldest	## How is this observation related to Koszul duality? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $X$ be a smooth variety, $\mathcal D$ the sheaf of algebraic differentail operators, $\Omega$ the algebraic deRham complex and $\mathcal M$ a quasi coherent $\mathcal O_X$-module. Now there is a bijection between $\mathcal D$-module structures on $\mathcal M$ and $\Omega$ dg-module structures on $\Omega \otimes_{\mathcal O_X} \mathcal M$. For example given a $\mathcal D$-module structure on $\mathcal M$ we can define the corresponding differential by imposing the rule $d(m)(X)=X.m$ on $\Omega\otimes_{\mathcal O_X} \mathcal M$. (X denotes a vectorfield and $m$ a local section of $\mathcal M$. A similar statement holds if one takes instead $\mathcal D=U(\mathfrak g)$ the universal envelope of a Lie-algebra and $\Omega=\bigwedge \mathfrak g^*$ the standard complex. Now in both of there cases $\mathcal D$ and $\Omega$ are Koszul-dual, meaning that there are equivalences between carefully defined versions of their "derived categories". Yet formulations of Koszul duality I am aware of, seem not to extend the above correspondence. Roughly speaking they are $\mathcal M \mapsto Hom(\Omega, \mathcal M)$ instead of $\mathcal M \mapsto \Omega \otimes \mathcal M$. So my questions are: What does the above observation have to do with Kosul duality? Is there a formulation of Koszul duality extending the above correspondence? - 1 Is it necessary to call the bijection "easy to see"? – S. Carnahan♦ Nov 25 2011 at 14:50 Maybe not, sorry. I automatically think of a $D$ modules as an integrable connections. And I internalized that a connection is integrable iff the map $M\rightarrow \omega^1 \otimes M$ can be extended to a differential. So with these two things in mind it is "easy to see", but I agree that it is not so obvious from other viewpoints. – Jan Weidner Nov 25 2011 at 16:30 ## 2 Answers As far as I remember this is explained (maybe in differet terms) in Kapranov's paper "On DG-modules over the de rham complex and the vanishing cycles functor", Lecture Notes in Mathematics, 1991, Volume 1479. You could also have a look at the discussion below this entry of the everything seminar. I think the kind of statement you are expecting ("equivalences between carefully defined versions of their "derived categories") are of a type you can find in this paper. In any way, the equivalence is given by a $\otimes$-Hom adjunction. I might have misunderstood something but I don't really understand where is the problem with having $Hom(\Omega,-)$ and $\Omega\otimes -$. EDIT: more precisely, I have the feeling that the statement is the following. If $A$ is Koszul and if $B$ is its Koszul dual then $Hom(A,-)$ and $A\otimes-$ defines an equivalence of DG categories between • $B$-modules • the category $\mathcal P_A$ as it is defined e.g. in Block's paper. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There are, basically, two answers to your question that I am aware of. The first one amounts to saying that there is no difference between tensoring with $\Omega$ and taking $Hom$ from $\Omega$. More precisely, the difference between these things corresponds to the difference between the left and right $\mathcal D$-modules: the former are supposed to be tensored with $\Omega$, while for the latter one takes $Hom$ from $\Omega$. Indeed, the $\mathcal O$-algebra $\Omega$ being Frobenius, tensoring with it only differs from taking the $Hom$ from it by the twist with the line bundle of top forms and the homological shift. This twist just transforms left $\mathcal D$-modules into right $\mathcal D$-modules. The three functors form a commutative triangle (up to the shift). Similarly, if your Lie algebra $\mathfrak g$ is finite-dimensional, tensoring $\mathfrak g$-modules with the cohomological or the homological standard complex of $\mathfrak g$ is almost the same functor, the twist with the one-dimensional $\mathfrak g$-module of top exterior forms on $\mathfrak g$ and the homological shift being the only difference between the two. The second answer purports to construct the kind of duality that you are looking for in the more general situations when $\Omega$ is no longer Frobenius. The first question that you can ask yourself in this case is, what would be the functor in the opposite direction, i.e., the adjoint functor to tensoring with $\Omega$ over $\mathcal O$? The answer is, it is the functor $Hom_{\mathcal O}(\mathcal D,{-})$. The latter functor looks somewhat problematic when the variety $X$ is not affine, as the internal $Hom$ from a noncoherent quasi-coherent sheaf may be not a well-behaved operation. Perhaps this problem can be dealt with, but at the moment I do not know how to do it. When $X$ is affine, however, the theory of derived $D$-$\Omega$ duality for the functors $\Omega\otimes_{\mathcal O}{-}$ and $Hom_{\mathcal O}(\mathcal D,{-})$ is developed in Appendix B to my AMS Memoir "Two kinds of derived categories, Koszul duality, and comodule-contramodule correspondence", http://arxiv.org/abs/0905.2621. In the case of a (possibly infinite-dimensional) Lie algebra $\mathfrak g$, this is done in Section 6.6 of the same paper. Basically, you want to look on the module-contramodule side of the commutative triangle of "Koszul triality", as presented in my paper. In particular, the appropriate category of $\Omega$-modules that you want to consider (when $\Omega$ is infinite-dimensional, e.g., $\Omega=\bigwedge\mathfrak g^\ast$ and $\mathfrak g$ is infinite-dimensional) is that of contramodules (modules with the infinite summation operations). And the appropriate version of the derived category of $\Omega$-modules is the contraderived category. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 66, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9254325032234192, "perplexity_flag": "head"}
http://theoreticalatlas.wordpress.com/2011/03/17/hgtqgr-part-iiia-workshop/	# Theoretical Atlas He had bought a large map representing the sea, / Without the least vestige of land: / And the crew were much pleased when they found it to be / A map they could all understand. March 17, 2011 ## HGTQGR – Part IIIa (Workshop) Posted by Jeffrey Morton under categorification, conferences, gauge theory, higher gauge theory, talks, tqft 1 Comment Now for a more sketchy bunch of summaries of some talks presented at the HGTQGR workshop. I’ll organize this into a few themes which appeared repeatedly and which roughly line up with the topics in the title: in this post, variations on TQFT, plus 2-group and higher forms of gauge theory; in the next post, gerbes and cohomology, plus talks on discrete models of quantum gravity and suchlike physics. ## TQFT and Variations I start here for no better reason than the personal one that it lets me put my talk first, so I’m on familiar ground to start with, for which reason also I’ll probably give more details here than later on. So: a TQFT is a linear representation of the category of cobordisms – that is, a (symmetric monoidal) functor $nCob \rightarrow Vect$, in the notation I mentioned in the first school post. An Extended TQFT is a higher functor $nCob_k \rightarrow k-Vect$, representing a category of cobordisms with corners into a higher category of k-Vector spaces (for some definition of same). The essential point of my talk is that there’s a universal construction that can be used to build one of these at $k=2$, which relies on some way of representing $nCob_2$ into $Span(Gpd)$, whose objects are groupoids, and whose morphisms in $Hom(A,B)$ are pairs of groupoid homomorphisms $A \leftarrow X \rightarrow B$. The 2-morphisms have an analogous structure. The point is that there’s a 2-functor $\Lambda : Span(Gpd) \rightarrow 2Vect$ which is takes representations of groupoids, at the level of objects; for morphisms, there is a “pull-push” operation that just uses the restricted and induced representation functors to move a representation across a span; the non-trivial (but still universal) bit is the 2-morphism map, which uses the fact that the restriction and induction functors are bi-ajdoint, so there are units and counits to use. A construction using gauge theory gives groupoids of connections and gauge transformations for each manifold or cobordism. This recovers a form of the Dijkgraaf-Witten model. In principle, though, any way of getting a groupoid (really, a stack) associated to a space functorially will give an ETQFT this way. I finished up by suggesting what would need to be done to extend this up to higher codimension. To go to codimension 3, one would assign an object (codimension-3 manifold) a 3-vector space which is a representation 2-category of 2-groupoids of connections valued in 2-groups, and so on. There are some theorems about representations of n-groupoids which would need to be proved to make this work. The fact that different constructions can give groupoids for spaces was used by the next speaker, Thomas Nicklaus, whose talk described another construction that uses the $\Lambda$ I mentioned above. This one produces “Equivariant Dijkgraaf-Witten Theory”. The point is that one gets groupoids for spaces in a new way. Before, we had, for a space $M$ a groupoid $\mathcal{A}_G(M)$ whose objects are $G$-connections (or, put another way, bundles-with-connection) and whose morphisms are gauge transformations. Now we suppose that there’s some group $J$ which acts weakly (i.e. an action defined up to isomorphism) on $\mathcal{A}_G(M)$. We think of this as describing “twisted bundles” over $M$. This is described by a quotient stack $\mathcal{A}_G // J$ (which, as a groupoid, gets some extra isomorphisms showing where two objects are related by the $J$-action). So this gives a new map $nCob \rightarrow Span(Gpd)$, and applying $\Lambda$ gives a TQFT. The generating objects for the resulting 2-vector space are “twisted sectors” of the equivariant DW model. There was some more to the talk, including a description of how the DW model can be further mutated using a cocycle in the group cohomology of $G$, but I’ll let you look at the slides for that. Next up was Jamie Vicary, who was talking about “(1,2,3)-TQFT”, which is another term for what I called “Extended” TQFT above, but specifying that the objects are 1-manifolds, the morphisms 2-manifolds, and the 2-morphisms are 3-manifolds. He was talking about a theorem that identifies oriented TQFT’s of this sort with “anomaly-free modular tensor categories” – which is widely believed, but in fact harder than commonly thought. It’s easy enough that such a TQFT $Z$ corresponds to a MTC – it’s the category $Z(S^1)$ assigned to the circle. What’s harder is showing that the TQFT’s are equivalent functors iff the categories are equivalent. This boils down, historically, to the difficulty of showing the category is rigid. Jamie was talking about a project with Bruce Bartlett and Chris Schommer-Pries, whose presentation of the cobordism category (described in the school post) was the basis of their proof. Part of it amounts to giving a description of the TQFT in terms of certain string diagrams. Jamie kindly credited me with describing this point of view to him: that the codimension-2 manifolds in a TQFT can be thought of as “boundaries in space” – codimension-1 manifolds are either time-evolving boundaries, or else slices of space in which the boundaries live; top-dimension cobordisms are then time-evolving slices of space-with-boundary. (This should be only a heuristic way of thinking – certainly a generic TQFT has no literal notion of “time-evolution”, though in that (2+1) quantum gravity can be seen as a TQFT, there’s at least one case where this picture could be taken literally.) Then part of their proof involves showing that the cobordisms can be characterized by taking vector spaces on the source and target manifolds spanned by the generating objects, and finding the functors assigned to cobordisms in terms of sums over all “string diagrams” (particle worldlines, if you like) bounded by the evolving boundaries. Jamie described this as a “topological path integral”. Then one has to describe the string diagram calculus – ridigidy follows from the “yanking” rule, for instance, and this follows from Morse theory as in Chris’ presentation of the cobordism category. There was a little more discussion about what the various properties (proved in a similar way) imply. One is “cloaking” – the fact that a 2-morphism which “creates a handle” is invisible to the string diagrams in the sense that it introduces a sum over all diagrams with a string “looped” around the new handle, but this sum gives a result that’s equal to the original map (in any “pivotal” tensor category, as here). Chronologically before all these, one of the first talks on such a topic was by Rafael Diaz, on Homological Quantum Field Theory, or HLQFT for short, which is a rather different sort of construction. Remember that Homotopy QFT, as described in my summary of Tim Porter’s school sessions, is about linear representations of what I’ll for now call $Cob(d,B)$, whose morphisms are $d$-dimensional cobordisms equipped with maps into a space $B$ up to homotopy. HLQFT instead considers cobordisms equipped with maps taken up to homology. Specifically, there’s some space $M$, say a manifold, with some distinguished submanifolds (possibly boundary components; possibly just embedded submanifolds; possibly even all of $M$ for a degenerate case). Then we define $Cob_d^M$ to have objects which are $(d-1)$-manifolds equipped with maps into $M$ which land on the distinguished submanifolds (to make composition work nicely, we in fact assume they map to a single point). Morphisms in $Cob_d^M$ are trickier, and look like $(N,\alpha, \xi)$: a cobordism $N$ in this category is likewise equipped with a map $\alpha$ from its boundary into $M$ which recovers the maps on its objects. That $\xi$ is a homology class of maps from $N$ to $M$, which agrees with $\alpha$. This forms a monoidal category as with standard cobordisms. Then HLQFT is about representations of this category. One simple case Rafael described is the dimension-1 case, where objects are (ordered sets of) points equipped with maps that pick out chosen submanifolds of $M$, and morphisms are just braids equipped with homology classes of “paths” joining up the source and target submanifolds. Then a representation might, e.g., describe how to evolve a homology class on the starting manifold to one on the target by transporting along such a path-up-to-homology. In higher dimensions, the evolution is naturally more complicated. A slightly looser fit to this section is the talk by Thomas Krajewski, “Quasi-Quantum Groups from Strings” (see this) – he was talking about how certain algebraic structures arise from “string worldsheets”, which are another way to describe cobordisms. This does somewhat resemble the way an algebraic structure (Frobenius algebra) is related to a 2D TQFT, but here the string worldsheets are interacting with 3-form field, $H$ (the curvature of that 2-form field $B$ of string theory) and things needn’t be topological, so the result is somewhat different. Part of the point is that quantizing such a thing gives a higher version of what happens for quantizing a moving particle in a gauge field. In the particle case, one comes up with a line bundle (of which sections form the Hilbert space) and in the string case one comes up with a gerbe; for the particle, this involves associated 2-cocycle, and for the string a 3-cocycle; for the particle, one ends up producing a twisted group algebra, and for the string, this is where one gets a “quasi-quantum group”. The algebraic structures, as in the TQFT situation, come from, for instance, the “pants” cobordism which gives a multiplication and a comultiplication (by giving maps $H \otimes H \rightarrow H$ or the reverse, where $H$ is the object assigned to a circle). There is some machinery along the way which I won’t describe in detail, except that it involves a tricomplex of forms – the gradings being form degree, the degree of a cocycle for group cohomology, and the number of overlaps. As observed before, gerbes and their higher versions have transition functions on higher numbers of overlapping local neighborhoods than mere bundles. (See the paper above for more) ## Higher Gauge Theory The talks I’ll summarize here touch on various aspects of higher-categorical connections or 2-groups (though at least one I’ll put off until later). The division between this and the section on gerbes is a little arbitrary, since of course they’re deeply connected, but I’m making some judgements about emphasis or P.O.V. here. Apart from giving lectures in the school sessions, John Huerta also spoke on “Higher Supergroups for String Theory”, which brings “super” (i.e. $\mathbb{Z}_2$-graded) objects into higher gauge theory. There are “super” versions of vector spaces and manifolds, which decompose into “even” and “odd” graded parts (a.k.a. “bosonic” and “fermionic” parts). Thus there are “super” variants of Lie algebras and Lie groups, which are like the usual versions, except commutation properties have to take signs into account (e.g. a Lie superalgebra’s bracket is commutative if the product of the grades of two vectors is odd, anticommutative if it’s even). Then there are Lie 2-algebras and 2-groups as well – categories internal to this setting. The initial question has to do with whether one can integrate some Lie 2-algebra structures to Lie 2-group structures on a spacetime, which depends on the existence of some globally smooth cocycles. The point is that when spacetime is of certain special dimensions, this can work, namely dimensions 3, 4, 6, and 10. These are all 2 more than the real dimensions of the four real division algebras, $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$ and $\mathbb{O}$. It’s in these dimensions that Lie 2-superalgebras can be integrated to Lie 2-supergroups. The essential reason is that a certain cocycle condition will hold because of the properties of a form on the Clifford algebras that are associated to the division algebras. (John has some related material here and here, though not about the 2-group case.) Since we’re talking about higher versions of Lie groups/algebras, an important bunch of concepts to categorify are those in representation theory. Derek Wise spoke on “2-Group Representations and Geometry”, based on work with Baez, Baratin and Freidel, most fully developed here, but summarized here. The point is to describe the representation theory of Lie 2-groups, in particular geometrically. They’re to be represented on (in general, infinite-dimensional) 2-vector spaces of some sort, which is chosen to be a category of measurable fields of Hilbert spaces on some measure space, which is called $H^X$ (intended to resemble, but not exactly be the same as, $Hilb^X$, the space of “functors into $Hilb$ from the space $X$, the way Kapranov-Voevodsky 2-vector spaces can be described as $Vect^k$). The first work on this was by Crane and Sheppeard, and also Yetter. One point is that for 2-groups, we have not only representations and intertwiners between them, but 2-intertwiners between these. One can describe these geometrically – part of which is a choice of that measure space $(X,\mu)$. This done, we can say that a representation of a 2-group is a 2-functor $\mathcal{G} \rightarrow H^X$, where $\mathcal{G}$ is seen as a one-object 2-category. Thinking about this geometrically, if we concretely describe $\mathcal{G}$ by the crossed module $(G,H,\rhd,\partial)$, defines an action of $G$ on $X$, and a map $X \rightarrow H^$ into the character group, which thereby becomes a $G$-equivariant bundle. One consequence of this description is that it becomes possible to distinguish not only irreducible representations (bundles over a single orbit) and indecomposible ones (where the fibres are particularly simple homogeneous spaces), but an intermediate notion called “irretractible” (though it’s not clear how much this provides). An intertwining operator between reps over $X$ and $Y$ can be described in terms of a bundle of Hilbert spaces – which is itself defined over the pullback of $X$ and $Y$ seen as $G$-bundles over $H^$. A 2-intertwiner is a fibre-wise map between two such things. This geometric picture specializes in various ways for particular examples of 2-groups. A physically interesting one, which Crane and Sheppeard, and expanded on in that paper of [BBFW] up above, deals with the Poincaré 2-group, and where irreducible representations live over mass-shells in Minkowski space (or rather, the dual of $H \cong \mathbb{R}^{3,1}$). Moving on from 2-group stuff, there were a few talks related to 3-groups and 3-groupoids. There are some new complexities that enter here, because while (weak) 2-categories are all (bi)equivalent to strict 2-categories (where things like associativity and the interchange law for composing 2-cells hold exactly), this isn’t true for 3-categories. The best strictification result is that any 3-category is (tri)equivalent to a Gray category – where all those properties hold exactly, except for the interchange law $(\alpha \circ \beta) \cdot (\alpha ' \circ \beta ') = (\alpha \cdot \alpha ') \circ (\beta \circ \beta ')$ for horizontal and vertical compositions of 2-cells, which is replaced by an “interchanger” isomorphism with some coherence properties. John Barrett gave an introduction to this idea and spoke about “Diagrams for Gray Categories”, describing how to represent morphisms, 2-morphisms, and 3-morphisms in terms of higher versions of “string” diagrams involving (piecewise linear) surfaces satisfying some properties. He also carefully explained how to reduce the dimensions in order to make them both clearer and easier to draw. Bjorn Gohla spoke on “Mapping Spaces for Gray Categories”, but since it was essentially a shorter version of a talk I’ve already posted about, I’ll leave that for now, except to point out that it linked to the talk by Joao Faria Martins, “3D Holonomy” (though see also this paper with Roger Picken). The point in Joao’s talk starts with the fact that we can describe holonomies for 3-connections on 3-bundles valued in Gray-groups (i.e. the maximally strict form of a general 3-group) in terms of Gray-functors $hol: \Pi_3(M) \rightarrow \mathcal{G}$. Here, $\Pi_3(M)$ is the fundamental 3-groupoid of $M$, which turns points, paths, homotopies of paths, and homotopies of homotopies into a Gray groupoid (modulo some technicalities about “thin” or “laminated” homotopies) and $\mathcal{G}$ is a gauge Gray-group. Just as a 2-group can be represented by a crossed module, a Gray (3-)group can be represented by a “2-crossed module” (yes, the level shift in the terminology is occasionally confusing). This is a chain of groups $L \stackrel{\delta}{\rightarrow} E \stackrel{\partial}{\rightarrow} G$, where $G$ acts on the other groups, together with some structure maps (for instance, the Peiffer commutator for a crossed module becomes a lifting $\{ ,\} : E \times E \rightarrow L$) which all fit together nicely. Then a tri-connection can be given locally by forms valued in the Lie algebras of these groups: $(\omega , m ,\theta)$ in $\Omega^1 (M,\mathfrak{g} ) \times \Omega^2 (M,\mathfrak{e}) \times \Omega^3(M,\mathfrak{l})$. Relating the global description in terms of $hol$ and local description in terms of $(\omega, m, \theta)$ is a matter of integrating forms over paths, surfaces, or 3-volumes that give the various $j$-morphisms of $\Pi_3(M)$. This sort of construction of parallel transport as functor has been developed in detail by Waldorf and Schreiber (viz. these slides, or the full paper), some time ago, which is why, thematically, they’re the next two speakers I’ll summarize. Konrad Waldorf spoke about “Abelian Gauge Theories on Loop Spaces and their Regression”. (For more, see two papers by Konrad on this) The point here is that there is a relation between two kinds of theories – string theory (with $B$-field) on a manifold $M$, and ordinary $U(1)$ gauge theory on its loop space $LM$. The relation between them goes by the name “regression” (passing from gauge theory on $LM$ to string theory on $M$), or “transgression”, going the other way. This amounts to showing an equivalence of categories between [principal $U(1)$-bundles with connection on $LM$] and [$U(1)$-gerbes with connection on $M$]. This nicely gives a way of seeing how gerbes “categorify” bundles, since passing to the loop space – whose points are maps $S^1 \rightarrow M$ means a holonomy functor is now looking at objects (points in $LM$) which would be morphisms in the fundamental groupoid of $M$, and morphisms which are paths of loops (surfaces in $M$ which trace out homotopies). So things are shifted by one level. Anyway, Konrad explained how this works in more detail, and how it should be interpreted as relating connections on loop space to the $B$-field in string theory. Urs Schreiber kicked the whole categorification program up a notch by talking about $\infty$-Connections and their Chern-Simons Functionals . So now we’re getting up into $\infty$-categories, and particularly $\infty$-toposes (see Jacob Lurie’s paper, or even book if so inclined to find out what these are), and in particular a “cohesive topos”, where derived geometry can be developed (Urs suggested people look here, where a bunch of background is collected). The point is that $\infty$-topoi are good for talking about homotopy theory. We want a setting which allows all that structure, but also allows us to do differential geometry and derived geometry. So there’s a “cohesive” $\infty$-topos called $Smooth\infty Gpds$, of “sheaves” (in the $\infty$-topos sense) of $\infty$-groupoids on smooth manifolds. This setting is the minimal common generalization of homotopy theory and differential geometry. This is about a higher analog of this setup: since there’s a smooth classifying space (in fact, a Lie groupoid) for $G$-bundles, $BG$, there’s also an equivalence between categories $G-Bund$ of $G$-principal bundles, and $SmoothGpd(X,BG)$ (of functors into $BG$). Moreover, there’s a similar setup with $BG_{conn}$ for bundles with connection. This can be described topologically, or there’s also a “differential refinement” to talk about the smooth situation. This equivalence lives within a category of (smooth) sheaves of groupoids. For higher gauge theory, we want a higher version as in $Smooth \infty Gpds$ described above. Then we should get an equivalence – in this cohesive topos – of $hom(X,B^n U(1))$ and a category of $U(1)$-$(n-1)$-gerbes. Then the part about the “Chern-Simons functionals” refers to the fact that CS theory for a manifold (which is a kind of TQFT) is built using an action functional that is found as an integral of the forms that describe some $U(1)$-connection over the manifold. (Then one does a path-integral of this functional over all connections to find partition functions etc.) So the idea is that for these higher $U(1)$-gerbes, whose classifying spaces we’ve just described, there should be corresponding functionals. This is why, as Urs remarked in wrapping up, this whole picture has an explicit presentation in terms of forms. Actually, in terms of Cech-cocycles (due to the fact we’re talking about gerbes), whose coefficients are taken in sheaves of complexes (this is the derived geometry part) of differential forms whose coefficients are in $L_\infty$-algebroids (the $\infty$-groupoid version of Lie algebras, since in general we’re talking about a theory with gauge $\infty$-groupoids now). Whew! Okay, that’s enough for this post. Next time, wrapping up blogging the workshop, finally. About these ads Like Loading... ### One Response to “HGTQGR – Part IIIa (Workshop)” 1. January 20, 2012 at 4:38 pm [...] Nickolaus spoke about the same kind of -equivariant Dijkgraaf-Witten models as at our workshop in Lisbon, so I’ll refer you back to my earlier post on that. However, speaking of [...] %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 123, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9223195910453796, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/70922/on-the-average-of-continuous-functions-f-mathbbr2-rightarrow0-1/70940	On the average of continuous functions $f:\mathbb{R}^2\rightarrow[0,1]$ Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is it true that if the average of a continuous function $f:\mathbb{R}^2\rightarrow[0,1]$ over a unit circle centered around $(x,y)$ is $f(x,y)$ for all $(x,y)\in\mathbb{R}^2$, then $f$ is necessarily constant? - Sorry, I mistyped my question. I meant f takes values in [0,1]! – Nimr Jul 21 2011 at 19:56 Note there exists function with this poperty. Holomorphic function are suitable if you allow complex valued functions and real harmonic functions in the case before. There are no real valued holomorphic function except constant functions, there are no bounded harmonic functions, so you're finished if you can show that f is indeed harmonic. – Marc Palm Jul 21 2011 at 20:06 With the restriction to [0,1], I can't see the link with harmonic functions anymore. – Nimr Jul 21 2011 at 20:09 No, i just suggest. If the integral formula holds, then it is sufficient to show that this implies that $f$ is harmonic. Then you can argue that since $f$ is bounded and harmonic, it must be constant. – Marc Palm Jul 21 2011 at 20:13 1 @George: although not solving the current problem, it is still quite interesting that $f(x,y)=e^{1.88044 x} \cos(6.947506 x)$ has the unit-circle averaging property, even though it is not harmonic! – Gerald Edgar Jul 22 2011 at 23:38 show 6 more comments 2 Answers Yes, any such $f$ is constant. In fact, if we relax the condition so that $f$ is only required to be bounded below, but not above, then it is still true that $f$ is constant. This can be proven by martingale theory, as can the statement that harmonic functions bounded below are constant (Liouville's theorem). Let $X_1,X_2,\ldots$ be a sequence of independent random random variables uniformly distributed on the unit circle, set $S_n=\sum_{m=1}^nX_m$ and let $\mathcal{F}_n$ be the sigma-algebra generated by $X_1,X_2,\ldots,X_n$. Then, $S_n$ is a random walk in the plane, and is recurrent. Your condition is equivalent to $\mathbb{E}[f(S_{n+1})\vert\mathcal{F}_n]=f(S_n)$. That is, $f(S_n)$ is a martingale. It is a standard result that a martingale which is bounded below converges to a limit, with probability one. However, as $S_n$ is recurrent, this only happens if $f$ is constant almost everywhere. By continuity of $f$, it must be constant everywhere. For the same argument applied to functions $f\colon\mathbb{Z}^2\to\mathbb{R}$, see Byron Schmuland's answer to this math.SE question. In general, for a continuous function $f\colon\mathbb{R}^2\to\mathbb{C}$, if $f(x,y)$ is the average of $f$ on the unit circle centered at $(x,y)$ then it does not follow that $f$ is harmonic. So, we cannot prove the result directly by applying Liouville's theorem. As an example (based on the comments by Gerald Edgar and by me), consider $f(x,y)=\exp(ax)$. The average of $f$ on the unit circle centered at $(x,y)$ is $$\frac{1}{2\pi}\int_0^{2\pi}f(x+\cos t,y+\sin t)\,dt=\frac{1}{2\pi}f(x,y)\int_0^{2\pi}e^{a\cos t}\,dt=f(x,y)I_0(a).