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First off, I would not call 4% beyond a reasonable doubt. In several places that would entail a significant chance of the state taking an innocent person's life. As a modern society we should expect and demand a lot better accuracy than than that. There would be no such thing as commercial aviation if 96% of airplane flights ended in a safe landing. Secondly, it appears that you are confusing predictivity, specificity, and sensitivity. You also aren't looking at the whole picture. I'll illustrate with some admittedly fictitious numbers. I have tried to make these numbers realistic. 4.06% of the people sent to jail in this example are truly innocent. 4% of those who go to trial and are truly innocent. 5% of the people swept up by the justice system end up having a trial. The vast majority (94.9% here) of the people swept up by the justice system are truly guilty. Suppose some unspecified number people in some area have been accused of committing some crime of period of time. Of those people, 2772 either accept a plea or go to trial. I'll focus on those 2772 people. I am ignoring those against whom charges were initially levied but later dropped. Some of those 2772 people will have been truly innocent of the crime while the rest will have been truly guilty. Some (most in my example, and most in real life) of these people will be offered a plea. Some will accept that plea; others will not. The trial outcome is of course innocent or guilty. The following confusion matrix presents my made-up numbers: [math] \begin{matrix} & \text{Found innocent} & \text{Found guilty} & \text{Pled Guilty} & \text{Total} \\ \text{Truly innocent} & 30 & 4 & 107 & 141 \\ \text{Truly guilty} & 8 & 96 & 2527 & 2631 \\ \text{Total} & 38 & 100 & 2634 & 2772 \\ \end{matrix} [/math] The confusion matrix for these people who went to trial is formed by the looking at only the first and second columns of the above confusion matrix. Note very well: What constitutes a type I versus a type II error depends on the null hypothesis. With a null hypothesis of innocence (a person is innocent unless proven guilty), a type I error results when a truly guilty person is found to be innocent while a type II error results when a truly innocent person is found to be guilty. The state (police+prosecution) use a different null hypothesis. As far as the state is concerned, the rule is "guilty until proven innocent". With 94.9% of the people swept up by the state truly being guilty, Bayes' law says assuming guilt makes for a very good prior. To the state, a type I error results when a truly innocent person is sent to jail while a type II error occurs when a guilty person is set free. [math] \begin{matrix} & \text{Found innocent} & \text{\bf{Found guilty}} & \text{Total} \\ \text{Truly innocent} & 30 & \boldsymbol{4} & 34 \\ \text{Truly guilty} & 8 & \boldsymbol{96} & 104 \\ \text{Total} & 38 & \boldsymbol{100} & 138 \\ \end{matrix} [/math] The emboldened column represents your 96%/4% number. This 96% figure is negative predictivity (positive predictivity from the state's POV) and is the best statistic that can be drawn from this confusion matrix. Other statistics are specificity, 92.2%; accuracy, 91.3%, sensitivity, 88.2%, and positive predictivity, 78.9%. The low positive predictivity is OK here in my mind. It is in general better to let a guilty person go free than it is to send an innocent person to jail. The low sensitivity is a bit more problematic. The trial system is not all that sensitive to innocence. Collecting the "found guilty" and "pled guilty" columns in the original 2×3 confusion matrix into one row yields yet another a binary confusion matrix: [math] \begin{matrix} & \text{Found innocent} & \text{\bf{Sent to jail}} & \text{Total} \\ \text{Truly innocent} & 30 & \boldsymbol{111} & 141 \\ \text{Truly guilty} & 8 & \boldsymbol{2623} & 2631 \\ \text{Total} & 38 & \boldsymbol{2734} & 2772 \\ \end{matrix} [/math] The emboldened column portrays the status of the 2734 people who were sent to jail. Of those 2734 people, 111, or 4.06%, are truly innocent. That is the 4.06% figure that Pangloss deemed "acceptable". This overall confusion matrix has a very high specificity of 99.7%. On the other hand, it has an incredibly low sensitivity of 21.3%. The system as a whole is quite insensitive to innocence. That the people are better off is not always that good a metric. While safety is important, it also leads to things like the Patriot Act. Unfortunately, this is not true. Trying to reduce type I errors often leads to an increase in type II errors, and vice versa. The only way to reduce both is to obtain better information -- and to use that improved information in an intelligent manner. For example, DNA testing truly does represent better information. Other high tech forensic evidence is not so good; some is apparently negative information.

