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Fibonacci with new seed numbers just gives you the golden ratio again. It isn't the initial values that makes the Fibonacci ratios converge to phi, but rather the particular recursion rule F[sub]n[/sub] = F[sub]n-1[/sub] + F[sub]n-2[/sub]. To get a different ratio for the limit, you need a different recursion. For example:
A[sub]n[/sub] = 2A[sub]n-1[/sub] + A[sub]n-2[/sub] implies A[sub]n+1[/sub]/A[sub]n[/sub] ---> 1 + √2
B[sub]n[/sub] = 6B[sub]n-1[/sub] - 2B[sub]n-2[/sub] implies B[sub]n+1[/sub]/B[sub]n[/sub] ---> 3 + √7
As an interesting tidbit, trying to explore what MIF intended by his furthest from fractions, I looked at | nx - round(nx) | for various inputs x to the animation and integers n. What I found was that the lowest value of n for which that distance drops below approximately 0.03 seems to be the number of "seeds" on the outside when animation stops +/- 1 when run on input x.
( It is also interesting to note that when x = golden ratio, the values of n for which the distance hits a new record low are 1, 2, 3, 5, 8, 13, ... . Not really surprising, but interesting just the same.)
Any number you enter actually will be a fraction. The real question is why some fractions work better than others, and this relates to which irrational numbers they approximate. It appears that algebraic numbers in general do better than others, but the reason isn't clear to me.
Another number that works well (though not quite as well) is sqrt(2) - 1 = 0.414.
Sqrt(3), sqrt(5), sqrt(7) are not as good, but still tighter than most numbers.
It is straight off that page, just a little below the animation (next to the big phi).
On 14, The intersection of two arithmetic series is the arithmetic series of the least common multiple. In this case, 6. Since the first series stops at 200, and the second at 240, the question becomes: how many multiples of 6 are there <= 200, which is [200/6] = 33.
I think to understand it, I would need to know a bit more about how the animation works. How fast do the seeds grow as you move out? how far do you move out with each new seed. I suspect that with these parameters, you can put together a polynomial equation whose roots include the golden ratio, but also something close to this number.
It seems to work pretty much the same no matter how many more digits you add. If you assume repeating 56, then it is the rational number 31/198.
(Speaking of which,
But the Golden Ratio (its symbol is the Greek letter Phi, shown at left) is an expert at not being any fraction. It is an Irrational Number (meaning you cannot write it as a simple fraction), but more than that ... it is as far as you can get from being near any fraction.
SAY WHAT??? Technically this is true. It isn't rational, and it is just as far from rational numbers as all other irrational numbers (distance = 0).
I have no doubt that you know better than this statement. So I am curious what you really meant by it.)
Check out the blunder I made in that trigonometry thread. Know what I did, I screamed, cried, kicked a few boxes, cursed at some people, held my head in bucket of cold water for 30 seconds and now I am back. Reinvested, and ready to blunder again.
With that remark, I know you are true mathematician!
PatternMan -
IQ tests are a very limited measure of intelligence. True intelligence has many aspects and people tend to be better in some aspects than in others. Yet IQ is a single number. It cannot adequately measure this. That Feynmann got 125 on an intelligence test is a indictment of the test rather than of him. And intelligence itself is not a great predicter of success in any field, even mathematics. I have met mathematicians that I can say with good certainty are less intelligent than I am. Yet they are practicing mathematicians with good accomplishments, whereas beyond my dissertation, I have never accomplished anything in the field.
The question of how far you want to go and in which direction is of course up to you. Don't pursue a degree in anything unless you feel a real desire to do it. But also, don't let ideas about "job availability" steer you away from doing what you love, either. I cannot speak for other professions, but there are good jobs available for those proficient in mathematics, physics, or engineering. Even though my job is engineering, my first boss told me that he preferred to hire those with mathematics or physics backgrounds because engineers generally wanted to do design work, and thus were less likely to stay in weight engineering. Mathematics is applicable everywhere, and those who hire mathematicians for various jobs often find them more valuable than those specifically educated for the job, because we are trained to think logically about issues in general, and therefore spot possibilities that those trained for the job never dreamed of. That was certainly the case for me. (The reasons I couldn't find any job upon graduating had far more to do with me than with a lack of jobs.)