$$ Here, $I_0(a)$ is the modified Bessel function of the first kind. Whenever $I_0(a)=1$ then $f$ satisfies the required property. This is true for $a=0$, in which case $f$ is constant, but there are also nonzero solutions such as $a\approx1.88044+6.94751i$. In that case $f$ satisfies the required property but is not harmonic. - You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. These functions are called harmonic functions. One the simplest examples is $f(x,y)=xy$. More generally, the real part of any holomorphic function $\mathbb C\to \mathbb C$ is a harmonic function $\mathbb C\to \mathbb R$. Added later: As mentioned by Gerald, harmonic functions are characterized by the property that $$\int_0^1f(z+re^{2\pi\theta})d\theta=f(z),\qquad \forall r\ge 0,\quad \forall z\in \mathbb C.$$ I don't know whether that property for $r=1$ implies that property for all $r\ge 0$. Partial answer to the edited question: If you require the function to be bounded, then I think that yes, that should force it to be constant. Liouville's theorem states that any bounded holomorphic function $\mathbb C\to \mathbb C$ is constant. There is also a version of Liouville's theorem for harmonic functions, so yes: the function is constant. Gap in the argument: ▹ why is the function harmonic? - He said unit circle. Harmonic functions have this property for all circles. Why do you say $xy$ is simplest? Shy not just $x$? – Gerald Edgar Jul 21 2011 at 19:26 I think the question was, if $f: \mathbb{C} \rightarrow \mathbb{R}$ with $\int_0^{1} f( z + e^{2 \pi i \theta}) d \theta = f(z)$ implies that $f$ is constant. I am not seeing this being answered here. Can you please elaborate your answer? – Marc Palm Jul 21 2011 at 19:49 @Gerald: You're right I missed the simplest one. @pm: It is not true that $\int_0^1f(z+e^{2\pi\theta})d\theta=f(z)$ $\forall z$ implies $f$ is constant. I have provided a counterexample. – André Henriques Jul 21 2011 at 19:58 Sorry, I mistyped my question. I meant f takes values in [0,1]! – Nimr Jul 21 2011 at 19:58 Dear André Henriques, I misread that you had implied that the formula implies that $f$ is indeed harmonic, but you only stated that the formula holds for harmonic functions and answered the question. Sorry for this. But I guess now the question is more interesting, since there are no bounded nonconstant harmonic functions=) – Marc Palm Jul 21 2011 at 20:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 53, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9223282337188721, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/110417/probability-theory-random-number-hit-from-a-pool	# Probability Theory - random number hit from a pool With the given data: After picking 30 random natural integers in a pool of X-natural-numbers (numbers do not disappear from the pool after picking). The probability of NOT picking 1 pre-defined specific number should be around a probability of 0,3%. This is what I'm trying to calculate: The size of the pool What I want to know: The formula - ## 2 Answers It is not entirely clear what the problem is, so I will make an interpretation that I hope is the one you intend. Out of habit, I will use $n$ instead of $X$. We have a pool of $n$ distinct natural numbers, or indeed any $n$ distinct objects. We suppose that $A$ is one of these objects. We pick from this pool $30$ times, each time replacing the object that we picked, so that the composition of the pool does not change. The probability that in $30$ trials, we never pick $A$, is $0.003$. We want to know the number $n$ of objects in the pool. The probability that, on any individual pick, we get the object $A$, is $\dfrac{1}{n}$. So the probability that we don't get $A$ is $1-\dfrac{1}{n}$. The probability that this happens $30$ times in a row is $$\left(1-\frac{1}{n}\right)^{30}.\qquad\qquad (\ast)$$ So we want to solve the equation $$\left(1-\frac{1}{n}\right)^{30}=0.003.$$ This equation can be solved in various ways, including "trial and error." In our particular situation, trial and error is a very good way. If we play with the calculator a bit, using the formula $(\ast)$, we find that if $n=5$, the probability of never getting $A$ is about $0.0012379$, while if $n=6$, the probability is about $0.0042127$. There is no integer $n$ such that the probability is exactly $0.003$. We get closest with $n=6$. We now describe a more systematic way of solving our equation. Take the logarithm of both sides. I will use logarithm to the base $10$, though I would prefer the natural logarithm (base $e$). We obtain $$30\log\left(1-\frac{1}{n}\right)=\log(0.003).$$ Calculate. We get $$\log\left(1-\frac{1}{n}\right)\approx -0.084096.$$ Recall that if $y=\log x$ then $x=10^y$. We conclude that $$\left(1-\frac{1}{n}\right)\approx 0.823956,$$ which gives $n=5.6803994$. Of course, that is not right, $n$ must be an integer. If we let $n=5$, the probability we never get $A$ is quite a bit less than $0.003$, while if $n=6$, the probability we never get $A$ is greater than $0.003$. Remark: You might be interested in numbers other than your special $30$ and $0.003$. More generally, suppose that we pick $k$ times, and we want the probability of never getting $A$ to be $p$. Then we need to solve the equation $$\left(1-\frac{1}{n}\right)^{k}=p.$$ Like in our concrete case, we can find the appropriate value of $n$ by using logarithms. In general, like in our concrete case, there will not be an integer $n$ that gives probability exactly $p$. Again, we use logarithms to the base $10$, though any base will do. We get $$k\log\left(1-\frac{1}{n}\right)=\log p,$$ and therefore $$\log\left(1-\frac{1}{n}\right)=\frac{\log p}{k},$$ and therefore $$1-\frac{1}{n}=10^{\frac{\log p}{k}}.$$ Solving for $n$, we obtain $$n=\frac{1}{1-10^{\frac{\log p}{k}}}.$$ Suppose that instead of your $0.003$, we let $p=0.95$. Let $k=30$. Using the above formula, we get $n\approx 585.4$. This could have taken some time to reach by trial and error. - thank you into positive infinity for your 2 answers! P.S.: my precious 0.3% -> 99.7% -> is Sigma 3 love – Terence Feb 17 '12 at 21:05 @Terence: Yes, the $0.3$% sounded implausible for a real problem, but you had a big NOT, so I calculated on the basis of $0.003$. With $0.997$ I get about $9985.5$. – André Nicolas Feb 17 '12 at 22:38 you were correct in everything, even about the 0.3%. But if you wondered where I got the 0.3% from: it's from 1 minus 3 times Sigma (normal distribution), I will be applying this in a real life software situation where I need this kind of certainty (~99,7%) – Terence Feb 17 '12 at 23:25 long story short ;) you interpreted my question correctly with the values – Terence Feb 17 '12 at 23:26 1 @Terence: One must be careful: nothing and nobody in the real world is truly normal. Joke aside, there is a serious point. You are using the normal distribution as a model, or as an approximation to the "real" distribution. In (say) the left tail, the normal approximation may be very good in that the true probability is close to $0$, and the normal gives an answer close to $0$. But the ratio of the tail probabilities may be quite far from $1$. Because of poor tail fit between truth and normal approximation in the ratio sense, the calculated $n$ might be quite far away from the truth. – André Nicolas Feb 17 '12 at 23:45 The chance of not picking the predefined integer in our thirty picks $= \left( \dfrac {x-1} {x}\right) ^{30} = 0.3$. Now solve for x. - isnt the 0.3 in your answer 30%? – Terence Feb 17 '12 at 18:56 well u have 0,3% in the question so i was n't sure what u meant and i assumed u meant 0.3. Solving with 0.3 i think the answer for x turns out to be 25. If you are interested i can post the calculation. – Hardy Feb 17 '12 at 19:04 30% or 0.3 is not a very unnatural number for this probability in the question. – Hardy Feb 17 '12 at 19:06 $\left( 1-\dfrac {1} {x}\right) ^{30}=0.3$ <=> $30\ln \left( 1-\dfrac {1} {x}\right) =\ln \left( 0.3\right)$ <=> $1-\dfrac {1} {x}=e^{\left( \dfrac {\ln _{0}.3} {30}\right) }$ <=> $x=\dfrac {1} {1-e^{\left( \dfrac {\ln _{0}.3} {30}\right) }}$ – Hardy Feb 17 '12 at 19:21 $x=\dfrac {1} {1-0.9606} = 25.420867390196038892702305462344$ – Hardy Feb 17 '12 at 19:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 64, "mathjax_display_tex": 10, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9309329986572266, "perplexity_flag": "head"}
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However, in my opinion, projective spaces seem intuitively less ... 3answers 46 views ### Fourier transform Could anyone explain to me how do we change Fourier transform equation from this [Wiki - look at the top of the page]: \mathcal{F}(x) = \int\limits_{-\infty}^{\infty} \mathcal{G}(k)\, e^{-2\pi i ... 1answer 48 views ### How do we distinguish “walks” or “paths”? For example, let $G(V,E)$ be a graph such that $V=\{v_1,v_2\}$ and $E=\{(v_1,v_2)\}$. And let $s_1:\{1,2\}\rightarrow V$ be a walk such that $s_1(1)=v_1$ and $s_1(2)=v_2$. And let ... 1answer 41 views ### What is this matrix called? Let $G=(V,E)$ be a finite graph where $V$ has $n$ elements so that $V=\{v_1,...,v_n\}$. Now, define $a_{ij}$ to be 1 if $(v_i,v_j)\in E$ and 0 otherwise. What is this $n\times n$ matrix $(a_{ij})$ ... 2answers 45 views ### Question about power of sets If two sets are finite and they have the same power, can we say that the two sets are equivalent? 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And help me show that this claim: A space $X$ is compact iff every infinite set in $X$ has a point of complete ... 3answers 73 views ### $(\Bbb R \to \Bbb R : x\mapsto x^2)\equiv(\Bbb R \to \Bbb{R}_{\geq 0} : x \mapsto x^2) \not\equiv (\Bbb C \to \Bbb C:x\mapsto x^2)$ Consider the following functions: $f:\Bbb R \to \Bbb R : x\mapsto x^2$ $g:\Bbb R \to \Bbb{R}_{\geq 0} : x \mapsto x^2$ $h:\Bbb C \to \Bbb C:x\mapsto x^2$ I'm quite sure that $h$ is not equal to $f$ ... 2answers 89 views ### Associativity with one operation or two (or more) operations It seems to me there are different 'types' of associative law that are all said to simply have the property of associativity. For example this term is applied if we are only considering one operator ... 1answer 69 views ### Metrics with infinite distances. I've been wondering about the spaces $\Bbb R\cup\{-\infty,+\infty\}$ and $\Bbb C\cup\{\infty\}.$ Is there a useful generalization of the definition of a metric they satisfy? I thought it would be ... 1answer 52 views ### Are bicategories and lax 2-categories the same? My question is that whether the definition of bicategories is the same as the definition of lax 2-categories. I heard that they are both week versions of 2-categories. Are they the same? If not, how ... 0answers 38 views ### Definition: Plotting a 2D surface I would very much appreciate a definition clarification. Suppose we have 2 functions of a variable $x$: $f(x)$ and $g(x)$ And we have that $$f'^2+g=c$$ where $c$ is a constant What does it mean to ... 2answers 40 views ### Understanding the definition of monotonically monolithic A collection $\mathcal{N}$ of subsets of $X$ is called an external network of $A \subset X$, when for every $x \in A$ and every neighbourhood $U$ of $x$, there exists some $N \in \mathcal{N}$ such ... 1answer 54 views ### “[T]ransversely isotropic and mirror-symmetric (space group:$D_{\infty h}$)”, its orbifold notation? I am trying to understand this frieze pattern $D_{\infty h}$ aka its orbifold notation. This describes spider's silk. The authors call it a space group, some sort of generalization from orbifolds. ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 109, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.932108998298645, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/199476-transferring-coefficients-equations.html	1Thanks • 1 Post By Prove It # Thread: 1. ## Transferring Coefficients in equations Hey all, I'm new here and I'm relearning the basic rules of algebra and I've come across a problem: (this is the simplified equation just for demonstration purposes, I've also used an image not knowing how to represent fractions here) 1st: I thought when transferring a value from one side to the other, you had to change it from negative to positive and vice-versa. Does this rule not apply because the -91 is a coefficient? 2nd: Is it a general rule that a negative over a negative can be put positive? I've tried both 168/91 AND -168/-91 and they both give 1.85 (rounded). Thanks for any help on this. I know this looks real simple but I've forgotten most algebra "rules". I've got so many more questions, but I guess this one is the most important for today without spamming your board. 2. ## Re: Transferring Coefficients in equations Originally Posted by neurohype Hey all, I'm new here and I'm relearning the basic rules of algebra and I've come across a problem: (this is the simplified equation just for demonstration purposes, I've also used an image not knowing how to represent fractions here) 1st: I thought when transferring a value from one side to the other, you had to change it from negative to positive and vice-versa. Does this rule not apply because the -91 is a coefficient? 2nd: Is it a general rule that a negative over a negative can be put positive? I've tried both 168/91 AND -168/-91 and they both give 1.85 (rounded). Thanks for any help on this. I know this looks real simple but I've forgotten most algebra "rules". I've got so many more questions, but I guess this one is the most important for today without spamming your board. 1. If you want to transform an equation you have to do exactly the same on both sides of the equation without changing the "equalness". Since you wanted to get 1x you used the the property that $\frac aa = 1, a \ne 0$. So you divided the LHS by (-91) and therefore you must divide the RHS by (-91) too. 2. If you combine two equal signs the result is positive. If you combine two unequal signs the result is negative. 3. Btw: Your result can be simplified to $x = \frac{168}{91}=\frac{7 \cdot 24}{7 \cdot 13}=\frac{24}{13}$ 3. ## Re: Transferring Coefficients in equations Originally Posted by earboth [quoting for notification in case not suscribed to thread] Thanks alot! This is very appreciated. It's great to know that the two equal signs giving a positive number is a "rule of thumb". About the division by -91 on the RHS because it was done on the left side, I guess I must have gotten that confused with transferring say constants from one side to the other. I'm still a little confused having just jumped back into math 12 years later (other than basic math problems I deal with when programming) so I don't have all the math terms correctly either yet. Just a last question, when you say = 1,a , what does the comma stand for? From what I can logically understand, you're listing 1 as the lowest common number because anything divided by itself is equal to 1, AND so therefore a/a could be simplified as a, is that it? I'm just confused about the comma, I just figure it stands to seperate list items. 4. ## Re: Transferring Coefficients in equations Originally Posted by neurohype Thanks alot! This is very appreciated. It's great to know that the two equal signs giving a positive number is a "rule of thumb". About the division by -91 on the RHS because it was done on the left side, I guess I must have gotten that confused with transferring say constants from one side to the other. I'm still a little confused having just jumped back into math 12 years later (other than basic math problems I deal with when programming) so I don't have all the math terms correctly either yet. Just a last question, when you say = 1,a , what does the comma stand for? From what I can logically understand, you're listing 1 as the lowest common number because anything divided by itself is equal to 1, AND so therefore a/a could be simplified as a, is that it? I'm just confused about the comma, I just figure it stands to seperate list items. The comma is not being used as a mathematical symbol, it's being used for its English purpose, to take a breath. He's saying that a/a = 1 provided that a is not 0. 5. ## Re: Transferring Coefficients in equations Thanks! #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9620020985603333, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/38680/can-an-algebraic-number-on-the-unit-circle-have-a-conjugate-with-absolute-value-d/38692	## Can an algebraic number on the unit circle have a conjugate with absolute value different from 1? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm fearful about putting this forward, because it seems the answer should be elementary. Certainly, the Weak Approximation Theorem allows every system of simultaneous inequalities among archimedean absolute values to be satisfied. But equality combined with inequality? - ## 8 Answers Yes. Take $$\alpha=\sqrt{2-\sqrt{2}}+i\sqrt{\sqrt{2}-1}.$$ Neither of the conjugates $$\sqrt{2+\sqrt{2}}\pm \sqrt{\sqrt{2}+1}$$ have absolute value 1. It is impossible, however, if $\mathbb{Q}(\alpha)/\mathbb{Q}$ is abelian, since then all automorphisms commute with complex conjugation. This was all stolen from Washington's Cyclotomic Fields book. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Yes. E.g. consider the two complex roots of $x^2-(1-\sqrt{2})x+1.$ They are algebraic integers of degree 4 on the unit circle, but their algebraic conjugates are the roots of $x^2-(1+\sqrt{2})x+1$, so they are irrational real numbers. - There was a similar question on math.stackexchange: http://math.stackexchange.com/questions/4323/are-all-algebraic-integers-with-absolute-value-1-roots-of-unity/4332#4332 and my answer below is partly taken from that. For any non-real algebraic number x inside C, its absolute value \|x\| is algebraic, so x/\|x\| is an algebraic number on the unit circle other than $\pm 1$. Most algebraic numbers do not have all their conjugates of equal absolute value, so it's quite easy to generate examples of what you ask for. There is a constraint on the minimal polynomial over Q of an algebraic number on the unit circle other than $\pm 1$. Let x be an algebraic number with absolute value 1. Then x and its complex conjugate x* = 1/x have the same minimal polynomial. Writing f(T) for the minimal polynomial of x over Q, with degree n, the polynomials T^nf(1/T) and f(T) are irreducible over Q with common root x, so the polynomials are equal up to a scaling factor: T^nf(1/T) = cf(T). Setting T = 1, f(1) = cf(1). Assuming x is not rational (i.e., x is not 1 or -1), f has degree greater than 1, so f(1) is nonzero and thus c = 1. Therefore T^nf(1/T) = f(T), so f(T) has symmetric coefficients (the coefficients of $T^i$ and $T^{n-i}$ are equal). In particular, its constant term is 1. Moreover, the roots of f(T) come in reciprocal pairs (since 1 and -1 are not roots), so n is even. Partial conclusion: an algebraic number in C other than 1 or -1 which has absolute value 1 has even degree over Q and its minimal polynomial has constant term 1. In particular, if x is an algebraic integer then it must be a unit in the number field it generates. There are no examples of algebraic integers with degree 2 and abs. value 1 that are not roots of unity since a real quadratic field has no elements on the unit circle besides 1 and -1 and the units in an imaginary quadratic field are all roots of unity (and actually are only 1 and -1 except for Q(i) and Q(w)). Thus the smallest degree x could have over Q is 4 and there are examples with degree 4: the polynomial T^4 - 2T^3 - 2T + 1 has two roots on the unit circle and two real roots (one between 0 and 1 and the other greater than 1). In Victor's answer, the two quadratic polynomials he writes down have product T^4 - 2T^2 + T^2 - 2T + 1. In Cam's answer, the algebraic integers he writes down have minimal polynomial T^8 - 12T^6 + 6T^4 - 12T^2 + 1. Note these polynomials are all symmetric, as they must be by the argument I gave above. Fact: polynomials f(T) of even degree n = 2m with symmetric coefficients are the same as polynomials of the form f(T) = T^mg(T + 1/T), where g has degree m. (This is a nice little exercise by induction.) Moreover, roots of f on the unit circle other than 1 or -1 correspond in a 2-to-1 way to real roots of g in the interval (-2,2) by $e^{i\theta} \mapsto e^{i\theta} + e^{-i\theta} = 2\cos \theta$. (Note $e^{-i\theta}$ is also a root of $f$ on the unit circle and it leads to the same real root of $g$ since $\cos(-\theta) = \cos(\theta)$.) This is how you can count the number of different roots of $f(T)$ on the unit circle: convert it into $g(T)$ and count the real roots of $g(T)$ in $(-2,2)$ using sign changes. For example, taking for f(T) Lehmer's polynomial of degree 10 in cfranc's answer, the corresponding polynomial g(T) is T^5 + T^4 - 5T^3 - 5T^2 + 4T + 3. (Although f(T) has symmetric coefficients, there is no reason g(T) needs to have symmetric coefficients.) This polynomial g has all real roots and 4 of them are in (-2,2), so Lehmer's polynomial has 24 = 8 roots on the unit circle. - I'm a little confused: several times you use the "fact" that 1 is not a root of f, but plenty of other algebraic numbers on the unit circle have 1 as a root of their minimal polynomial, like the cube roots of unity. These have odd degree and the coefficients have non-symmetric minimal polynomial. Did you want some other assumptions on the algebraic number under consideration? – Noah Stein Sep 14 2010 at 18:25 2 @Noah: The minimal polynomial of the (nontrivial) cube roots of unity is the quadratic $x^2+x+1$ and does not have 1 as a root. More generally, the only number whose minimal polynomial has 1 as a root is 1. – Andreas Blass Sep 14 2010 at 19:29 Thanks -- I forgot to turn my brain on. Now the argument makes much more sense. – Noah Stein Sep 14 2010 at 19:40 Another nice example is Lehmer's number $\lambda \simeq 1.17628$, which is the largest real root of the monic irreducible polynomial $x^{10} + x^9 - x^7-x^6-x^5-x^4-x^3+x+1$. This polynomial has a second real root inside the unit circle, and the remaining roots lie on the unit circle. The Mahler measure of a monic irreducible polynomial is the absolute value of the product of the roots with norm $\geq 1$. The Mahler measure of an algebraic integer is the Mahler measure of its minimal polynomial. It is believed that Lehmer's number has minimal Mahler measure. For more on Lehmer's number check out this "What is" paper. - 1 Something wrong here. The conjugates of a root of unity are roots of unity of the same order (the roots of one cyclotomic polynomial). Therefore either $\lambda$ is of modulus one (clearly not the case), or the polynomial is not irreducible (over ${\mathbb Q}$). I opt for the third possibility: you did not mean 'root of unity', but only 'complex number on the unit circle' – Denis Serre Sep 14 2010 at 14:14 Yup, sorry. Fixed now. – cfranc Sep 14 2010 at 14:27 One interesting set of example is Salem numbers. These are real algebraic integer all of whose conjugates has absolute value less than or equal to one and at least one of the conjugates lies on the unit circle. So you have a scenario of, one which flew over the cuckoo's nest, at least one in the edge and the rest are all in the nest (caged ?) . http://en.wikipedia.org/wiki/Salem_number Also if all its conjugates lies on the unit circle then it has to be a root of unity, its known as Kronecker Theorem. http://mathoverflow.net/questions/33169/good-effective-version-of-kroneckers-theorem - One more comment on this frequently answered question: Let $a_0 x^{2n} + a_1 x^{2n-1} + \cdots + a_{n} x^{n} + \cdots a_1 x + a_0$ be a palindromic polynomial with real coefficients and $2k$ isolated roots on the unit circle. Then any sufficiently small perturbation of this polynomial to another real palindromic polynomial also has $2k$ roots on the unit circle. Proof: Notice that $e^{\pm i \theta}$ is a root of this polynomial if and only if $a_0 \cos (n \theta) + a_1 \cos ((n-1) \theta) + \cdots + a_{n/2-1} \cos \theta + a_0/2=0$. Write $f(\theta)$ for the right hand side of this equation. Our hypothesis is that $f$ has $k$ isolated roots, $\theta_1$, $\theta_2$, ..., $\theta_k$. Then we can find $\epsilon>0$ and disjoint intervals $(a_i, b_i)$ around each $\theta_i$ such that (1) We have $f((a_i, b_i)) \supseteq (-\epsilon, \epsilon)$. (2) On $(a_i, b_i)$, we have $\|f'\|<\epsilon$ (3) Off of the $(a_i, b_i)$, we have $\|f\|>\epsilon/2$. Then, if our perturbation is small enough that $f$ and $f'$ change by less than $\epsilon/4$ everywhere, then the perturbed $f$ still has one root in each $(a_i, b_i)$, and no roots elsewhere. QED Why do I point this out? Take any of the above examples and perturb its coefficients slightly, while keeping them rational and palindromic. Then you get another example! This observation destroys most attempts to classify such polynomials by number theoretic criteria. - Nice. Since I sort my answers on "newest," most of the "above" examples are actually below.... – Gerry Myerson Sep 15 2010 at 0:18 I try to remember not to use the words above and below on MO for this reason, but it's a hard habit to break. – David Speyer Sep 15 2010 at 0:20 1 Writing the palindromic polynomial of degree 2n as x^n*g(x+1/x), those 2k distinct roots in x on the unit circle correspond in a 2-to-1 way to k distinct real roots of g(y) in [-2,2], so your comment amounts to saying that any small perturbation of a polynomial with real coefficients doesn't change the number of its distinct real roots in an (open) interval, and that is more intuitive. However, couldn't there be a problem at endpoints -2 and 2 for g (corr. to roots 1 and -1 for f), as a perturbation may move those roots of g outside [-2,2], so f may lose the roots 1 and -1. What do you think? – KConrad Sep 15 2010 at 0:43 First of all, I acknowledge that I didn't address the endpoints. I claim that a palindromic polynomial of even degree can't have single roots at $1$ or $-1$! If it had a single root at $1$, then the constant term would be negative the leading term, not equal to it. And, once you know that there is no root at $1$, if there were a root at $-1$ then it would have odd degree. – David Speyer Sep 15 2010 at 1:21 Ah, right, I forgot about the distinctness of roots at endpoints. So your argument reduces exactly to the issue of perturbing a real polynomial with distinct roots in (-2,2). – KConrad Sep 16 2010 at 4:31 If $f(x)$ is a monic polynomial in ${\mathbb Z}[x]$ and all roots have absolute value 1 (i.e. lie on the circle $S^1$ of radius 1), then all roots are roots of unity (satisfy $x^k=1$). Indeed, if $x_1,...,x_n$ are the roots of $f$, then for every integer $k$, $x_1^k,...,x_n^k$ are all roots of some monic polynomial $f_k$ of degree $n$ over $\mathbb Z$. The coefficients of $f_k$ are (by Vieta's formulas) $\pm$ elementary symmetric polynomials of $x_1^k,...,x_n^k$. Since the absolute values of $x_i^k$ are 1, the coefficients of $f_k$ are bounded independently of $k$. Hence there are finitely many different polynomials $f_k$, so $x_i^k=x_i^l$ for some $k\ne l$ and $i=1,2,...,n$, hence $x_i^{l-k}=1$. Now if you take any algebraic number on $S^1$ which is not a root of unity, some of its conjugates will not belong $S^1$. - One more example. Let $\alpha_1$ be the real, and $\alpha_2$ and $\alpha_3$ the nonreal, roots of $x^3-x-1$. Then the six numbers $\alpha_i/\alpha_j$, $i\ne j$, are a complete set of conjugates, exactly two of which ($\alpha_2/\alpha_3$ and $\alpha_3/\alpha_2$) lie on the unit circle. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 83, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9278438687324524, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Quantum_superposition	Quantum superposition This article . Please improve it by verifying the claims made and adding inline citations. Statements consisting only of original research may be removed. (March 2010) Quantum mechanics Introduction Glossary · History Background Fundamental concepts Experiments Formulations Equations Interpretations Advanced topics People Quantizations Mechanics Interpretations - Theorems Mathematical Formulations Quantum superposition is a fundamental principle of quantum mechanics that holds that a physical system—such as an electron—exists partly in all its particular, theoretically possible states (or, configuration of its properties) simultaneously; but, when measured or observed, it gives a result corresponding to only one of the possible configurations (as described in interpretation of quantum mechanics). Mathematically, it refers to a property of solutions to the Schrödinger equation; since the Schrödinger equation is linear, any linear combination of solutions to a particular equation will also be a solution of it. Such solutions are often made to be orthogonal (i.e. the vectors are at right-angles to each other), such as the energy levels of an electron. In other words, the overlap of the states is nullified, and the expectation value of an operator is the expectation value of the operator in the individual states, multiplied by the fraction of the superposition state that is "in" that state (see also eigenstates). An example of a directly observable effect of superposition is interference peaks from an electron wave in a double-slit experiment. Another example is a quantum logical qubit state, as used in quantum information processing, which is a linear superposition of the "basis states" $\|0 \rangle$ and $\|1 \rangle$. Here $\|0 \rangle$ is the Dirac notation for the quantum state that will always give the result 0 when converted to classical logic by a measurement. Likewise $\|1 \rangle$ is the state that will always convert to 1. Concept The principle of quantum superposition states that if a physical system may be in one of many configurations — arrangements of particles or fields — then the most general state is a combination of all of these possibilities, where the amount in each configuration is specified by a complex number. For example, if there are two configurations labelled by 0 and 1, the most general state would be $c_0 \mid 0 \rangle + c_1 \mid 1 \rangle$ where the coefficients are complex numbers describing how much goes into each configuration. The principle was described by Paul Dirac as follows: The general principle of superposition of quantum mechanics applies to the states [that are theoretically possible without mutual interference or contradiction] ... of any one dynamical system. It requires us to assume that between these states there exist peculiar relationships such that whenever the system is definitely in one state we can consider it as being partly in each of two or more other states. The original state must be regarded as the result of a kind of superposition of the two or more new states, in a way that cannot be conceived on classical ideas. Any state may be considered as the result of a superposition of two or more other states, and indeed in an infinite number of ways. Conversely any two or more states may be superposed to give a new state... The non-classical nature of the superposition process is brought out clearly if we consider the superposition of two states, A and B, such that there exists an observation which, when made on the system in state A, is certain to lead to one particular result, a say, and when made on the system in state B is certain to lead to some different result, b say. What will be the result of the observation when made on the system in the superposed state? The answer is that the result will be sometimes a and sometimes b, according to a probability law depending on the relative weights of A and B in the superposition process. It will never be different from both a and b [i.e, either a or b]. The intermediate character of the state formed by superposition thus expresses itself through the probability of a particular result for an observation being intermediate between the corresponding probabilities for the original states, not through the result itself being intermediate between the corresponding results for the original states.[1] Anton Zeilinger, referring to the prototypical example of the double-slit experiment, has elaborated regarding the creation and destruction of quantum superposition: "[T]he superposition of amplitudes ... is only valid if there is no way to know, even in principle, which path the particle took. It is important to realize that this does not imply that an observer actually takes note of what happens. It is sufficient to destroy the interference pattern, if the path information is accessible in principle from the experiment or even if it is dispersed in the environment and beyond any technical possibility to be recovered, but in principle still ‘‘out there.’’ The absence of any such information is the essential criterion for quantum interference to appear.[2] Theory Examples For an equation describing a physical phenomenon, the superposition principle states that a combination of solutions to a linear equation is also a solution of it. When this is true the equation is said to obey the superposition principle. Thus if state vectors f1, f2 and f3 each solve the linear equation on ψ, then ψ = c1 f1 + c2 f2 + c3 f3 would also be a solution, in which each c is a coefficient. The Schroedinger equation is linear, so quantum mechanics follows this. For example, consider an electron with two possible configurations, up and down. This describes the physical system of a qubit. $c_1 \mid \uparrow \rangle + c_2 \mid \downarrow \rangle$ is the most general state. But these coefficients dictate probabilities for the system to be in either configuration. The probability for a specified configuration is given by the square of the absolute value of the coefficient. So the probabilities should add up to 1. The electron is in one of those two states for sure. $p_\text{up} = \mid c_1 \mid^2$ $p_\text{down} = \mid c_2 \mid^2$ $p_\text{up or down} = p_\text{up} + p_\text{down} = 1$ Continuing with this example: If a particle can be in state up and down, it can also be in a state where it is an amount 3i/5 in up and an amount 4/5 in down. $\|\psi\rangle = {3\over 5} i \|\uparrow\rangle + {4\over 5} \|\downarrow\rangle.$ In this the probability for up is $\left\|\;\frac{3i}{5}\;\right\|^2=\frac{9}{25}$. The probability for down is $\left\|\;\frac{4}{5}\;\right\|^2=\frac{16}{25}$. Note that $\frac{9}{25}+\frac{16}{25}=1$. In the description, only the relative size of the different components matter, and their angle to each other on the complex plane. This is usually stated by declaring that two states which are a multiple of one another are the same as far as the description of the situation is concerned. Either of these describe the same state for any nonzero $\alpha$ $\|\psi \rangle \approx \alpha \|\psi \rangle$ The fundamental law of quantum mechanics is that the evolution is linear, meaning that if state A turns into A′ and B turns into B′ after 10 seconds, then after 10 seconds the superposition $\psi$ turns into a mixture of A′ and B′ with the same coefficients as A and B. For example, if we have the following $\mid \uparrow \rangle \to \mid \downarrow \rangle$ $\mid \downarrow \rangle \to \frac{3i}{5} \mid \uparrow \rangle + \frac{4}{5} \mid \downarrow \rangle$ Then after those 10 seconds our state will change to $c_1 \mid \uparrow \rangle + c_2 \mid \downarrow \rangle \to c_1 \left( \mid \downarrow \rangle + c_2 \left(\frac{3i}{5} \mid \uparrow \rangle + \frac{4}{5} \mid \downarrow \rangle \right)\right)$ So far there have just been 2 configurations, but there can be infinitely many. In illustration, a particle can have any position, so that there are different configurations which have any value of the position x. These are written: $\|x\rangle$ The principle of superposition guarantees that there are states which are arbitrary superpositions of all the positions with complex coefficients: $\sum_x \psi(x) \|x\rangle$ This sum is defined only if the index x is discrete. If the index is over $\reals$, then the sum is not defined and is replaced by an integral instead. The quantity $\psi(x)$ is called the wavefunction of the particle. If we consider a qubit with both position and spin, the state is a superposition of all possibilities for both: $\sum_x \psi_+(x)\|x,\uparrow\rangle + \psi_-(x)\|x,\downarrow\rangle \,$ The configuration space of a quantum mechanical system cannot be worked out without some physical knowledge. The input is usually the allowed different classical configurations, but without the duplication of including both position and momentum. A pair of particles can be in any combination of pairs of positions. A state where one particle is at position x and the other is at position y is written $\|x,y\rangle$. The most general state is a superposition of the possibilities: $\sum_{xy} A(x,y) \|x,y\rangle \,$ The description of the two particles is much larger than the description of one particle — it is a function in twice the number of dimensions. This is also true in probability, when the statistics of two random things are correlated. If two particles are uncorrelated, the probability distribution for their joint position P(x, y) is a product of the probability of finding one at one position and the other at the other position: $P(x,y) = P_x (x) P_y(y) \,$ In quantum mechanics, two particles can be in special states where the amplitudes of their position are uncorrelated. For quantum amplitudes, the word entanglement replaces[citation needed] the word correlation, but the analogy[which?] is exact. A disentangled wave function has the form: $A(x,y) = \psi_x(x)\psi_y(y) \,$ while an entangled wavefunction does not have this form. This section requires expansion. (July 2012) Analogy with probability In probability theory there is a similar principle. If a system has a probabilistic description, this description gives the probability of any configuration, and given any two different configurations, there is a state which is partly this and partly that, with positive real number coefficients, the probabilities, which say how much of each there is. For example, if we have a probability distribution for where a particle is, it is described by the "state" $\sum_x \rho(x) \|x\rangle$ Where $\rho$ is the probability density function, a positive number that measures the probability that the particle will be found at a certain location. The evolution equation is also linear in probability, for fundamental reasons. If the particle has some probability for going from position x to y, and from z to y, the probability of going to y starting from a state which is half-x and half-z is a half-and-half mixture of the probability of going to y from each of the options. This is the principle of linear superposition in probability. Quantum mechanics is different, because the numbers can be positive or negative. While the complex nature of the numbers is just a doubling, if you consider the real and imaginary parts separately, the sign of the coefficients is important. In probability, two different possible outcomes always add together, so that if there are more options to get to a point z, the probability always goes up. In quantum mechanics, different possibilities can cancel. In probability theory with a finite number of states, the probabilities can always be multiplied by a positive number to make their sum equal to one. For example, if there is a three state probability system: $x \|1\rangle + y \|2\rangle + z \|3\rangle \,$ where the probabilities $x,y,z$ are positive numbers. Rescaling x,y,z so that $x+y+z=1 \,$ The geometry of the state space is a revealed to be a triangle. In general it is a simplex. There are special points in a triangle or simplex corresponding to the corners, and these points are those where one of the probabilities is equal to 1 and the others are zero. These are the unique locations where the position is known with certainty. In a quantum mechanical system with three states, the quantum mechanical wavefunction is a superposition of states again, but this time twice as many quantities with no restriction on the sign: $A\|1\rangle + B\|2\rangle + C\|3\rangle = (A_r + iA_i) \|1\rangle + (B_r + i B_i) \|2\rangle + (C_r + iC_i) \|3\rangle \,$ rescaling the variables so that the sum of the squares is 1, the geometry of the space is revealed to be a high dimensional sphere $A_r^2 + A_i^2 + B_r^2 + B_i^2 + C_r^2 + C_i^2 = 1 \,$. A sphere has a large amount of symmetry, it can be viewed in different coordinate systems or bases. So unlike a probability theory, a quantum theory has a large number of different bases in which it can be equally well described. The geometry of the phase space can be viewed as a hint that the quantity in quantum mechanics which corresponds to the probability is the absolute square of the coefficient of the superposition. Hamiltonian evolution The numbers that describe the amplitudes for different possibilities define the kinematics, the space of different states. The dynamics describes how these numbers change with time. For a particle that can be in any one of infinitely many discrete positions, a particle on a lattice, the superposition principle tells you how to make a state: $\sum_n \psi_n \|n\rangle \,$ So that the infinite list of amplitudes $\scriptstyle (... \psi_{-2},\psi_{-1},\psi_0,\psi_1,\psi_2 ...)$ completely describes the quantum state of the particle. This list is called the state vector, and formally it is an element of a Hilbert space, an infinite dimensional complex vector space. It is usual to represent the state so that the sum of the absolute squares of the amplitudes add up to one: $\sum \psi_n^\psi_n = 1$ For a particle described by probability theory random walking on a line, the analogous thing is the list of probabilities $(...P_{-2},P_{-1},P_0,P_1,P_2,...)$, which give the probability of any position. The quantities that describe how they change in time are the transition probabilities $\scriptstyle K_{x\rightarrow y}(t)$, which gives the probability that, starting at x, the particle ends up at y after time t. The total probability of ending up at y is given by the sum over all the possibilities $P_y(t_0+t) = \sum_x P_x(t_0) K_{x\rightarrow y}(t) \,$ The condition of conservation of probability states that starting at any x, the total probability to end up somewhere must add up to 1: $\sum_y K_{x\rightarrow y} = 1 \,$ So that the total probability will be preserved, K is what is called a stochastic matrix. When no time passes, nothing changes: for zero elapsed time $\scriptstyle K{x\rightarrow y}(0) = \delta_{xy}$, the K matrix is zero except from a state to itself. So in the case that the time is short, it is better to talk about the rate of change of the probability instead of the absolute change in the probability. $P_y(t+dt) = P_y(t) + dt \sum_x P_x R_{x\rightarrow y} \,$ where $\scriptstyle R_{x\rightarrow y}$ is the time derivative of the K matrix: $R_{x\rightarrow y} = { K_{x\rightarrow y}(dt) - \delta_{xy} \over dt} \,$. The equation for the probabilities is a differential equation which is sometimes called the master equation: ${dP_y \over dt} = \sum_x P_x R_{x\rightarrow y} \,$ The R matrix is the probability per unit time for the particle to make a transition from x to y. The condition that the K matrix elements add up to one becomes the condition that the R matrix elements add up to zero: $\sum_y R_{x\rightarrow y} = 0 \,$ One simple case to study is when the R matrix has an equal probability to go one unit to the left or to the right, describing a particle which has a constant rate of random walking. In this case $\scriptstyle R_{x\rightarrow y}$ is zero unless y is either x+1,x, or x−1, when y is x+1 or x−1, the R matrix has value c, and in order for the sum of the R matrix coefficients to equal zero, the value of $R_{x\rightarrow x}$ must be −2c. So the probabilities obey the discretized diffusion equation: ${dP_x \over dt } = c(P_{x+1} - 2P_{x} + P_{x-1}) \,$ which, when c is scaled appropriately and the P distribution is smooth enough to think of the system in a continuum limit becomes: ${\partial P(x,t) \over \partial t} = c {\partial^2 P \over \partial x^2 } \,$ Which is the diffusion equation. Quantum amplitudes give the rate at which amplitudes change in time, and they are mathematically exactly the same except that they are complex numbers. The analog of the finite time K matrix is called the U matrix: $\psi_n(t) = \sum_m U_{nm}(t) \psi_m \,$ Since the sum of the absolute squares of the amplitudes must be constant, $U$ must be unitary: $\sum_n U^_{nm} U_{np} = \delta_{mp} \,$ or, in matrix notation, $U^\dagger U = I \,$ The rate of change of U is called the Hamiltonian H, up to a traditional factor of i: $H_{mn} = i{d \over dt} U_{mn}$ The Hamiltonian gives the rate at which the particle has an amplitude to go from m to n. The reason it is multiplied by i is that the condition that U is unitary translates to the condition: $(I + i H^\dagger dt )(I - i H dt ) = I \,$ $H^\dagger - H = 0 \,$ which says that H is Hermitian. The eigenvalues of the Hermitian matrix H are real quantities which have a physical interpretation as energy levels. If the factor i were absent, the H matrix would be antihermitian and would have purely imaginary eigenvalues, which is not the traditional way quantum mechanics represents observable quantities like the energy. For a particle which has equal amplitude to move left and right, the Hermitian matrix H is zero except for nearest neighbors, where it has the value c. If the coefficient is everywhere constant, the condition that H is Hermitian demands that the amplitude to move to the left is the complex conjugate of the amplitude to move to the right. The equation of motion for $\psi$ is the time differential equation: $i{d \psi_n \over dt} = c^* \psi_{n+1} + c \psi_{n-1}$ In the case that left and right are symmetric, c is real. By redefining the phase of the wavefunction in time, $\psi\rightarrow \psi e^{i2ct}$, the amplitudes for being at different locations are only rescaled, so that the physical situation is unchanged. But this phase rotation introduces a linear term. $i{d \psi_n \over dt} = c \psi_{n+1} - 2c\psi_n + c\psi_{n-1}$ which is the right choice of phase to take the continuum limit. When c is very large and psi is slowly varying so that the lattice can be thought of as a line, this becomes the free Schrödinger equation: $i{ \partial \psi \over \partial t } = - {\partial^2 \psi \over \partial x^2}$ If there is an additional term in the H matrix which is an extra phase rotation which varies from point to point, the continuum limit is the Schrödinger equation with a potential energy: $i{ \partial \psi \over \partial t} = - {\partial^2 \psi \over \partial x^2} + V(x) \psi$ These equations describe the motion of a single particle in non-relativistic quantum mechanics. Quantum mechanics in imaginary time The analogy between quantum mechanics and probability is very strong, so that there are many mathematical links between them. In a statistical system in discrete time, t=1,2,3, described by a transition matrix for one time step $\scriptstyle K_{m\rightarrow n}$, the probability to go between two points after a finite number of time steps can be represented as a sum over all paths of the probability of taking each path: $K_{x\rightarrow y}(T) = \sum_{x(t)} \prod_t K_{x(t)x(t+1)} \,$ where the sum extends over all paths $x(t)$ with the property that $x(0)=0$ and $x(T)=y$. The analogous expression in quantum mechanics is the path integral. A generic transition matrix in probability has a stationary distribution, which is the eventual probability to be found at any point no matter what the starting point. If there is a nonzero probability for any two paths to reach the same point at the same time, this stationary distribution does not depend on the initial conditions. In probability theory, the probability m for the stochastic matrix obeys detailed balance when the stationary distribution $\rho_n$ has the property: $\rho_n K_{n\rightarrow m} = \rho_m K_{m\rightarrow n} \,$ Detailed balance says that the total probability of going from m to n in the stationary distribution, which is the probability of starting at m $\rho_m$ times the probability of hopping from m to n, is equal to the probability of going from n to m, so that the total back-and-forth flow of probability in equilibrium is zero along any hop. The condition is automatically satisfied when n=m, so it has the same form when written as a condition for the transition-probability R matrix. $\rho_n R_{n\rightarrow m} = \rho_m R_{m\rightarrow n} \,$ When the R matrix obeys detailed balance, the scale of the probabilities can be redefined using the stationary distribution so that they no longer sum to 1: $p'_n = \sqrt{\rho_n}\;p_n \,$ In the new coordinates, the R matrix is rescaled as follows: $\sqrt{\rho_n} R_{n\rightarrow m} {1\over \sqrt{\rho_m}} = H_{nm} \,$ and H is symmetric $H_{nm} = H_{mn} \,$ This matrix H defines a quantum mechanical system: $i{d \over dt} \psi_n = \sum H_{nm} \psi_m \,$ whose Hamiltonian has the same eigenvalues as those of the R matrix of the statistical system. The eigenvectors are the same too, except expressed in the rescaled basis. The stationary distribution of the statistical system is the ground state of the Hamiltonian and it has energy exactly zero, while all the other energies are positive. If H is exponentiated to find the U matrix: $U(t) = e^{-iHt} \,$ and t is allowed to take on complex values, the K' matrix is found by taking time imaginary. $K'(t) = e^{-Ht} \,$ For quantum systems which are invariant under time reversal the Hamiltonian can be made real and symmetric, so that the action of time-reversal on the wave-function is just complex conjugation. If such a Hamiltonian has a unique lowest energy state with a positive real wave-function, as it often does for physical reasons, it is connected to a stochastic system in imaginary time. This relationship between stochastic systems and quantum systems sheds much light on supersymmetry. Experiments and applications Successful experiments involving superpositions of relatively large (by the standards of quantum physics) objects have been performed.[3] • A "cat state" has been achieved with photons.[4] • A beryllium ion has been trapped in a superposed state.[5] • A double slit experiment has been performed with molecules as large as buckyballs.[6] • An experiment involving a superconducting quantum interference device ("SQUID") has been linked to theme of the thought experiment: " The superposition state does not correspond to a billion electrons flowing one way and a billion others flowing the other way. Superconducting electrons move en masse. All the superconducting electrons in the SQUID flow both ways around the loop at once when they are in the Schrödinger’s cat state."[7] • A piezoelectric "tuning fork" has been constructed, which can be placed into a superposition of vibrating and non vibrating states. The resonator comprises about 10 trillion atoms.[8] • An experiment involving a flu virus has been proposed.[9] • The Heisenberg uncertainty principle states that at any given time, an electrons speed and location cannot be determined, it will already have changed. In quantum computing the phrase "cat state" often refers to the special entanglement of qubits wherein the qubits are in an equal superposition of all being 0 and all being 1; i.e., $\| \psi \rangle = \frac{1}{\sqrt{2}} \bigg( \| 00...0 \rangle + \|11...1 \rangle \bigg)$. Formal interpretation Applying the superposition principle to a quantum mechanical particle, the configurations of the particle are all positions, so the superpositions make a complex wave in space. The coefficients of the linear superposition are a wave which describes the particle as best as is possible, and whose amplitude interferes according to the Huygens principle. For any physical property in quantum mechanics, there is a list of all the states where that property has some value. These states are necessarily perpendicular to each other using the Euclidean notion of perpendicularity which comes from sums-of-squares length, except that they also must not be i multiples of each other. This list of perpendicular states has an associated value which is the value of the physical property. The superposition principle guarantees that any state can be written as a combination of states of this form with complex coefficients. Write each state with the value q of the physical quantity as a vector in some basis $\psi^q_n$, a list of numbers at each value of n for the vector which has value q for the physical quantity. Now form the outer product of the vectors by multiplying all the vector components and add them with coefficients to make the matrix $A_{nm} = \sum_q q \psi^{*q}_n \psi^q_m$ where the sum extends over all possible values of q. This matrix is necessarily symmetric because it is formed from the orthogonal states, and has eigenvalues q. The matrix A is called the observable associated to the physical quantity. It has the property that the eigenvalues and eigenvectors determine the physical quantity and the states which have definite values for this quantity. Every physical quantity has a Hermitian linear operator associated to it, and the states where the value of this physical quantity is definite are the eigenstates of this linear operator. The linear combination of two or more eigenstates results in quantum superposition of two or more values of the quantity. If the quantity is measured, the value of the physical quantity will be random, with a probability equal to the square of the coefficient of the superposition in the linear combination. Immediately after the measurement, the state will be given by the eigenvector corresponding to the measured eigenvalue. It is natural to ask why "real" (macroscopic, Newtonian) objects and events do not seem to display quantum mechanical features such as superposition. In 1935, Erwin Schrödinger devised a well-known thought experiment, now known as Schrödinger's cat, which highlighted the dissonance between quantum mechanics and Newtonian physics, where only one configuration occurs, although a configuration for a particle in Newtonian physics specifies both position and momentum. In fact, quantum superposition results in many directly observable effects, such as interference peaks from an electron wave in a double-slit experiment. The superpositions, however, persist at all scales, absent a mechanism for removing them. This mechanism can be philosophical as in the Copenhagen interpretation, or physical. Recent research indicates that chlorophyll within plants appears to exploit the feature of quantum superposition to achieve greater efficiency in transporting energy, allowing pigment proteins to be spaced further apart than would otherwise be possible.[10][11] If the operators corresponding to two observables do not commute, they have no simultaneous eigenstates and they obey an uncertainty principle. A state where one observable has a definite value corresponds to a superposition of many states for the other observable. References 1. P.A.M. Dirac (1947). The Principles of Quantum Mechanics (2nd edition). Clarendon Press. p. 12. 2. Zeilinger A (1999). "Experiment and the foundations of quantum physics". Rev. Mod. Phys. 71: S288–S297. 3. Scholes, Gregory; Elisabetta Collini, Cathy Y. Wong, Krystyna E. Wilk, Paul M. G. Curmi, Paul Brumer & Gregory D. Scholes (4 February 2010). "Coherently wired light-harvesting in photosynthetic marine algae at ambient temperature". Nature 463 (7281): 644–647. Bibcode:2010Natur.463..644C. doi:10.1038/nature08811. PMID 20130647. 4. Moyer, Michael (September 2009). "Quantum Entanglement, Photosynthesis and Better Solar Cells". Scientific American. Retrieved 12 May 2010.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 83, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9241814613342285, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/59853-something-do-reduction-formulae.html	# Thread: 1. ## Something to do with Reduction Formulae...? Hello everybody, I have got into a terrible mess with a question. In my question, I am asked to derive the reduction formulae for integrating (sin x)^m * (cos x)^n, which are on the bottom of this webpage: Reduction Formulas I keep nearly getting there, but not quite near enough! I think I must be doing the reduction part wrong. Any help would be greatly appreciated!! Jessica. 