That was from page 1, so I assume you have been beaten up quite a bit regarding that statement. So, piling on, That is a terrible statistic! What if 4.06% of all plane flights ended with a fatal crash? 4.06% automobile trips involved a collision? 4.06% of all vaccinations resulted in a serious adverse reaction? A 4.06% failure rate is acceptable for things that are completely inconsequential such as infant mortality in light bulbs. It isn't even acceptable for things that are extremely risky such as flying into orbit the Space Shuttle. A 4.06% false positive rate for the government ruining someone's life via a false conviction is IMHO utterly unacceptable.

Whether climate is chaotic is an open issue. Whether weather is chaotic is not an open issue. All it takes is the flap of a butterfly's wing in Brazil to eventually to set off a tornado in Texas. Turbulence is an integral part of weather. This alone makes weather inherently chaotic. Weather is predictable for a week or so at most.

The question "why do clouds exist?" is intimately tied with the question "why do we have weather?". We have weather because The Earth's equatorial regions receive a lot more solar energy than do the poles. That means the equatorial regions are hotter than the poles, and that in turn means that the atmosphere is going to move around to transfer energy from the equatorial regions to the poles. The Earth's surface is about 71% water, 29% land. Air over the oceans will naturally be fairly humid. When that humid air blows over land all kinds of havoc will break loose. Humid air is less dense than is dry air at the same temperature and pressure. When humid air confronts dry air that humid air will rise. The packet of rising humid air will expand (atmospheric pressure decreases with increasing altitude) and cool off (conservation of energy) as it rises. While the absolute humidity of the packet of rising humid air won't change, the relative humidity will. The temperature of the rising packet of humid air will eventually fall below the dew point. A cloud forms when the packet of rising humid air reaches that critical altitude. Warm air is less dense than is cool air at the same pressure. Combine warm moist air with cool dry air and things can become quite violent. Now the clouds are fueled by humidity and temperature differences. These phenomena can happen very fast. Diffusion is a comparatively slow process.

Substituting [math]x=a\pm\sqrt{a^2-b}[/math] in [math]x^2-2ax+b=0[/math] yields 0=0, a tautology. That [math]x^2-2ax+b=0[/math] has two distinct real roots is a given. From this, triclino correctly derived the condition [math]4a^2-4b>0[/math]. Stated more succinctly, [math]a^2-b>0[/math]. This reduces the problem to Given [math]a^2-b>0[/math], show that [math]6a^4-2a^3-2a^2b+a^2-6a^2b+2ab+2b^2-b\geq 0[/math]. As shyvera already noted, the first step to proving this is to factor the left hand side into something of the form [math](a^2-b)(\cdots)[/math].

Some rules that will be of use here: If u≥0 and v≥0 then u+v≥0. If u≥0 and v≥0 then uv≥0. Where do you get these tedious but trivial problems?

No. It works fine. Use implicit differentiation.

The Sun is about 93 million miles away from Earth. If the atmosphere was perfectly transparent, the intensity of sunlight after traveling an extra 62 miles (the altitude at which ones earns their astronaut wings) is reduced by a paltry 0.0002%. One problem with that analysis: That is a mighty big if. The atmosphere is not perfectly transparent. The sky is blue, after all. The atmosphere, excluding clouds, reflects about 6% of the incoming radiation right back into space and absorbs another 16%. Sunlight on a sunny day at sea level is 22% less intense than it is up on the Space Station.

[math]x^2+y^2=a^2[/math] is not a function of y in terms of x or of x in terms of y, and yet at all but four points on the circle one can compute dy/dx and dx/dy -- and dx/dy = 1/(dy/dx). Things become problematic in higher dimensions working with partial derivatives.

I agree with the first sentence. The last sentence, not so much. Evolution is a slow and inefficient process. The reason there is so much biological diversity on the Earth now is because evolution had a long, long time (3.8 billion years or so) to overcome the inherent slowness and inefficiency of the process. Evolutionary pressure does not necessarily come from the environment. It can come from within. Sexual selection pressure can arise in what some view to be maladaptations. Deer and their oversized antlers, birds and their ostentatious plumage, humans and their bad boys/nasty girls. Even when evolutionary pressure does come from the environment, it can be the wrong part of the environment ("wrong" in the sense of a gods-eye view of the best direction a species should be heading to ensure long-term survival). Evolution does not have a gods-eye view. It has no long-term view, no intelligence, a limited tool set. It creeps across the evolutionary landscape. Evolution blindly and stupidly responding to immediate evolutionary pressure, and even then only to the pressure it is equipped to respond to. When it can respond, the response can be stupid. Species can go into all-out interspecies arms race with the end result being a Pyrrhic victory. The opposite extreme of an arms race is cooperation, and this too can be problematic. Avocados depend on ground sloths to disperse their fruit. One problem with that: ground sloths went extinct with the other megafauna. I definitely disagree with that statement. Overspecialization and sexual selection can seal the fate of a species. Genetic drift can push a species down a wrong path. You are giving evolution more credit than it is due. It is slow, inefficient, stupid, and blind. 99.9% of all species that have existed are extinct.