More generally, doctorates in any field can be useful. About a year or two after starting this job, I was sent a survey. Some college student was studying the usefulness of doctorates to business. I don't know what the outcome of that was, but what impressed me was the list of doctorates it was sent to. Though my company has thousands of salaried employees, and I had met only a small number of them, I recognized almost every name on the list. Most of them I knew as being the crucial people in their particular groups. Doctorates were evidently well worth the hire for my company.
Yes. I am curious if the test designer intended that point or the other one. Probably the other, since it has integer coordinates. Either way, they apparently missed that there are two solutions.
I have a BS in Physics and a PhD in Mathematics specializing in Differential Geometry. I started in Physics, which had been my intended direction since 7th grade. But I was always good at math, and once I got through the classes expected for all physics majors, I decided to take the few extra needed to pick up a minor in Math. However, my introduction to real analysis course completely changed my understanding of mathematics. I learned the rigor of true mathematics, and took delight in learning the inner workings of all the things that I had only vague ideas about before. I enjoyed it enough, I took another year and got a double major in Physics and Math instead. Afterwards, an opportunity to get my Masters in Math opened. I followed it and the doctorate after.
Like any PhD program, it was long slog, and I was burned out on the subject by the time I reached the end of it. That, little reputation for my alma mater, and a cyclic glut of new doctorates vs available post-docs (these things go up and down in cycles) meant that I had no success in finding a position after graduating. Instead I spent a few years in manufacturing before an opportunity opened for my present job as a Weight Engineer at an aircraft company (weight engineers track the distribution of weight in the aircraft, which is needed for aerodynamics, stress, loads, and flutter analysis). At first, while a step up from manufacturing, this seemed like a poor fit for my education, but in fact has proven to challenging, enjoyable, and has opened many new avenues of investigation for me.
Like you, I originally considered higher mathematics to be mostly without practical benefit. I have since developed a strong disagreement with that concept. Most of modern technology works by principles that people originally studied as a lark, without believing that it would ever have practical application. Instead, had they not studied it, our lives would be far different, and harder, than they are now. The thing is, you don't know in advance what will prove useful, and what won't. But generally, anything has an application. Hardy used to brag that nothing he did had a practical application (yes, he considered that something to brag about). But since his day, every theory he developed has found practical application.
Whether Physics or Math or Engineering, or some other field is the best fit for you, I don't know. But don't be too quick to choose. When you get to college, look around. See what you find the most intriguing. Then follow it.
if they intersect at all, there must be 4 points of intersection.
Or 2 points, but that requires that the circle radius exactly match one of the ellipse semi-axes, which is not the case here.
But, niharika_kumar, I doubt you were intended to solve this problem algebraicly. If you have already studied the equations of circles and ellipses, then it is likely the test preparer hoped for you to recognize them, and then be able to do exactly what Bob did in his first answer. Realize that the circle and ellipse had to intersect in either 0, 2, or 4 points. Then by examination of the radii and axes to realize it had to be 4.
But he or she also made it simple enough to solve for x[sup]2[/sup] and y[sup]2[/sup], and to notice that neither one is negative, meaning that two real values of x and of y exist that solve them, from which 2x2 = 4 points can be constructed.
It is not necessary, or even desirable, to actually figure out exactly what x and y are to solve this problem.
Position, not font size, indicates exponentiation.
10
10still means 10[sup]10[/sup] despite the font sizes (this is the only way I've been able to show a superscript of the same size font - every other method appears to be disabled).
Use of Log as a function really isn't distinct from use of it as an operator, They are the same thing. It is a matter of how you look at it. Since you obviously have encountered functions yet, I wouldn't worry about it. You'll catch on later.