2. Originally Posted by j_clough Hello everybody, I have got into a terrible mess with a question. In my question, I am asked to derive the reduction formulae for integrating (sin x)^m * (cos x)^n, which are on the bottom of this webpage: Reduction Formulas I keep nearly getting there, but not quite near enough! I think I must be doing the reduction part wrong. Any help would be greatly appreciated!! Jessica. Hello, What exactly did you do ? integrate by parts with u=cos^(n-1) (x) and v'=cos(x)sin^m (x) this gives : $I_{m,n}=\frac{\cos^{n-1}(x) \sin^{m+1}(x)}{m+1}+\frac{n-1}{m+1} \int \left(\sin(x) \cos^{n-2}(x)\right)\left(\sin^{m+1}(x)\right) ~ dx$ $I_{m,n}=\frac{\cos^{n-1}(x) \sin^{m+1}(x)}{m+1}+\frac{n-1}{m+1} \int \cos^{n-2}(x) \sin^m (x) \sin^2(x) ~ dx$ Use the identity $\sin^2(x)=1-\cos^2(x)$ to find $I_{m,n-2}$ and $I_{m,n}$ in the right hand side of the equation =)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 5, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9602250456809998, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/229606/ideals-of-the-ring-mathbbf-qx-xn-1	# Ideals of the ring $\mathbb{F}_q[X]/(X^n-1)$ I need little help in proving the following result : Consider the ring $R:=\mathbb{F}_q[X]/(X^n-1)$, where $\mathbb{F}_q$ is a finite field of cardinality $q$ and $n\in\mathbb{N}$. Then any ideal $I$ of $R$ is principle and can be written as $I=(g(X))$, such that $g(X)\|(X^n-1)$. - 2 Any ideas, thoughts, self work, insights...? – DonAntonio Nov 5 '12 at 11:46 ## 2 Answers If $\varphi\colon S\to T$ is an epimorphism of rings with $1$, and if $I$ is an ideal of $T$, then $J=\varphi^{-1}(I)$ is an ideal of $S$. Now if $S$ is PID, that is, if every ideal is generated by a single element, then $J=xS$, for some $x\in S$. So $I=\varphi(J)=\varphi(s)I$, is also principle. Apply the above to $\varphi\colon \mathbb{F}_q[X]\to R$ and you'll get that every ideal of $R$ is principle. - Hint: Think about $R=\Bbb F_q[X]/(X^n-1)$ as the ring of polynomials of degree $<n$, and multiplication is 'modulo $(X^n-1)$', meaning that $X^n=1$ is the rule to use in $R$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 26, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9583637714385986, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/107782/continuously-deform-2-torus-with-a-line-through-one-hole-to-make-it-go-through-b	Continuously deform 2-torus with a line through one hole to make it go through both The problem is in this video at 18:00. If you have a 2-holed 2D torus with a line ($\mathbb{A}^1$) going through one of its holes. How do you deform it to make the line go through both holes. - 1 The two-holed "torus" is not $S^1\times S^1$. It could be argued to be (homotopy) equivalent to the wedge sum $S^1\wedge S^1$ for the purpose of this exercise, though. – Henning Makholm Feb 10 '12 at 13:01 Whoops.. you are right.. I edited that out! Thanks. – aelguindy Feb 10 '12 at 13:16 1 Answer This would be easiest to describe with a series of pictures but I don't have the time to draw them at the moment. Imagine your 2-holed torus is actually a 1-holed torus with another handle attached to one side along two disks. Now move one of these disks so that it circumambulates the hole of the central torus, moving around the perimeter until it comes back to near where it started. Now both of the torus's holes are penetrated by the line, kind of like two concentric circles which are tangent at a point. If you want you can push one of these circles up a bit so that the line goes through each hole at a different height. I hope this helps! Update: here is a hand-drawn sequence of pictures which I took a photo of with my cellphone. - If I understand correctly, what you do involves passing a part of the surface through itself. That's not allowed. Did I misunderstand you? – aelguindy Feb 10 '12 at 19:50 2 @aelguindy: nope I'm not moving the surface through itself. I am sliding the end of one handle along the surface of the central torus. This is an isotopy. Which part of what I said looks like moving the surface through itself? – Grumpy Parsnip Feb 10 '12 at 22:39 1 I'll draw a picture. – Grumpy Parsnip Feb 10 '12 at 22:43 Awesome! Thank yo very much! – aelguindy Feb 11 '12 at 0:17	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9530931115150452, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/241726/proving-that-this-family-of-angles-cannot-be-trisected	# proving that this family of angles, cannot be trisected Given a field $\Bbb Q \subset K \subset \Bbb C$. One can prove that $\beta \in \Bbb C$ is constructible over $K$ iff the galois group of the minimal polynomial over $K$, $m_{\beta}(x)\in K[x]$ is a 2-group. Thus given $\alpha \in \Bbb C$ with $\|\alpha\|=1$, one want to prove if it's possible to construct $\alpha^{\frac{1}{3}}$ in other words if it's possible to trisect the angle given by $\alpha$. Then using the above, one has to consider the polynomial $x^3-\alpha \in Q(\alpha)=K$, clearly the minimal polynomial of $\alpha$ divides $x^3-\alpha$ and the galois group of a cubic could be $S_3$ or $\Bbb Z_3$ none of them is a 2-group. So the criteria is the following: The important of all this Given $\alpha \in \Bbb C$ with $\|\alpha\|=1$ it's $\underline{not}$ possible to trisect the angle given by $\alpha$ (construct $\alpha^{\frac{1}{3}}$) if $x^3-\alpha \in \Bbb Q(\alpha)$ irreducible. The problem: Given $\alpha \in \Bbb C$ such that $\|\alpha\|=1$ and trascendental. Then it's not possible trisect $\alpha$. To prove this using the lemma it's just enough to prove that $x^3-\alpha$ is irreducible over $\Bbb Q(\alpha)$. If $\beta$ is a root, then $\beta^3 = \alpha$ the other roots are given by $\beta , \beta \zeta_3 , \beta \zeta_3^2$. How supposing that it's reducible I can I prove that this contradicts the fact that $\alpha$ is trascendental? - ## 1 Answer We prove an easy weaker result. Let $\alpha$ be a real transcendental. Then $x^3-\alpha$ is irreducible over $\mathbb{Q}(\alpha)$. Suppose to the contrary that $x^3-\alpha$ is reducible over $\mathbb{Q}(\alpha)$. Then the real cube root $\alpha^{1/3}$ of $\alpha$ is in $\mathbb{Q}(\alpha)$. Thus there are polynomials $P$ and $Q$ with rational coefficients such that $$\alpha^{1/3}=\frac{P(\alpha)}{Q(\alpha)}.$$ Without loss of generality we may assume that the constant terms of $P$ and $Q$ are not both $0$. Take the cube of both sides. We obtain $\alpha Q^3(\alpha)=P^3(\alpha)$. This contradicts the fact that $\alpha$ is transcendental, since $P^3(x)-xQ^3(x)$ cannot be identically $0$. To show that $P^3(x)-xQ^3(x)$ is not identically $0$, note this is clear if the constant term of $P$ is non-zero. If the constant term of $P$ is $0$, then the constant term of $Q$ is non-zero. But then the coefficient of $x$ in $xQ^3(x)$ is non-zero, while the coefficient of $x$ in $Q^3(x)$ is $0$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 57, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.932698667049408, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/194642/ellipse-equation-parameters?answertab=votes	# Ellipse equation parameters If I have an ellipse expressed by: $ax^2 + 2cxy + by^2 = constant$ what does this expression equal to: $1/ \sqrt{ab - c^2}$ with respect to the ellipse ? Thanks matlabit - is it an ellipse or an ellipsoid? $z$ can't be constant for an ellipsoid. – ajay Sep 12 '12 at 11:23 which equation?? – Avatar Sep 12 '12 at 12:25 Where you write "equation", this is an expression, not an equation. Also note that you don't need the parentheses inside the square root. – joriki Sep 12 '12 at 13:22 ## 2 Answers [Note:] As David pointed out, I'm assuming the constant is $1$; if not, divide through by the constant. This is the product of the two semi-axes, and thus except for a factor of $\pi$ it is the area of the ellipse. Write the left-hand side as $$\pmatrix{x&y}\pmatrix{a&c\\c&b}\pmatrix{x\\y}\;.$$ We can diagonalize the matrix to bring this into the form $$\pmatrix{x'&y'}\pmatrix{\lambda_1&0\\0&\lambda_2}\pmatrix{x'\\y'}\;,$$ where $\lambda_1$ and $\lambda_2$ are the eigenvalues of the matrix and the inverse squares of the semi-axes. Their product is the determinant $ab-c^2$. See also Wikipedia. - You are assuming matlabit's "constant" is 1. – David Speyer Sep 12 '12 at 14:31 Why not divide the equation by ${\rm constant}$ to make it $a x^2+2 c x y + b y^2 = 1$ without any loss of generality. Then for a rotated ellipse with major radius $r_1$ and minor radius $r_2$ and orientation angle $\theta$ the coefficients are $$a = (\frac{1}{r_1^2}-\frac{1}{r_2^2}) \cos^2\theta+\frac{1}{r_2^2}$$ $$b = (\frac{1}{r_2^2}-\frac{1}{r_1^2}) \cos^2\theta+\frac{1}{r_1^2}$$ $$c = (\frac{1}{r_1^2}-\frac{1}{r_2^2}) \sin\theta \cos\theta$$ and the said quantity $$\boxed{ \frac{1}{\sqrt{a b-c^2}} = r_1 r_2 }$$ which appears to be the area over $\pi$. Note the equation of the ellipse is best expressed by the quadratic form of the conic section tensor $C$ $$\boldsymbol{x}^\top C \boldsymbol{x} = 0$$ $$\begin{pmatrix} x & y & 1 \end{pmatrix} \begin{pmatrix} a & c & 0\\c & b & 0\\0 & 0 & \text{-}1 \end{pmatrix} \begin{pmatrix} x\\y\\1 \end{pmatrix} = a x^2+2 c x y + b y^2 -1 = 0$$ or in rotated coordinates $x'=x \cos\theta-y\sin\theta$, $y'=x \sin\theta+x\cos\theta$ $$\begin{pmatrix} x' & y' & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{r_1^2} & 0 & 0\\0 & \frac{1}{r_2^2} & 0\\0 & 0 & \text{-}1 \end{pmatrix} \begin{pmatrix} x'\\y'\\1 \end{pmatrix} = \frac{x'^2}{r_1^2}+\frac{y'^2}{r_2^2}-1 = 0$$ - Thank You guys, both answers were really helpful! – matlabit Sep 13 '12 at 8:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8965073823928833, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/783/is-dr-quantums-double-slit-experiment-video-scientifically-accurate/785	# Is Dr Quantum's Double Slit Experiment video scientifically accurate? I'm fascinated by the fundamental questions raised by the Double Slit Experiment at the quantum level. I found this "Dr Quantum" video clip which seems like a great explanation. But is it scientifically accurate? - At the same time you can also copy paste the comments from area51 ... – Cedric H. Nov 14 '10 at 18:10 – Wikis Nov 14 '10 at 18:12 1 The video is qualitatively accurate but has a quantitative lapse, which, however, is also widely present in the diagrams of the general literature on the topic. Roughly, if the two single slit distributions are as cleanly separated as indicated in the video, then the double slit interference pattern would not be as 'broad' as indicated. Precisely, the double slit pattern must always lie between the square of the sum of the square roots of the single slit patterns and the square of the difference of the square roots of the single slit patterns. But that's asking a lot from a cartoon video! – Gordon N. Fleming Nov 17 '10 at 23:07 ## 3 Answers A bad thing about the video is how they explained the part where you try to observe which slit the electron goes through. They made it sound more mysterious than it really is. What we have to ask ourselves is: what does it mean to observe an electron? What does it mean to observe anything? If we want to look at something, we need light. We see things because light is reflected off objects and our eyes collect this light which is then interpreted by our brains. If we want to see which slit the electron goes through, we shine light upon it, but this fundamentally alters the experiment. Small particles are very sensitive to perturbations and shining light on an electron is a big perturbation. Now, and this is technical, the Heisenberg uncertainty relation tells you how much the electron's path will be perturbed by the light. The path is more perturbed as the energy of the photon is greater, but to determine the position of the electron accurately, you need high energy according to Heisenberg. High energy means perturbing the electron a lot and as a consequence destroy the interference pattern. So, you might want to give up on accuracy to avoid perturbing the path of the electron too much, but if you do that, the Heisenberg relation will show you that you have to diminish the energy of the photon so much that you will not be able to locate the electron anymore. The interference pattern on the other hand will reappear. More details can be found in the Feynman Lectures, Volume 3, Chapter 1. - very thorough and clear answer, thank you. – Wikis Nov 14 '10 at 19:34 1 And what about the delayed choice quantum eraser? That experiment cancels out any probability for the measurement devices to affect the result. It's only the choice of the observer - keep the slit data or not - which determines the result. In that experiment the detectors can be turned ON all the time, it's only the delayed choice eraser which gets turned on and off (observer's choice) AFTER the photon has already went through the slits (or one of them). – Martin May 4 '12 at 19:45 1 I don't get what observers have to do with it? Just make a more elaborate measurement device that makes the "choice" and the delayed choice quantum eraser results will stay the same. Or do you mean to say that the measurement device would then have to be counted as an observer? What is an observer and what isn't? It's a very shaky foundation to build fundamental physics on the existance of observers when we know that there have not always been observers in history. – Raskolnikov May 5 '12 at 9:02 This answer is completely WRONG! As others have pointed out, how do you then explain delayed choice quantum erasure? The video is fairly accurate. I'm aghast that this incorrect answer has been upvoted so much. – Dheeraj V.S. Jan 27 at 4:31 Maybe you could explain to me what the delayed choice quantum erasure changesto the problem? Because just saying it makes a difference doesn't quite cut it. – Raskolnikov Jan 27 at 17:09 Note that the Dr. Quantum video is from the pseudoscience film "What the Bleep Do We Know?", which takes the following approach: 1. Use examples of quantum physics to show the viewer that the universe is far more weird and complex than our basic human perception/intuition suggests. 2. Attempt to convince the viewer that if quantum weirdness is real, then the weirdness of someone channeling a 35,000 year old warrior-spirit named Ramtha is also real. 3. Profit. Despite its flaws, the Dr. Quantum video on its own isn't terrible. But I think the source of scientific information should be taken into consideration when assessing its accuracy. There's a similar but better video here: http://www.youtube.com/watch?v=UMqtiFX_IQQ - I get this feeling that this is Jack Sarfatti as a digital avatar. If you have ever encountered him he has all sorts of strange ideas about things. Cedric’s comment about Area 51 sort of tipped me off, for Sarfatti has all sorts of UFO ideas. This little video is correct, but lapses into mysteriousness at the end. The process of measurement of a system in a superposition is to replace that superposition with an entanglement. We might think of it as a process where the phase associated with the “waviness” of the superposition of a system is transferred to a nonlocal property of this system with another. Consider a two slit experiment where a photon wave function interacts with a screen. The wave vector is of the form $$\|\psi\rangle~=~e^{ikx}\|1\rangle~+~e^{ik’x}\|2\rangle$$ as a superposition of states for the slits labeled $1$ and $2$. The normalization is assumed. The state vector is normalized as $$\langle\psi\|\psi\rangle~=~1~=~\langle 1\|1\rangle~+~\langle 2\|2\rangle~+~ e^{i(k’~+~k)x}\langle 1\|2\rangle~+~ e^{-i(k’~+~k)x}\langle 2\|1\rangle$$ The overlaps $\langle 1\|2\rangle$ and $\langle 2\|1\rangle$ are multiplied by the oscillatory terms which are the interference probabilities one measures on the photoplate. We now consider the classic situation where one tries to measure which slit the photon traverses. We have a device with detects the photon at one of the slit openings. We consider another superposed quantum state. This is a spin space that is $$\|\phi\rangle~=~\frac{1}{\sqrt {2}}(\|+\rangle~+~\|-\rangle).$$ This photon quantum state becomes entangled with this spin state. So we have $$\|\psi,\phi\rangle~=~e^{ikx}\|1\rangle\|+\rangle~+~e^{ik’x}\|2\rangle\|-\rangle$$ which means if the photon passes through slit number 1 the spin is + and if it passes through slit 2 the spin is in the – state. Now consider the norm of this state vector $$\langle\psi,\phi\|\psi,\phi\rangle~=~\langle 1\|1\rangle\langle +\|+\rangle~+~\langle 2\|2\rangle\langle-\|-\rangle~+~ e^{i(k’~+~k)x}\langle 1\|2\rangle\langle +\|-\rangle~+~ e^{-i(k’~+~k)x}\langle 2\|1\rangle\langle-\|+\rangle.$$ The spin states $\|+\rangle$ and $\|-\rangle$ are orthogonal and thus $\langle +\|-\rangle$ and $\langle-\|+\rangle$ are zero. This means the overlap or interference terms are removed. In effect the superposition has been replaced by an entanglement. This analysis does not tell us which state is actually measured, but it does tell us how the interference term is lost due to the entanglement of the system we measure with an instrument quantum state. So one does not need to invoke an outright collapse to illustrate how a superposition is lost. How the actual state is obtained is some matter of debate. We might think of there being some other system which now measures this spin state. So with the $\|\pm\rangle$ states we now entangle another system with two states. Yet it is clear this does not help much, for we could do this inductively “forever” and get presumably no closer to finding out which state obtains. However, maybe this third state could be a heftier spin, or an angular momentum in this case, say a spinning buckyball cooled to some low temperature. The buckyball could enter into an entanglement, as various quantum properties of these have been observed. What has this accomplished? The path integral for the entire entangled system is now narrowed closer to a classical path. We have a bit of quantum superposition properties here, but “barely.” Now we need to measure the buckyball’s rotational state. This carries us up to an even larger system and …, well we have the Schrodinger cat issue. However, some sort of asymmetry enters into the picture with the buckyball which puts the buckyball in a .7 to .3 probability ratio of being either angular momentum configuration. Further entangling this reduces the probability ratio further to .9 to .1 and the so forth. Somehow entanglement phase is being transferred completely out of the picture or into the environment (or demolished) , which then gives this state reduction in a measurement. From the perspective of a quantum path integral the set of paths are reduced to an ever narrow set of paths which corresponds to the outcome. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.951682448387146, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/70777/list	## Return to Answer 4 added 3 characters in body Artie, here is a sketch of an argument that I think should work. Definition (just in case someone needs this): A threefold singularity is called cDV or comopound Du Val or cDV if a general hyperplane section through the singular point has Du Val singularities (a.k.a. rational double point). Remark Since a rational double point surface singularity is Gorenstein, it follows that so is a cDV singularity. A typical cDV singularity is a one-parameter deformation of a rational double point. A terminal threefold singularity is a cyclic quotient of an isolated cDV singularity. Say $X'\to X$ is such that $X'$ has cDV singularities. Suppose we already know the statement for cDV singularities. Let $Y''$ denote a resolution of $Y'=X'\times_X Y$. I think $Y'$ should be smooth over a general point of the exceptional locus of $f:Y\to X$, so it should only have isolated singularities, so a general point of the exceptional locus of $f$ is covered by the codimension $2$ part of the exceptional locus of $Y''\to X$. In other words, if we know that that is covered by rational curves, then we are done. (Admittedly I did not work out this part, but it seems reasonable). Now for the cDV case, it should be easy. A general hyperplane section through the singular point has Du Val singularities, so the exceptional locus over that point can only contain rational curves forming a tree. I don't know if this is written down anywhere, I just made it up. (So it might not be air-tight). The same statement for $X$ smooth is known as Abhyankar's lemma and is (1.3) in Kollár-Mori. Interestingly, it is used in the proof of Bend-and-Break (although only for surfaces) which is used in the Cone Theorem, at least the original proof. And I think B&B is actually enough to prove what you want, so you don't need the Cone Theorem, although that is of course also a big gun. Addendum To answer Artie's question in the comments: The proof of this characterisation of terminal threefold singularities is actually not easy. However, it seems to me that it is not that far from what you want to be proven, so if that can't be proven easily, then probably neither can your statement. Here are some thoughts to support that pseudo-claim: 1) The cyclic cover part is easy, since one takes the index-one cover, which will be an index-one terminal singularity. I will assume this from this point. 2) A terminal singularity is rational and hence CM, so an index-one terminal singularity is Gorenstein. Therefore any hyperplane section is also Gorenstein. 3) In the situation of the question a general hyperplane section through the singular point may not be resolved by $f$, but if it were, then the exceptional divisor of that resolution would be just the exceptional locus of $f$. That would imply that the exceptional locus of the resolution of the hyperplane section consists of rational curves and since it is equal to the exceptional locus of $f$, if it contained a loop, it would imply that $R^1f_\mathscr O_Y\neq 0$ (I think I can prove this. It is relatively easy, but not super-easy.) In other words, since $X$ has rational singularities it follows that so does this general hyperplane section. That implies that it has Du Val singularties and hence $X$ has (isolated) cDV. Conclusion Even if the addendum is not a full proof of the used characterisation of terminal threefold singularities, it seems to suggest that you question is about as hard as that is. 3 added 2052 characters in body Definition (just in case someone needs this): A threefold singularity is called cDV or comopound Du Val if a general hyperplane section through the singular point has Du Val singularities (a.k.a. rational double point). Remark Since a rational double point surface singularity is Gorenstein, it follows that so is a cDV singularity. A typical cDV singularity is a one-parameter deformation of a rational double point. A terminal 3-fold threefold singularity is a cyclic quotient of an isolated cDV singularity. Say $X'\to X$ is such that $X'$ has cDV singularities. Suppose we already know the statement for cDV singularities. Let $Y''$ denote a resolution of $Y'=X'\times_X Y$. I think $Y'$ should be smooth over a general point of the exceptional locus of $f:Y\to X$, so it should only have isolated singularities, so a general point of the exceptional locus of $f$ is covered by the codimension $2$ part of the exceptional locus of $Y''\to X$. In other words, if we know that that is covered by rational curves, then we are done. (Admittedly I did not work out this part, but it seems reasonable). which is used in the Cone Theorem, at least the original proof. And I think B&B is actually enough to prove what you want, you don't need the Cone Theorem, although that is of course also a big gun. AddendumTo answer Artie's question in the comments: The proof of this characterisation of terminal threefold singularities is actually not easy. However, it seems to me that it is not that far from what you want to be proven, so if that can't be proven easily, then probably neither can your statement. Here are some thoughts to support that pseudo-claim: 1) The cyclic cover part is easy, since one takes the index-one cover, which will be an index-one terminal singularity. I will assume this from this point. 2) A terminal singularity is rational and hence CM, so an index-one terminal singularity is Gorenstein. Therefore any hyperplane section is also Gorenstein. 3) In the situation of the question a general hyperplane section through the singular point may not be resolved by $f$, but if it were, then the exceptional divisor of that resolution would be just the exceptional locus of $f$. That would imply that the exceptional locus of the resolution of the hyperplane section consists of rational curves and since it is equal to the exceptional locus of $f$, if it contained a loop, it would imply that $R^1f_\mathscr O_Y\neq 0$ (I think I can prove this. It is relatively easy, but not super-easy.) In other words, since $X$ has rational singularities it follows that so does this general hyperplane section. That implies that it has Du Val singularties and hence $X$ has (isolated) cDV. Conclusion Even if the addendum is not a full proof of the used characterisation of terminal threefold singularities, it seems to suggest that you question is about as hard as that is. 2 added 3 characters in body Artie, here is a sketch of an argument that I think should work. A terminal 3-fold singularity is a cyclic quotient of an isolated cDV singularity. Say $X'\to X$ is such that $X'$ has cDV singularities. Suppose we already know the statement for cDV singularities. Let $Y''$ denote a resolution of $Y'=X'\times_X Y$. I think $Y'$ should be smooth over a general point of the exceptional locus of $f:Y\to X$, so it should only have isolated singularities, so a general point of the exceptional divisor locus of $f$ is covered by the codimension $2$ part of the exceptional divisor locus of $Y''\to X$. In other words, if we know that that is covered by rational curves, then we are done. (Admittedly I did not work out this part, but it seems reasonable). Now for the cDV case, it should be easy. A general hyperplane section through the singular point has Du Val singularities, so the exceptional locus over that point can only contain rational curves forming a tree. I don't know if this is written down anywhere, I just made it up. (So it might not be air-tight). The same statement for $X$ smooth is known as Abhyankar's lemma and is (1.3) in Kollár-Mori. Interestingly, it is used in the proof of Bend-and-Break (although only for surfaces) which is used in the Cone Theorem, at least the original proof. And I think B&B is actually enough to prove what you want, you don't need the Cone Theorem, although that is of course also a big gun. 1 Artie, here is a sketch of an argument that I think should work. A terminal 3-fold singularity is a cyclic quotient of an isolated cDV singularity. Say $X'\to X$ is such that $X'$ has cDV singularities. Suppose we already know the statement for cDV singularities. Let $Y''$ denote a resolution of $Y'=X'\times_X Y$. I think $Y'$ should be smooth over a general point of the exceptional locus of $f:Y\to X$, so it should only have isolated singularities, so a general point of the exceptional divisor is covered by the codimension $2$ part of the exceptional divisor of $Y''\to X$. In other words, if we know that that is covered by rational curves, then we are done. (Admittedly I did not work out this part, but it seems reasonable). Now for the cDV case, it should be easy. A general hyperplane section through the singular point has Du Val singularities, so the exceptional locus over that point can only contain rational curves forming a tree. I don't know if this is written down anywhere, I just made it up. (So it might not be air-tight). The same statement for $X$ smooth is known as Abhyankar's lemma and is (1.3) in Kollár-Mori. Interestingly, it is used in the proof of Bend-and-Break (although only for surfaces) which is used in the Cone Theorem, at least the original proof. And I think B&B is actually enough to prove what you want, you don't need the Cone Theorem, although that is of course also a big gun.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9618566036224365, "perplexity_flag": "head"}