Here is a way to compute the number of ways to generate all possible sums of m tosses of a die with n faces. Use the generating function, which for a fair die of n faces is [math]g(z) = z+z^2+\cdots+z^n[/math]. Next compute [math]g(z)^m[/math], where m is the number of tosses. Finally, read off the polynomial coefficients. For example, the number of ways to get a total of three is the coefficient of [math]z^3[/math] in [math]g(z)^m[/math]. Here is a perl script that computes and prints this: #!/usr/bin/perl -w use strict; sub mult_poly($$); my ($tosses, # Number of tosses of the die $faces, # Number of faces on the die $gen, # Generating polynomial g(z) $pow); # g(z)^m # Get command line arguments. Usage is <script_name> [#tosses [#faces]] $tosses = scalar @ARGV ? shift : 2; $faces = scalar @ARGV ? shift : 6; # Construct the generating function, g(z)=z+z^2+...i+z^n, n=#faces. $gen = [0, (1) x $faces]; # Compute g(z)^m, m=#tosses $pow = [1]; $pow = mult_poly $pow, $gen foreach (1 .. $tosses); # Print results. printf "%3s %9s\n", qw(N Count); printf +("%3d %9d\n", $_, $pow->[$_]) foreach ($tosses .. $#$pow); # Sanity check; the printed sum should be n^m. my $s = 0; map {$s += $_} @$pow; printf "Sum %9d\n", $s; # Multiply two polynomials. sub mult_poly ($$) { my ($p, $q) = @_; my $prod = [(0) x ($#$p + $#$q + 1)]; for (my $ip = 0; $ip <= $#$p; $ip++) { for (my $iq = 0; $iq <= $#$q; $iq++) { $prod->[$ip+$iq] += $p->[$ip] * $q->[$iq]; } } return $prod; } Merged post follows: Consecutive posts mergedThe above might overkill if you don't want all possible combos. If all you want the number of ways to achieve a specific sum s given t tosses of a die with f faces, Ooops. What I posted for this individual problem doesn't work. Deleted. For now use the algorithm in the first half of this multi-post.

2005 is not several years. 2005 was the all-time record in terms of number of tropical storms (they had to use Greek letters), all-time record in terms of number of hurricanes, and one short of the all-time record in terms of number of major hurricanes. 2008 was no slouch either, with 16 named storms, 8 hurricanes, and 5 major hurricanes. Last summer was a dud primarily because it was an El Nino summer. That El Nino is now gone. We are heading for a La Nina. If the transition occurs sooner than expected, look out for a particularly nasty end to the 2010 hurricane season. Be careful what you wish for. You just might receive it. There are two ways that NOAA's forecast can be wrong. If NOAA's forecast does turn out to be wrong, I suspect it will because they guessed low rather than high. Conditions are ripe for a particularly nasty hurricane season. Reason #1: Hurricane seasons go through a long cycle of 20-30 years of high activity followed by 20-30 years of low activity, then back to high activity. We are 15 years into the high activity side of the cycle. There is no reason to think things have switched to the low activity side 5 to 15 years earlier than expected. Reason #2: Atlantic sea surface temperatures in May are very telling regarding the strength of the upcoming hurricane season. This year they are not just abnormally high. They are off the charts high. Reason #3: That El Nino that made last year's hurricane season a complete dud is now dead. El Nino 3.4 plummeted from +0.8C to -0.2C during the month of May. Another 0.3C drop and we are in a full-blown La Nina. Just as El Ninos tend to stifle Atlantic hurricanes, La Ninas make conditions ripe for Atlantic hurricanes. So, if you live near the Atlantic or Gulf coast, make sure you keep your cars at least half full of gas. Have your escape route planned well ahead of time, but also have alternative routes in mind (those freeways can turn into parking lots). Buy some camping equipment, some water, some iodine to sterilize the water you forgot to buy, and maybe some MREs (el yucko). A propane stove and lantern and plenty of propane can make weeks without power a lot more bearable. Buy a generator so you don't have to toss the contents of your fridge and freezer. Added plus: you can eat steak instead of MREs. Stock up on some canned food. It's not as good as steak but it's not nearly as disgusting as are MREs.

triclino, you are never going to learn if you do not try. That is why students are given homework problems. So, a homework problem for you: The reals are a field (i.e., they have operators + and * with the standard axioms for those operators). What makes the reals special is that they are a totally order field. This means there is an operator ≥ whose characteristics include, for a,b,c in R, If a ≥ b and b ≥ c then a ≥ c. a ≥ b or b ≥ a. If a ≥ b then a + c ≥ b + c. Given the above axioms prove that The sum of two non-negative reals is non-negative. Hint: In other words, prove that if a≥0 and b≥0 then a+b≥0. The sum of a finite set of non-negative reals is non-negative. Hint: Use the above result.