Circumflexes are a common way of denoting exponentiation in computer programs. The notation carries over into other fields. A variant some languages use is **, so 2^3 = 2**3 = 2[sup]3[/sup].
You can occasionally not have an explicit operator symbol. A common example, as you may be aware, is multiplication: 2x means 2 times the value of x, even though there is no operator symbol between them. As long as it is possible to differentiate between the two operands, juxtaposition (i.e, positioned next to each other) is a valid operation indicator.
(There is an even more common example. Another operator that we use all the time and denote by juxtaposition as well. But you may have a hard time recognizing it, because people don't normally think of it as an operation. I'll leave it to you to guess what it is.)
The first step is not to learn rules, but to learn why the rules hold. Some things are simply a matter of convention (such as working from the top down on power towers instead of the bottom up). But most mathematical rules have to be the way they are, such as -(-x) = x. When you understand why it has to be that way, then you have mastered the subject. And then you don't have to remember tons of rules, because you can usually figure out the rules again when you need them.
For -(-x) = x, this follows from two things:
1. the definition of the opposite: -x is the unique number which, when added to x, gives 0: x + (-x) = 0.
2. the commutivity of addition: if x + (-x) = 0, then (-x) + x = 0.
That is, x is the unique number which, when added to (-x), gives 0. Thus -(-x), has to be x.
(Okay - I fudged that by putting "unique" in the definition, which is really a property that needs to be proven of the opposite.)
Applying that to your example: -(-(-(-2))) = -(-(2)) = -(-2) = 2.
Please, subscripts are too useful as contravariant indices to be giving them another interpretation. To do so would just seed confusion (just as the combined use of superscripts to denote exponentiation, differentiation, repeated composition, and covariant indices can already be confusing). The traditional notation for logarithm works just fine.
Rather than considering LOG to be some sort of binary operator, we generally consider it to be a parameterized family of functions: ln is one function, log[sub]10[/sub] is another, log[sub]2[/sub] is a third, etc. From a strict logical standpoint this is the same thing as an operator, but it does tend to shape how we approach it, and how we denote it. Thus we want it to look like a function.
Font size has nothing to do with operation. We just prefer a smaller font for superscripts and subscripts in general for readability reasons. A full-sized font for them would either interfere with the surrounding lines are require more space between them.
There are 10 kinds of people in the world:
Those who understand Binary and those who don't.
Borrowed from xkcd:
There are 10 kinds of people in the world:
Those who understand binary, those who don't, and those who have heard of trinary.
I have heard numerous people say that they never used algebra again after graduating. Those in more technical fields use basic algebra, but never anything higher. It has been claimed that higher mathematics exists only in academia.
I have found just the opposite to be true. My job (weight engineer for an aircraft company) has given me cause to invert a 7x7 matrix (if I had really gone whole hog on the problem, I would have had to invert a 12x12 matrix). It has required me to find eigenvectors and eigenvalues. About 3 or 4 years ago, I came across an application where using complex numbers simplified a calculation more than any other approach (calculating gear axle location from a measurement of the "shock absorber"). I have had need for differential and integral calculus. I have solved differential equations, developed numerical methods, explored esoteric spaces, all just to do my job.
Most of my collegues never go beyond algebra or a little analytic geometry (and trig is alwasys useful). But it isn't because the opportunities for use of higher mathematics are not there. It is because they are not as familiar with them, so they don't see the opportunities. (And because, since I am available, many of those jobs get passed to me.) Instead, they fall to experimentation to solve similar issues.
Of course I also engage in recreational mathematics, or else I wouldn't be here. Anytime I find myself with nothing else to occupy my time, I pull up a problem in my head and muse upon it. Attempting to find a human proof of the 4 color theorem is one of my favorites. Of course, I haven't found one, but I have come across a number of intriguing results. (The 4 color theorem is famous for a computer proof of it that was produced in 1976 - the first well-known hypothesis to be proven by computer.) I haven't been truly bored in decades.
Except that these are sets of real numbers, not integers, so you can't list all the members of them (not saying that bobbym is wrong - he is just giving examples of intervals of integers, not intervals of real numbers).