http://mathoverflow.net/questions/71731?sort=oldest	Is there a path'' between any two fiber functors over the same field in Tannakian formalism? Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I will take the approach of this question: http://mathoverflow.net/questions/23860/tannaka-formalism-and-the-etale-fundamental-group and think of the etale fundamental group as Tannakian formalism for $\mathbb{F}_1$. Then our "Tannakian category" is the category of finite etale covers, and each fiber functor is a functor from this category to $Sets$ (thought of as finite dimensional spaces over $\mathbb{F}_1$). For the etale fundamental group, it is true that for any two fiber functors (given by two different geometric points) there is a path'' between them. Meaning: there is a natural isomorphism between these two functors. My question is whether this independence of the basepoint'' result applies to Tannakian formalism as well: Is it true that for any two fiber functors $H_1, H_2: \mathcal{C}\rightarrow Vec_K$, there is a natural isomorphism $H_1 \cong H_2$? - 2 Answers I don't think this is true in general. The point is that that there are non-isomorphic groups with equivalent categories of representations; since the category of representations together with the fiber functor determines the group, this gives a counterexample. This happens under the following circumstances; the following construction was first given, I believe, in Giraud's book on non-abelian cohomology. Suppose that $G$ is an affine algebraic group over $\mathop{\rm Spec} K$ and $P \to \mathop{\rm Spec}K$ is a $G$-torsor. Call $H$ the group scheme of automorphisms of $P$ as a torsor; then $P$ becomes an $(H, G)$-bitorsor, that is, admits commuting actions of $G$ on the right and of $H$ on the left, and is a torsor for both. Conversely, if $P \to \mathop{\rm Spec}K$ is an $(H, G)$-bitorsor, then $H$ is the automorphism group scheme of $P$ as a $G$-torsor. Then the categories of representations of $G$ and $H$ are isomorphic. This follows, essentially, from descent theory; if $V$ is a representation of $G$, then the quotient $(P \times_{\mathop{\rm Spec}K} V)/G$ is a vector space on $K$ with an action of $H$; the inverse functor is obtained by exchanging $G$ and $H$ (and right and left actions). My favorite example of this is the following: if $q$ and $q'$ are non-degenerate quadratic forms in $n$ variable, the orthogonal groups $\mathrm O(q)$ and $\mathrm O(q')$ have equivalent categories of representations (although they are not isomorphic, in general). The bitorsor is the functor of isometries of $q$ and $q'$. On the other hand, if $K$ is algebraically closed the fiber functors are indeed isomorphic; if memory serves me well, this is in Deligne's paper in Grothendieck's Festschrift, but I don't have it here and can't check right now. - 1 This is indeed in Deligne's Festschrift article. One can also look at Breen's article in the Motives volume. What Deligne shows is that, for any Tannakian category, any two fiber functors are isomorphic over a field extension. This was the `missing' piece in Saavedra's thesis. – Keerthi Madapusi Pera Jul 31 2011 at 17:24 I guess that makes sense. Does this mean that $\mathbb{F}_1$ behaves like an algebraically closed field?! I must say that I'm growing very fond of this Tannakian formalism business. – James D. Taylor Jul 31 2011 at 17:25 You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The obstruction to the existence of such an isomorphism is a (bi)torsor, that has been studied in various real-life situations. An example extracted from On the relation between Nori Motives and Kontsevich Periods Annette Huber, Stefan Müller-Stach http://arxiv.org/abs/1105.0865 "As already explained by Kontsevich, singular cohomology and algebraic de Rham cohomology are both fiber functors on the same Tannaka category of motives. By general Tannaka formalism, there is a pro-algebraic torsor of isomorphisms between them. The period pairing is nothing but a complex point of this torsor." Basically, by tannaka duality, you can build a counter-example out of any couple of non-isomorphic objects of a gerbe. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "mathjax_display_tex": 0, "mathjax_asciimath": 3, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9270952343940735, "perplexity_flag": "head"}
http://en.m.wikibooks.org/wiki/The_Science_of_Programming/The_Simplest_Things	# The Science of Programming/The Simplest Things If you have studied Chapter 4 of CME, you will have learned the power rule for differentiating simple polynomial functions, known as terms. A term has the form: ` $a x ^ n$` with a being known as the coefficient of the term and n being known as the exponent of the term. Here is the power rule for terms: ` $\frac{d(a x^n)}{dx} = a n x^{n - 1}$` As example, consider the polynomial: ` $y = 2 x^3$` If the power rule is correct, then the derivative should be: ` $\frac{dy}{dx} = 6 x^2$` Let's test. We start out by defining a function to represent the polynomial: ``` function y(x) { 2 * (x ^ 3); } ``` As before, the dependent variable becomes the name of the function and the independent variable becomes the formal parameter. Before we continue, now would be a good time to learn about using Sway with files, since typing into the interpreter can be rather painful for function definitions. Thus for the rest of this book, even though interactions with the interpreter are shown, you should be storing Sway programs in files and running Sway programs from files. The derivative, or $\frac{dy}{dx}$ can be written as this function: ``` function dy/dx(x) { 6 * (x ^ 2); } ``` Is there any way to test whether or not the power rule is correct? Recall our revised ratio function from the previous chapter: ``` function ratio(x,dx,y) { var dy = y(x + dx) - y(x); dy / dx; } ``` As you should remember, the ratio it computes is an approximation for the derivative at a given x. Note that we've changed the names of the formal parameters, w to x, dw to dx, and h to y, to reflect the fact that we are working with polynomials now.[1] We believe that ratio (for small dx) produces a number that should be close to what the function dy/dx produces. Let's define a testing function: ``` function test(x) { var dx = 0.000001; println("for x = ",x); inspect(ratio(x,dx,y)); inspect(dy/dx(x)); println(); } ``` Given a value for x, the test function will print out the two results. Time for some nomenclature. When we use our ratio function, we are finding the derivative numerically. That is to say, we use a small number for dx, find the resulting dy, and then compute the ratio $\frac{dy}{dx}$. On the other hand, using the power rule (as embodied in the function dy/dx), we find the derivative symbolically since we never actually compute an actual ratio. ``` sway> test(10); for x = 10 ratio(x,dx,y) is 600.00005965 dy/dx(x) is 600 sway> test(20); for x = 20 ratio(x,dx,y) is 2400.0001231 dy/dx(x) is 2400 sway> test(30); for x = 30 ratio(x,dx,y) is 5400.0001837 dy/dx(x) is 5400 ``` For this test at least, we see that there is good agreement between our numeric and symbolic solutions. Clearly, though, the symbolic solution is preferred because: • it is exact • it is easier to compute You may be asking yourself, can we write a program to find a derivative symbolically? ## A Shortcoming to Our Approach It certainly would be nice if we could use the symbolic approach to finding derivatives instead of our numeric approach using ratios. The problem is that our function representing a term: ``` function y(x) { 2 * (x ^ 3); } ``` has the coefficient 2 and the exponent 3 hard-wired. If we are given such a function but aren't privy to the internal details (such as the value of the hard-wired exponent), we cannot create the derivative function since we need to know the value of the exponent to do so and have no way to get at it. We need to contrive a way to retrieve the exponent from the function so we can use it to construct the derivative function. The next section is to give you an exposure to one approach for extracting the components of a function. The approach uses objects. If you don't know what an object is or how to create one in Sway, please read about using objects in the The Sway Reference Manual now. OK, now that you are familiar with objects, we can begin to use objects to write a program that finds a derivative symbolically. ↑Jump back a section ## Polynomial Objects and the Power Rule Instead of using a function to represent the polynomial y, we are going to use an object instead. From studying the reference manual, you know that an object is simply an environment and an environment is simply a table of variables and their values. You also know that you use a function to create an object and that a function that creates an object is conventionally known as a constructor. Finally, you know that to define a constructor function, you have that function return the predefined variable this. Here's a constructor for a simple polynomial object. We name the constructor term: ``` function term(a,n) { function value(x) { a * (x ^ n); } this; } ``` Note that the constructor includes an internal function for computing the value of y given an x, just as before. We call that function by using the dot operator. ``` sway> var y = term(2,3); OBJECT: <OBJECT 1671> sway> y . value(4); INTEGER: 128 sway> y . value(5); INTEGER: 250 ``` Indeed, $2 * 4 ^ 3$ is truly 128 and $2 * 5 ^ 3$ is truly 250. Now comes the good part! Storing a polynomial term as an object rather than a function allows us to extract the coefficient a and the exponent n: ``` sway> y . a; INTEGER: 2; sway> y . n; INTEGER: 3; ``` We can even pretty print the object y to see all its fields: ``` sway> pp(y); <OBJECT 2561>: context: <OBJECT 749> dynamicContext: <OBJECT 749> callDepth: 1 constructor: <function term(a,n)> this: <OBJECT 2561> value: <function value(x)> a: 2 n: 3 OBJECT: <OBJECT 2561> ``` We see, among other fields, the correct values for a and n. Later on, you will understand the meaning and purpose of the predefined fields: context, dynamicContext, and constructor. The this field you already know. We want access to a and n because the power rule needs them to compute the derivative polynomial. Now we can write an expression that allows us to compute the derivative: ``` sway> var coeff = y . a; sway> var exp = y . n; ``` ``` sway> var z = term(coeff * exp,exp - 1); OBJECT: <OBJECT 2783> ``` ``` sway> z . a; INTEGER: 6 ``` ``` sway> z . n; INTEGER: 2 ``` ``` sway> z . value(4) INTEGER: 96 ``` Of course, we could write a function to do this task for us: ``` function powerRule(t) { var coeff = t . a; var exp = t . n; term(coeff * exp,exp - 1); } ``` Note that the powerRule function takes a term object as an argument and returns a new term object that represents the derivative. This is because the new term has a coefficient of a * n and an exponent of n - 1, just like the power rule dictates. Let's check and make sure our power rule works as intended: ``` sway> var y = term(2,3); sway> var z = powerRule(y); sway> y . value(4); INTEGER: 128 sway> z . value(4); INTEGER: 96 ``` Indeed it does! Moreover, pretty printing z shows us the correct values for a and n: ``` sway> pp(dy/dx); <OBJECT 2922>: context: <OBJECT 749> dynamicContext: <OBJECT 2808> callDepth: 2 constructor: <function term(a,n)> this: <OBJECT 2922> value: <function value(x)> a: 6 n: 2 OBJECT: <OBJECT 2922> ``` You may not have noticed it, but we have a little lack of symmetry. While the term object has an internal function for computing its value, we use an external function, powerRule to compute its derivative. When we program using objects, we use internal functions when possible. So let's rewrite the term function so that it can compute its own derivative: ``` function term(a,n) { function value(x) { a * (x ^ n); } function diff() { term(a * n,n - 1); } this; } ``` We name this new internal function, diff, to indicate taking the differential of our object. Note how we have dispensed with the variables coeff and exp inside the diff function and just used a and n directly. Finally, we add a third internal function to term. This one is used to visualize our object in a more concise way than using the pp function. By convention, we will call this function toString: ``` function term(a,n) { function value(x) { a * (x ^ n); } function diff() { term(a * n,n - 1); } function toString() { string(a) + "x^" + string(n); } this; } ``` The toString function converts both the coefficient and the exponent to strings and then concatenates those strings together to form a string representation of the term: ``` sway> y = term(2,3); sway> z = y . diff(); ``` ``` sway> y . toString() STRING: "2x^3" ``` ``` sway> x . toString() STRING: "6x^2" ``` Notice how much easier it is to pick out the coefficient and the exponent of a term by using the toString function. Another way to formulate the toString function takes advantage of the fact that if you add a string and a number (with the string on the left hand side of the plus sign, the number is automatically converted to a string. That, with the fact that Sway combines mathematical operators from left-to-right, lets us remove the string conversion of n: ``` function toString() { string(a) + "x^" + n; } ``` If we use an empty string to start the expression, we can remove the first string conversion as well: ``` function toString() { "" + a + "x^" + n; } ``` ↑Jump back a section ## A Whole System of Objects From here on out, we are going to develop a system for finding the derivative (and eventually, the integral) of many kinds of mathematical objects (limited only by our imagination and attention span). Each object in our system will do (at least) three things. The first is to compute its value at a given spot. The second is to compute its derivative (and eventually, its integral). The third is to compute its visualization. So all our mathematical objects will have value, diff, and toString internal functions or methods. [2] Even the simplest real-world mathematical objects such as numbers and variables will need to be re-created as Sway objects. That way, we can always take the derivative of an object without having to ask first if taking the derivative makes sense. ↑Jump back a section ## Questions All formulas are written using Grammar School Precedence. 1. Why does the Sway implementation of $2 x^3$ have parentheses: 2 * (x ^ 3)? 2. Using the powerRule function, can you take the derivative of a derivative? Explain. 3. Write a function that uses the power rule function to answer the following questions on p. 58 of Thompson: 1, 2, 4, 6, 7. Name your function p58. Your function should calculate, then print the answers to the questions. 4. Using pencil and paper, work exercises 3, 8, 9, and 10 on p. 58 of Thompson. ↑Jump back a section ## Footnotes 1. In general, it does not matter what names are used for the formal parameters, so we could have left the ratio function untouched and it would still work exactly the same. 2. In the world of object-oriented programming (which is another way of saying programming with objects), these internal functions are called methods. We will use the term method from now one. Just remember, a method is simply an internal function. ↑Jump back a section	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 9, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9123979210853577, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-math-topics/8755-another-nasty-integral-2-print.html	Another nasty integral Printable View Show 40 post(s) from this thread on one page • December 31st 2006, 07:16 AM CaptainBlack Quote: Originally Posted by CaptainBlack Rough outline: Given a sufficiently smooth function f(x) we have that: $\lim_{x \to \infty}f(x) + \lim_{x \to - \infty}f(x) = \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon \|x\|}$ 1. Every function can be written as the sum of an odd function and an even function. 2. The result is trivially true for a sufficiently well behaved odd function. Both sides are zero for a sufficiently well behaved odd function. Quote: 3. The result is a result of the fundamental theorem of calculus, and the Laplace transform of a derivative for an even function zero at the origin(sufficiently well behaved). For an even function $f(x)$ such that $f(0)=0$ $\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon \|x\|}=2 \lim_{\epsilon \to 0^+} \epsilon \int_{0}^{\infty} dx f(x) e^{-\epsilon \|x\|}=2 \lim_{\epsilon \to 0^+} \left[ \mathcal{L}f' \right] (\epsilon)+f(0)=$ $2 \lim_{\epsilon \to 0^+} \left[ \mathcal{L}f' \right] (\epsilon)$ .............. $=2\int_{0}^{\infty}f'(x) dx=2\left[\lim_{x \to \infty}f(x) - \lim_{x \to 0+}f(x)\right]$ but as $f$ is even (and sufficiently well behaved) the last limit above is $0$, and $\lim_{x \to \infty}f(x)=\lim_{x \to -\infty}f(x)$ For a constant we have: $\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx\ k\ e^{-\epsilon \|x\|}=2 \lim_{\epsilon \to 0^+} \epsilon \int_{0}^{\infty} dx\ k\ e^{-\epsilon \|x\|}=$ $2 \lim_{\epsilon \to 0+} \epsilon\ \mathcal{L}k(\epsilon)=2 \lim_{\epsilon \to 0^+} \epsilon\ \frac{k}{\epsilon}=2\ k<br />$, which is the required result in this case. (Note we could have done this for an arbitrary even function by keeping track of the f(0) terms rather than setting them to zero, then we would not have had to handle the constant term separately) Quote: 4. Hence the result is true for a sufficiently well behaved function. I can amplify the details if you need (though I doubt that I will feel like going for full rigour) RonL RonL • December 31st 2006, 10:55 AM CaptainBlack Quote: Originally Posted by CaptainBlack Rough outline: 1. Every function can be written as the sum of an odd function and an even function. Obvious but here goes, for any function $f(x)$ on $\mathbb{R}$ define the even and odd functions $g(x)$ and $h(x)$ as: $g(x)=\left[f(x)+f(-x)\right]/2$ $h(x)=\left[f(x)-f(-x)\right]/2$, then: $f(x)=g(x)+h(x)$ I hope this is all very familiar from other contexts RonL • December 31st 2006, 10:58 AM CaptainBlack Quote: Originally Posted by CaptainBlack Rough outline: Given a sufficiently smooth function f(x) we have that: $\lim_{x \to \infty}f(x) + \lim_{x \to - \infty}f(x) = \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon \|x\|}$ We need that if (sufficiently well behaved) functions $f_1(x)$ and $f_2(x)$ satisfy the above relation then so does any linear combination of them. This is obvious and I will not prove it, but it needs stating for the demonstration to go through when we split a function into its odd and even components. RonL • December 31st 2006, 01:54 PM CaptainBlack Quote: Originally Posted by CaptainBlack Both sides are zero for a sufficiently well behaved odd function. For an even function $f(x)$ $\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon \|x\|}=2 \lim_{\epsilon \to 0^+} \epsilon \int_{0}^{\infty} dx f(x) e^{-\epsilon \|x\|}=2 \lim_{\epsilon \to 0^+} \left[ \mathcal{L}f' \right] (\epsilon)$ .............. $=2\int_{0}^{\infty}f'(x) dx=2\left[\lim_{x \to \infty}f(x) - \lim_{x \to 0+}f(x)\right]$ but as $f$ is even (and sufficiently well behaved) the last limit above is $0$, and $\lim_{x \to \infty}f(x)=\lim_{x \to -\infty}f(x)$ There is a slight error here, in that I have assumed that the even function is zero at $x=0$, which is not necessarily true. Instead supose we decompose the original function into the sum of a constant, and odd function and an even function which is zero at the origin (an alternative is to leave the decomposition into odd and even, and kept track of the f(0) terms in the even function analysis when they would have cancelled out). The relation is satisfied by the constant, the odd function and the even function zero at the origin, so the result holds. I have gone back and modified the previous posts to reflect this change. RonL Show 40 post(s) from this thread on one page All times are GMT -8. The time now is 01:48 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 28, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9370825290679932, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/22897/fields-with-trivial-automorphism-group/22898	## Fields with trivial automorphism group ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Is there a nice characterization of fields whose automorphism group is trivial? Here are the facts I know. 1. Every prime field has trivial automorphism group. 2. Suppose L is a separable finite extension of a field K such that K has trivial automorphism group. Then, if E is a finite Galois extension of K containing L, the subgroup $Gal(E/L)$ in $Gal(E/K)$ is self-normalizing if and only if L has trivial automorphism group. (As pointed out in the comments, a field extension obtained by adjoining one root of a generic polynomial whose Galois group is the full symmetric group satisfies this property). 3. The field of real numbers has trivial automorphism group, because squares go to squares and hence positivity is preserved, and we can then use the fact that rationals are fixed. Similarly, the field of algebraic real numbers has trivial automorphism group, and any subfield of the reals that is closed under taking squareroots of positive numbers has trivial automorphism group. My questions: 1. Are there other families of examples of fields that have trivial automorphism group? For instance, are there families involving the p-adics? [EDIT: One of the answers below indicates that the p-adics also have trivial automorphism group.] 2. For what fields is it true that the field cannot be embedded inside any field with trivial automorphism group? I think that any automorphism of an algebraically closed field can be extended to any field containing it, though I don't have a proof) [EDIT: One of the answers below disproves the parenthetical claim, though it doesn't construct a field containing an algebraically closed field with trivial automorphism group]. I suspect that $\mathbb{Q}(i)$ cannot be embedded inside any field with trivial automorphism group, but I am not able to come up with a proof for this either. [EDIT: Again, I'm disproved in one of the answers below]. I'm not even able to come up with a conceptual reason why $\mathbb{Q}(i)$ differs from $\mathbb{Q}(\sqrt{2})$, which can be embedded in the real numbers. ADDED SEP 26: All the questions above have been answered, but the one question that remains is: can every field be embedded in a field with trivial automorphism group? Answering the question in general is equivalent to answering it for algebraically closed fields. - 1 Extension by any single root of a general polynomial has trivial automorphism group. – Ryan Thorngren Apr 28 2010 at 20:48 1 @Georges: by "general" I think he just means something like "irreducible and separable, of degree at least 3, and the Galois group of the splitting field is the full symmetric group". So for example over Q this produces a lot of examples. – Kevin Buzzard Apr 28 2010 at 22:05 3 I just want to point out that one of the OP's questions remains wide open: can every field be embedded in a field with trivial automorphism group? I don't even myself know the answer for $\mathbb{C}$. – Pete L. Clark Apr 29 2010 at 6:09 1 One conceptual difference between $\mathbb{Q}(i)$ and $'\mathbb{Q}( \sqrt{2})$ is that a field in which $-1$ is a square cannot be ordered. Since $'\mathbb{Q}( \sqrt{2})$ has real embeddings it can obviously be ordered. There is a somewhat related invariant called the level of of the field, which if I remember correctly is the least number of summands needed to express -1 as a sum of squares. In the cases above the levels are 1 and $\infty$ respectively. If we had chosen $\sqrt{-2}$ instead the level would have been 2. – K.J. Moi May 3 2010 at 8:11 1 For the question Pete points out, see the references in mathoverflow.net/questions/61058 – dke Apr 8 2011 at 17:24 show 7 more comments ## 4 Answers As Robin as pointed out, for all primes $p$, $\mathbb{Q}_p$ is rigid, i.e., has no nontrivial automorphisms. It is sort of a coincidence that you ask, since I spent much of the last $12$ hours writing up some material on multiply complete fields which has applications here: Theorem (Schmidt): Let $K$ be a field which is complete with respect to two inequivalent nontrivial norms (i.e., the two norms induce distinct nondiscrete topologies). Then $K$ is algebraically closed. Corollary: Let $K$ be a field which is complete with respect to a nontrivial norm and not algebraically closed. Then every automorphism of $K$ is continuous with respect to the norm topology. (Proof: To say that $\sigma$ is a discontinuous automorphism is to say that the pulled back norm $\sigma^*\|\| \ \|\|: x \mapsto \|\|\sigma(x)\|\|$ is inequivalent to $\|\| \ \|\|$. Thus Schmidt's theorem applies. In particular this applies to show that $\mathbb{Q}_p$ and $\mathbb{R}$ are rigid, since every continuous automorphism is determined by its values on the dense subspace $\mathbb{Q}$, hence the identity is the only possibility. (It is possible to give a much more elementary proof of these facts, e.g. using the Ostrowski classification of absolute values on $\mathbb{Q}$.) At the other extreme, each algebraically closed field $K$ has the largest conceivable automorphism group: `$\# \operatorname{Aut}(K) = 2^{\# K}$`: e.g. Theorem 80 of http://math.uga.edu/~pete/FieldTheory.pdf. There is a very nice theorem of Bjorn Poonen which is reminiscent, though does not directly answer, your other question. For any field $K$ whatsoever, and any $g \geq 3$, there exists a genus $g$ function field $K(C)$ over $K$ such that $\operatorname{Aut}(K(C)/K)$ is trivial. However there may be other automorphisms which do not fix $K$ pointwise. There is also a sense in which for each $d \geq 3$, if you pick a degree $d$ polynomial $P$ with $\mathbb{Q}$-coefficients at random, then with probability $1$ it is irreducible and $\mathbb{Q}[t]/(P)$ is rigid. By Galois theory this happens whenever $P$ is irreducible with Galois group $S_d$, and by Hilbert Irreducibility the complement of this set is small: e.g. it is "thin" in the sense of Serre. Addendum: Recall also Cassels' embedding theorem (J.W.S. Cassels, An embedding theorem for fields, Bull. Austral. Math. Soc. 14 (1976), 193-198): every finitely generated field of characteristic $0$ can be embedded in $\mathbb{Q}_p$ for infinitely many primes $p$. It would be nice to know some positive characteristic analogue that would allow us to deduce that a finitely generated field of positive characteristic can be embedded in a rigid field (so far as I know it is conceivable that every finitely generated field of positive characteristic can be embedded in some Laurent series field $\mathbb{F}_q((t))$, but even if this is true it does not have the same consequence, since Laurent series fields certainly have nontrivial automorphisms). - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This is not a family but instead an interesting example ... Let $\mathcal{U}$ be a nonprincipal ultrafilter over the set $\mathbb{P}$ of primes and consider the corresponding ultraproduct $K = \prod_{\mathcal{U}}\mathbb{F}_{p}$ of the fields of prime order $p$. If $CH$ holds, then $K$ always has $2^{2^{\aleph_{0}}}$ automorphisms ... but none of the nontrivial ones is easily seen by the naked eye. There is a good reason for this. Shelah has recently shown that it is consistent that there exists a nonprincipal ultrafilter $\mathcal{U}$ such that $K = \prod_{\mathcal{U}}\mathbb{F}_{p}$ has no nontrivial automorphisms. - Aren't all but two automorphisms of $\mathbb{C}$ hard to see as well by the naked eye? – Zsbán Ambrus Apr 29 2010 at 8:57 3 Very much so! And in a Solovay model $L(\mathbb{R})$, there are only two automorphisms of $\mathbb{C}$. But this is a model in which the Axiom of Choice fails. I find it interesting that it is consistent that $ZFC$ that the nontrivial automorphisms of $\prod_{\mathcal{U}} \mathbb{F}_{p}$ are not just hard to see ... they aren't even there! – Simon Thomas Apr 29 2010 at 15:22 In the paper below Shelah, among many other things, gives constructions of real closed fields with no nontrivial automorphisms that are not subfields of the reals. S. Shelah, Models with second order properties. IV. A general method and eliminating diamonds -- Annals Pure and Applied Logic 25 (1983) 183-212 - There are examples involving the $p$-adics: for instance $\mathbb{Q}_p$ itself has trivial automorphism group. Indeed as $\mathbb{Q}(i)$ embeds in $\mathbb{Q}_p$ when $p\equiv1$ (mod 4) then $\mathbb{Q}(i)$ does embed in a field with trivial automorphism group. Indeed this is the case for all number fields (finite extensions of $\mathbb{Q}$). Now for an example of an algebraically closed field, $K$, an extension $L$ of $K$ and an automorphism $\tau$ of $K$ not extending to one of $L$. Let $K$ be the algebraic closure of $\mathbb{Q}$, considered as a subfield of $\mathbb{C}$ and let $\tau$ be complex conjugation. Let $L=K(x,\sqrt{x^3+ax+b})$ be the function field of an elliptic curve $E$ over $K$. Each automorphism of $L$ takes $K$ to itself. Suppose the $j$-invariant of $E$ is $i$ (considered as an element of $K$). Then any automorphism of $L$ taking $K$ to itself must fix $i$, and so cannot restrict to $\tau$ on $K$. - Could you sketch a proof that the p-adics have trivial automorphism group? I'm guessing it's the same reason as the reals, but how do you show that any automorphism must "respect" the valuation? – Vipul Naik Apr 28 2010 at 21:27 2 One shows, by fair means or foul, that an automorphism of $\mathbb{Q}_p$ must be continuous. For instance it suffices to prove that an automorphism preserves $\mathbb{Z}_p$. – Robin Chapman Apr 28 2010 at 21:32 15 The point is that various subsets of Q_p are algebraically defined. For example 1+pZ_p is precisely the elements of Q_p which have n'th roots for all n prime to p (or 1+4Z_p if p=2). Hence 1+p^nZ_p is algebraically-defined, so p^nZ_p is algebraically-defined, so "small" is algebraically-defined, and now you're home. – Kevin Buzzard Apr 28 2010 at 22:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 93, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9365200400352478, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/151586/infinite-degree-algebraic-field-extensions/151602	Infinite Degree Algebraic Field Extensions In I. Martin Isaacs Algebra: A Graduate Course, Isaacs uses the field of algebraic numbers $$\mathbb{A}=\{\alpha \in \mathbb{C} \; \| \; \alpha \; \text{algebraic over} \; \mathbb{Q}\}$$ as an example of an infinite degree algebraic field extension. I have done a cursory google search and thought about it for a little while, but I cannot come up with a less contrived example. My question is What are some other examples of infinite degree algebraic field extensions? - 3 One example which is quite natural is the field given by adjoining all roots of unity or all radicals to $\mathbb Q$. But I must say, I don't like this question: It is not that hard to give you any number of infinite algebraic field extensions (take a suitable collection of polynomials and consider the splitting field), but it seems like there is very little to be learned from such an exercise in thinking up any examples. – Sam May 30 '12 at 13:50 3 @MTurgeon $\pi$ is transcendental over $\mathbb{Q}$, hence $\mathbb{Q}(\pi)/\mathbb{Q}$ is not algebraic. – Holdsworth88 May 30 '12 at 14:07 9 How are the algebraic numbers a contrived example? They are the largest algebraic extension of $\mathbb{Q}$! Any infinite algebraic extension lies in them! – Qiaochu Yuan May 30 '12 at 14:08 @Holdsworth88 I misread the requirements for this infinite degree extension. – M Turgeon May 30 '12 at 14:28 5 Answers Another simple example is the extension obtained by adjoining all roots of unity. Since adjoining a primitive $n$-th root of unity gives you an extension of degree $\varphi(n)$ and $\varphi(n)=n-1$ when $n$ is prime, you get algebraic numbers of arbitrarily large degree when you adjoin all roots of unity. - 8 And, as is well-known, this is the largest abelian algebraic extension – M Turgeon May 30 '12 at 14:30 The field of algebraic numbers is important, as is the field of real algebraic numbers. There are plenty of other examples of the same nature. The field of Euclidean constructible numbers is an extension field of the rationals, of infinite degree over the rationals, that comes up "naturally." - $\mathbb Q[\sqrt 2, \sqrt 3, \sqrt 5, \cdots]$, obtained by adjoining the square root of the primes, is an example because if you use just $n$ primes, you get an extension of degree $2^n$. - 5 – Qiaochu Yuan May 30 '12 at 14:10 1 – lhf May 30 '12 at 14:12 So I suppose $\mathbb{Q}(\sqrt{m_1},\sqrt{m_2},...)