Of course. It is trivial. You are the one who needs help here, why don't you try it?

Show your work.

There are lots of well-known techniques when the problem is linear. Linear programming, for example, addresses the problem of maximizing a linear function of a set of N variables [math]f(x) = \sum_{i=1}^N a_i x_i[/math] subject to a set of M linear constraints [math]\sum_{j=1}^N c_{i,j} x_j \le b_j,\quad i=1,2,\cdots,M[/math] Quadratic programming extends the cost function to a quadric form. The constraints are still linear in quadratic programming. There are a number of well-known techniques for solving quadratic programming problems. In general, however, the answer is there are no general techniques. (There aren't any general techniques for computing anti-derivatives either. There are nonetheless a lot of tricks.) Just because there are no general techniques does not mean you have to give up. This particular problem is fairly easy. As Mr. Skeptic hinted, try substitute a single variable for the product xy. What does that do to the problem?

If, needless to say, you have no idea how to start, why are you doing these problems? You need to do a better assessment of your skills. You are apparently biting off more than you can chew. Some questions: What are you studying? The name of the course, the book, etc. Why are you studying? Is it because you have to in order to get through school, because you want to improve your knowledge, or some other reason? Are you studying in a structured setting (e.g. a class in some) or an unstructured setting (e.g., studying on your own)?

Just because you do not understand something does not mean it is wrong, triclino.

Do you understand that a function f(x) that is quadratic in some variable x will necessarily have the same sign for all real x if the roots are not real-valued? This is very basic stuff, triclino.

I would say "prove it".