(a, b) = set of all real numbers x such that a < x and x < b = { x | a < x and x < b}
(a, b] = set of all real numbers x such that a < x and x ≤ b = { x | a < x and x ≤ b}
[a, b) = set of all real numbers x such that a ≤ x and x < b = { x | a ≤ x and x < b}
[a, b] = set of all real numbers x such that a ≤ x and x ≤ b = { x | a ≤ x and x ≤ b}
(In all of those, it is assumed that a < b. Some people allow a = b, in which case the first three are empty, and [a, a] = {a}, a singleton set containing just a. Some people also allow b < a, in which case all four sets are empty. Usually, though intervals are only given with a < b.)
So the set (0, 1) contains 0.1, 0.01, 0.001, 0.0001, etc., no matter how many 0s occur before the 1. I also contains 0.9, 0.99, 0.999, etc (for any finite number of 9s), because all of the these numbers are > 0, but < 1. It does not contain either 0 or 1.
The set (0, 1] contains any number in the open interval (0,1), but also contains 1 (but still not 0).
Similarly, the set [0, 1) contains everything on (0,1), but also includes 0, but not 1.
Finally, [0, 1] contains 0, 1, and all the numbers in (0,1).
The next difficulty is that sometimes we want everything less than a number, or everything greater than a number. For these, we replace a with -∞ or b with ∞. But since ∞ is not a real number, it cannot be a part of the set, so it is only given with an open side:
(-∞, b) = the set of all real numbers x such that x < b = { x | x < b}
(-∞, b] = the set of all real numbers x such that x ≤ b = { x | x ≤ b}
(a, ∞) = the set of all real numbers x such that a < x = { x | a < x}
[a, ∞) = the set of all real numbers x such that a ≤ x = { x | a ≤ x}
(-∞, ∞) = the set of all real numbers.
It has gone too far. Sure, the first year that Dumbledore had stolen the house cup from Slytherin by a last minute awarding of large numbers of points to his favorites, nobody minded except Slytherin. Who cares if a bunch of slimey gits have their faces rubbed in the mud? But every year since, it has been the same. Another awarding of huge points to Harry and his friends for evermore ludicrous reasons. The Gryffindors didn't even try anymore. They just sat back and waited to see what excuse Dumbledore would use to give the cup to them. This year, when your own house of Hufflepuff had a 400 point lead over Gryffindor, he awarded 405 points to Harry "for parting his hair in a most amusing fashion." Even the other teachers had the grace to look embarassed.
You could stand it no longer. So you stood in protest of this favoritism. Everyone except the Gryffindors have joined you. "Unjust, am I?", Dumbledore roared. "We shall see about that! Since you accuse me of dishonest judging, let us settle this with a challenge! Yes, yes. You shall serve as champion for the Hufflepuffs, and Harry shall serve as champion for the Gryffindors! Snape! Go to my office and bring the Urns of Peano here." As you wait, Dumbledore sits back with a sly smile. You know he has a trick up his sleeve. Presently, Snape returns with three large urns, each with a dispenser at the bottom and a short track. For one of the urns, the track contains balls numbered from 0 to 6. The tracks for the other urns are empty. Dumbledore explains, "These urns have a marvelous enchantment upon them. They can each contain an infinite number of these balls. Indeed, this urn already contains a ball for every natural number." Suddenly, the History Professor Binns interupted, "But 0 is not a natural number! The natural numbers are the numbers of the Greeks!". "Please! Do pull yourself out of your stuffy old books!", replied the Arithmancy Professor, Septima Vector. "0 obviously belongs in the Natural numbers! You can't start counting without it!" The argument became more raucous, until Dumbledore was forced to expel them from the hall.