$, where the $m_i$ are square-free coprime integers would also be an example then. – Holdsworth88 May 30 '12 at 14:13 1 @QiaochuYuan: Dear Qiaochu, While you are right that this result is not obvious, it is not that hard; it follows from the most basic algebraic number theory (more precisely, computations of discriminants for quadratic extensions). Regards, – Matt E Jun 1 '12 at 18:35 How about the following example: for any field $k$, consider the field extension $\cup_{n\geq 1} k(t^{2^{-n}})$ of the field $k(t)$ of rational functions. This extension is algebraic and of infinite dimension. The idea behind is quite simple. But I admit it require some work to define the extension rigorously. - Let $\{n_1,n_2,...\}$ be pairwise coprime, nonsquare positive integers. Then $\mathbb{Q}(\sqrt{n_1},\sqrt{n_2},...)$ is an algebraic extension of infinite degree. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 19, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9184883236885071, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/23043-optimization-problem.html	# Thread: 1. ## Optimization Problem Here is the problem, A straight piece of wire is to be cut into two pieces. One piece is to be formed into a circle and the other piece into a square. Where should the wire be cut so that the sum of the areas of the circle and square will be greatest? The least? I know the square area formula x=side, x(squared). The same stuff for the circle, x=circumference, x=2(Pi)r I do not know the Length of the wire, but that length is a constant away (right?, I will assume this for now) so I will use “L”. So the side of the square will be x, the circumference will be L-x Stuck right here, not sure what to do next 2. Originally Posted by BClary A straight piece of wire is to be cut into two pieces. One piece is to be formed into a circle and the other piece into a square. Where should the wire be cut so that the sum of the areas of the circle and square will be greatest? The least? I know the square area formula x=side, x(squared). The same stuff for the circle, x=circumference, x=2(Pi)r I do not know the Length of the wire, but that length is a constant away (right?, I will assume this for now) so I will use “L”. So the side of the square will be x, the circumference will be L-x Stuck right here, not sure what to do next Good start, except that the perimeter of the square will be x (not the side). So the side of the square will be x/4, and the radius of the circle will be (L–x)/2π. That means that the total area enclosed will be $(x/4)^2 + \pi\Bigl({\textstyle\frac{L-x}{2\pi}}\Bigr)^2$. Now use calculus to find when that is a maximum or a minimum. 3. ## Ty wow, thats for the great start, looks like I am just a "little" algebra away now 4. ## What now Batman??? Optimization Problem Cont..... Okay, found the algebra easier if I made the square "L-x" and let the circle be "x" so the sum of the two areas would be Circle + Square = 2Pi(x/2Pi)(squared)+((L-x)/4)(squared) I think that is right. So the next step should be to take a derivative, SO I will do the left hand side first, add the right hand later (this can get ugly looking quickly here): Left hand side: A'(x)=[(Pix(squared))'(4Pi(square))-Pix(squared)(4Pi(squared)]/16Pi(squared) [(2xPi)(4Pi(squared))-0]/(16Pi(fourth)) 8xPi(third)/16Pi(fourth) x/2Pi for the left hand side right hand now: A'(x)=((L-x)/4)(Squared) 2(L-x)(-1)(16)-0/16 -2(L-x)/16 -(L-x)/8 Combined back: x/2Pi + -L+x/8 (4+xPi-Lpi)/8Pi so set top to zero 4+xPi-LPi=o Stuck again, been at this for hours, I know I need to find both max and min, but the L is causing me great problems. The original problem does not address L or ask for it, could I use a simple value of 1 "one" for it, as the ratio of any cut of the wire would apply to all lengths. I need to walk away for a moment, bang my head against a solid object, multiple times (no I will not take a derivative to find v of that), and hope someone reads this a gives me the push in the right direction Signed Soon to have a headache 5. Edit: I figured I should add an overview to make this easier to follow since it's so long. OVERVIEW 1. Set up the problem (define values and default relationships) 2. Solve for area of a square in terms of length of it's wire 3. Solve for area of a circle in terms of length of it's wire 4. Add them together to get a formula for total area 5. Find where the slope of the area = 0 6. Test that point. 7. Test the end points of your domain to see where they lie 8. Wrap up the process into a conclusion SETUP THE PROBLEM the length of the wire is L the length of the piece that will make the square is s the length of the piece that will make the circle is c c+s=L c=L-s FIND THE AREA OF THE SQUARE So, now you need to find an equation to make the square, you know the circumference of a square is 4 times one side, and that is equal to the length of your piece of wire which will make the square. So one side of the square will be 1/4 of s, or s/4. And since the Area of a square is the one side squared, you know that the area of the square is $(\frac{s}{4})^{2}$ which is $\frac{s^{2}}{16}$ FIND THE AREA OF THE CIRCLE Your piece of wire b will become the circumference of the circle, and the circumference of a circle is $2\pi r$ so $c=2\pi r$ and so $r=\frac{c}{2\pi}$. And you know the area of the circle is $\pi r^{2}$ so substitute the value of r into the equation for area and get $\pi(\frac{c}{2\pi})^{2}$ Which can be simplified to $\pi(\frac{c^{2}}{4\pi^{2}})$ Which simplifies to to $\frac{c^{2}}{4\pi}$ Now, you don't want two variables (s and c) and we know that c=L-s (L is not a variable, it is a constant). so we substitute that into our equation. $\frac{(L-s)^{2}}{4\pi}$ and simplify: $\frac{L^{2}-2Ls+s^{2}}{4\pi}$ FIND THE EQUATION FOR THE SUM OF THE SQUARE AND THE CIRCLE This is easy, just add the area of the square to the area of the circle $\frac{s^{2}}{16}+\frac{L^{2}-2Ls+s^{2}}{4\pi}$ Common denominator: $\frac{\pi s^{2}}{16\pi}+\frac{4(L^{2}-2Ls+s^{2})}{4(4\pi)}$ Simplify: $\frac{\pi S^{2} + 4L^{2}-8Ls+4s^{2}}{16\pi}$ And turn this into an equation, by realizing it is equal to our area, A, in terms of s (s is the only variable in this equation since we substituted out the b and L is a constant) $A(s)=\frac{\pi s^{2} + 4L^{2}-8Ls+4s^{2}}{16\pi}$ CONSIDER A=area in terms of s, so the value of A will give us our total area, and the graph of A will show us our area as it changes according to a. Now, we want the point where area is greatest. At that point, we know that it will be the highest on the graph of area. And we know that because it is highest, it must be higher than all the points around it, which means the slope of A is increasing up to it, and decreasing after it, and equal to 0 at that point. By the same thinking, wherever the area is minimized, the slope will be decreasing before that piont and increasing after it, and equal to zero on it. So wherever the slope is equal to zero is a potential maximum/minimum value of area. So lets find the slope and set it to zero, then test the points around it to see what the slope is doing. So we want to find the equation for the slope of this line, and then set that equal to zero. The derivative is the equation for the slope of the line, so lets find the derivative. FIND THE DERIVATIVE Initial equation: $A\prime(s)=\frac{\pi s^{2} + 4L^{2}-8Ls+4s^{2}}{16\pi}$ Rewrite: $A\prime(s)=\frac{1}{16\pi}(\pi s^{2} + 4L^{2}-8Ls+4s^{2})$ Differentiate (remember that L is a constant, and the derivative of a constant is zero) $A\prime(s)=\frac{1}{16\pi}(2\pi s -8L+8s)$ FIND THE ZERO OF THE SLOPE Now we have the equation of our slope, lets find where it is equal to zero as that will be a potential maximum value: $0=\frac{1}{16\pi}(2\pi s -8L+8s)$ Divide out the irrelevant constant $0=2\pi s -8L+8s$ Divide out a 2 $0=\pi s -4L+4s$ Factor out an s $0=s(\pi +4) -4L$ Add 4L $4L=s(\pi +4)$ Divide by the coefficient of s $\frac{4L}{\pi +4}=s$ CONSIDER Now we have a value for s where the change in the total area of our circle and square is zero, but lets make sure that it is a maximum. Because all of the terms we are using are arbitrary, (s, c, L) it is very difficult to choose actual points to plug into our first derivative, so what we can instead do is find the second derivative, plug our point into the second derivative and see whether it is concave up or down. If it is concave down, then the open end is down, so it must be increasing up to that point and decreasing after it, making it a maximum. If the open end is up, it looks like a parabola, and we can see that our point is at the bottom, making it a minimum. FIND THE SECOND DERIVATIVE First derivative $A\prime(s)=\frac{1}{16\pi}(2\pi s -8L+8s)$ Differentiate: $A\prime\prime(s)=\frac{1}{16\pi}(2\pi +8)$ Simplify $A\prime\prime(s)=\frac{\pi +4}{8\pi}$ CONSIDER: This tells us that Now we can see that the second derivative is ALWAYS positive, this means that the slope is always concave up, so our s value must be a minimum. This tells us that when $s=\frac{4L}{\pi +4}$ our area is at a minimum. Well what do we do now? what about our maximum? There was only 1 value where the prime was equal to zero, so where does our other point come from? Well, we only have 2 other critical numbers, the two endpoints of our graph, where the domain begins and ends. Now we know we can't have a length less than zero, so s must be > zero. And we can't have a length of s that is greater than the length of the wire, so s is < to L. This means that the domain is 0 < s < L Now we have two more critical values to check out. Whichever is higher must be our maximum value. CHECK THE AREA OF OUR TWO ENDPOINTS Initial equation $A(s)=\frac{\pi s^{2} + 4L^{2}-8Ls+4s^{2}}{16\pi}$ Check area of s=zero $A(0)=\frac{\pi 0^{2} + 4L^{2}-8L0+40^{2}}{16\pi}$ Simplify $A(0)=\frac{4L^{2}}{16\pi}$ Simplify $A(0)=\frac{L^{2}}{4\pi} \approx .079599L^{2}$ Check area of s=L $A(L)=\frac{\pi L^{2} + 4L^{2}-8L*L+4L^{2}}{16\pi}$ Simplify $A(L)=\frac{\pi L^{2}}{16\pi}$ Simplify $A(L)=\frac{L^{2}}{16} = .0625L^{2}$ CONSIDER Now our two endpoints are each in terms of L^2, and we know that L is positive, so we can just compare their coefficients and see which is greater: .079599 > .0625 Therefore our area is maximized when s equals 0 CONCLUSION We know our area is minimized when $s=\frac{4L}{\pi +4}$ And maximized when $s=0$ So to get the greatest area, do not cut the wire at all and instead use it to make only a circle, and to get the least area, cut the wire $\frac{4}{\pi+4}$ of the length and use that to make a square, and the remaining bit to make a circle. ---- There was a lot in there, I hope I didn't make any errors :/ 6. ## TY - Angel you are If you are not a teacher, you should be, great explaination, its earily, but I will review and review, thank you	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 39, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.927713930606842, "perplexity_flag": "head"}
http://mathoverflow.net/questions/22189/what-is-your-favorite-strange-function/102367	## What is your favorite “strange” function? [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) There are many "strange" functions to choose from and the deeper you get involved with math the more you encounter. I consciously don't mention any for reasons of bias. I am just curious what you consider strange and especially like. Please also give a reason why you find this function strange and why you like it. Perhaps you could also give some kind of reference where to find further information. As usually: Please only mention one function per post - and let the votes decide :-) - 2 m.reddit.com/r/math/comments/9txzv/… – Regenbogen Apr 22 2010 at 14:21 2 Counterexamples in analysis has some nice ones: books.google.nl/… – skupers Aug 21 2010 at 18:40 show 3 more comments ## 37 Answers Characteristic p commutative algebra leads naturally to the construction of various continuous functions on [0,1]^m that have beautiful self-similarity properties; for explication and some pictures see: ````Pedro Teixeira, Syzygy gap fractals--I, arXiv 1008.0583 ```` - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. How about the function given by the Banach-Tarski paradox? This maps a ball into two copies of the same size ball, and is composed of isometries on subsets of $\mathbb{R}^3$. - Nonconstant continuous locally recurrent functions are quite unintuitive. A real-valued function is locally recurrent on $\mathbb R$ if for every $x_0\in\mathbb R$ and every deleted neighborhood $N(x_0)$ of $x_0$, there exists $x\in N(x_0)$ for which $f(x)=f(x_0)$. Thus in some sense a nonconstant continuous locally recurrent function looks everywhere like $x\sin(1/x)$ at $x=0$. See papers in the American Math. Monthly of Bush (1962), Marcus (1963), and Mauldon (1965). - The formula for the nth term in the Fibonacci sequence $F_{n} = \cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1+\sqrt{5}}{2}\right)^n-\cfrac{1}{\sqrt{5}}\cdot\left(\cfrac{1-\sqrt{5}}{2}\right)^n$ This is interesting because it is a non-recursive expression for the Fibonacci sequence and also because it involves the golden ratio. - Just a simple construction to illustrate Nate Eldredge's answer about functions with dense graphs. Pick any $\mathbb{R}$-vector space E with a norm. On E, choose a non-continuous linear form $L: E \to \mathbb{R}$; now this can only be done if $\dim(E)=\infty$, of course. Then, pick y such that $L(y)=1$, and let $T: E \to E$ be defined by $Tx=x-L(x)y$. Then obviously T maps E onto the kernel of L; it is not difficult to prove that $\ker (L)$ must be dense in E for any non-continuous L (the two conditions are even equivalent), and thus the graph of T must be dense in $E \times E$. - How about a function f: f(f(x)) = exp(x). - I like the beauty and mysticism of Euler's identity: $$f(\theta) = e^{i\theta} = \cos\theta + i \sin\theta$$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9154516458511353, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/120853/list	## Return to Question 3 contains vs is contained in Let $G$ be a connected, simply connected, semi-simple algebraic group defined and split over a local non-arch field $k$ with integer ring $R$. Let $B$ be the corresponding reduced building. Fix an apartment $A$ and corresponding root system $\Phi$ and split torus $T$; then for each $x \in A$ and $r>0$, we have a corresponding Moy-Prasad filtration subgroup $G_{x,r}$. Our hypotheses give $G_x=G_{x,0}$. Define $\Omega_{A,r}$ as the set $${ y \in A : \forall \alpha \in \Phi, \vert \alpha(x-y) \vert \leq r }.$$ Is it true that $T(R)G_{x,r} = \cap_{y\in \Omega_{A,r}} G_y$ ? (With $r>0$ the product gives a group which contains is contained in the given intersection. Equality seems to be about whether all hyperplanes $H_{\alpha,r}$ meet $\Omega_{A,r}$.) Let $A(x)$ be the set of all apartments of $B$ containing $x$. Let $Z$ be the (finite) center of $G$. Is it true that $ZG_{x,r} = \cap_{A \in A(x)}\cap_{y\in \Omega_{A,r}}G_y$ ? Equiv by (1): Is $ZG_{x,r} =\cap_{g \in G_x} (gT(R)g^{-1})G_{x,r}$? This is part of the larger question: I would like to understand the stabilizers of certain subsets of $B$ (which are not contained within any apartment $A$). Pointers to any literature much appreciated. 2 Improved formatting. Let $G$ be a connected, simply connected, semi-simple algebraic group defined and split over a local non-arch field $k$ with integer ring $R$. Let $B$ be the corresponding reduced building. Fix an apartment $A$ and corresponding root system $\Phi$ and split torus $T$; then for each $x \in A$ and $r>0$, we have a corresponding Moy-Prasad filtration subgroup $G_{x,r}$. Our hypotheses give $G_x=G_{x,0}$. Define $\Omega_{A,r}$ as the set $${ y \in A : \forall \alpha \in \Phi, \vert \alpha(x-y) \vert \leq r }.$$ Is it true that $T(R)G_{x,r} = \cap_{y\in \Omega_{A,r}} G_y$ ? (With $r>0$ the product gives a group which contains the given intersection. Equality seems to be about whether all hyperplanes $H_{\alpha,r}$ meet $\Omega_{A,r}$.) Let $A(x)$ be the set of all apartments of $B$ containing $x$. Let $Z$ be the (finite) center of $G$. Is it true that $ZG_{x,r} = \cap_{A \in A(x)}\cap_{y\in \Omega_{A,r}}G_y$ ? Equiv by (1): Is $ZG_{x,r} =\cap_{g \in G_x} (gT(R)g^{-1})G_{x,r}$? This is part of the larger question: I would like to understand the stabilizers of certain subsets of $B$ (which are not contained within any apartment $A$). Pointers to any literature much appreciated. 1 # When is a Moy-Prasad filtration subgroup the stabilizer of a subset of the building (up to center)? Let $G$ be a connected, simply connected, semi-simple algebraic group defined and split over a local non-arch field $k$ with integer ring $R$. Let $B$ be the corresponding reduced building. Fix an apartment $A$ and corresponding root system $\Phi$ and split torus $T$; then for each $x \in A$ and $r>0$, we have a corresponding Moy-Prasad filtration subgroup $G_{x,r}$. Our hypotheses give $G_x=G_{x,0}$. Define $\Omega_{A,r}$ as the set $${ y \in A : \forall \alpha \in \Phi, \vert \alpha(x-y) \vert \leq r }.$$ 1. Is it true that $T(R)G_{x,r} = \cap_{y\in \Omega_{A,r}} G_y$ ? (With $r>0$ the product gives a group which contains the given intersection. Equality seems to be about whether all hyperplanes $H_{\alpha,r}$ meet $\Omega_{A,r}$.) Let $A(x)$ be the set of all apartments of $B$ containing $x$. Let $Z$ be the (finite) center of $G$. 1. Is it true that $ZG_{x,r} = \cap_{A \in A(x)}\cap_{y\in \Omega_{A,r}}G_y$ ? Equiv by (1): Is $ZG_{x,r} =\cap_{g \in G_x} (gT(R)g^{-1})G_{x,r}$? This is part of the larger question: I would like to understand the stabilizers of certain subsets of $B$ (which are not contained within any apartment $A$). Pointers to any literature much appreciated.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 75, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.916766881942749, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/3439/mpt-adding-constraint-on-minimum-asset-weight?answertab=active	# MPT: Adding constraint on minimum asset weight I'm new to finance in general, and recently read about Modern Portfolio Theory. Now I'm wondering how to add the following constraint on asset weights: • Each asset weight $w_i$ should either be $w_i = 0$, or it should be $0.05 <= w_i <= 1.0$ (With 0.05 as the lower bound just being an example.) From doodling a bit, it looks to me as if that would give a non-convex problem, and thus the usual optimization approaches won't work. Can someone point me into the right direction on how to efficiently solve this problem for a large number of assets? Edit: Alternatively, I could reformulate the additional constraint as • For each asset weight $0.05 <= w_i <= 1.0$ • For each asset there is an indicator $I_i \in {0, 1}$ • The combined weight of the selected assets must be 1: $\sum I_i w_i = 1$ What optimization technique is suitable for this problem? - This is actually a standard linear constraint that just about any basic portfolio optimizer can handle (assuming you don't have any integer constraints, like maximum number of names). Is there a particular software package that you're using? – michaelv2 May 4 '12 at 20:25 – rodion May 4 '12 at 20:55 I'm not sure what you're saying is as simple as what michael is saying. He is solving a traditional problem that uses linear inequality constraints and appears to be correct. If I'm not mistaken, what you're saying is technically a mixed integer problem. I think CPLEX is a popular solver to use. You'd have to look at some examples on how to set it up, but it's similar to what michael did. – John May 4 '12 at 21:31 I think I would express the third bullet as it is done traditionally, without the I. The trick is the first constraint needs to be re-specified so it's like an either/or constraint. You need to make it so that it is between 0.05% and 1% or equal to zero. – John May 4 '12 at 21:35 – michaelv2 May 4 '12 at 22:04 show 1 more comment ## 2 Answers You can use a branch-and-cut algorithm (this is what mixed-integer solvers use). The idea is to solve the problem recursively, by considering subproblems in which the constraints are $w_i=0$ for some stocks and $0.05 \leq w_i \leq 1$ for others. This gives $2^n$ convex optimization problems, and you want the best solution among them. Even when $n$ is small, that too much, but you can arrange those optimization problems in a tree, and prune large parts of it, as follows. The root of the tree has constraints of the form $0 \leq w_i \leq 1$ (and gives a bound on the value of the optimal portfolio). Its children add constraints on the first stock: $w_1=0$ for the first child, $0.05\leq w_1 \leq 1$ for the second. The grand-children similarily add constraints on the second stock, and so on. The fully constrained problems we are interested in are the leaves of this tree. First solve the problem at the root of the tree and one leaf: this gives you two values that bound the value of the optimal portfolio. Then, search the tree, depth-first, but discard a subtree without exploring it if its value is worse than the best leaf found so far. - Thanks a lot! The bounds for w_i you mention are a bit different from my question, but I understand the concept. – rodion May 5 '12 at 10:38 I made an edit but was not really sure about "Its children add..." so I reverted it. – Bob Jansen May 5 '12 at 14:51 @BobJansen: I have made the change you suggested, to match the constraints in the question. – Vincent Zoonekynd May 5 '12 at 15:03 A very basic implementation using the quadprog package in R would look something like the following: ````library(quadprog) library(MASS) # -------------------------------------------------------- # Generate a set of random returns for a covariance matrix # -------------------------------------------------------- set.seed(100) n <- 100 # number of assets m <- 200 # number of states of the world rho <- 0.7 sigma <- 0.2 mu <- .10 Cov <- matrix(rhosigmasigma, ncol=n, nrow=n) diag(Cov) <- rep(sigma*sigma, n) S <- 1 + matrix(mvrnorm(m, rep(mu, n), Sigma=Cov), ncol=n) Dmat <- var(S) # -------------------------------------------------------- # Setup quadratic problem # -------------------------------------------------------- # The weights must sum to 1 Amat <- matrix(1, n) # x >= 5% bLo <- rep(0, n) bvec <- c(1, bLo) Amat <- cbind(Amat, diag(n)) # x <= 5% bHi <- rep(.05, n) bvec <- c(bvec, -bHi) Amat <- cbind(Amat, -diag(n)) dvec <- rep(0, nrow(Amat)) meq <- 1 # the first column of Amat is an equality constraint sol <- solve.QP(Dmat=Dmat, dvec=dvec, Amat=Amat, bvec=bvec, meq) ```` - That only considers the constraint $0 \leq w_i \leq 0.05$, not the non-convex constraint $w_i=0$ or $0.01 \leq w_i \leq 0.05$''. – Vincent Zoonekynd May 4 '12 at 23:15 Quite right, I completely mis-read that first part. I'll re-formulate the answer using Rglpk and post an edit shortly... – michaelv2 May 4 '12 at 23:36 1 The constraints $w_i = 0$ or $0.01 \leq w_i \leq 0.05$'' can be rewritten as $0.01 n_i \leq w_i \leq 0.05 n_i$ by adding binary variables $n_i \in \{0,1\}$. The constraints are still linear, but the objective function is still quadratic: we would need a mixed integer quadratic solver -- Rglpk is only a (mixed integer) linear solver... – Vincent Zoonekynd May 5 '12 at 12:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 2, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9085209965705872, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/108231-basis-dimension-matrices.html	Thread: 1. Basis and dimension of Matrices Hello, first time posting on the forum. Been here a lot in my past couple of semesters with homework help. I've browsed as much as I could for help on this problem but I haven't made any advancement. Any help would be greatly appreciated. The problem reads Find a basis and the dimension of the subspace W of V = $M_{2,2}$, spanned by $A = \left(\begin{array}{cc}1&-5\\-4&2\end{array}\right)$ $B = \left(\begin{array}{cc}1&1\\-1&5\end{array}\right)$ $C = \left(\begin{array}{cc}2&-4\\-5&7\end{array}\right)$ $D = \left(\begin{array}{cc}1&-7\\-5&1\end{array}\right)$. TBH I barely have any idea where to start here. I understand bases and I can do them when they involve 1x1 vectors, but these are 2x2 and have no idea how to set it up. TIA DP 2. Originally Posted by illness Hello, first time posting on the forum. Been here a lot in my past couple of semesters with homework help. I've browsed as much as I could for help on this problem but I haven't made any advancement. Any help would be greatly appreciated. The problem reads Find a basis and the dimension of the subspace W of V = $M_{2,2}$, spanned by $A = \left(\begin{array}{cc}1&-5\\-4&2\end{array}\right)$ $B = \left(\begin{array}{cc}1&1\\-1&5\end{array}\right)$ $C = \left(\begin{array}{cc}2&-4\\-5&7\end{array}\right)$ $D = \left(\begin{array}{cc}1&-7\\-5&1\end{array}\right)$. TBH I barely have any idea where to start here. I understand bases and I can do them when they involve 1x1 vectors, but these are 2x2 and have no idea how to set it up. TIA DP You have to find out how many, and what, matrices here are linearly independent and then those matrices span W. For example, and as a little help, note that C = A + B ==> C is lin. dependent on A,B and thus we can count it out. Check now what happens if we take A,B,D... Tonio 3. I'm not quite sure I follow. It's just I don't know what to do with the matrices. To be a little clearer, as far as I understand with vectors and such, is that you set one free variable =1 and the rest 0 and thats your basis. <- taking the solution from Ax=0. I'm sure the process is different because unless the approach is something along the lines of c1A + c2B + c3C + c4D = 0, how do i play around with the matrices into something to work with? Dp	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9723381996154785, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-equations/130621-linear-equation-integrating-factor-print.html	# linear equation with integrating factor Printable View • February 24th 2010, 03:47 PM collegestudent321 linear equation with integrating factor Hello, I have been trying this problem for about an hour now and I just can't seem to figure it out: dA/dt + 2A/(50+t) = 3 I found that the integrating factor = (50+t)^2 and so d/dt (50+t)^2(A) = 3(50+t)^2 but after this my problem falls apart... I can't seem to figure out how to go from there... I tried integrating both sides and factoring but that was not the right answer... Any help would be greatly appreciated! Thank you! • February 24th 2010, 06:18 PM arbolis Quote: Originally Posted by collegestudent321 Hello, I have been trying this problem for about an hour now and I just can't seem to figure it out: dA/dt + 2A/(50+t) = 3 I found that the integrating factor = (50+t)^2 Until now, everything's fine. Quote: Originally Posted by collegestudent321 and so d/dt (50+t)^2(A) = 3(50+t)^2 I'm not really sure what you did here. Maybe Latex could help me to understand. Quote: but after this my problem falls apart... I can't seem to figure out how to go from there... I tried integrating both sides and factoring but that was not the right answer... Any help would be greatly appreciated! Thank you! From it, multiply both side by the IF. Then integrate with respect to t. You'll reach $(50+t)^2y=3\int (50+t)^2 dt$. • February 24th 2010, 08:20 PM Prove It Quote: Originally Posted by collegestudent321 Hello, I have been trying this problem for about an hour now and I just can't seem to figure it out: dA/dt + 2A/(50+t) = 3 I found that the integrating factor = (50+t)^2 and so d/dt (50+t)^2(A) = 3(50+t)^2 but after this my problem falls apart... I can't seem to figure out how to go from there... I tried integrating both sides and factoring but that was not the right answer... Any help would be greatly appreciated! Thank you! $\frac{dA}{dt} + \frac{2A}{50 + t} = 3$ $(50 + t)^2\frac{dA}{dt} + 2(50 + t)A = 3(50 + t)^2$ $\frac{d}{dt}[(50 + t)^2A] = 3(50 + t)^2$ $(50 + t)^2A = \int{3(50 + t)^2\,dt}$ $(50 + t)^2A = (50 + t)^3 + C$ $A = 50 + t + \frac{C}{(50 + t)^2}$. • February 25th 2010, 04:21 AM collegestudent321 wow... I would have never guessed that. Thank you so much! • February 25th 2010, 07:26 AM arbolis Quote: Originally Posted by collegestudent321 wow... I would have never guessed that. Thank you so much! I suggest you to read the second post of this thread: http://www.mathhelpforum.com/math-he...