Capn's proof is correct, triclino. Your attitude is a huge hindrance to your learning. Since you insist on being spoon-fed, here are some big servings. Method 1 By inspection, [math]a^2+ab+b^2 = 3\left(\frac{a+b}2\right)^2 + \left(\frac{a-b}2\right)^2[/math] The two terms on the right are non-negative for all real a and b. The sum of two non-negative values is non-negative, and therefore a^2+ab+b^2 is non-negative, or in other words [math]a^2+ab+b^2 \ge 0[/math] Method 2 Viewing a^2+ab+b^2 as a quadratic in a, the zeros are given by [math]a = \frac{-b \pm \sqrt{-3b^2}}2[/math] When b=0 the two solutions are degenerate and are zero. In this case, the expression a^2+ab+b^2 degenerates to a^2, and this is non-negative for all real a. When b≠0, the two solutions are imaginary. Thus a^2+ab+b^2 has the same sign for all real a. At a=0, a^2+ab+b^2 degnerates b^2, which is positive for all non-zero real b. Thus [math]a^2+ab+b^2 \ge 0[/math] Method 3 By inspection, [math]a^2+ab+b^2\ge 0[/math] is equivalent to [math]a^2+b^2\ge -ab[/math]. This is trivially true when ab≥0. This leaves the case ab<0 to be investigated. Define [math] \aligned \alpha &\equiv \max(|a|, |b|) \\ \beta &\equiv \min(|a|, |b|) \endaligned [/math] In terms of these new variables, [math]a^2+b^2=\alpha^2+\beta^2[/math] and in the case where [math]ab<0[/math], [math]-ab = \alpha\beta[/math]. Note that by definition, [math]\alpha\ge\beta[/math]. Multiplying both sides by [math]\alpha[/math] does not change the inequality since [math]\alpha>0[/math]. Thus [math]\alpha^2\ge\alpha\beta[/math]. Since [math]\beta \ge 0[/math] by definition, [math]\alpha^2+\beta^2\ge \alpha^2[/math]. As greater than is transitive, so [math]\alpha^2+\beta^2\ge \alpha\beta[/math]. In terms of the original variables, [math]a^2+b^2\ge -ab[/math] when ab<0. Combining the cases ab≥0 and ab>0 yields [math]a^2+b^2\ge -ab[/math] for all real a, b, and thus [math]a^2+ab+b^2 \ge 0[/math] Method 4 By inspection, [math]a^2+ab+b^2\ge 0[/math] is equivalent to [math](a+b)^2\ge ab[/math]. This is trivially true when ab≤0. This leaves the case ab>0 to be investigated. Define [math] \aligned \alpha &\equiv \max(|a|, |b|) \\ \beta &\equiv \min(|a|, |b|) \endaligned [/math] In terms of these new variables, [math](a+b)^2=(\alpha+\beta)^2[/math] and in the case where [math]ab>0[/math], [math]ab = \alpha\beta[/math]. Since [math]\alpha\ge\beta[/math] by definition, [math]\alpha^2+\beta^2>\alpha^2\ge\alpha\beta[/math] and thus [math]a^2+ab+b^2 \ge 0[/math] Method 5 Viewing a^2+ab+b^2 as a quadric in a and b, [math]a^2 + ab + b^2 = \bmatrix a & b \endbmatrix \boldsymbol A \bmatrix a \\ b \endbmatrix [/math] where [math]\boldsymbol A = \bmatrix 1 & 1/2 \\ 1/2 & 1 \endbmatrix[/math] The eigenvalues of the matrix A are 3/2 and 1/2 (verification left as an exercise to the reader). The matrix A is thus positive definite and thus serves as a metric on the space [math](a,b)\in\mathbb R^2[/math]. Thus [math]a^2+ab+b^2 \ge 0[/math] Method 6 Method 5 suggests expressing a and b in terms of the eigenvectors of the matrix A, which are [math]\bmatrix 1/\surd 2 & 1/\surd 2\endbmatrix^T[/math] and [math]\bmatrix 1/\surd 2 & -1/\surd 2\endbmatrix^T[/math]. By inspection, [math] \aligned a &\equiv \frac{a+b}2 + \frac{a-b}2 \\ b &\equiv \frac{a+b}2 - \frac{a-b}2 \endaligned [/math] Thus [math] \aligned a^2 &=\left(\frac{a+b}2\right)^2+ 2\left(\frac{a+b}2\right)\left(\frac{a-b}2\right)+ \left(\frac{a-b}2\right)^2 \\ b^2 &=\left(\frac{a+b}2\right)^2- 2\left(\frac{a+b}2\right)\left(\frac{a-b}2\right)+ \left(\frac{a-b}2\right)^2 \\ ab &= \left(\frac{a+b}2\right)^2 - \left(\frac{a-b}2\right)^2 \endaligned [/math] and hence [math] a^2+ab+b^2 =3\left(\frac{a+b}2\right)^2 + \left(\frac{a-b}2\right)^2 [/math] See method 1.

I am not going to spoon feed you the answers to your homework-like questions, triclino. That won't help you learn. I am instead trying to help you solve the problem mostly on your own with a little nudge here and there. Suppose I told you that the eigenvalues of the matrix [math]\bmatrix 1 & 1/2 \\ 1/2 & 1 \endbmatrix[/math] are 1/2 and 3/2 and therefore [math]a^2+ab+b^2 \ge 0[/math] because [math]a^2 + ab + b^2 = \bmatrix a & b \endbmatrix \bmatrix 1 & 1/2 \\ 1/2 & 1 \endbmatrix \bmatrix a \\ b \endbmatrix[/math] Did that help you? (No, it didn't.) Spoon-feeding someone the answers is generally far easier than trying to help that person learn.

It would be good if you paid attention to that valuable advice. You do not have to assume [math]a^2+ab+b^2\ge 0[/math] to prove [math](a+b)^2\ge ab[/math]. These two statements say exactly the same thing. This means of course that if you do prove the former you have automatically proven the latter. However, if you can prove the latter you have automatically proven the former. I thought you might have an easier time proving the latter statement.

Baloney. Let's start with [math]a^2+ab+b^2 \ge 0[/math]. Adding 0 to both sides of the inequality does not change the inequality, so [math]a^2+ab+b^2 +ab -ab \ge 0 + 0[/math]. (Note that I write 0 in the form [math]ab-ab[/math] on the left hand side.) The left hand side is [math](a+b)^2 - ab[/math], and thus the original inequality is equivalent to [math](a+b)^2 - ab \ge 0[/math]. Adding [math]ab[/math] to both sides results in [math](a+b)^2 \ge ab[/math]. There is a simpler way to obtain the same result. Start again with [math]a^2+ab+b^2 \ge 0[/math]. Adding [math]ab[/math] to both sides yields [math]a^2+ab+b^2 + ab \ge ab[/math], or [math](a+b)^2 \ge ab[/math].

No, for two reasons. 1. This looks too much like homework. You need to figure this out on your own. 2. You already did it in post #9.