"Getting back to business. The maker of these urns included 0. Now, if you will please take a ball and drop it back into the top of the urn?". You remove the ball marked 1. The higher numbered balls all roll down, allowing a ball marked 7 to imerge from the dispenser. You drop the 1 ball into the top. Ball 7 rolls back into the dispenser, and balls 2 through 6 move back up the track. Ball 1 appears, passing though the other balls as rolls back to its original spot. Dumbledore continues "Another enchantment on the urns causes the lowest numbered balls to alway be displayed. Now as you can see, the other two urns are empty. The middle urn will be assigned to Hufflepuff, while the Gryffindors will have the urn on the right. The challenge is this. You, Hufflepuff, will take two balls at a time from the full urn and place them in the Hufflepuff urn. Each time that you do so, Harry will take a single ball from the Hufflepuff urn and add it to the Gryffindor urn. You will continue until the original urn is empty." Hagrid pops up. 'Surely, ye cannat be thinkin' of 'aving 'arry 'ere spend the rest o' 'is days movin' little balls around. Even I knows that'd take forever!" "Ah, do not fear!", Dumbledore replies, "for I have an answer. I shall cast Zeno's Infinite Acceleration upon them. All our brave champions must do is to fix their plans of action in their heads as I cast the spell. The spell will cause each repetition of these actions to proceed at twice the pace of the previous one. As Professor Vector could explain if she were still here, this will cause the entire contest to finish in only a short time."
Professor McGonagall suddenly appears quite agitated. She quickly hurries to Dumbledore and whispers to him. "Hmmm, yes. I see. Still, it will be most entertaining to watch!" Looking at her face, he goes on, "No, I suppose not." Turning to the assembly, Dumbledore says "Minerva has reminded me that if we don't want our two champions to erupt into balls of fire, a further step will be needed." Before he can continue, Snape speaks up, "Headmaster, it so happens that I have two potions of Caratheodory's Entropic Isolation upon my person. If our contestants do not waste their time, they might be able to finish before the potion wears off. Possibly, anyway." "Splendid", Dumbledore replies, "well, then, if our champions will just step forward, we can be underway!". Harry looks confused. "Professor Dumbledore, if I take one ball out of every two in Hufflepuff, how would I ever win? We will always have the same number of balls!" "Ahh, Harry. I am afraid your task is harder than that! Did I forget to explain? Gryffindor must do more than just match Hufflepuff in balls. If, when we are finished, there is even a single ball left in the Hufflepuff urn, then Hufflepuff will be declared the winner! Let them accuse me of favoring Gryffindor now!"
Harry is clearly panicked, but Hermione leans over and starts whispering to him. Suddenly he relaxes and a smile spreads over his face. Clearly there is more to this contest than it seems. You wonder, how does Dumbledore expect Harry to win? And how can you defeat him?
You'd better add "monic" to your description of f, or else it can't be proved, since for any a, af(x) is another polynomial with the same roots.
So is the refractive index of B relative to C be more than unity or less than unity?
Since sine is increasing and 45 > 40, the index for B must be smaller than the index for C.
When smaller numbers are divided by larger numbers, the ratio is < 1.
Bob has gone into greater detail, but I had already provided the answer.
The polynomial is f(t) = (3+2t)[sup]2[/sup](1-t[sup]2[/sup]) - 1. It has turning points at roughly -3/2, -7/8, 3/8 (the later two are only approximate). Since t = sin x, it only extends in value from -1 to 1. So the -3/2 turning point is of no interest. f(-1) = f(1) = -1. f(3/8) ~= 12.5, f(-7/8) ~= -6.5. So there are 2 roots of the polynomial in the interval of interest: One between -7/8 and 3/8, and another between 3/8 and 1.
I don't have time now, but you could almost certainly estimate them very closely using Newton's method, then take the inverse sin to find the values of x between 0 and 10 that yield them.
Yeah. When I did it, I accidently dropped the square from the (3+2sin x) factor, and just got a cubic. Still couldn't solve it easily, though.
Given that the answers are in decimal, I wouldn't be surprised if this was intended to be solved numerically. I would suggest using Newton's method rather than trial and error, or even graphical solutions. It might still be easier to use Newton's method to solve the quartic. I doubt Ferrari's method is would be any more illuminating.