-tutorial.html. If you would have never guessed how to proceed after getting the IF, it means you didn't really understand the method itself. Chris' explanation of the method in 3 lines explain it very well. Have a look. All times are GMT -8. The time now is 04:40 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 7, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9723857045173645, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/170346-finding-nth-derivative.html	# Thread: 1. ## Finding the nth derivative Find the nth derivative of $f(x) = x^n$ by calculating the first few derivatives and observing the pattern that occurs. So, here's what I've done so far: $f'(x) = nx^{x-1})$ $f''(x) = n(n-1)x^{n-2}$ $f''(x) = (n^2-n)x^{n-2}$ $f'''(x) = (n^3-2n^2+2n)x^{n-3}$ $f^{(4)}(x) = (n^4 - 5n^3 + 8n^2 - 6n)x^{n-4}$ The pattern of the exponents is clear to me, but the pattern of the coefficients is not. Can anybody help me? Thanks. 2. Maybe letting them as such and not expanding will help you? $f'(x) = nx^{n-1}$ $f''(x) = n(n-1)x^{n-2}$ $f'''(x) = n(n-1)(n-2)x^{n-3}$ $f''''(x) = n(n-1)(n-2)(n-3)x^{n-4}$ 3. Originally Posted by joatmon Find the nth derivative of $f(x) = x^n$ by calculating the first few derivatives and observing the pattern that occurs. So, here's what I've done so far: $f'''(x) = (n^3-2n^2+2n)x^{n-3}$ $f^{(4)}(x) = (n^4 - 5n^3 + 8n^2 - 6n)x^{n-4}$ The pattern of the exponents is clear to me, but the pattern of the coefficients is not. Can anybody help me? Thanks. $f'=nx^{n-1}$ $f''=n(n-1)x^{n-2}$ $f'''=n(n-1)(n-2)x^{n-3}$ $f^{(4)}=n(n-1)(n-2)(n-3)x^{n-3}$ So, looks like maybe the coefficients are following some kind of reverse factorial... Something like $f^{(k)}=\frac{n!}{(n-k)!}x^{n-k}$ I could be wrong because I have to go and I haven't checked my work here, but maybe... 4. First, consider the nth derivative of $x^m$ (for clarity). Finding the first few derivatives indicate that we've: $\displaystyle (x^m)^{(n)} = x^{m-n}\prod_{k=0}^{n-1}(n-k)$. This can be proven by induction. Put $m = n$ to get: $\begin{aligned}\displaystyle (x^n)^{(n)} = x^{n-n}\prod_{k=0}^{n-1}(n-k) = \prod_{k=0}^{n-1}(n-k) = n!\end{aligned}$ 5. All of these replies are helpful. Unknown008, that helps a lot to not multiply them out. I can see the sequence now clearly. What I don't know is how to write this in mathematical notation. The other two posts are a little beyond my abilities. I don't understand the notation used by CoffeeMachine, and I'm not sure that I agree with VonNemo's (although it might be right). Can anybody help me with how to write this out? Thanks. 6. You might not be familar with the factorial notation, which is: $n! = n(n-1)(n-2)(n-3)...(n-(n-1))$ For example, $5! = 5\times4\times 3 \times 2 \times 1$ In the derivative, you put n! in the numerator, but you get everything till 1, but that's not in the derivative! So you divide by another factorial, such that the factorial in the denominator will cancel out anything that should not be in the numerator. Example, you want 9 x 8 x 7 x 6 But you have 9! What you do, is divide by 5!, so that: $\dfrac{9!}{5!} = \dfrac{9\times8\times7\times6\times5\times4\times3 \times2\times1}{5\times4\times 3 \times 2 \times 1}$ What is unneeded after 6 is cancelled out. 7. I'm somewhat familiar with factorial notation, but not to the extent that I feel comfortable writing this like you have done. Here is what I am going to do: $\frac{d^ky}{dx^k} = n(n-1)(n-2)...(n-k+1) x^{n-k}$ Is the Leibnitz notation correct? They are throwing this notation at us in the book and in lecture, but never really telling us how to write it for ourselves. Thanks. 8. Originally Posted by joatmon Is the Leibnitz notation correct? If $y = x^n$, then what you've found is the kth derivative $y$. So if that's what you wanted, then it's absolutely fine. But wasn't your question, as stated in the OP, to find the nth derivative of $x^n$? If so, then you need to replace $k$ by $n$ in the equation you have found for the kth derivative of $y$, and then note that $x^{n-n} = 1$. The product $n(n-1)(n-2)...(n-k+1)$ evaluates to $n!$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 35, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.963980495929718, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Acute_angle	# Angle (Redirected from Acute angle) "Oblique angle" redirects here. For the cinematographic technique, see Dutch angle. ∠, the angle symbol in Unicode is U+2220 In geometry, an angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle.[1] Angles are usually presumed to be in a Euclidean plane or in the Euclidean space, but are also defined in non-Euclidean geometries. In particular, in spherical geometry, the spherical angles are defined, using arcs of great circles instead of rays. Angle is also used to designate the measure of an angle or of a rotation. This measure is the ratio of the length of a circular arc to its radius. In the case of a geometric angle, the arc is centered at the vertex and delimited by the sides. In the case of a rotation, the arc is centered at the center of the rotation and delimited by any other point and its image by the rotation. The word angle comes from the Latin word angulus, meaning "a corner". The word angulus is a diminutive, of which the primitive form, angus, does not occur in Latin. Cognate words are the Greek ἀγκύλος (ankylοs), meaning "crooked, curved," and the English word "ankle". Both are connected with the Proto-Indo-European root *ank-, meaning "to bend" or "bow".[2] Euclid defines a plane angle as the inclination to each other, in a plane, of two lines which meet each other, and do not lie straight with respect to each other. According to Proclus an angle must be either a quality or a quantity, or a relationship. The first concept was used by Eudemus, who regarded an angle as a deviation from a straight line; the second by Carpus of Antioch, who regarded it as the interval or space between the intersecting lines; Euclid adopted the third concept, although his definitions of right, acute, and obtuse angles are certainly quantitative.[3] ## Measuring angles The size of a geometric angle is usually characterized by the magnitude of the smallest rotation that maps one of the rays into the other. Angles that have the same size are sometimes called congruent angles. In some contexts, such as identifying a point on a circle or describing the orientation of an object in two dimensions relative to a reference orientation, angles that differ by an exact multiple of a full turn are effectively equivalent. In other contexts, such as identifying a point on a spiral curve or describing the cumulative rotation of an object in two dimensions relative to a reference orientation, angles that differ by a non-zero multiple of a full turn are not equivalent. The measure of angle θ is the quotient of s and r. In order to measure an angle θ, a circular arc centered at the vertex of the angle is drawn, e.g. with a pair of compasses. The length of the arc s is then divided by the radius of the arc r, and possibly multiplied by a scaling constant k (which depends on the units of measurement that are chosen): $\theta = k \frac{s}{r}.$ The value of θ thus defined is independent of the size of the circle: if the length of the radius is changed then the arc length changes in the same proportion, so the ratio s/r is unaltered. ### Units Units used to represent angles are listed below in descending magnitude order. Of these units, the degree and the radian are by far the most commonly used. Angles expressed in radians are dimensionless for the purposes of dimensional analysis. Most units of angular measurement are defined such that one turn (i.e. one full circle) is equal to n units, for some whole number n. The two exceptions are the radian and the diameter part. For example, in the case of degrees, n = 360. A turn of n units is obtained by setting k = n/(2π) in the formula above. (Proof. The formula above can be rewritten as k = θr/s. One turn, for which θ = n units, corresponds to an arc equal in length to the circle's circumference, which is 2πr, so s = 2πr. Substituting n for θ and 2πr for s in the formula, results in k = nr/(2πr) = n/(2π).) • The cycle (or turn, full circle, revolution, or rotation) is one full circle. A turn can be subdivided in centiturns and milliturns. A turn is abbreviated $\tau$ or rev or rot depending on the application, but just r in rpm (revolutions per minute). 1 turn = 360° = 2π rad = 400 grad = 4 right angles. • The quadrant is 1/4 of a turn, i.e. a right angle. It is the unit used in Euclid's Elements. 1 quad. = 90° = π/2 rad = 1/4 turn = 100 grad. In German the symbol ∟ has been used to denote a quadrant. • The sextant (angle of the equilateral triangle) is 1/6 of a turn. It was the unit used by the Babylonians,[4] and is especially easy to construct with ruler and compasses. The degree, minute of arc and second of arc are sexagesimal subunits of the Babylonian unit. 1 Babylonian unit = 60° = π/3 rad ≈ 1.047197551 rad. θ = s/r rad = 1 rad. • The radian is the angle subtended by an arc of a circle that has the same length as the circle's radius (k = 1 in the formula given earlier). One turn is 2π radians, and one radian is 180/π degrees, or about 57.2958 degrees. The radian is abbreviated rad, though this symbol is often omitted in mathematical texts, where radians are assumed unless specified otherwise. When radians are used angles are considered as dimensionless. The radian is used in virtually all mathematical work beyond simple practical geometry, due, for example, to the pleasing and "natural" properties that the trigonometric functions display when their arguments are in radians. The radian is the (derived) unit of angular measurement in the SI system. • The diameter part (occasionally used in Islamic mathematics) is 1/60 radian. One "diameter part" is approximately 0.95493°. • The astronomical hour angle is 1/24 of a turn. Since this system is amenable to measuring objects that cycle once per day (such as the relative position of stars), the sexagesimal subunits are called minute of time and second of time. Note that these are distinct from, and 15 times larger than, minutes and seconds of arc. 1 hour = 15° = π/12 rad = 1/6 quad. = 1/24 turn ≈ 16.667 grad. • The point, used in navigation, is 1/32 of a turn. 1 point = 1/8 of a right angle = 11.25° = 12.5 grad. Each point is subdivided in four quarter-points so that 1 turn equals 128 quarter-points. • The binary degree, also known as the binary radian (or brad), is 1/256 of a turn.[5] The binary degree is used in computing so that an angle can be efficiently represented in a single byte (albeit to limited precision). Other measures of angle used in computing may be based on dividing one whole turn into 2n equal parts for other values of n.[6] • The degree, denoted by a small superscript circle (°), is 1/360 of a turn, so one turn is 360°. One advantage of this old sexagesimal subunit is that many angles common in simple geometry are measured as a whole number of degrees. Fractions of a degree may be written in normal decimal notation (e.g. 3.5° for three and a half degrees), but the "minute" and "second" sexagesimal subunits of the "degree-minute-second" system are also in use, especially for geographical coordinates and in astronomy and ballistics: • The grad, also called grade, gradian, or gon, is 1/400 of a turn, so a right angle is 100 grads. It is a decimal subunit of the quadrant. A kilometre was historically defined as a centi-grad of arc along a great circle of the Earth, so the kilometer is the decimal analog to the sexagesimal nautical mile. The grad is used mostly in triangulation. • The mil is approximately equal to a milliradian. There are several definitions ranging from 0.05625 to 0.06 degrees (3.375 to 3.6 minutes), with the milliradian being approximately 0.05729578 degrees (3.43775 minutes). In NATO countries, it is defined as 1/6400th of a circle. Its value is approximately equal to the angle subtended by a width of 1 metre as seen from 1 km away (2π / 6400 = 0.0009817… ≒ 1/1000). • The minute of arc (or MOA, arcminute, or just minute) is 1/60 of a degree = 1/21600 turn. It is denoted by a single prime ( ′ ). For example, 3° 30′ is equal to 3 + 30/60 degrees, or 3.5 degrees. A mixed format with decimal fractions is also sometimes used, e.g. 3° 5.72′ = 3 + 5.72/60 degrees. A nautical mile was historically defined as a minute of arc along a great circle of the Earth. • The second of arc (or arcsecond, or just second) is 1/60 of a minute of arc and 1/3600 of a degree. It is denoted by a double prime ( ″ ). For example, 3° 7′ 30″ is equal to 3 + 7/60 + 30/3600 degrees, or 3.125 degrees. • The hexacontade is a unit of 6° that Eratosthenes used, so that a whole turn was divided into 60 units. • The Babylonians sometimes used the unit pechus of about 2° or 2½°. ### Positive and negative angles Although the definition of the measurement of an angle does not support the concept of a negative angle, it is frequently useful to impose a convention that allows positive and negative angular values to represent orientations and/or rotations in opposite directions relative to some reference. In a two dimensional Cartesian coordinate system, an angle is typically defined by its two sides, with its vertex at the origin. The initial side is on the positive x-axis, while the other side or terminal side is defined by the measure from the initial side in radians, degrees, or turns. With positive angles representing rotations toward the positive y-axis and negative angles representing rotations toward the negative y-axis. When Cartesian coordinates are represented by standard position, defined by the x-axis rightward and the y-axis upward, positive rotations are anticlockwise and negative rotations are clockwise. In many contexts, an angle of −θ is effectively equivalent to an angle of "one full turn minus θ". For example, an orientation represented as − 45° is effectively equivalent to an orientation represented as 360° − 45° or 315°. However, a rotation of − 45° would not be the same as a rotation of 315°. In three dimensional geometry, "clockwise" and "anticlockwise" have no absolute meaning, so the direction of positive and negative angles must be defined relative to some reference, which is typically a vector passing through the angle's vertex and perpendicular to the plane in which the rays of the angle lie. In navigation, bearings are measured relative to north. By convention, viewed from above, bearing angle are positive clockwise, so a bearing of 45° corresponds to a north-east orientation. Negative bearings are not used in navigation, so a north-west orientation corresponds to a bearing of 315°. ### Alternative ways of measuring the size of an angle There are several alternatives to measuring the size of an angle by the corresponding angle of rotation. The grade of a slope, or gradient is equal to the tangent of the angle, or sometimes the sine. Gradients are often expressed as a percentage. For very small values (less than 5%), the grade of a slope is approximately the measure of an angle in radians. In rational geometry the spread between two lines is defined at the square of sine of the angle between the lines. Since the sine of an angle and the sine of its supplementary angle are the same any angle of rotation that maps one of the lines into the other leads to the same value of the spread between the lines. ### Astronomical approximations Astronomers measure angular separation of objects in degrees from their point of observation. • 1° is approximately the width of a little finger at arm's length. • 10° is approximately the width of a closed fist at arm's length. • 20° is approximately the width of a handspan at arm's length. These measurements clearly depend on the individual subject, and the above should be treated as rough approximations only. ## Identifying angles In mathematical expressions, it is common to use Greek letters (α, β, γ, θ, φ, ...) to serve as variables standing for the size of some angle. (To avoid confusion with its other meaning, the symbol π is typically not used for this purpose.) Lower case roman letters (a, b, c, ...) are also used. See the figures in this article for examples. In geometric figures, angles may also be identified by the labels attached to the three points that define them. For example, the angle at vertex A enclosed by the rays AB and AC (i.e. the lines from point A to point B and point A to point C) is denoted ∠BAC or $\widehat{\rm BAC}.$ Sometimes, where there is no risk of confusion, the angle may be referred to simply by its vertex ("angle A"). Potentially, an angle denoted, say, ∠BAC might refer to any of four angles: the clockwise angle from B to C, the anticlockwise angle from B to C, the clockwise angle from C to B, or the anticlockwise angle from C to B, where the direction in which the angle is measured determines its sign (see Positive and negative angles). However, in many geometrical situations it is obvious from context that the positive angle less than or equal to 180 degrees is meant, and no ambiguity arises. Otherwise, a convention may be adopted so that ∠BAC always refers to the anticlockwise (positive) angle from B to C, and ∠CAB to the anticlockwise (positive) angle from C to B. ## Types of angles ### Individual angles Acute (a), obtuse (b), and straight (c) angles. • Angles smaller than a right angle (less than 90°) are called acute angles ("acute" meaning "sharp"). • An angle equal to 1/4 turn (90° or π/2 radians) is called a right angle. Two lines that form a right angle are said to be normal, orthogonal, or perpendicular. • Angles larger than a right angle and smaller than a straight angle (between 90° and 180°) are called obtuse angles ("obtuse" meaning "blunt"). • Angles equal to 1/2 turn (180° or π radians) are called straight angles. • Angles larger than a straight angle but less than 1 turn (between 180° and 360°) are called reflex angles. • Angles equal to 1 turn (360° or 2π radians) are called full angles or a perigon. • Angles that are not right angles or a multiple of a right angle are called oblique angles. The names, intervals, and measured units are shown in a table below: \| \| \| \| \| \| Units \| Interval \| \|---------\|----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\|-----------------------------------------------------------------------------\| \| Name \| acute \| right angle \| obtuse \| straight \| reflex \| perigon \| \| \| \| \| \| \| \| \| \| Turns \| $(0,\tfrac{1}{4})$ \| $\tfrac{1}{4}$ \| $(\tfrac{1}{4},\tfrac{1}{2})$ \| $\tfrac{1}{2}$ \| $(\tfrac{1}{2},1)$ \| $1$ \| \| Radians \| $(0,\tfrac{1}{2}\pi)$ \| $\tfrac{1}{2}\pi$ \| $(\tfrac{1}{2}\pi,\pi)$ \| $\pi$ \| $(\pi,2\pi)$ \| $2\pi\,$ \| \| Degrees \| (0,90)° \| 90° \| (90,180)° \| 180° \| (180,360)° \| 360° \| ### Equivalence angle pairs • Angles that have the same measure (i.e. the same magnitude) are said to be equal or congruent. An angle is defined by its measure and is not dependent upon the lengths of the sides of the angle (e.g. all right angles are congruent). • Two angles which share terminal side, however differ in size by an integer multiple of a turn, are called coterminal angles. • A reference angle is the acute version of any angle determined by repeatedly subtracting or adding 180 degrees, and subtracting the result from 180 degrees if necessary, until a value between 0 degrees and 90 degrees is obtained. For example, an angle of 30 degrees has a reference angle of 30 degrees, and an angle of 150 degrees also has a reference angle of 30 degrees (180-150). An angle of 750 degrees has a reference angle of 30 degrees (750-720).[7] ### Intersecting angle pairs • Two angles opposite each other, formed by two intersecting straight lines that form an "X"-like shape, are called vertical angles or opposite angles or vertically opposite angles. These angles are equal in measure. • Angles that share a common vertex and edge but do not share any interior points are called adjacent angles. • Alternate angles, corresponding angle, interior angles and exterior angles are associated with a transversal of a pair of lines by a third. ### Combine angle pairs The complementary angles a and b (b is the complement of a, and a is the complement of b). Here the sum of the reflex angle and the acute angle makes an explementary angle. Here, a and b are supplementary angles. • Two angles that sum to one right angle (90°) are called complementary angles. The difference between an angle and a right angle is termed the complement of the angle. • Two angles that sum to a straight angle (180°) are called supplementary angles. The difference between an angle and a straight angle (180°) is termed the supplement of the angle. • Two angles that sum to one turn (360°) are called explementary angles or conjugate angles. ### Polygon related angles • An angle that is part of a simple polygon is called an interior angle if it lies on the inside of that simple polygon. A concave simple polygon has at least one interior angle that exceeds 180°. In Euclidean geometry, the measures of the interior angles of a triangle add up to π radians, or 180°, or 1/2 turn; the measures of the interior angles of a simple quadrilateral add up to 2π radians, or 360°, or 1 turn. In general, the measures of the interior angles of a simple polygon with n sides add up to [(n − 2) × π] radians, or [(n − 2) × 180]°, or (2n − 4) right angles, or (n/2 − 1) turn. • The angle supplementary to the interior angle is called the exterior angle. It measures the amount of rotation one has to make at this vertex to trace out the polygon. If the corresponding interior angle is a reflex angle, the exterior angle should be considered negative. Even in a non-simple polygon it may be possible to define the exterior angle, but one will have to pick an orientation of the plane (or surface) to decide the sign of the exterior angle measure. In Euclidean geometry, the sum of the exterior angles of a simple polygon will be one full turn (360°). • Some authors use the name exterior angle of a simple polygon to simply mean the explementary (not supplementary!) of the interior angle.[8] This conflicts with the above usage. ### Plane related angles • The angle between two planes (such as two adjacent faces of a polyhedron) is called a dihedral angle. It may be defined as the acute angle between two lines normal to the planes. • The angle between a plane and an intersecting straight line is equal to ninety degrees minus the angle between the intersecting line and the line that goes through the point of intersection and is normal to the plane. ## Angles between curves The angle between the two curves at P is defined as the angle between the tangents A and B at P The angle between a line and a curve (mixed angle) or between two intersecting curves (curvilinear angle) is defined to be the angle between the tangents at the point of intersection. Various names (now rarely, if ever, used) have been given to particular cases:—amphicyrtic (Gr. ἀμφί, on both sides, κυρτός, convex) or cissoidal (Gr. κισσός, ivy), biconvex; xystroidal or sistroidal (Gr. ξυστρίς, a tool for scraping), concavo-convex; amphicoelic (Gr. κοίλη, a hollow) or angulus lunularis, biconcave.[9] ## Dot product and generalisation In the Euclidean plane, the angle θ between two vectors u and v is related to their dot product and their lengths by the formula $\mathbf{u} \cdot \mathbf{v} = \cos(\theta)\ \\|\mathbf{u}\\|\ \\|\mathbf{v}\\|.$ This formula supplies an easy method to find the angle between two planes (or curved surfaces) from their normal vectors and between skew lines from their vector equations. ## Inner product To define angles in an abstract real inner product space, we replace the Euclidean dot product ( · ) by the inner product $\langle\cdot,\cdot\rangle$, i.e. $\langle\mathbf{u},\mathbf{v}\rangle = \cos(\theta)\ \\|\mathbf{u}\\|\ \\|\mathbf{v}\\|.$ In a complex inner product space, the expression for the cosine above may give non-real values, so it is replaced with $\operatorname{Re}(\langle\mathbf{u},\mathbf{v}\rangle) = \cos(\theta)\ \\|\mathbf{u}\\|\ \\|\mathbf{v}\\|.$ or, more commonly, using the absolute value, with $\|\langle\mathbf{u},\mathbf{v}\rangle\| = \cos(\theta)\ \\|\mathbf{u}\\|\ \\|\mathbf{v}\\|.$ The latter definition ignores the direction of the vectors and thus describes the angle between one-dimensional subspaces $\operatorname{span}(\mathbf{u})$ and $\operatorname{span}(\mathbf{v})$ spanned by the vectors $\mathbf{u}$ and $\mathbf{v}$ correspondingly. ## Angles between subspaces The definition of the angle between one-dimensional subspaces $\operatorname{span}(\mathbf{u})$ and $\operatorname{span}(\mathbf{v})$ given by $\|\langle\mathbf{u},\mathbf{v}\rangle\| = \cos(\theta)\ \\|\mathbf{u}\\|\ \\|\mathbf{v}\\|$ in a Hilbert space can be extended to subspaces of any finite dimensions. Given two subspaces $\mathcal{U},\mathcal{W}$ with $\operatorname{dim}(\mathcal{U}):=k\leq \operatorname{dim}(\mathcal{W}):=l$, this leads to a definition of $k$ angles called canonical or principal angles between subspaces. ## Angles in Riemannian geometry In Riemannian geometry, the metric tensor is used to define the angle between two tangents. Where U and V are tangent vectors and gij are the components of the metric tensor G, $\cos \theta = \frac{g_{ij}U^iV^j} {\sqrt{ \left\| g_{ij}U^iU^j \right\| \left\| g_{ij}V^iV^j \right\|}}.$ ## Angles in geography and astronomy In geography, the location of any point on the Earth can be identified using a geographic coordinate system. This system specifies the latitude and longitude of any location in terms of angles subtended at the centre of the Earth, using the equator and (usually) the Greenwich meridian as references. In astronomy, a given point on the celestial sphere (that is, the apparent position of an astronomical object) can be identified using any of several astronomical coordinate systems, where the references vary according to the particular system. Astronomers measure the angular separation of two stars by imagining two lines through the centre of the Earth, each intersecting one of the stars. The angle between those lines can be measured, and is the angular separation between the two stars. Astronomers also measure the apparent size of objects as an angular diameter. For example, the full moon has an angular diameter of approximately 0.5°, when viewed from Earth. One could say, "The Moon's diameter subtends an angle of half a degree." The small-angle formula can be used to convert such an angular measurement into a distance/size ratio. ## Notes 1. Sidorov, L.A. (2001), "Angle", in Hazewinkel, Michiel, , Springer, ISBN 978-1-55608-010-4 2. Slocum, Jonathan (2007), Preliminary Indo-European lexicon — Pokorny PIE data, University of Texas research department: linguistics research center, retrieved 2 Feb., 2010 3. Chisholm 1911; Heiberg 1908, pp. 177–178 4. J.H. Jeans (1947), The Growth of Physical Science, p.7; Francis Dominic Murnaghan (1946), Analytic Geometry, p.2 5. ooPIC Programmer's Guide (archived) www.oopic.com 6. Angles, integers, and modulo arithmetic Shawn Hargreaves blogs.msdn.com 7. Chisholm 1911; Heiberg 1908, p. 178 ## References • Heiberg, Johan Ludvig (1908). In Heath, T. L.. Euclid. The thirteen books of Euclid's Elements 1. Cambridge University press. • This article incorporates text from a publication now in the public domain: Chisholm, Hugh, ed. (1911). "Angle". (11th ed.). Cambridge University Press.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 31, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9019116759300232, "perplexity_flag": "middle"}
http://nrich.maths.org/2360/note?nomenu=1	## Lots of Lollies If you are a teacher, click here for a version of the problem suitable for classroom use, together with supporting materials. Otherwise, read on ... Frances and Rishi were given a bag of lollies. They shared them out evenly and had one left over. Just as they had finished sharing them their friends Kishan, Hayley and Paul came along. They wanted some lollies too so the children shared them out again between all of them. This time they had two lollies left over. How many lollies could there have been in the bag? Once you've had a chance to think about it, click below to see how three different groups of pupils began working on the task. Sarah, Danielle and Sally said: "We noticed that $17$ works as when there are only two of them they get $8$ each, with one left over. But when their friends come along they get three each with $2$ left over. We also notice that $7$ works and $27$ works, as well as $107$." Poppy began like this: If the two children end up with one lolly it must be an odd number of lollies. Then three more children come making the total number of children $5$. Say they had $1$ lolly each when they shared them, the number of lollies would be $7$ because $1$ times $5$ is $5$ add on $2$ for the left over ones and it makes seven. If we carry this on to $10$ lollies each it shows: $1$ lolly each - $7$ lollies $2$ lollies each - $12$ lollies $3$ lollies each - $17$ lollies $4$ lollies each - $22$ lollies $5$ lollies each - $27$ lollies $6$ lollies each - $32$ lollies $7$ lollies each - $37$ lollies $8$ lollies each - $42$ lollies $9$ lollies each - $47$ lollies $10$ lollies each - $52$ lollies Here is the start of Phoebe and Alice's work: Can you take each of these starting ideas and develop it into a solution? ### Why do this problem? This problem requires children to apply their knowledge of factors and multiples, and is a good way of making the link between sharing, division and multiples/factors. It may also be used to introduce learners to the fact that a problem can have more than one solution and that the solutions can be generalised. It can be approached in many different ways so can be a useful context in which to talk about different ways of recording and different methods of solving problems. ### Possible approach You could introduce this challenge by acting it out. Invite two children to the front of the class and ask everyone to imagine that you have a bag of lollies. Explain that you give them out to the two volunteers so that they have the same number each, but there is one left in the bag. You could mime giving out some lollies so that everyone gets the idea. Then invite three more children to the front. Mime gathering the lollies back into the bag and then distributing them equally again, this time explaining that you have two left over. Pose the question "I wonder how many lollies could have been in the bag?". Give the class a few minutes to consider, individually, how they might go about tackling the problem, then pair them up and suggest that they talk to their partner about their ideas so far. Try to stand back and observe, and resist the temptation to make helpful suggestions! Allow pairs to work on the task so that you feel they have made some progress, but do not worry if they have not completed it or if they report being stuck. The aim at this stage is for everyone to 'get into' the problem and work hard on trying to solve it, but not necessarily to achieve a final solution. At a suitable time, hand out this (doc pdf ) to pairs. Suggest to the class that when they've finished or can't make any further progress, they should look at the sheet showing three approaches used by children working on this task. Pose the question, "What might each do next? Can you take each of their starting ideas and develop them into a solution?". You may like pairs to record their work on large sheets of paper, which might be more easily shared with the rest of the class in the plenary. Allow at least fifteen minutes for a final discussion. Invite some pairs to explain how the three different methods might be continued. You may find that some members of the class used completely different approaches when they worked on the task to begin with, so ask them to share their methods too. You can then facilitate a discussion about the advantages and disadvantages of each. Which way would they choose to use if they were presented with a similar task in the future? Why? (You might find it helpful to adapt this Smart Notebook file for use on the interactive whiteboard. Thank you to Gemma for giving us permission to include it here.) ### Key questions How many children are there altogether when they share the lollies the second time? What is the smallest number of lollies there could be? 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http://mathhelpforum.com/pre-calculus/94620-dreaded-word-problems.html	# Thread: 1. ## Dreaded Word Problems So I was doing my homework quite merrily until I go to the end where I ran into word problems.. (why do they always stick them at the end) Can you guys point me into the right direction? Q#1. The minute hand on a clock is 6 cm long and the hour hand is 4 cm long. Give an expression for d , the distance between the tips of the hands, in terms of a , the angle between the hands. What Does "d"=? My Plan of Attack: I drew a lame picture of the clock hands neatly labeled. I noticed you can form a triangle and thus use Pythagorean to find the distance between the 2 points. But Im not sure which side is the longer side? Do you guys think it should be $6^2+4^2=c^2?$ 2nd, I think, once I know the 3 side lengths, I can use either Cosine law or Sine law to find the angle.. And what is the question asking when they are asking for an expression for "d"? Q#2. At noon, ship A passes 8 km west of a harbour, heading due south at 10 km/h. At 9 a.m., ship B had left the harbour, sailing due south at 5 km/h. Give an expression for d , the distance between the two ships, in terms of t , the number of hours after noon. My Plan of Attack: Confused. I have no idea why I can't handle word problems. TIA as always 2. Originally Posted by mvho So I was doing my homework quite merrily until I go to the end where I ran into word problems.. (why do they always stick them at the end) Can you guys point me into the right direction? Q#1. The minute hand on a clock is 6 cm long and the hour hand is 4 cm long. Give an expression for d , the distance between the tips of the hands, in terms of a , the angle between the hands. What Does "d"=? My Plan of Attack: I drew a lame picture of the clock hands neatly labeled. I noticed you can form a triangle and thus use Pythagorean to find the distance between the 2 points. But Im not sure which side is the longer side? Do you guys think it should be $6^2+4^2=c^2?$ 2nd, I think, once I know the 3 side lengths, I can use either Cosine law or Sine law to find the angle.. And what is the question asking when they are asking for an expression for "d"? Join the tips of the minute hand and hour hand, You have a triangle: 1 side is 6cm long, second side is 4cm long and third side is d cm. Since you know the angle between the known sides:Apply cosine law. 3. Yes, That is what I thought but I was more confused at what they actually want me to find? And does it matter if the unknown side is the longer side or shorter side? I am not clear about that as well. Do you guys think it should be $6^2+4^2=c^2 OR, c^2+4^2=6^2?$ 4. Originally Posted by mvho Q#2. At noon, ship A passes 8 km west of a harbour, heading due south at 10 km/h. At 9 a.m., ship B had left the harbour, sailing due south at 5 km/h. Give an expression for d , the distance between the two ships, in terms of t , the number of hours after noon. See figure. Ship A started from A at noon. After time t, it is at C hence, AC=10t Ship B started at harbour(B) at 9am. At noon, it was at D. BD=???? At time t after noon, it is at E. DE=???? Attached Thumbnails 5. Originally Posted by mvho Yes, That is what I thought but I was more confused at what they actually want me to find? And does it matter if the unknown side is the longer side or shorter side? I am not clear about that as well. Do you guys think it should be $6^2+4^2=c^2 OR, c^2+4^2=6^2?$ Neither! You only use the Pythagorean Theorem for right triangles, and for the most part, the triangles formed by the two hands plus the distance between them at the tips are not right triangles. You have a triangle where you know two of the sides, and the angle between them is a. The unknown side is d. As malaygoel said, just use the Law of Cosines: $\begin{aligned}<br /> d^2 &= 4^2 + 6^2 - 2(4)(6)\cos a \\<br /> d^2 &= 52 - 48\cos a \\<br /> d &= \sqrt{52 - 48\cos a}<br /> \end{aligned}$ See diagram below. 01 Attached Thumbnails 6. Thanks everyone Ship A started from A at noon. After time t, it is at C hence, AC=10t Ship B started at harbour(B) at 9am. At noon, it was at D. BD=???? At time t after noon, it is at E. DE=???? So, BD=5t and AC=10t Is it also true that the length from point A to point B is 8km since the question said it passes 8k west? If this is the case, I can use pythagorean since a rectangle would be a right triangle. I think I should have: $d=\sqrt{8^2+5t^2}$ Since I know that AC=5t and BD=5t so if I minus 10t-5t, I'll know the length. #### Search Tags View Tag Cloud Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9583150148391724, "perplexity_flag": "middle"}
http://cstheory.stackexchange.com/questions/2386/inspirational-talk-for-final-year-high-school-pupils/2414	# Inspirational talk for final year high school pupils I am often asked by my department to give talks to final year high school pupils about the more mathematical elements of computer science. I do my best to pick topics from TCS which might inspire their interest (which mostly involves something to do with the Halting problem) but would love hear other people's ideas/successes/failures. The remit is that these are pupils who are considering applying for a CS undergraduate degree at a decent university but may be more attracted by maths or another one of the sciences. I find that the usual topics of shortest path algorithms or faster sorting methods don't really work any more to pique their interest. - 11 I'm thinking this should be CW ? – Suresh Venkat♦ Oct 22 '10 at 19:43 Is this really TCS research level question?! – Mohammad Al-Turkistany Oct 22 '10 at 21:32 18 @turkistany: Yes. Selling the importance of research is an essential part of doing that research. It is also a part where many theoreticians are weak. To paraphrase Feynman, we don't really understand TCS unless we can explain in to bright high school students. – Aaron Sterling Oct 22 '10 at 22:32 9 @turkistany: Yes, yes, a thousand times yes. – JɛﬀE Oct 23 '10 at 17:41 1 @JeffE, Ok, Ok,..., infinite number of times OK. I get now :) – Mohammad Al-Turkistany Oct 23 '10 at 19:26 show 1 more comment ## 13 Answers There is a neat way to introduce zero-knowledge proofs to students, which I think is originally due to Oded Goldreich (please correct me if I'm wrong). You have a red ball and a green ball, which poor colorblind Charlie believes are the same color. You want to convince Charlie that you can tell the difference between the red ball and green ball, and you want to do this in a way that Charlie does not learn which is red and which is green. (You want to prove something is true, in such a way that no one else can turn around and claim a proof of that something as their own.) How can you do this? Or is it impossible? One protocol is the following. Charlie puts a ball in each hand, then chooses to either switch the two balls behind him, or not. Then he presents the two balls again. If you can always detect whether he switched the two balls or not, then Charlie is increasingly convinced that you can tell the difference between them. If Charlie does this shuffle at random and you really can't tell the difference between the colors, then you will only guess correctly with probability $1/2$. After $k$ trials, Charlie should be convinced that you can tell the difference with probability at least $1-1/2^k$. Now while Charlie becomes increasingly convinced that you can tell the difference, he frustratingly never learns which ball is red and which one is green. - 2 Presenting ZK proofs is a very good choice. Another example which I think will be understandable to students is graph coloring. – Kaveh♦ Oct 23 '10 at 5:57 2 There's a cool ZK sudoku demo from Moni Naor's page. – Suresh Venkat♦ Oct 23 '10 at 7:07 1 That is very nice. Thank you. – Raphael Oct 23 '10 at 17:15 While Goldreich has contributed a lot to this field, ZK proofs are originally due to Goldwasser, Micali, and Rackoff. PS: The color-blind-convincing protocol is actually due to Goldreich (see http://www.wisdom.weizmann.ac.il/~oded/poster03.html). – Sadeq Dousti Dec 15 '10 at 20:16 1 @Sadeq: I am sure Ryan meant that ZKP for ball color with a color blind prover is due to Goldreich :) – Sasho Nikolov Jun 15 '11 at 21:29 A good source for education purposes in general is CS unplugged, which has lots of neat CS ideas translated into high-school and middle-school activities. - That is a very good link thanks. The most remarkable thing about it is that it is aimed at middle school children. I doubt there is a single middle school in the UK that teaches anything like it, sadly. – Raphael Oct 24 '10 at 13:01 The Teacher's Edition book looks more suitable for primary-school and middle-school children, rather than for high-school students. – Alessandro Cosentino Oct 25 '10 at 3:57 One of the most appealing aspects TCS is how it uses abstract mathematical ideas for day-to-day practical applications. A presentation can focus on the abstract ideas that lie one step behind what they see daily on the Internet: Shortest paths becomes exciting once they are put in the context of friends-of-friends on Facebook. More graph algorithms can ride on Pagerank; Amazon recommendations raise the challenge of machine learning; and buying-stuff on the Internet is certainly a good lead for the public-key crypto. - 4 Also, any StarCraft player is aware of the importance of a good shortest path algorithm. And I guess that high school students are still playing videogames (do they ?). – Sylvain Peyronnet Oct 23 '10 at 10:24 1 They're definitely playing video games. – Daniel Apon Oct 24 '10 at 13:59 I think almost any topic in computer science can be used for giving an interesting talk, but some are better suited, the more important part is the presentation. ### Fun Side of Computer Science I have used various games from Combinatorial Games Theory, mainly from Richard Guy's "Fair Games" and Elwyn R. Berlekamp, John H. Conway, and Richard K. Guy's "Winning Ways for your Mathematical Plays" (wiki). They are fun, and you can play them in the class with them and let them find the right way to play it, give some hints so at the end they find the way to win the game. These games are probably more suitable for younger students. There are other fun topics in Computer Science where you can pick a problem which is more suitable for your audience and use it to engage them. ### Philosophical Side of Computer Science There many topics in theoretical computer science which are related to philosophy and the big questions. From Gödel's incompleteness theorem to zero-knowledge proofs, security, privacy, algorithmic game theory, P vs NP, machine learning, ... I would not go into details, just demonstrate that the problems are interesting, they are more than just computer science, they are related to big questions. (Take a look at Scott Aaronson's Quantum Computing Since Democritus and Great Ideas In Theoretical Computer Science lectures). Don't make them feel like the topic is dead (i.e. all questions are answered), make them feel that the area is alive, there has been progress but there are still big challenges ahead, and it is a journey to an undiscovered land. ### Technological Side of Computer Science Talk about the computer science behind technologies. There are so many topic that one can choose here, familiar technologies from video-games to Google search, machine translation, vision, ... technologies that everyone uses each day, or even unfamiliar ones. Talk about in progress and next generation technologies, about the impact they have had on our lives, and how they have improved it. Talk about research going on in big famous companies (like Google, Microsoft, Apple, IBM, ...) and products they develop. Talk about big problems of our time and what effect computer science has on them. ### Mathematical Side of Computer Science This is good for students who are already interested in mathematics, those interested in the pure and exact side, but without combining it with other theme mentioned above it won't be as effective to other students. I would go with a big question and at some point mention start talking about mathematical problems involved. ### Interdisciplinary Side of Computer Science Computer Science is probably one of the most interdisciplinary subjects, there are some connection with almost any other subject, humanistic (sociology, linguistics, economics, philosophy, ...), natural sciences (mathematics, physics, ...), biology, medical sciences, art, engineering (electronics, mechanics, ...), ... anything! Whatever topic you are interested in, there is something in computer science that is related to it! As Scott said, Every Other Major Sucks By Comparison :). ### All of Them You can also try to mention all of the themes I have mentioned above. I haven't tried this, and I am not sure how effective it would be. You have to transfer the feeling and make the point, and it takes sometime. One other options is to mention all of them briefly at the start (or the end) and then go on with one of them, and tell them that they can contact you to get more informations about the other ones if they are interested. ### some comments Whatever you are going to talk about, you should be enthusiastic about it. It is going to be much more difficult to interest them in a topic which is not really interesting to yourself. Tell them about your own reasons for selecting computer science. And don't be boring. - 1 +1 for "make them feel that the area is alive". – András Salamon Oct 24 '10 at 14:04 I've used two talks quite successfully with both high school students and entering freshmen. 1. Origami. I lead off with the 5-point star problem (this works well in american contexts, because of the connection to the american flag) and let students try to figure out how to make a five-point star with folding + 1 cut. I talk about the "resource" (cutting) and how algorithm design is about working with limited resources. Then I talk about other origami questions and applications in the real world (heart valves, NASA telescopes, crumple zones in cars). 2. Sorting pancakes: there's a beautiful connection between sorting pancakes and genome rearrangement, and I actually made stacks of pancakes from foam for students to play with. Works great, and lets me talk about algorithms, gene sequencing, Bill Gates (!), and other fun things. - Thanks very much. Those do look great fun. – Raphael Nov 15 '10 at 13:51 Cryptography is always something that captures the mind of younger (and I personally hope older) individuals. I had friends who wanted to be nurses' assistants, hockey players, businessmen and politicians and friends (who despite their higher goals) took jobs as grocery baggers and cart pushers, construction workers and kennel assistants - all of whom invented and broke each others' (admittedly naive and simple) codes. In particular, the existence of public key cryptography is usually pretty easy to explain if one takes the route of RSA. One might also list some of the important results without proofs or constructions - Zero-Knowledge proofs and Homomorphic Encryption are bound to juice the geek factor for what it's worth. Forward Error Correction and Error Detection codes are also very cool and if done right can be taught to a curious audience. To make them easier to digest, you could mention the "universality" of the index of coincidence - that all our spoken language and writing have small redundancies and exaggerations that help us to communicate in the noisy channel of a room containing shuffling bags, feet and humming air conditioners. Finally, I also would suggest doing a simple introduction to complexity theory - something along the lines of my answer to A Dinner-table description of Theoretical Computer Science. - The New Turing Omnibus by A. K. Dewey has 66 so-called excursions in computer science. It covers topics such as analysis of algorithms, AI, complexity theory, theory of computation, cryptography, computer graphics and so forth. Every topic is written in a rather condensed form, capturing some landmark result in computer science. This book could provide some inspiration. Another possibility is to allow students to get their hands dirty via something like Google's Code-in program. It's a bit like Google's Summer of Code, but, you know, for kids. Perhaps showing some of the amazing coding projects students can be involved in is one possible way of piquing interest. - Of course, the book is from 1993 (I think) and thus a bit old school. – Dave Clarke♦ Oct 22 '10 at 20:07 2 Yes there is a problem with trying to excite them about the future if one is referring to a book written before they were born :) – Raphael Oct 22 '10 at 20:33 In my opinion, to be sexy to high school students you need to be some kind of magician. That's why I think that randomized algorithms are very good as a student attractor. For instance property testing is really something intriguing, and also something that can be explained (not the technicalities, but the idea) to anyone. PCP is also magic, but I guess that this is out of reach... - I had once given a talk about PCP to talented high school students, of course without proving it, but showing its applications to hardness of approximation and giving the general 'feel' of the theorem. I think they liked it, so it isn't that much out of reach (but they had listened to some talks about approximation algorithms before, without this I think they wouldn't grab the motivation of the theorem). – Karolina Sołtys Oct 23 '10 at 21:01 Here is a very nice article on coding theory aimed at high school students by Michael Mitzenmacher: http://www.eecs.harvard.edu/~michaelm/FUTUREOFCS/codes-mitzenmacher.pdf - 2 this is an excellent survey – Suresh Venkat♦ Oct 27 '10 at 19:54 1 Thanks for that. – Raphael Oct 27 '10 at 20:07 2 This seems to part of a book that's a work-in-progress. Michael Mitzenmacher's blog post (mybiasedcoin.blogspot.com/2008/04/theorycs-book.html) has a link to that, which also has a very nice expository chapter (cs.princeton.edu/~chazelle/pubs/algorithm.html) on algorithms by Bernard Chazelle. That chapter isn't mathematics per se, but it is rich in mathematical ideas. – Cong Han Oct 27 '10 at 21:44 My answer is not directly connected with TCS, but it can show that math can be beautiful and useful. You could make a speech about how to get reliable data about how many students were cheating on the exam. If you asked them directly then You wouldn't get reliable data. The idea of how to get reliable data is very simple. First You tell every student to think about some integer number, then You say: - If it was odd number write down whether You like green color or not. You can choose any other simple question, but You have to know, from some other survey, what percentage of people answer yes on this question. - If it was even number write down whether You were cheating or not. About 50% of students are going to answer on first question, and the other 50% are going to answer on second question. Now it is very easy to estimate how many students were cheating. Example: If 40% of answers were yes, and You know that 30% of people like green color then You know that about 50% of students were cheating. - I think this is closely related to http://cstheory.stackexchange.com/questions/1471/dinner-table-description-of-theoretical-computer-science As I posted there I feel that algorithmics relate the best to every-day problems and can therefore motivate TCS very well. ("How long one google search would take if they would search in the same way you look up telephone numbers") - 1 Hello Raphael! The main difference I feel is that these are all mathematically inclined students making an active choice about what to do with their future. The problem we have had in recruitment, which may be peculiar to the UK, is that high school teaches them that CS is neither for great intellectuals nor for people who want to change the world. I have 20 minutes to redress this misconception :) – Raphael Oct 24 '10 at 12:55 That is right (also in Germany) and there might be some differences in attitude but the amount of CS specific knowledge present might be about the same as for the dinner table folks. I would agree that you have wrap the package differently for the other audience but I would choose the same content. – Raphael Oct 24 '10 at 14:59 According to me, "computer science" is the "science of all sciences" :) What is "science"? We get data from nature, and we try to construct a model that explains the data. Also, we assume implicitly that nature is not arbitrary. The laws of nature must have a concise expression, the data must satisfy some symmetries, etc. But this is exactly a learning problem! The data is generated by some process that is promised to be of "low complexity", and our task is to reconstruct a description of the process. Our understanding of such problems is at such a primitive level that it's your duty to work on them! :) Even our understanding of the seemingly simpler problem of whether the output of a black-box process is equivalent to some fixed function is far from complete. For example, suppose that we are promised that the black-box is evaluating a function that can be computed by a small-depth arithmetic circuit (this is easy to explain to high-schoolers), and we want to find out whether the box is computing the zero function. We don't know if this can be done in the lifetime of the universe for functions on reasonable sized domains! Cue to start talking about arithmetic complexity theory, the chasm at depth 4, the role of randomness in computation, what's known if we reduce # of multiplication gates, etc. etc. ... - In the Algorithms in the Field workshop a month ago in DIMACS, Graham Cormode was arguing in favor of teaching sketching techniques from streaming algorithms to undergraduates. Moses Charikar said that they do teach them in Princeton, I think @Suresh Venkat also mentioned he teaches things like the Misra-Gries algorithm for heavy hitters. I think some basic streaming results would be great for high school students too: they rely on basic but important math tricks, the problem formulations are like puzzles, and the solutions feel like magic, and magic is a great way to inspire high school students. You can make sure to emphasize the dramatic difference between the scale of the problem and the amount of resources you can use. A silly example: suppose that you can ask every person entering or leaving JFK airport their zipcode. Can you keep track how many different zipcodes have been represented in JFK during month, only using a notepad? - yup. this is a good example – Suresh Venkat♦ Jun 17 '11 at 17:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9553534388542175, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/51464-inequality-proof-print.html	# inequality proof Printable View • September 30th 2008, 09:25 PM lllll inequality proof Using the triangle inequality show $a$ and $b$ are in $\mathbb{R}$ with $a \neq b$, then $\exists$ open intervals $U$ centered at $a$ and $V$ centered at b, both with radius $\epsilon=\frac{1}{2}\|a-b\|$ so far all I have is $\|a-b\|=2\epsilon$ $\|\|a\|-\|b\|\| \leq \|a-b\|$ now I'm not sure if what I'm doing is right: $-\epsilon<\|\|a\|-\|b\|\| < \epsilon \rightarrow \|\|a\|-\|b\|\| < 2\epsilon \rightarrow \|\|a\|-\|b\|\| < \|a-b\|$ ? • September 30th 2008, 11:09 PM CaptainBlack Quote: Originally Posted by lllll Using the triangle inequality show $a$ and $b$ are in $\mathbb{R}$ with $a \neq b$, then $\exists$ open intervals $U$ centered at $a$ and $V$ centered at b, both with radius $\epsilon=\frac{1}{2}\|a-b\|$ You do realise that as this stands it is not the question that you think it is, don't you? Do you want to show that: For all $a,b \in \mathbb{R},\ a \ne b$, then the two disjoint open intervals $U$ centered at $a$ and $V$ centered at $b$, both with radius $\epsilon=\frac{1}{2}\|a-b\|$ are disjoint? RonL • October 3rd 2008, 11:03 PM lllll Quote: Originally Posted by CaptainBlack Do you want to show that: For all $a,b \in \mathbb{R},\ a \ne b$, then the two disjoint open intervals $U$ centered at $a$ and $V$ centered at $b$, both with radius $\epsilon=\frac{1}{2}\|a-b\|$ are disjoint? RonL I completely overlooked that part of the question. But I did found a prove for this statement but am still lost as to how it was done, even after asking my prof. This is what I found: $U=(x-\epsilon, \ x+\epsilon), \ V = (y-\epsilon, \ y+\epsilon)$ $U \bigcap V = \emptyset$ suppose $z$ is in $U \bigcap V$ $\|x-y\| \leq \|x-z\|+\|z-y\| < \epsilon + \epsilon = 2\epsilon = \|x-y\|$ which is apparently a contradiction. If anyone could clarify this, it would help me out a lot. • October 3rd 2008, 11:08 PM CaptainBlack Quote: Originally Posted by lllll I completely overlooked that part of the question. But I did found a prove for this statement but am still lost as to how it was done, even after asking my prof. This is what I found: $U=(x-\epsilon, \ x+\epsilon), \ V = (y-\epsilon, \ y+\epsilon)$ $U \bigcap V = \emptyset$ suppose $z$ is in $U \bigcap V$ $\|x-y\| \leq \|x-z\|+\|z-y\| < \epsilon + \epsilon = 2\epsilon = \|x-y\|$ which is apparently a contradiction. If anyone could clarify this, it would help me out a lot. The bit that is missing is: $\|x-y\|=\|x-z+z-y\|=\|(x-z)-(y-z)\|$ Now by the triangle inequality: $\|x-y\|=\|(x-z)-(y-z)\| \le \|x-z\|+\|y-z\|$ etc. RonL • October 4th 2008, 03:22 PM lllll Quote: Originally Posted by CaptainBlack The bit that is missing is: $\|x-y\|=\|x-z+z-y\|=\|(x-z)-(y-z)\|$ Now by the triangle inequality: $\|x-y\|=\|(x-z)-(y-z)\| \le \|x-z\|+\|y-z\|$ etc. RonL That solves a part of the proof, but where is the contradiction? It lies is the fact that $z$ is assumed to be in $U \cap V$ so lies in $U$ and in $V$ so $\|x-z\|<\epsilon,$ and $\|y-z\|<\epsilon$, so: $\|x-y\|<2 \epsilon$ But $\|x-y\|=2 \epsilon$, so we have: $\|x-y\|<\|x-y\|$ which is a contradiction. RonL All times are GMT -8. The time now is 06:24 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 56, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9720219373703003, "perplexity_flag": "head"}

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